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4. 7. EXERCISES
159
able to generalize results previously discussed by P. R. Montmort (1708). Also, De Moivre (1730) derived Theorem 4.3 for si = 1, and mi = m, i = 1, 2, ... , n. After De Moivre, C. Jordan (1867) derived expressions analogous to those of Theorem 4.4 and its Corollary 4.4. Ch. Jordan (1926, 1927a,b, 1934, 1939a,b) obtained Theorems 4.4 and 4.6 and their Corollaries 4.4 and 4.6 and presented several applications. More general formulations of the inclusion and exclusion principle were given by M. Frechet (1940, 1943), K. L. Chung (1941, 1943a,b,c) and L. Takacs (1967b), who furnished a very extensive reference list. Further, L. Takacs (1991) derived a limiting form of Theorem 4.4 as the number n of sets (events) increases to infinity. The work of C. E. Bonferroni (1936) is considered as the foundation of the inequalities in Theorem 4.5. The menages problem was presented and solved by E. Lucas (1891); before that under a different formulation this problem had been examined by A. Cayley (1878a,b) and T. Muir (1878). The expression for the reduced menages number given in Example 4.5 is due to J. Touchard (1934, 1953). A general method of enumeration of permutations with restricted positions was developed by I. Kaplansky and J. Riordan in a series of papers. For details, see Chapters 7 and 8 of the book on combinatorial analysis by J. Riordan (1958).
4.7
EXERCISES
1. Calculate the number offourdigit university campus telephone numbers in which each of the digits 1, 2 and 3 appear at least once.
2. Calculate the number of positive integers less than or equal to 1000 that are neither perfect squares nor perfect cubes. 3. Suppose five pairs of gloves are to be distributed to five people by a child. Calculate the number of different distributions of two gloves to each of the five people with no one getting a matching pair. 4. Suppose that four gloves are randomly chosen from a drawer containing five different pairs of gloves. Calculate the number of selections that include at least one pair of gloves. 5. Calculate the number of 6combinations of the set W4 = {w1 , w2 , w3 , w4} with repetition and the restriction that the element Wi is allowed to appear, at most, i times, i = 1, 2, 3, 4. 6. In the base 5 number system each number is represented by an ordered sequence of the digits from the set {0, 1, 2, 3, 4} with repetition. For example, the numbers 625, 859 and 3125, which are expressed in terms of
THE PRINCIPLE OF INCLUSION AND EXCLUSION
160
powers of 5 as 625 = 1·5 4 +0·5 3 +0·5 2+0·5 1+0·5°, 859 = 1·5 4 +1·5 3 +4·5 2+ 1·5 1+4·5° and 3125 = 1·55 +0·5 4 +0·5 3 +0·5 2 +0·5 1+0·5°, are represented by the ordered sequences (1, 0, 0, 0, 0), (1, 1, 4, 1, 4) and (1, 0, 0, 0, 0, 0). Note that each of 2500 integers that are greater than or equal to 54 = 625 and less that 55 = 3125 may be represented by a 5permutation of the set {0, 1, 2, 3, 4} with repetition, in which the first position is occupied by digit 1. Calculate the number of integers greater than or equal to 625 and less than 3125 for which only three of the five digits {0, 1, 2, 3, 4} appear in the last four positions of their 5nary sequence representation. 7. Show that the number of positive integers, less than or equal to 70n, that are not divisible by any of the prime numbers 2, 5 and 7 equals 24n. 8. Show that the number of positive integers, less than or equal to 100, that have not repeated prime factors equals 61.
9. Let s 1, s2, ... , sr, be positive integers, relatively prime, and n a positive integer. Show that the number N(n;s 1,s 2, ... ,sr) of positive integers, less than or equal to n, that are not divisible by any of the numbers s1, s2, ... , Sr is given by
where, in the inner sum, the summation is extended over all kcombinations {i 1, i2, ... , ik} of the r indices { 1, 2, ... , r} and [x] denotes the integral part of x. 10. The sieve of Eratosthenes. Let a 1 = 2, a 2 = 3, a 3 = 5, ... , be the sequence of primes (in increasing order) and let E(s), s > 0, be the number of prime numbers that are less than or equal to s. Show that
E(r) E(Vr) = Sn,owith n
= E( vr)
Sn,1
+ Sn,2 · · · + (1)nSn,n,
and
Sn,o = r 1, Sn,k =
L[
r
ail ai2 ... aik
] , k = 1, 2, ... , n,
where the summation is extended over all kcombinations {i 1, i2, ... , ik} of then indices {1,2, ... ,n}. 11. (Continuation). Show that the number E(100) of prime numbers less than or equal to 100 equals 25.
12. Show that
4. 7. EXERCISES
161
by a suitable combinatorial interpretation of the binomial coefficients and application of the inclusion and exclusion principle. 13. Prove the identity
by a suitable combinatorial interpretation of the binomial coefficients and application of the inclusion and exclusion principle. 14. The Galilei problem. Let G(n, k) be the number of different outcomes of a throw of n distinguishable dice, in each of which the sum of the numbers on the upmost faces of the dice equals k. Show that
G(n,k) =
~(1Y(~) e~:; 1), s = [(kn)/6].
15. Let U(n, k, r) be the number of kpermutations (j 1 , h, ... , jk) of the set {1, 2, ... , n }, with repetition, in which]!+ h + · · · + jk ~ r. Show that
U(n, k, r)
= ~( 1)
8
G)
(r k ns), m
= [(r k)jn]
and conclude that
16*. Runs of consecutive elements in combinations. Let C(n, k, s) be the number of kcombinations of the set { 1, 2, ... , n} which include no run of s consecutive integers. Show that
17*. Runs of consecutive elements in combinations of circularly ordered elements. Let B(n, k, s) be the number of (circular) kcombinations of the set { 1, 2, ... , n} of n integers, displayed on a circle (whence n and 1 are consecutive), which include no run of s consecutive points. Show that [k/s]
B(n, k, s) = ""' ( 1Y (n k) _n_ (n rs). ~ r n rs n k r=O
162
THE PRINCIPLE OF INCLUSION AND EXCLUSION
18*. Let C(n, k; a 1 , b1 , ... , akl, bkl) be the number of kcombinations {i 1,i 2, ... ,ik}, it< i2 < ··· < ik, of {1,2, ... ,n}, with differences dm = im+l  im, m = 1, 2, ... , k 1, satisfying the inequalities am < dm :S bm, m = 1, 2, ... , k 1, where am < bm, m = 1, 2, ... , k 1, are given positive integers. Show that C(n,k;at,bl,··· ,akt,bkd kl = 1)j
L( L (n Tkl (em,: Cm
2
+ · · · + CmJ),
J=O
where Tkl = a1 + a2 + · · · + akt, Cm = bm am, m = 1, 2, ... , k 1, and in the inner sum the summation is extended over all ]combinations {m 1,m2,··· ,mj} of the k1 indices {1,2, ... ,k1}. In the particular case am= r, bm = s, m = 1,2, ... ,k 1, setting C(n,k;r,s, ... ,r,s) C(n, k; r, s), conclude that
=
C(n,k;r,s)
= ~(1)i
e;
1)
(n
(k
1)~ j(s r)).
19*. Let B(n, k; a1, b1, ... , ak, bk) be the number of kcombinations {i1,i2,··· ,ik}, it< i2 < ··· < ik, of {1,2, ... ,n}, with differences dm = im+l  im, m = 1, 2, ... , k  1, satisfying the inequalities am < dm :S bm, m = 1, 2, ... , k  1, and span d = ik  it satisfying the inequality ak < n d :S bk, where am < bm, m = 1, 2, ... , k, are given positive integers. Show that
where Tkl =at+ a2 + ··· + akt, Cm = bm am, m = 1,2, ... ,k 1, and in the inner sums the summation is extended over all jcombinations {m 1,m2, ... ,m1 } of the k 1 indices {1,2, ... ,k 1}. In the particular case am = r, bm = s, m = 1, 2, ... , k 1, and ak = u, bk = w setting
4. 7. EXERCISES
163
B(n, k;; r, s, ... , r, s, u, w)
=B(n, k; r, s, u, w), conclude that
. _ ~ _ J(k1)n(k1)(ru)j(sr) B(n,k,r,s,u,w) ~( 1) . (k 1) "( ) J
J=O
x(n_
(k 1)r ~
~( _ 1)i
f;:o
x(n
n 
(kj

r  u  J s r
u j(s r))
1) n (k 1)(r w) j(s r) n(k1)rwj(sr)
(k 1)r ~ w j(s r)).
20. The menages problem at a straight table. Let 2n!Ln be the number of ways of seating n married couples at a straight table so that men and women alternate and no man is next to his wife. Show that
21. (Continuation). Let Mn and Ln be the reduced menages number for a circular and a straight table, respectively. Show that
(n 2)Mn = n(n 2)Mnl (n 1)Ln = (n 2

+ nMn2 + 4( 1)n+l,
n 1)Lnl
+ nLn2 + 2( 1t+l.
22. A modified menages problem. Let Hn be the number of ways of seating n married couples around a circular table so that no man is next to his wife. Show that
23. (Continuation). Let Wn be the number of ways of seating n married couples at a straight table so that no woman is next to her husband. Show that
164
THE PRINCIPLE OF INCLUSION AND EXCLUSION
24. Show that the number Dn,k of the terms in the development of a determinant of order n that contain k diagonal elements is given by
n! Dn,k
= k!
(1)J L .,, J
nk
k = 0,1, ... ,n.
j=O
25. Consider a series of k throws of a die and let Q(k, r) be the number of different outcomes, in each of which r :::; 6 specified faces appear. Show that
26. Let f"h = {(wl,w2,···,wk): wi = (ui,vi), ui,vi = 1,2, ... ,6, i = 1, 2, ... , k} be the set of different outcomes of a series of k throws of a pair of distinguishable dice and let Ak be the subset of f"h containing the outcomes, in each of which all six pairs (u, u), u = 1, 2, ... , 6 appear. Show that
27. Permutations with restricted repetition. Let U1 (n, k) be the number of kpermutations of n with repetition and the restriction that each element appears at least once. Show that U,(n, k)
= n!S(k,n) = i)1)nj (~)/, j=O
J
where S(k, n) is the Stirling number of the second kind.
28. Combinations with restricted repetition. Let Er,s(n, k) be the number of kcombinations of n with repetition and the restriction that each element appears at least r and at most s times. Show that
29. Consider 2n elements that belong to n different kinds, with two like elements of each kind, and let Qn be the number of permutations of these elements in which no two like elements are consecutive. Show that
4. 7. EXERCISES
165
30. Consider an urn containing n balls numbered from 1 ton. Assume that, in a drawing, r numbers are simultaneously drawn from the urn and returned to it before the next drawing. An stuple drawing consists of s consecutive drawings. If L(n, r, s) denotes the number of stuple drawings, in each of which all then numbers are drawn (each of them at least once), show that
L(n, r, s) =
~( 1)nk (~)G)
5
31. Let D(n, r, s) be the number of divisions (A 1 , A2, ... , Ar) of a finite set W, with N(W) = n, in r subsets with N(Aj) 2: s, i = 1, 2, ... , r. Show that D(n, r, s
for s
+ 1) =
~( 1)j G)~:?;: D(n sj, r j, s),
= 0, 1, ... , [n/r] 1 and conclude that D(n, r, 1) = r!S(n, r) = t ( 1)j
(~) (r j)n, J
j=O
where S(n, r) is the Stirling number of the second kind. 32. (Continuation). Consider the sum of multinomial coefficients
S2(n,r) = 11 ~ ~ (
r.
n
k1, k2, ... , kr1
where the summation is extended over all ki k1 + k2 + · · · + kr1 + kr = n. Show that
> 2,
) , i
1,2, ...
,r,
with
t(1)j(~)s(nj,rj).
S2(n,r) =
J
j=O
The number S2(n, r) is called associated Stirling number of the second kind. 33. Let C(n, r; s)
= ~~ L
(;J (:J ···(;J,
where the summation is extended over all ki 2: 1, i = 1, 2, ... , r, with k1 + k2 + · · · + kr = n. Show that C(n,r;s)
= ~ t(1ti(~)(sj)n· j=O
J
The number C(n, r; s) is called generalized factorial coefficient.
166
THE PRINCIPLE OF INCLUSION AND EXCLUSION
34. (Continuation). Let
where the summation is extended over all ki 2: 2, i = 1, 2, ... , r, with kt + k2 + · · · + kr = n. Show that
C2(n, r; s) =
i)
1)j (~) sJC(n j, r j; s). J
j=O
The number C 2 (n,r; s) is called associated generalized factorial coefficient. 35*. Frechet inequality . Let At, A2 , ... , An be subsets of a finite set il and Sn,O = N(il), Sn,r = LN(Ai,Ai2 ···Air)! r = 1,2, ... ,n, where the summation is extended over all rcombinations {it, i 2 , .•• , ir} of then indices {1,2, ... ,n}. Show that Sn,r
I(;)~ Sn,rt
I (r:
). r = 1,2, ... ,n. 1
36*. (Continuation). Gumbel inequality. Show that
for r = 2,3, ... ,n. 37*. Mobius function on partially ordered and locally finite sets. A set P = {. . . , x, y, z, ... } supplied with an order relation, x ~ y (x is less than or is included in or is ahead of y), which is valid for some pairs of elements (x, y) and with an equality relation, x = y, so that (1) x ~ x for every element x in P (2) if x ~ y and y ~ z, then x ~ z (3) if x ~ y andy ~ x, then x = y is called partially ordered. Note that the notation y 2: x may be used as an alternative form of x ~ y. Also x < y (or y > x) if x ~ y (or y 2: x) and x f:. y. A partially ordered set P is called locally finite if the number of elements in every interval [x, y] = { z: x ~ z ~ y} is finite. Let F be the set of real functions f(x, y) for x, y elements in P such that f(x, y) = 0 if x 1:. y (that is x > y or x and y are not comparable). The Kronecker delta function defined by J(x, y) = 1 if x
=y
and J(x, y)
=0
if x
f:.
y
4. 7. EXERCISES
167
and the zeta function defined by
=1
((x,y)
if x::; y and ((x,y)
=0
otherwise,
belong to the set F. (a) Let f be a function in F. Prove that there exist functions g and h in F such that
t5(x,y)
=
L
f(x,z)g(z,y), t5(x,y)
=
L
h(x,z)f(z,y),
if and only if f(x, x) :j:. 0 for every element x in P. (b) Let g(x, y) and h(x, y) be solutions of the equations in (a). Prove that h(x, y) = g(x, y) for all elements x, y in P. This common solution g is called the inverse of f. (c) The inverse of the zeta function, denoted by J.L(x, y) for x, y elements in P is called the Mobius function of P. Therefore
J.L(x,x)
=1
for every element x in P and
J.L(X, y) =for x
L
J.L(x, z), J.L(x, y) =
L
J.L(z, y)
< y fixed points in P.
38*. (Continuation). Mobius inversion formula. Let P be a partially ordered and locally finite set with a zero element 0. If f(x) and g(x) are functions defined for every element x in P and J.L(X, y) is the Mobius function of P, prove that the relation
g(x)
=L
f(y)
y'5,x
implies the relation
f(x) =
L g(y)J.L(Y, x) y<x
and vice versa. 39*. (Continuation). Let P be the set of the subsets of a finite set S with respect to the order relation~ (is included, is a subset of). Prove that the Mobius function of P is given by
J.L(X, Y)
= (1)N(Y)N(X)'
X~ Y,
where N(X) and N(Y) are the numbers of elements of S that belong to X and Y, respectively.
168
THE PRINCIPLE OF INCLUSION AND EXCLUSION
40*. (Continuation). Consider n subsets A1, A2, ... , An of a finite set D and let S = { 1, 2, ... , n}. To each element w E D, there corresponds a set of indices J <; S such that w E A1 for every j E J. Let f(K) be the number of elements of n that belong only to the intersection of the sets A 1 , j ¢ K forK a subset of S. Then g(I) = LKci f(K) is the number of elements of n that belong (not necessarily exclusively) to the intersection of the sets Aj, j ¢ I. Prove that
f(K) =
L (1)N(K)N(I)g(I), JCK
where N(K) and N(I) are the numbers of elements of S that belong to K and I, respectively. Deduce the number Nn,o of elements of D that do not belong to any of the n sets.
ChapterS PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
5.1
INTRODUCTION
One of the classical combinatorial problems in the theory of probability is the famous problem of coincidences {probleme des recontres), which constitutes in the computation of the ways of putting n cards, numbered from 1 ton, in a series so that, in a given number of cards their position coincides with their number. This problem was initially formulated in the particular case n = 13 by Montmort (16781719) and then in the general case by De Moivre (1667 1754), whose solution essentially constitutes an application of the inclusion and exclusion principle. Later, several reformulations and generalizations of this problem were followed. In this chapter targeting to the problem of coincidences, the number of permutations of the first n positive integers { 1, 2, ... , n} (a) with a given number of fixed points (with respect to their natural order ( 1, 2, ... , n)) and (b) of a given rank are computed by applying the inclusion and exclusion principle. In the same way, the computation of the permutations of the first n positive integers with a given number of pairs of successive points is carried out.
5.2
PERMUTATIONS WITH FIXED POINTS
DEFINITIONS.l Let (j 1 , h, ... , in) beapermutationoftheset {1, 2, ... , n}. The point ir is called fixed point of this permutation if ir = r, r = 1, 2, ... , n.
170
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
Note that, according to this definition, the problem of coincidences constitutes in the computation of the permutations of the set {1, 2, ... , n}, with a given number of fixed points. The permutations without a fixed point are particularly interesting and deserve special attention. A particular name is given to such a permutation. DEFINITION S.2 A permutation (j1, h, ... , ]n) of the set { 1, 2, ... , n} is called a derangement if it does not have a fixed point, that is, if Jr =j:. r for all r=1,2, ... ,n.
These notions are illustrated in the following simple example. ExampleS./ Consider the permutations (j 1 , )2, }3, j 4 ) of the set { 1, 2, 3, 4}. Clearly, these 24 permutations can be classified according to the fixed points they have as follows. (a) The derangements are the following 9: (2, 1, 4, 3), (2, 4, 1, 3), (2, 3, 4, 1), (3, 1, 4, 2), (3, 4, 1, 2), (3,4,2,1), (4,1,2,3), (4,3,1,2), (4,3,2,1).
(b) The permutations with one fixed point are the following 8: (1,3,4,2), (1,4,2,3), (3,2,4,1), (4,2,1,3), (2,4,3,1), (4,1,3,2), (3,1,2,4), (2,3,1,4).
(c) The permutations with two fixed points are the following 6: (1,2,4,3), (1,4,3,2), (1,3,2,4), (4,2,3,1), (3,2,1,4), (2,1,3,4).
(d) There are no permutations with three fixed points, while
(1, 2, 3, 4) is the only permutation with four fixed points.
0
The following two theorems are concerned with the number of derangements. THEOREMS./ The number Dn of derangements of the set {1, 2, ... , n} equals n
(1)k
Dn=n!L~
(5.1)
k=O
and satisfies the recurrence relation
Dn=nDnI+(l)n, n=1,2, ... , Do=l.
(5.2)
5.2. PERMUTATIONS WITH FIXED POINTS
171
PROOF Consider the set {} of the permutations (j 1, )2, ... , Jn) of the set { 1, 2, ... , n} and let Ar be the subset of permutations (j 1, )2, ... , Jn) for which the point Jr is a fixed point, r = 1, 2, ... , n. Then, the number Dn of derangements of the set { 1, 2, ... , n} is given by
Clearly, for any selection of k indices {i 1, i 2, ... , ik} out of the n indices { 1, 2,
... ,n}, vk
= N(Ai,Ai, · · · Aik) = (n k)!,
k
= 1, 2, ...
,n,
which is the number of permutations of the n  k nonfixed points. This expression implies the exchangeability of the sets A 1, A2, ... , An and so, according to Corollary 4.3,
Using this expression, recurrence relation (5.2) is readily deduced: n1 Dn = n(n 1)!
L
(k~)
k
+ (1t = nDn1 + (1)n.
k=O
As regards the initial value of this recurrence relation, it is clear that D 1 which is equivalent to D 0 = 1. I
0,
THEOREM 5.2 Euler's recurrence relation The number Dn of derangements of the set { 1, 2, ... , n} satisfies the recurrence relation Dn
= (n
1){Dn1 + Dn2}, n
= 2, 3, ...
, Do
= 1,
D1
= 0.
(5.3)
PROOF Note that the last position of a derangement (j 1, )2, ... , Jn) of the set {1, 2, ... , n} can be occupied by any of the n  1 numbers {1, 2, ... , n  1}. Assume that Jn = k, k =/: n. Then, the derangements (j 1 , )2, ... , Jnl) of the set { 1, 2, ... , k  1, k + 1, ... , n }, can be distinguished according to whether n does or does not occupy the kth position. If ik = n, then the number of these derangements is equal to the number Dn_ 2 of the derangements (j1,)2, ... ,Jk1,ik+1, ... ,in1) of then 2 numbers {1, 2, ... , k 1, k + 1, ... , n  1}. If ik =1 n, then the number of these derangements equals the number of permutations (ii, h, . .. , in I) of the n  1 numbers { 1, 2, ... , k1,k+1, ... ,n},withjr =f:.r,r= 1,2, ... ,k1,k+1, ... ,n1,jk =f:.n.
172
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
The latter number equals the number Dn1 of the derangements Ut, j2, ... , in1) of then 1 numbers { 1, 2, ... , n 1}. Consequently
Dn = (n 1){Dn1
+ Dn2},
n = 2,3, ... , Do= 1, Dt = 0,
I
which is the required recurrence relation.
REMARK 5.1 The similarity of recurrence relations (5.2) and (5.3) to the corresponding recurrence relations for the factorials,
n! = n(n 1)!, n = 1, 2, ... , 0! = 1, n! = (n 1){(n 1)! + (n 2)!}, n = 2, 3, ... , 0! = 1, 1! = 1, led Whitworth to call the number Dn nsubfactorial. Using recurrence relation (5.2) or (5.3), a table of subfactorials can be constructed. Table 5.1 gives the I numbers Dn for n = 0, 1, ... , 10.
Table 5.1 0
Subfactorials Dn 1 0
2
3 2
4 0
5 44
6 265
7
8
9
10
1854
148331
133496
1334961
The next theorem is concerned with the number of permutations with a given number of fixed points.
THEOREM5.3 The number Dn,k of pennutations of the set {1, 2, ... , n} with k fixed points equals Dn,k
=
n! k!
nk
L
· (1)3
.!J
j=O
=
(n) k
(5.4)
Dnk·
PROOF Consider the set D of the permutations (it, i2, ... , in) of the set { 1, 2, ... , n} and let Ar be the subset of permutations Ut ,i2, . . . , j n) for which the point ir is a fixed point, r = 1, 2, ... , n. Then, the number Dn,k equals the number of elements of the set D that are contained in k among the n exchangeable sets A 1 , A 2 , ... , An. This number, according to Corollary 4.4 and since Vr = N(A; 1 A; 2 • • • Ai,) = (n r)!, is given by
n nk . ( nDn,k= ( k ) ~(1)1 . J=O
J
k) (nkj)!=k!L::.,. n! nk ( 1 ) · 1
J=O
J
5.2. PERMUTATIONS WITH FIXED POINTS
Further Dn,k =
(~) (n k)!
173
I: (.? J
j=O
and so, using (5.1), the second part of (5.4) is deduced.
I
Example 5.2 The classical problem of coincidences Consider an urn containing n balls numbered from 1 to n and assume that all the balls are successively drawn one after the other without replacement. The draw of the jth ball at the jth trial, j = 1, 2, ... , n, is called a coincidence. Calculate the number of series of drawings of the n balls that result in k coincidences. Clearly, to each series of drawings of then balls, there corresponds a permutation (j 1 , }2, ... , in) of the set {1, 2, ... , n} and to each coincidence there corresponds a fixed point of the permutation. Therefore the number of series of drawings of the n balls that result in k coincidences equals
n! nk (1)i
Dn,k =
kr L
. j=O
.J· 1
,
D
the number of permutations of the set { 1, 2, ... , n} with k fixed points.
REMARK 5.2 The number Dn,k is known in the bibliography as the coincidence number , a name due to the preceding example. Using the relation Dn,k = (~)Dnk. Table 5.1 of subfactorials and Table 2.1 of binomial coefficients, a table of the numbers Dn k can be constructed. Table 5.2 gives the I coincidence numbers Dn,k fork= 1', 2, ... , n, n = 1, 2, ... , 9.
Table 5.2 Coincidence Numbers Dn,k
k
1
2
I 0 3 8 45 264 I855 14832 133497
0 6 20 I35 924 7420 66744
3
4
5
6
I 0 I5 70 630 5544
I 0 21 112 1134
0 28 168
8
9
1 0 1 36 0
1
7
n I 2 3 4 5 6 7 8 9
I
1 0 lO
40 315 2464 22260
I
174
5.3
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
RANKS OF PERMUTATIONS
DEFINITION 5.3
If
ir f:. r,
Let (ii, h, ... , in) be a permutation ofthe set { 1, 2, ... , n }.
r
= 1, 2, ...
, k 1,
ik = k,
2~ k
~
n,
the permutation is said to be of rank k. In panicular, if j 1 = 1. the permutation is said to be of rank 1, while if ir f:. r, r = 1, 2, ... , n, it is said, by convention, to be of rank n + 1.
Note that several problems in combinatorics and in probability theory, known as problems of duration of trials (drawings), reduce to the enumeration of the permutations of the set { 1, 2, ... , n} of a given rank. Example5.3 The 24permutations (ii, h, iJ, j 4 ) of the set {1, 2, 3, 4} can be classified according to their rank as follows. (a) The permutations of rank 1 are the following 6: (1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2). (b) The permutations of rank 2 are the following 4: (3,2,1,4), (3,2,4,1), (4,2,1,3), (4,2,3,1). (c) The permutations of rank 3 are the following 3: (2,1,3,4), (2,4,3,1), (4,1,3,2). (d) The permutations of rank 4 are the following 2:
(2, 3, 1, 4), (3, 1, 2, 4). (e) The permutations of rank 5 are the following 9: (2, 1, 4, 3), (2, 4, 1, 3), (2, 3, 4, 1), (3, 1, 4, 2), (3, 4, 1, 2), (3,4,2,1), (4,1,2,3), (4,3,1,2), (4,3,2,1). Note that the permutations of rank 5 are, by convention, the derangements of the set{1,2,3,4}. D The next theorem is concerned with the number of permutations of a given rank.
5.3. RANKS OF PERMUTATIONS
175
THEOREM5.4 The number Rn,k of permutations of the set { 1, 2, ... , n} of rank k equals k1
Rn,k = ~( 1)
8
(k ~ 1) (n s 1)!,
(5.5)
fork = 1, 2, ... , nand Rn,n+l = Dn. Further, it satisfies the recurrence relation
Rn,k = Rn,k1  Rnl,k1, k = 2, 3, ... , n, n = 2, 3, ... , with Rn,l = (n 1)!, Rn,k = 0, k
(5.6)
> n + 1.
PROOF Consider the set D of the permutations (j1,h, ... ,Jn) of the set { 1, 2, ... , n} and let A,. be the subset of permutations (j 1, )2, ... , Jn) for which the pointj,. is a fixed point, r = 1, 2, ... , n. The number Rn,k equals the number of elements of D of rank k with respect to the n ordered and exchangeable sets A 1, A2, ... , An. Since V 8 = N(Ai,Ai 2 • • • Ai.) = (n s)!, s = 1, 2, ... , n, this number, according to Corollary 4.6, is k1
Rn,k = ~(1)
8
(k ~ 1) (n s 1)!,
fork= 1, 2, ... , n. Clearly Rn,n+l = N(A~ A~··· A~) = Dn. Further, expression (5.5), on using Pascal's triangle, can be written as 1
Rn,k = (n 1)! + ~(1) k
8 {
(k ~ 2) + (k =12)} (n s 1)! 8
and so
(k ~ 2) (ns1)!+~(1) kl (k =2) (ns1)! 1 (k s 2) (ns1)!~(1)' (k i 2) (ni2)!, =~(1Y k2
Rn,k=~(1) 8 k2
8
8
k2
.
which, by virtue of (5.5), implies recurrence relation (5.6).
I
Example 5.4 Duration of a series of drawings Consider a lotteryurn containing n balls numbered from 1 to n and assume that balls are drawn one after the other without replacement. This series of drawings is terminated with occurrence of the first coincidence. Calculate the number of the series of drawings with a given duration of k drawings.
176
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
The required number, according to the Definition 3.1 of the rank of a permutation, is equal to the number of permutations of { 1, 2, ... , n} of rank k which, according to Theorem 3.1, is kl
Rn,k = 2)1)
5
(k 1) (n s 1)!,
k
:
= 1,2, ...
,n.
0
s=O
REMARK 5.3 The number Rn,k has been studied by M. Frechet ( 1943); J. Riordan (1958) called it rank number. Using recurrence relation (5.6), Table 5.3 of the numbers Rn,k. fork = 1, 2, ... , n + 1, n = 1, 2, ... , 8, is constructed. I
Table 5.3 Rank Numbers Rn,k
k
I
2
3
4
5
6
7
8
9
265 265 2119
1854 1854
14833
n I
2 3 4 5 6 7 8
I 0 1 0 2 I 6 4 24 18 120 96 720 600 5040 4320
1 I
3 14 78 504 3720
2 2 11 64
426 3216
9 9 53 362 2790
44 44
309 2428
5.4 PERMUTATIONS WITH SUCCESSIONS DEFINITION5.4 If
Let (j 1 , h, ... , iJ) beapermutationoftheset{1, 2, ... , n}.
ir+l
=
ir + 1,
1~ r
~
n  1,
then the pair Ur, ir+ 1 ) is called a succession of the permutation. Example5.5 Consider the permutations (]I, h, h, j 4 ) of the set { 1, 2, 3, 4}. These 24 permutations can be classified according to their successions as follows.
5.4. PERMUTATIONS WITH SUCCESSIONS
177
(a) The permutations with the no succession are the following 11: (1, 3, 2, 4), (1, 4, 3, 2), (2, 1, 4, 3), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 4, 2), (3,2,1,4), (3,2,4,1), (4,1,3,2), (4,2,1,3), (4,3,2,1).
(b) The permutations with one succession are the following 9: (1,2,4,3), (3,1,2,4), (4,3,1,2), (2,3,4,1), (4,2,3,1), (1, 4, 2, 3), (3, 4, 2, 1), (1, 3, 4, 2), (2, 1, 3, 4).
(c) The permutations with two successions are the following 3: (2,3,4,1), (3,4,1,2), (4,1,2,3).
(d) The permutation (1,2,3,4)
is the only permutation with three successions.
0
The next theorem is concerned with the number of permutations without a succession. THEOREM5.5 The number Sn of permutations of the set { 1, 2, ... , n} without a succession is equal to
nt Sn = (n  1)! _L) 1) k n
~k
(5.7)
k=O
and satisfies the recurrence relation
nSn
= (n 2 1)Snt (1)n, n = 2,3, ...
, S1
= 1.
(5.8)
PROOF Consider the set fl of the permutations (j 1 , }2, . .. , Jn) of the set { 1, 2, ... , n} and let Ar be the set of these permutations for which the pair Ur, Jr+d is a succession, r = 1, 2, ... , n1. Then the number Sn of permutations of the set {1, 2, ... , n} without a succession is given by Sn = N(A~ A~··· A~_ 1 ). Further, the sets A1 , A 2 , ••• , An 1 are exchangeable and vo
= N(fl) = n!,
vk
= N(Ai,Ai
2 •
··AiJ
= (n
k)!, k
= 1,2, ...
,n 1.
Indeed, in the case where the set of indices {it, i 2 , ••• , ik}, with 1 ~ it < i 2 < · · · < ik ~ n, consists of consecutive integers, ir = i + r  1, r = 1, 2, ... , k, 1 :S i ~ n k, the k successions constitute a run p = (ji, ji+ 1 , ... , Ji+d· Thus,
178
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
the remaining n  k + 1 integers {h, h, ... , ii1, ii+k+1, ... , in}. along with the run p, constitute a set of n  k elements and the number N (Ai, Ai 2 • • • Aik) equals (n  k) !, the number of permutations of this set. If the set of indices I= {i 1,i 2, ... ,ik}, with 1 :::; i 1 < i 2 < · · · < ik :::; n, does not consist of consecutive integers, then it is partitioned in 8 subsets 11 , 12 , ... , Is. where I c. with N (Ic) = me. c = 1, 2, ... , 8, is either a onepoint set or a set of consecutive integers and m 1 + m 2 + · · · + m 8 = k. In this case, the k successions constitute 8 runs, P1, p2, ... , p 8 , of consecutive integers, which are formed by ( m1 + 1) + (m 2 + 1) + · · · + (ms + 1) = k + 8 integers. Thus the remaining n k 8 integers along with the 8 runs constitute a set of n  k elements and the number N (Ai, Ai 2 • • • Aik) is equal to (n  k) !, the number of permutations of this set. Therefore, according to Corollary 4.3, the number Sn of permutations of the set { 1, 2, ... , n} without a succession is
Further, multiplying both members of the identity
n(nk)
k!
=
(n+1)(nk1) 1 1 k! + k! + (k 1)!
by ( 1)k(n 1)! and summing fork = 0, 1, ... , n 1, we get the expression 1 2 nn k nn k 1 n(n1)!L(1)k~=(n 2 1)(n2)!L(1)k k! +(1)n 1 ,
k=O
k=O
which, by virtue of(5.7), implies (5.8).
I
REMARK 5.4 The number Sn can be expressed in terms of the number Dr. of rsubfactorial. Indeed, rewriting expression (5.7) as n1
Sn = n(n 1)!
L
(l)k
~ (n 1)!
k=O n
=
( 1) k
n1
L
k=1
(  1)k (k _ )!
1
n 1 (  1)J
n!L ~ +(n1)!L ,.,J k=O
j=O
and, using (5.1), we get the relation Sn = Dn + Dn1• n = 1, 2, ... , which, combined with the recurrence relation (5.3), yields for Sn the recurrence relation
Sn = (n 1)SnI + (n 2)Sn2, n = 3, 4, ... , S1 = S2 = 1.
(5.9)
Using either of the recurrence relations (5.8) or (5.9), a table of the numbers Sn can be constructed. Table 5.4 gives the numbers Sn for n = 1, 2, ... , 10. I
5.4. PERMUTATIONS WITH SUCCESSIONS
Table 5.4
179
The Numbers Sn 2
3 3
4 11
5 53
6 509
8 16687
7 2119
9 148329
10 1468457
The next theorem is concerned with the number of permutations with a given number of successions.
THEOREM5.6 The number Sn,k of permutations of the set { 1, 2, ... , n} with k successions is equal to
_ (n1)!n~l( )inkj _ (n1)s S n,kk! ~ 1 '! k nk· j=O
(5.10)
J
PROOF Consider the set fl of the permutations (jt, j 2 , ••• , in) of the set { 1, 2, ... , n} and let Ar be the set of these permutations for which the pair Ur, ir+I) is a succession, r = 1, 2, ... , n  1. Then the number Sn,k equals the number of elements of the set fl, which are contained ink among the n 1 sets A 1 , A2 , ... , AnI· As it was shown in the proof of Theorem 5.5, the sets At, A2, ... , AnI are exchangeable and
Vo
= N(fl) = n!,
Vr
= N(Ai, Ai2 · · · Ai.) = (n r)!,
r
= 1, 2, ...
, n 1.
Thus, according to Corollary 4.4, the number Sn,k is given by
Sn,k =
(n ~
1)
n~l (1)j (n; 1)(n k j)!
(n1)!n~l( 1 )inkj k!
~
j=O
'!
= (n1)s
k
J
with the last equality implied by virtue of (5.7).
.
nk,
I
Using (5.10), Table 5.4 of the numbers Sn and Table 2.1 of binomial coefficients, a table of succession numbers Sn,k can be constructed. Table 5.5 gives the numbers Sn,k for k = 1, 2, ... , n 1, n = 2, 3, ... , 10.
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
180
Table 5.5
k n 2 3 4 5 6 7 8 9 10
Succession Numbers Sn,k
1
2
3
1 2 I 9 3 18 44 265 110 1854 795 14833 6489 59332 133496 1334961 600732
4
1 4 I 30 5 220 45 1855 385 17304 3710 177996 38934
5
6
7
8 9
1 6 1 63 1 7 616 84 8 1 6678 924 108 9
1
Example 5.6 Caravan in the desert A caravan of n camels travels across a desert in a journey that lasts for several days. The cameldrivers find it boring to see the same camel in front of them every day. The calculation of the number of permutations of the camels so that the next day each camel follows a different camel is then of interest. For this purpose, let us consider the initial order of the camels and number them from I to n. Then the required number is equal to the number of permutations of the set { 1, 2, ... , n} that do not include any ofthe n 1 pairs: ( 1, 2), (2, 3), ... , (n 1, n). According to Theorem 5.5, this number equals
Sn
= (n
n1
(
1)! L:)1)k n ~!
k)l ·.
k=O
If, more generally, the interest is in the number of permutations of the n camels so that the next day each of k camels follows the same camel, then, according to Theorem 5.6, this number equals _
S . = (n n,k
1
nk1
1). "" ( )in kl ~ 1 .
j=O
_
_
k .1
.
J.
D
J.
5.5
CIRCULAR PERMUTATIONS WITH SUCCESSIONS Consider the positive integers 1, 2, ... , n displayed on a circle (h, h, ... ,
in), where i1
follows in· Thus a circularly ordered permutation, called a cir
5.5.
CIRCULAR PERMUTATIONS WITH SUCCESSIONS
181
cular permutation of the set {1, 2, ... , n }, is formed. Note that, to each circular permutation (it, i 2 , . . . , in), there correspond n linear permutations (il, i2, .. · ,in), (h, iJ, ... , in, i!), ... , (in,il, ... , ind and so the number of circular permutations of the set {1, 2, ... , n} is equal to n!/n = (n 1)!. In the case of the circular permutations of the set { 1, 2, ... , n}, in addition to the pairs (1, 2), (2, 3), ... , (n 1, n), the pair (n, 1) is also a succession. The next theorem is concerned with the number of circular permutations without a succession. THEOREM5.7 The numberCn of circularpennutations of the set {1, 2, ... , n} without a succession equals n1
(_
1)k Cn = n! ( ; (n k)k!
+ (1t.
(5.11)
PROOF Consider the set D of the circular permutations (i 1 , h, ... ,in) of the set { 1, 2, ... , n} and let Ar be the subset, of fl, of circular permutations in which the pair Ur, ir+I) constitutes a succession, r = 1, 2, ... , n 1. In addition let An be the subset, of D, of circular permutations in which the pair (in, it) constitutes a succession. Then the number Cn of circular permutations of the set {1, 2, ... , n} without a succession is given by Cn = N(A~ A~··· A~). Further, the sets At, A 2 , •.• , An are exchangeable and
vo Vk
= N(fl) = (n
1)!,
= N(A; 1 A; 2 ···A;k) = (n k 1)!, k = 1,2, ... ,n 1.
Indeed, as in the proof of Theorem 5.5, the cases in which the indices it, i 2 , •.• , ik are or are not consecutive integers are separately examined. In both cases, the number N(A; 1 A; 2 • • • A;k) is reduced to the number (n k 1)!, of circular permutation of n  k distinct elements. Also
which is the only circular permutation, ( 1, 2, ... , n), belonging in all n sets. Thus, by virtue of Corollary 4.3, the number Cn is given by Cn
= ~(1)k (~) (n k 1)! + (1)n n1
= n!
(_
1)k
2: (n k)k! + (1t,
k=O
which is the required formula.
I
182
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
REMARK 5.5
The number Cn can be expressed in terms of the number Dr, of rsubfactorial. Indeed, rewriting expression (5 .11) as
n1 _ )'"( )dnk)+k Cn ( n 1. 1 (n _ k)k!
6
+ ( 1)n
_ n1 ( 1)k n1 ( 1)k n  (n 1)! {; ~ + (n 1)! {; (n k)(k 1)! + (1)
=(n1)!~(1)k {(n1)!~ (1)i .. +(1)n1} ~ k! ~(n1J)J! k=O
J=O
and, using (5.1), we get
Cn+CnI=Dn1, n=3,4, ....
(5.12)
Writing this relation in the form
and summing for k = 3, 4, ... , n, it follows that
n L(1)nkck k=3
n
n
+ L(1)nkck1 = L(1)nkDk1· k=3
The lefthand side of this relation is equal to Cn
k=3
+ (1 )n 3 C 2 and, since C2
= 0,
n
Cn=L(1)nkDk1, n=3,4, .... k=3 Furthermore, note that (5.12), by using recurrence relations (5.2) and (5.3), yields the recurrence relations:
Cn = (n 2)Cn1 + (n 1)Cn2 + (1)n 1, n = 4, 5, ... , with
(5.13)
c2 = 0, c3 = 1, and Cn = (n 3)Cn1 + 2(n 2)Cn2 + (n 2)Cn3• n = 5, 6, ... , (5.14)
with C2 = 0, C3 = 1, C4 = 1. Using either of these recurrence relations, a table of the numbers Cn can be constructed. Table 5.6 gives the numbers Cn for n = 2, 3, ... , 12. I
5.5.
183
CIRCULAR PERMUTATIONS WITH SUCCESSIONS
Table 5.6 2 0
3
4 1
The Numbers Cn 5 8
6 36
7 229
8 1625
9 13208
10 120288
11 1214673
12 13469897
THEOREM5.8 The number Cn,k of circular pennutations of the set { 1, 2, ... , n} with k successions equals
Cn,k
n! {nk1 j;
= k!
(1)i (n _ k _ j)j!
+
(l)nk} (n _ k)!
=
(n)
k Cnk·
(5.15)
PROOF Consider the set fl of the circular permutations of the set { 1, 2, ... , n} and the n exchangeable sets A 1 , A 2 , ... , An, subsets of fl, defined in the proof of Theorem 5.7. Then, the number Cn,k equals the number of elements of n that are contained in k among the n subsets. Since v0 Vr
= N(Ai 1 A; 2
· ·
= N(fl) = (n
1)!,
·A;J = (n r 1)!, k = 1,2, ... ,n 1,
Vn = N(AI A2 ···An)= 1, it follows from Corollary 4.4 that
c•.• ~
mr~\1)1
("
~ k) (n k j 1)! + (!)··}
n! {nk1 (1)i (1)nk} (n) {; (nj)j! + (n k)! =
= k!
k
with the last equality implied by virtue of (5.11 ).
k Cnk,
I
Example 5. 7 A roundtable conference Consider n representatives participating in a roundtable conference scheduled to continue for several days. It is decided that every day they sit around the table so that each has to his right a different person. At their first seating, let us number the n representatives from I ton in the ordinary direction (counterclockwise), starting from any one of them. The problem of their second seating is reduced to the
184
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
calculation of the circular permutations of the set { 1, 2, ... , n} that do not include any of then pairs: (1, 2), (2, 3), ... , (n 1, n), (n, 1). According to Theorem 5.7, this number is n1
Cn = n!
L
(  1)k
(n k)k! + (1)n.
k=O
More generally, if the interest is on the number of ways of creating a second seating of the n representatives so that k of them have to their right the same person, then this number, according to Theorem 5.8, equals
n!{nk1 ( 1)i Cn,k = k!
~
(l)nk}
(n k j)j! + (n k)!
D ·
5.6 BIBLIOGRAPHIC NOTES The famous problem of coincidences (matches, recontres) was initially treated in the particular case of 13 cards by P. R. Montmort (1708) and J. Bernoulli (1714). Abraham De Moivre (1718) examined the general case of n cards by using the inclusion and exclusion principle. L. Euler (1751) derived a recurrence relation for the number of derangements. W. A. Whitworth (1867), in his book on the combinatorics of the games of chance, studied the matching problem and contributed to its popularization. P. MacMahon (1902, 1915, 1916) provided a generating function of the number of arrangements of the n cards with k matches. The number of permutations of a given rank has been examined by M. Frechet (1945) and J. Riordan (1958). Generalizations and applications of the matching problem have been discussed by several authors; the interested reader is referred to the bibliography furnished in the review article of D. E. Barton (1958). The two sections on the enumeration of the linear and circular permutations by successions were inspired by the problem of the caravan in the desert posed and solved by N.Y. Vilenkin (1971).
5. 7
EXERCISES 1. Let Dn be the nsubfactorial. Show that
IDnn!e 1 1<1/(n+1), n=1,2, ...
5. 7. EXERCISES
185
and thus conclude, for large n, the asymptotic expression
2. Let Dn,k be the number of permutations ofthe set {1, 2, ... , n} with k fixed points. Show that (a)
Dn,k = nDn1,k
+ (l)nk(~)
and (b)
Dn k = Dn1 k1 1
)
+ (n
1){Dn1 k + Dn2 k Dn2 k1}· )
I
I
3. Let Rn,k be the rank number and Dr rsubfactorial. Show that
nk
Rn,k
=L
(n
~ k)vk+r1,
= 1,2, ... , n.
k
r=O 4. Let Rn,k be the rank number and
Rn
n+1
n+1
k=1
k=1
= L Rn,k, En=
L
kRn,k·
Using the recurrence relation
Rn,k
= Rn,k1 Rn1,k1,
k
= 1, 2, ... , n,
with Rn, 1 = (n 1)!, Rn,n = Dn1, Rn,n+1 = Dn, show that
Rn1 = (n 1)!, En! = n! Dn, n = 1, 2, .... 5. Show that the rank number Rn,k satisfies the recurrence relation
Rn,k = (n k)Rn1,k
+ (k
1)Rn1,k1,
fork= 2, 3, ... , n 1, n = 3, 4, .... 6. Let En,r be the number of permutations of the set {1, 2, ... , n} without a fixed point in r predetermined positions. Show that
En,r
=
t( j=O
1) 1
(~) (n j)!. J
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS
186
Derive the recurrence relations
= En,rl  Enl,r1, r = 1, 2, ... , n, En,o = n!, = (n r)Enl,r + rEnl,r1, r = 1, 2, ... ,n 1,
En,r En,T
En,n n
= Dn,
= 2, 3, ....
and, if Rn,k is the rank number, conclude that
7. (Continuation). Let En,r,k be the number of permutations of the set { 1, 2, ... , n} with k fixed points in r predetermined positions. Show that
and conclude that En,r,l
= rRn,r·
8. (Continuation). Show that En,r,k
= (n k)En1,r,k + (k + 1)Enl,k+l·
9. (Continuation). Show that
where Dn is the nsubfactorial and, using the recurrence relation
conclude that En,r,k = (n r)En1,r,k
+ rEn1,r1,k·
10. Double coincidences. Let Qn be the number of ways of placing n letters and n invoices in n envelopes, one letter and one invoice in each, so that no envelope contains its corresponding letter and invoice. Show that
Qn
=
1
~( 1 )n (n k)!
n.~ k=O
k!
.
11. (Continuation). Show that Qn
= n 2 Qnl + n(n 1)Qn2 + (l)n, n = 2, 3, ... , Qo = 1,
Q1
= 0.
5. 7. EXERCISES
187
12. (Continuation). Let Qn,k be the number of ways of placing n letters and n invoices in n envelopes, one letter and one invoice in each, so that each of k envelopes contains its corresponding letter and invoice. Show that
)J(nkj)! Q n,k = n!~(k! L.....i 1 ., . J
j=O
13. Multiple coincidences. Consider r urns, each containing n balls numbered from 1 to n. Assume that a ball is drawn from each urn and, without replacement, this multiple drawing is repeated until all the balls are drawn. The drawing of the jth ball from each of the urns at the jth rtuple trial constitutes an rtuple coincidence. (a) Let Bn,r be the number of rtuple drawings without an rtuple coincidence. Show that
Bn,r =
i)
l)nj
(~) (jW. J
j=O
(b) If Bn,r,k is the number of rtuple drawings with k rtuple coincidences, show that
Bn,r,k =
(~)Bnk,r·
14. (Continuation). Show that the number Bn,r satisfies the relations
Bn,r
( = n n1 ~(1)ns 1 n~
_ ~ Bn,r~( 1) ns1 (n S
1) (sJt(s+ l)r1) (s.) (s + 1) 1
r
r
1
+ (1)n,
 Bn1,r
and proving that
ns1
L
j=O
. ( ns1 ) (1)1 . j!S(k,j;nj)=(s+1)k, J
where
S(k,j;u) =
~ t(1)i 1 (~)(i+u)k,
J. i=O
~
conclude the recurrence relations T
Bn,r = L(n)jS(r 1,j 1; n j j=1
+ 1)Bnj,r + (1)n,
188
PERMUTATIONS WITH FIXED POINTS AND SUCCESSIONS n+1
Bn,r = (nr 1)Bnl,r + L(n 1)jtS(r,j 1; n j
+ 1)Bnj,r·
j=2
The number S(k,j; u) is called noncentral Stirling number of the second kind. 15. A generalization of the problem of coincidences. Consider an urn containing s like series of balls with n balls, numbered from 1 to n, in each series. Assume that all the sn balls are drawn one after the other without replacement. (a) Let Gn,s be the number of drawings without a coincidence. Show that Gn,s
=
f)
1)j
(~) sj (sn j)!. J
j=O
(b) If Gn,k,s is the number of drawings with k coincidences, show that
16. (Continuation). Assume that the successive drawing of balls is terminated with the appearance of the first coincidence. If H n,k,s is the number of drawings with a given duration of k trials, show that Hn,k,s =
k1 . (k 1) . L( 1)1 . s1+
1
(sn j  1)!.
J
j=O
17. Let Sn,k be the number of permutations of the set {1, 2, ... , n} with k successions. Show that Sn+l,k = (k
for k = 1, 2, ... , n, n k ?:n.
+ 1)Sn,k+l + (n
= 2, 3, ... ,
with St,o
k)Sn,k
=
+ Sn,k1,
1, Sn,n1
=
1, Sn,k
= 0,
18. Let Cn,k be the number of circular permutations of the set {1, 2, ... , n} with k successions. Show that Cn,k
+ Cnl,k
 Cnl,k1
= Dnl,k,
where Dn,k is the coincidence number, and conclude that Cn,k = (n 2)Cn1,k
+ Cnl,k1 + (n
+(n  1)Cn2,kl
+
1)Cn2,k 1 (1)nk 1 ( n ~ ).
5. 7. EXERCISES
189
19. Permutations with a given number of transpositions. A permutation can always be represented as a product of circular permutations. A circular permutation of length 2 is called a transposition. Let Tn,k be the number of permutations of the set {1, 2, ... , n} which can be written as a product of k transpositions. Show that
Tn,k
=
n!(1)k
mk
k! 2 k
L
j=O
(1)i
uJ•
m
J
= [n/2].
20. (Continuation). Show that Tzr,k = 2rTzrl,k
+ (1 )
r
(2r)! k!(r _ k)! 2 k,
and Tzr+I,k
= (2r + 1)Tzr,k·
Chapter6 GENERATING FUNCTIONS
6.1
INTRODUCTION
The generating functions constitute an important means for a unified treatment of combinatorial and probabilistic problems. P. S. Laplace, their inventor, has first introduced them in the form of power series. Later, generating functions of a more general than the power series form have been used. Moreover, for their combinatorial uses, they are to be regarded, following E. T. Bell, as tools in the study of an algebra of sequences; thus, despite all appearances they belong to algebra and not to analysis. A generating function expanded generates a sequence of numbers ak,
k
= 0, 1, ... , the coefficients of the expansion, which in combinatorics ex
press the number of elements of a certain finite set. In this respect its study is a natural complement to the study of these numbers. In the next section, a justification of the use of generating functions in the derivation of the number of elements of certain finite sets is provided by a brief presentation of two simple combinatorial problems preceding the formal definitions. After introducing the basic forms of generating functions, several examples illustrating their use are discussed. Also, the general form of a generating function is introduced and exemplified. In order to facilitate the derivation and expansion of generating functions, the Maclaurin, Newton and Lagrange series are briefly discussed. A general method for constructing generating functions for combinations and permutations is presented and illustrated in a separate section. The factorial moments of a sequence ak, k = 0, 1, ... , of combinatorial numbers (or probabilities or masses) are closely connected with the general term of the inclusion and exclusion formula, which is one of the basic counting expressions. The inclusion of a section on the moment generating functions is thus justified. In ending this chapter, bivariate and multivariate generating functions are examined.
GENERATING FUNCTIONS
192
6.2 6.2.1
UNIVARIATE GENERATING FUNCTIONS Definitions and basic properties
Before proceeding to the formal definition and study of generating functions, let us examine two simple combinatorial problems that suggest the use of a certain form of generating functions. The first problem is concerned with the derivation of the number of kcombinations of n. Let us consider first three elements labelled x 1 , x 2 and x 3 and form the algebraic product
which, after executing the multiplication and arranging the terms in ascending powers oft, is equal to
Introducing the elementary symmetric functions with respect to the three variables x 1 , x 2 and x3,
it follows that
3
3
i=!
k=O
IT (1 + x;t) = L ak(x
1,
x 2 , x 3 )tk.
Note that the function ak = ak(x 1 , x 2 , x 3 ), k = 1, 2, 3, which is the coefficient of tk, contains one term for each kcombination of the three elements. Hence, the number C(3, k) of the kcombinations of 3 is given by the value of ak = ak(x 1 , x2, x 3 ) at the point (1, 1, 1), that is, C(3, k) =
= 1, i
and so, setting x;
ak(1,
1, 1),
0, 1, 2, 3,
= 1, 2, 3, we conclude that 3
(1
k =
+ t) 3 = L
3
ak(1,
k=O
1, 1)tk
=L
C(3, k)tk.
k=O
Similarly, in the case of n distinct elements labelled x 1 , x 2 , ... tain n
IT (1 + i=l
, Xn,
we ob
n
x;t) =
L k=O
ak(X!, x2, ... , Xn)tk,
(6.1)
6.2. UNIVARIATE GENERATING FUNCTIONS
193
with ak = ak(x 1, x 2 , •.. , xn), k = 0, 1, ... , n, the elementary symmetric functions, with respect to then variables x 1,x 2 , ... ,xn, which are defined by the sum ak=ak(XI,X2, ... ,xn)=LXi 1 Xi 2 ···Xik, k=O,l, ... ,n,
(6.2)
where the summation is extended over all kelement subsets {i 1,i 2 , ... ,ik} of the set of n indices {1, 2, ... , n }. It is clear that the symmetric function ak ( x 1, x 2 , ••• , xn) contains one term for each kcombination of n. Hence the number C(n, k) of the kcombinations of n is given by the value of ak(x1,x 2 , ... ,xn) at the point (1,1, ... ,1), C(n,k)=ak(1,1, ... ,1), k=0,1, ... ,n. Consequently, setting in (6.1) Xi= 1, i = 1, 2, ... , n, we conclude that n
n
k=O
k=O
(6.3)
The function A(t) = (1 + t)n, which generates, in the sense of (6.3), the sequence of the numbers C(n, k) = ak(1, 1, ... , 1), k = 0, 1, ... , n, may be called enumerating generating function of combinations of n distinct elements. Let us now examine, from the same viewpoint, the problem of the derivation of the number of the kpermutations of n. Since the products x 1x 2 and x 2 x 1 are not distinguished in the usual algebraic product, it is not possible to define a function analogous to (6.2) containing one term for each kpermutation of n and, consequently, it is impossible to define a function similar to (6.1). Nevertheless, an enumerating generating function of the number P(n, k) of the kpermutations of n is easy to find. Indeed, from (6.3), on using the relation C(n, k) = P(n, k)/k!, it follows that (1
+ t)n
n tk = LP(n, k) k!. k=O
(6.4)
The function E(t) = (1 + t)n, which generates, in the sense of (6.4), the sequence of the number P(n, k), k = 0, 1, ... , n, may be called enumerating generating function of permutations of n distinct elements. The preceding discussion of the enumerating generating functions for combinations and permutations suggests the study of the following generating functions.
DEFINITION 6.1 sum
Let ak, k
=
0, 1, ... , be a sequence of real numbers. The 00
A(t)
=
Laktk k=O
(6.5)
GENERATING FUNCTIONS
194
is called (ordinary) generating function, while the sum CXl tk E(t) = Lak k!
(6.6)
k=O
is called exponential generating function of the sequence ako k = 0, 1, ....
The existence of the generating functions A(t) and E(t) requires the absolute convergence of the corresponding series for t 0 < t < t 1 , with t 0 and t 1 real numbers. The algebra of the corresponding sequences of the (ordinary) generating functions is known as Cauchy algebra and of the exponential generating functions is known as Blissard or symbolic calculus. It is not necessary to burden this presentation of the generating functions with formal definitions and study of the relative algebras. Note only that the equality of two generating functions entails the equality of the coefficients of the powers oft of the same degree. The operations of summation and multiplication, as well as differentiation and integration with respect to t, of the generating functions can be used to derive relations between the corresponding coefficients. In order to clarify this, let us consider the generating functions CXl
CXl
CXl
A(t)=Laktk, B(t)=Lbktk, C(t)=Lcktk. k=O
k=O
k=O
A direct consequence of the equality of generating functions in the following pairs of relations is that the one relation entails the other: (a) Sum C(t)
= A(t) + B(t)
iff Ck
= ak + bk,
k
= 0, 1,....
(6.7)
k
= 0, 1,....
(6.8)
(b) Product k
C(t)
= A(t)B(t)
iff Ck
= Laibkj, j=O
Note that relations (6. 7) and (6.8) define the sum and the product of sequences in Cauchy algebra. The sequence ck, k = 0, 1, ... , defined by (6.8) is called convolution of the sequences ak, k = 0, 1, ... , and bkl k = 0, 1,.. . . Several applications of the product of generating functions (6.8) may be obtained by specifying one or two of the generating functions. Such an application is given in the following example. Example 6.1 Sum of terms of an arithmetic progression The sum Sk = 1 + 2 + · · · + k, k = 1, 2, ...
6.2. UNIVARIATE GENERATING FUNCTIONS
195
may be evaluated by using (6.8) as follows. Note first that this sum is of the form
where Co
= 0,
= 1 + 2 + · · · + k, k = 1, 2, ... , = k, bk = 1, k = 0, 1, ....
Ck := Sk ak
Thus, according to (6.8),
= A(t)~(t)
C(t)
=
and the evaluation of the sum Sk Ck of an arithmetic progression reduces to the derivation of the generating functions A(t) and ~(t). By the geometric series, it follows that
~(t)
00
00
k=O
k=O
= 2:)ktk = 2:: tk = (1 t) 1, It I < 1.
In order to evaluate the generating function 00
00
k=O
k=1
L aktk = L ktk,
A(t) = we differentiate the relation 00
L:tk k=O
= (1 t)1
and get 00
2:: ktk1 = (1 t)2. k=1
Therefore, 00
A(t) = t
L ktk 1 = t(1 t)
2
k=1
and
C(t)
= A(t)~(t) = t(1 t) 3.
The generating function C(t), on using Newton's negative binomial formula (see Theorem 3.4), with n = 3, is expanded into powers oft as
GENERATING FUNCTIONS
196
Thus
k(k+ I) 2
k = 1,2, ....
0
Example 6.2 The Catalan numbers Consider a product of n numbers in a specified order, y = x 1 x 2 · · · Xn, and let Cn be the number of ways of evaluating this product by successive multiplication of exactly two numbers at each step and without altering the initial order. Note that c2 = 1 : X}X2, c3 = 2 : X} (x2x3), (xlx2)XJ, c4 = 5 : x1 (x2(x3x4)), x1 ((x2x3)x4), (x1x2)(x3x4), (x1 (x2x3))x4, ((x1x2)x3)X4. In order to determine a recurrence relation for the numbers Cn, n = 2, 3, ... , note that the last multiplication, with which the evaluation of the product is completed, is y = YkYnko where Yk is the result of the product X1X2 · · · Xk and Ynk is the result of the product Xk+l xk+ 2 · · · Xn, with k = 1, 2, ... , n  1. The number of ways of evaluating the product Yk = x1 x 2 · · · xk is C k. while the number of ways of evaluating the product Ynk = xk+1Xk+2 · · · Xn is Cnko k = 1, 2, ... , n 1. According to the multiplication principle, the number of ways of evaluating the products Yk and Ynk is CkCnk. k = 1, 2, ... , n  1. Thus, by the addition principle, the total number Cn of ways of evaluating the product y is equal to Cn
= C1Cn1
+ C2Cn2 + · · · + CnlCl,
n = 2, 3, ....
C£, which, by virtue of C2 = 1,
In the particular case of n = 2, it holds C2 = entails cl = 1 and so the initial conditions are
Using the recurrence relation and the initial conditions, we construct Table 6.1 of Catalan numbers Cn for n = 1, 2, ... , 12.
Table 6.1 2
Catalan Numbers 3
4
2 5
5 14
6 42
7 132
8 429
9 1430
10 4862
II 16796
12 58786
In order to determine the generating function 00
C(t) =
L
Cntn,
n=l note that the righthand side of the recurrence relation is the convolution of the sequence an = Cn+l• n = 0, 1, ... with itself bn = Cn+h n = 0, 1, .... Thus,
6.2.
UNIVARIATE GENERATING FUNCTIONS
197
multiplying the recurrence relation by tn and summing for n = 2, 3, ... , by virtue of (6.8), we get
and so
C(t) t = [C(t)) 2 . Therefore [C(t)) 2 and CI(t) = Since C(O)
1

C(t)
2 (1 v'l=4t)'
+t =
C2(t) =
0 1
2 (1 + v'l=4t).
= 0, the solution C 2(t) is rejected because C 2(0) = 1 and so C(t)
= 21 (1 v'J=4t).
This generating function, on using Newton's general binomial formula (see Theorem 3.6), is expanded into powers oft as
Therefore
Cn
= (1)n122nl (1/2) n
= 1 · 3 · · · (2n 3) n!
= ( 1)n1 22n1 (1/2)n
n!
. nl . 2 · 4 · · · (2n 2) 2 2·4·(2n2)
=
(2n 2)! n!(n1)!
and
1 (2nCn = ;;: n _ l2) , n = 1, 2, ....
D
Let us now consider the exponential generating functions 00 tk 00 tk E(t) = Lak k!' F(t) = Lbk k!' G(t) k=O k=O
00
tk
= Lck k!" k=O
In this case we have (a) Sum
G(t) = E(t)
+ F(t)
iff Ck = ak
+ bk.
k = 0, 1,... .
(6.9)
GENERATING FUNCTIONS
198
(b) Product k
G(t)
= E(t)F(t)
iff Ck
= L (~)aibkj, j=O
k
= 0, 1, ....
(6.10)
J
Note that (6.9) is the same as its corresponding relation (6. 7) for the ordinary generating functions, while (6.10) differs from its corresponding relation (6.8) through the presence of the binomial coefficients. This difference suggests a basic device of the symbolic calculus, which may informally be presented as follows. The subscripts of a sequence ak, k = 0, 1, ... , are replaced by exponents transforming it into the sequence ak, k = 0, 1, .... Then ak is treated as an ordinary power (with a0 not necessarily equal to 1 and a1 not necessarily equal to a) until symbolic formulae are transformed back into ordinary formulae by transforming ak, k = 0, 1, ... , into ak, k = 0, 1, .... A reminder of such a transformation, of the form ak ak, should accompany a symbolic formula. Thus, with this device, the second relation of (6.10) is written as
=
ck=(a+b)k, ai:aj, br:br
(6.11)
and the exponential generating functions take the form
E(t) =eat, ak
=ak,
F(t)
= ebt,
bk
=bk, G(t) = ect, ck =Ck
and the first relation of (6.10) becomes
Expanding its first and last members and equating the coefficients of tk, (6.11) is deduced. Note that the derivative of E(t) =eat, ak ak, with respect tot, gives
=
d ( ) dt E t
= ae at ,
ak
= ak.

This expression, expanded into powers oft, takes the ordinary form
d oo tk oo tk oo tk oo r1  " ' kI  " ' k+1 k1  " '  " ' dt E(t) a ~a ~a ~ak+1kl~ar( k. . . r _ 1) 1. • k=O k=O k=O r=1 Applications of the symbolic calculus are given in the following examples.
Example 6.3 Exponential Bell numbers Let
6.2. UNIVARIATE GENERATING FUNCTIONS
199
and
F(t)
= exp{E(t) E(O)}.
(6.12)
The numbers bk. k = 0, 1, ... , the exponential generating function of which is connected with the generating function of the numbers, ak. k = 0, 1, ... , by the exponential relation (6.12) are called exponential Bell numbers. The derivation of a recurrence relation for the numbers bk, k = 0, 1, ... , can be efficiently carried out by using the symbolic calculus as follows. The defining relation (6.12) is written symbolically as ebt  exp(eat  a o, ) bk =  bk.
Differentiating it with respect to t, we get bebt = aeatexp(eat ao) = aeatebt = ae
= bk,
ar
= ar.
Expanding the first and last member of this relation into powers oft, k
00
k
00
L.... bk+l tk! _ " L....' a (a+ b) k tk!, bk
"' k=O
_ = bk,
a
= ar,
r 
k=O
and equating the coefficients of tk / k! in both sides, we conclude that k
bk+l =a( a+ b)k =
L j=O
and bk+I =
t
(~) aj+! bkj,
bk
=
bk, ar
=
ar
J
(~)aj+lbkj, k = 0, 1, ... , J
j=O
which is the required recurrence relation. In the particular case of ak = 1, k 0, 1, ... , b0 = 1, we have bk+!
=
t (~)bkj, j=O
k
= 0, 1, ...
, b0
=
= 1.
J
The Bell numbers bk. k = 0, 1, ... , in this particular case, express the number of partitions of a finite set of k elements (see Exercise 2.42). 0
Example 6.4 Exponential generating function of subfactorials Let Dn be the number of permutations (j 1 , h, ... , in) of the set { 1, 2, ... , n} without a fixed point, that is, Jr =/; r, r = 1, 2, ... , n. Dn is nsubfactorial (see Section 5.2). Let us determine the exponential generating function
GENERATING FUNCTIONS
200
Note that the nsubfactorial Dn is connected with the coincidence number Dn,k by the relation (see Theorem 5.3)
Dn,k=
(~)Dnk,
k=0,1, ...
,n,
n=0,1, ....
Consequently,
and if
then
Therefore But
n
Pn
= L Dn,k = n!,
oo
P(t)
k=O
tn
oo
= L Pn n! = L tn = (1 t) 1 n=O
n=O
and so
The ordinary and the exponential generating functions are particular classes of the following general form of a generating function. DEFINITION 6.2 Let ak, k = 0, 1, ... , be a sequence of real numbers and 9k(t), k = 0, 1, ... , be a sequence of linearly independent real .functions. The sum 00
G(t)
=L
ak9k(t)
(6.13)
k=O is called generating function of the sequence ak, k = 0, 1, ... , with respect to the sequence 9k(t), k = 0, 1, ....
The existence of the generating function G(t) requires the absolute convergence of the series for to < t < t 1 , with t 0 and t 1 real numbers. The
6.2. UNIVARIATE GENERATING FUNCTIONS
201
assumption of the linear independence of the sequence gk(t), k = 0, 1, ... assures the uniqueness of the definition of a generating function. The general form (6.13) of a generating function directly refers to algebra where it belongs. The most frequently used sequences of linearly independent functions in combinatorics and probability theory are gk (t) = tk, k = 0, 1, ... and gk(t) = tk jk!, k = 0, 1, ... , which correspond to the ordinary and the exponential generating functions, respectively. The sequences Yk(t) = (t)k, k = 0, 1, ... and gk(t) = (!), k = 0, 1, ... are also widely used. Their use requires particular names for the corresponding generating functions. Specifically, the following definitions, which are particular cases of Definition 6.2, are introduced. DEFINITION 6.3 sum
Let ak, k = 0, 1, ... , be a sequence of real numbers. The
00
F(t) =
L ak(t)k
(6.14)
k=O
is called factorial generating function, while the sum
B(t)=~akG)
(6.15)
is called binomial generating function of the sequence ako k
= 0, 1, ....
Example 6.5 Stirling numbers of the second kind Let S(n, k), k = 0, 1, ... , n, be the number of partitions of a finite set of n elements into k subsets. Determine the factorial generating function n
Yn(t) =
L S(n, k)(t)k. k=O
Note first that, to each partition { A 1 , A 2 , ... , Ak} of a finite set W n of n elements into k subsets, there correspond k! divisions (A}!, Ah, ... , AJk) of Wn into k nonempty subsets, which are obtained from it by permuting the k sets in all possible ways. Hence W(n, k) = k!S(n, k) is the number of divisions of a finite set of n elements into k nonempty subsets. Further, the number of divisions of a finite set of n elements into r subsets among which k are nonempty is given by (see Example 4.9)
(~)W(n,k) = (r)kS(n,k),
k
= 0,1, ...
,r, r
= 0, 1, ...
,n,
GENERATING FUNCTIONS
202
while, by Corollary 3.3, the total number of divisions of a finite set of n elements into r subsets equals rn. Consequently, by the addition principle, n
L S(n, k)(r)k = rn. k=O Since this formula is an identity for all r = 0, 1, ... , n, it follows that n
9n(t) =
L S(n, k)(t)k = tn k=O
holds for all real numbers t. The numbers S(n, k) are called Stirling numbers of the second kind. D
6.2.2 Power, factorial and Lagrange series The derivation of certain generating functions is facilitated by the use of some wellknown expansions of functions. Every function expanded into a power or a factorial series can be considered as a generating function. The Maclaurin, Newton and Lagrange series are presented (without proof) in this section. Maclaurin power series (6.16)
Note that, with
ak =
[:,~kkf(t)L=o'
k=0,1, ... ,
(6.16) is of the form (6.5), 00
f(t) =
L aktk,
(6.17)
k=O while for
bk
= [:tkk f(t)L=O,
k
= o, 1, ... ,
(6.16) is of the form (6.6), (6.18)
6.2. UNIVARIATE GENERATING FUNCTIONS
203
Example6.6 Let us consider the function f(t) = et. Its derivatives are
dk t t dtk e = e , k = 0, 1, .... Thus bk
dk ] = [ dket
t
= 1,
k
= 0, 1, ...
t=O
and, by (6.18), we find the expansion of the exponential function:
The derivatives of the function f(t)
dk dtk { log(1 t)}
= log(1 t) are
= (k 1)!(1 t)k,
k
= 1, 2, ....
Thus
ao ak =
= f(O) = log1 = 0,
[~!:tkk{log(1t)}L=o = ~'
k= 1,2, ...
and, by (6.17), we get the expansion of the logarithm: 00
log(1 t) =
L
tk
k"
k=I
Setting u = t, this expansion can also be written as k
00
log(1
+ u) =
L( 1)kI uk . k=I
The derivatives of the function
dk dtk (1
f (t)
= ( 1 + t )X are
+ ty~: = (x)k(1 + t)kx,
Thus
ak
=
x] 1 dk ( [ k! dtk 1 + t) t=O
(x)k = kl =
k
= 0, 1, ....
(X)
k ' k
and, by (6.17), we deduce the general binomial formula:
= 0, 1, ...
GENERATING FUNCTIONS
204
Example 6.7 Stirling numbers of the first kind Expanding the factorial oft of degree n, (t)n. into a Maclaurin's series and putting s(n, k) =
[~! !kk (t)n L=o,
k = 0, 1, ... , n,
we get n
(t)n = L
s(n, k)tk.
k=O
The coefficients s(n, k) are called Stirling numbers of the first kind. Note that the sequence s(n, k), k = 0, 1, ... , n, for fixed n, has generating function gn(t) = (t)n· D
Newton factorial series
where Llk f(t) denotes the difference of order k of the function f(t). Note that, with
ak =
[~!Llkf(t)L=o'
k=0,1, ... ,
(6.19) is of the form (6.14), ()()
(6.20)
f(t) = Lak(t)k. k=O
Setting
bk = [Llk j(t)]t=O, k = 0, 1, · ·. , (6.19) is of the form (6.15), (6.21)
Example 6.8 Let us consider the function f(t) = ( u is a fixed real number. Then
Ll(u
+ t)n
= (u
+ t)n, where n
+ t + 1)n (u + t)n
is a positive integer and u
= n(u
+ t)n1
6.2. UNIVARIATE GENERATING FUNCTIONS
205
and, in general,
.1k(u + t)n Thus
ak =
= (n)k(u + t)nk,
k
= 0, 1, ....
[~! Llk(u + t)n L=O = (~) (u)nk,
k = 0, 1, ...
and so, from (6.20), we deduce Vandermonde's formula:
+ t)n
(u
=
~ (~) (u)nk(t)k.
D
Lagrange series The Lagrange series appears in two basic forms: 00
t(t) = t(o)
dk1 (
d
+ :L: dtk1 gk(t) d/(t) [
)]
uk t=O
k=1
/J' u =
t (t) g
(6.22)
and
d ) f(t) ( 1 ud g(t) t
00
l
=L
k=O
[
dk ] k dk (l(t)f(t)) uk', u t t=O .
t = (). (6.23) gt
Note that these two series constitute a basic means for the solution of equations of implicit form and for the inversion of series. They are also used as generating functions with variables defined implicitly. More specifically, with
G(t) = f(t) ( 1 g!t)
!g(t))
I
I
bk = [!kk (l(t)f(t))] t=O,
(6.22) and (6.23) take the forms 00
00
k=O
k=O
which are of the general form, (6.13), of a generating function. Some interesting applications of such generating functions are given in the following examples. Example 6.9 Abel's formula Let g(t) = ezt and f(t) =ext. Then from (6.22) and since
GENERATING FUNCTIONS
206
dk (kg (t) d/(t) d )] [dtk1 1
dk1
t=O
= [ xe<x+kz)t dtk1
we deduce the expansion
=L
:! , k
00
ext
x(x
+ kz)kl
k=O
u
]
= x(x
+ kz)kl
t=O
= tezt.
'
(6.24)
Similarly, from (6.23), we get (6.25) The coefficients in (6.24),
_ x(x + kz)k 1 k A k (x, z ) k! ' = 0, 1, ... ' are called Abel polynomials. Expanding both members of the identity
into powers of u = tezt, using (6.24), we get
~ A.(x + y,z)u" ~ (t. A,(x,z)u') (~ A;(y,z)u') =
~ {~ Ak(x, z)Ank(Y, z)} un
and, consequently, n
An(x
+ y, z) = L Ak(x, z)An_k(y, z). k=O
Further, consider the identity
e<x+w)t
ewt
1  zt
1  zt
....,... = ext
and expand both its members into powers of u = tezt, using (6.24) and (6.25). We get the relation
6.2. UNIVARIATE GENERATING FUNCTIONS
207
which, according to (6.10), implies Abel's formula:
Putting y = w
+ nz, we find the expression (x +y)n =
t
(~)x(x + kz)kl(y kz)nk.
k=O
Note that Newton's binomial formula is a particular case of Abel's formula, corresponding to z = 0. 0
Example 6.10 Gould's formula Let f(t) = (1 + t)x and g(t) = (1 + t)Z. Then from (6.22) and since
= _ x (x+kz), x+kz
k
we deduce the expansion
Similarly, from (6.23), we get
(1 + w+l = Loo (X+ kz)uk, 1 + t zt k
U
= t(1 + t)z.
(6.27)
k=O
The coefficients in (6.26),
G (
k x, z
) = _x_ (x
x + kz
+ kz) k
= x(x
+ kzk!
1)kI
'
k
= 0, 1, ... '
are called Gould polynomials. Expanding both members of the identity
into powers of u = t( 1 + t) z, using (6.26), we get the relation
GENERATING FUNCTIONS
208
which, according to (6.8), implies n
Gn(x
+ y, z) = L Gk(x, z)Gn_k(y, z). k=O
Expanding both members of the identity
+ t)'~:+w+l
+ t)w+I
'' = (1 + w''(1
1 + t  zt
into powers of u = t(1
(1
1 + t  zt
+ t) z, using (6.26) and (6.27), we get the relation
which, according to (6.1 0), implies Gould's formula:
(x
+ w + nz)n
=
t (~)x(x
+ kz 1)kdw + nz kz)nk·
k=O
Putting y = w
+ nz, it yields
(x+y)n
=
t
G)x(x+kz1)kI(ykz)nk·
k=O
Note that Vandermonde's formula is a particular case of Gould's formula, correD sponding to z = 0.
6.3
COMBINATIONS AND PERMUTATIONS
Let us consider a finite set Wn = {w1,w2,··· ,wn} and the set Ck(Wn) of the kcombinations of n and attach the indicator Xj to the element Wj, j = 1, 2, ... , n. Then the homogeneous product of order k, x~ 1 2 • • • x~", with ri 2: 0, j = 1, 2, ... , n, and r 1 + r 2 + · · · + rn = k, uniquely determines the kcombination of n, {a 1 , a 2 , ... , ak}, ai E Wn, j = 1, 2, ... , k, in which ri 2: 0 elements are Wj, j = 1, 2, ... , n, with r 1 + r 2 + · · ·+ rn = k. Further, the homogeneous product sum
x;
(6.28) where the summation is extended over all ri 2: 0, j = 1, 2, ... , n, such that r 1 + r 2 + · · · + rn = k, exhibits all kcombinations of n. The advantage
6.3. COMBINATIONS AND PERMUTATIONS
209
of this correspondence is that the homogeneous product sum (6.28) can be obtained algebraically. Specifically, oo
n
II (1 +
Xjt
+ x]t2 + · · ·) = :~:::>k(Xl, X2, ... , Xn)tk.
(6.29)
k=O
j=l
Indeed, if the operations are executed by selecting the term tr; from the jth factor of the product and forming the product tr 1F 2 · .. tr" = tr1 +r2+··+rn, with r 1+r2 + · ·+r n = k, the kth power oft, tk, is turned up with coefficient equal to the sum of the homogeneous products of order k, x~ 1 x; 2 • • • x~", extended over all rj 2: 0, j = 1, 2, ... , n, such that r 1 + r 2 + · · · + rn = k. Note that, in the jth factor of the product in (6.29), (6.30)
x?
the term tr; indicates that the element Wj may appear in any combination rj times, rj = 0, 1, .... Moreover, any restrictions on the appearance of any elements of Wn modify the factors (6.30) accordingly. Thus, if the element Wj is allowed to appear in any combination at least mj and at most nj times, then (6.30) becomes (6.31)
for j = 1, 2, ... , n and expression (6.29) is modified accordingly. Putting Xj = 1, j = 1, 2, ... , n in the resulting expression, the generating function of the number of kcombinations of n is deduced. Note that, in the particular case of kcombinations of n (without repetitions), the homogeneous product sum of order k, (6.28), coincides with the elementary symmetric function of order k, (6.2), since rj = 1 for k indices, {i 1, i 2, ... , ik}, out of the n indices {1, 2, ... , n }, and rj = 0 for the remaining n k indices, {1,2, ... ,n} {i 1,i 2, ... ,ik}· Also, the factor (6.31) reduces to (1 + Xjt) and expression (6.29), modified accordingly, coincides with (6.1). Since the ordinary multiplication is commutative, whence x 1 x 2 and x 2 x 1 are not distinguishable, it is not possible to construct a function exhibiting all permutations of the set Pk(Wn), of the kpermutations of n. For this reason, to the homogeneous product of order k, x~ 1 x; 2 • • • x~", with rj 2: 0, i = 1, 2, ... , n, and r 1 + r 2 + · · · + rn = k, in addition to the kpermutation ofn, (a1,a2, ... ,ak), aj E Wn, j = 1,2, ... ,k, in which the first n 2:0 elements are w1 , the next r 2 2: 0 elements are w2 and finally the last rn 2: 0 elements are Wn, there correspond all its permutations, which (including the initial), according to Theorem 2.4, are equal to k!
I I I, k = r1 + r2 + · · · + r1.r2. · · · rn·
Tn·
GENERATING FUNCTIONS
210
In consequence, to the set Pk(Wn) of the kpermutations of n, we correspond the sum
9k(XI,X2,···,xn)=L
1 x~ 1 X~2 ···x~n,k=0,1, ... ,(6.32)
~!
1 T).T2· · · ·Tn.
where the summation is extended over all r 1 2 0, j = 1, 2, ... , n, such that
r 1 + r 2 + · · · + rn = k. This sum, according to the multinomial Theorem 3.7, equals
9k(XI,X2,··· ,xn) =(xi +x2 + ··· +xn)k and may be obtained algebraically as follows: (6.33)
since
~ xktk xt ~ k! e . k=O
Moreover, any restrictions on the appearance of any elements in permutations of Wn modify accordingly the corresponding factors,
x2t2
E(t,xj)=1+x1t+ ~! +···, j=1,2, ... ,n,
(6.34)
of the product in (6.33). Thus, if the element Wj is allowed to appear in any permutation at least m 1 and at most n 1 times, then (6.34) becomes
Em;,n; (t, Xj)
=
x'J'itmi m1!
xm;+ltm;+l
+
J
(mi
+ 1)!
xnitni
+ ... +
_J_ _
ni!
'
(6.35)
for j = 1, 2, ... , n and expression (6.33) is modified accordingly. Putting Xj = 1, j = 1, 2, ... , n in the resulting expression, the exponential generating function of the number of kpermutations of n is deduced.
Example 6.11 Combinations with repetition The generating function of the number of kcombinations of n with (unrestricted) repetition may be obtained as follows. The enumerator of the appearance of the jth element in any combination, since it is allowed to appear Tj times, with r1 = 0, 1, ... , is A0 (t,x 1 ) = 1 + x 1t + x]t2 + · · · = (1 x 1t) 1 , j = 1, 2, ... ,n.
Therefore, the enumerator of kcombinations of n is given by
IT Ao(t,xj) = n(l Xjt)n
j=l
n
j=l
oo
1
= Lhk(XJ,X2,· .. k=O
,xn)tk,
6.3. COMBINATIONS AND PERMUTATIONS
211
which, with x 1 = 1, j = 1, 2, ... , n, yields the generating function of the number of kcombinations of n with repetition:
The coefficient of tk, E(n,k)
= (n+~ 1 ),
gives the number of kcombinations of n with repetition (see Theorem 2.7).
D
Example 6.12 In the case of kcombinations of n with repetition, in which every element appears at least once, the enumerator of the appearance of the jth element, since it is allowed to appear Tj times, with r1 = 1, 2, ... , is A 1 (t, Xj)
= Xjt + x]t2 + · · · = Xjt(1 Xjt) 1 ,
j
= 1, 2, ... , n.
The generating function of the number of kcombinations of n with repetition, in which every element appears at least once, is deduced from n
n
II A1 (t, Xj) = II Xjt(1j=1
Xjt)
1,
j=1
by setting x 1 = 1, j = 1, 2, ... , n. Expanding the resulting generating function into powers oft, we get
A1(t)
= tn(l t)n = ~(1r (rn)tr+n = ~ (n + ~ 1)tr+n =
f: (~ =~) f: (~ =~) tk =
k=n
tk.
k=n
Therefore, the number of kcombinations of n with repetitions, in which every element appears at least once, is zero for k < n and
1)
k ( n1 fork 2: n.
D
Example 6.13 Combinations with limited repetition The generating function of the number of kcombinations of n with repetition and the restriction that each element appears, at most, s times is deduced from the
GENERATING FUNCTIONS
212
enumerator: n
n
II (1 + Xjt + x]t 2 + · · · + xjt") = II (1 xj+ t•+l )(1 x 1t) 1, 1
j=l
by putting xi = 1, j
j=l
= 1, 2, ... , n. Thus, An,s(t) = (1 t"+I)n(1 t)n.
The determination of the number E.(n, k), of kcombinations of n with repetition and the restriction that each element appears, at most, s times requires the expansion of the generating function into powers oft,
Note that, in the case k ::; s, the first factor of this product in addition to the first term, which is equal to 1, contains powers of t of order greater than k. Consequently, the coefficient of tk is equal to the coefficient of this power in the second factor of the product, that is, E.(n,k)
= ( n + kk
1)
, k::; s.
In the case of k 2: s + 1, the term tk is formed by multiplying the term tr(•+t) of the first factor with the term tkr(s+I) of the second factor. Consequently, the coefficient of tk is equal to the sum of the product of the coefficients of tr(•+I) from the first factor and the coefficients of tkr(s+l) from the second factor, for r = 0, 1, ... , n, whence
The next two particular cases are worth noticing. For s
An,l (t)
= (1 + tt =
t (~) k=O
and so
while, for s + oo, we have
tk,
= 1 we have
213
6.3. COMBINATIONS AND PERMUTATIONS
and so
D Example 6.14 Permutations with repetition The generating function of the number of kpermutations of n with (unrestricted) repetition is deduced from the function n
(
II
x2t2 1 +x·t+  12! ]
+ ... )
,
j=l
by putting xi = 1, j = 1, 2, ... , n. Hence, we get
Expanding it into powers oft, k
00
" k t Eo ( t ) = e nt = " L...n k! ,
k=O
we conclude the number of kpermutations of n with repetition equals U(n,k)=nk.
D
Example 6.15 The generating function of the number of kpermutations of n, with repetition, in which every element appears at least once, is deduced from the function n
II
( X
]·t +
) 3 x2t2 2! + " ·
,
j=l
by setting Xj = 1, j = 1, 2, ... , n. Expanding the resulting generating function into powers of t, we find
E(t) 1
= (t
+ ~~ + ... )
= ~(l)nr(~)
= (et
1)n
=
~(1)nr(~)ert
~rk~~ = ~ {~(ltr(~)rk} ~
Therefore, the number of kpermutations of n, with repetition, in which every element appears at least once, is given by
GENERATING FUNCTIONS
214
D
where S(k, n) is the Stirling number of the second kind.
Example 6.16 Permutations with limited repetition The generating function of the number of kpermutations of the set Wn = {w 1 ,w2 , ... ,wn}. with repetition, in which the element Wj is allowed to appear at most ki times, j = 1, 2, ... , n, where k1 + k2 + · · · + kn = r ;:::: k, is deduced from the function n
II
j=1
by setting xi
(
xk; tk;)
x2t2
1 1+x·t++ .. ·+11 2! k ! J
= 1, j = 1, 2, ... , n. Then we get the expression F(t)
=II n
(
1+t
j=1
t2 tk;) ++ ... +2! k·! J
'
which, after executing the multiplication and arranging the terms in ascending order of powers oft, takes the form
r
F(t)
=L
tk U(k1, k2, ... , kn; k) k! ·
k=O
The coefficient of tk / k! is the number of kpermutations of n with repetition, in which the jth element is allowed to appear at most kj times, j = 1, 2, ... , n. In the particular case of k = r, the number U(k 1, k2, ... , kn; r) M(k1, k2, ... , kn) of rpermutations of n with repetitions, in which the jth element is allowed to appear at most kj times, j = 1, 2, ... , n, where k1 + k 2 + · · ·+ kn = r, equals
=
M(k1, k2, ... , kn)
r!
= k 1·'k 2·1 ••• k n·1, r = k1 + k2 + · · · + kn,
since the coefficient oftr = tk 1 +k 2 + ·+kn is equal to the product
1 1 1 k1! k2! ... kn!
1
where 1/ k1 ! is the coefficient of the last term tk; of the jth factor of the generating function F(t), j = 1, 2, ... , n. In the general case, it similarly follows that
where the summation is extended overall 8J = 0, 1, ... , k 1 , j = 1, 2, ... , n, such that 81 + 82 + · · · + 8n = k. 0
6.4. MOMENT GENERATING FUNCTIONS
6.4
215
MOMENT GENERATING FUNCTIONS
Let us consider a sequence of real numbers ak> k = 0, 1,.... In combinatorics, the general term ak of such a sequence, usually, expresses a combinatorial number. Also, in probability theory and, in particular, when the total probability is concentrated at discrete points, such a sequence of probabilities defines a probability (density) function. In mechanics and particularly when the total mass is concentrated at discrete points, such a sequence of masses defines a density mass function. Among the characteristics of the sequence ak, k = 0,1, ... of special interest are its moments defined as follows. The (power) moment of order r of the sequence ak, k = 0, 1, ... , denoted by J.L~, is defined by ()()
J.L~=Lkrak, r=0,1, ... ,
(6.36)
k=O
while the factorial moment of order r of the same sequence, denoted by f.L( r), is defined by ()()
f.L(r)
= L(k)rak, r
= 0, 1, ....
(6.37)
k=O
It should be noted that the absolute convergence of the series (6.36) and (6.37) is an essential requirement for the existence of the moments J.L~ and J.L(r)· Note, also, that the factorial moment of order r, J.L(r)• is closely connected with the binomial moment of order r, b(r)• which is defined by (6.38)
whence b(r) = f.L(r)/r!, r = 0, 1, .... REMARK 6.1 The moments J.L~, J.L(r) and b(r) are particular cases of the weighted moment of order r, mr. defined by ()()
mr = L
Cr(k)ak, r = 0, 1, ... ,
k=O
where, for fixed r, cr(k), k = 0, 1, ... , is a sequence of weights. When ak, J.L~ = J.L( 1) is the mean and k = 0, 1, . . . is a sequence of probabilities, J.L
=
GENERATING FUNCTIONS
216
a2
= p,~  p,'? = IL( 2 )

/L(I) 
1Lfl) is the variance; when it is a sequence of
masses, p, is the center of gravity and a 2 is the moment of inertia.
I
The calculation of the moments p,~, 1L(r) and b(r)• r = 0, 1, ... , when the sequence ak, k = 0, 1, ... , is known and vice versa, is facilitated by using suitable generating functions. Such generating functions are the moment generating function, defined by (6.39)
and the binomial or factorial moment generating function, defined by (6.40)
These generating functions are connected with the ordinary generating functions of the sequence ak. k = 0, 1, ... , 00
A(t) = I>ktk.
(6.41)
k=O
Indeed, from (6.36) and (6.37), we have
and, using (6.41) with e1 instead oft, we get M(t) = A(e 1 ), A(u) = M(log u),
(6.42)
while, from (6.39) and (6.40), we find
and so B(t)
= A(1 + t),
A(u)
= B(u 1).
(6.43)
From (6.42) and (6.43) it follows that M(t)
= B(et 1),
B(t)
= M(log(1 + t)).
(6.44)
6.4. MOMENT GENERATING FUNCTIONS
217
Expanding the second of (6.43) into powers of u, using (6.40) and (6.41), we get
and thus 1 oo (1yk
= k! L
(r k)! f..L(r)· r=k Introducing the binomial moments b(r) = f..L(r)/r!, r = 0, 1, 2, ... , this expression takes the form ak
Note that the last expression is similar to (4.14), which expresses the number Nn,k, k = 0, 1, ... , n, of elements that are contained in k among n subsets, A1 , A2 , •.• , An, of a finite set D, in terms of the numbers Sn,r = L: N(Ai, Ai2 • • • AiJ, r = 0, 1, ... , n, where the summation is extended over all rcombinations, {h,i 2 , ... ,ir}, of then indices {1,2, ... ,n}. Example 6.17 Let us determine the factorial moments f..L(r)• r = 0, 1, ... , of the sequence of geometric probabilities Pk = pqk, k = 0, 1, ... , where 0 < p < 1 and q = 1  p. The probability generating function, on using the geometric series, is obtained as 00
00
A(t) = LPktk = p L(qt)k = _P_, 1  qt k=O k=O
ltl < 1/q.
Therefore, by the first of (6.43), the factorial moment generating function is given by 1
B(t) = 1 _ (qfp)t'
ltl < pfq
which, expanded into powers oft, B(t)
=
1
00
= "'(qjp)rtr, 1 (qfp)t ~
by virtue of (6.40), yields f..L(r)
= r!(qfpY,
r
= 0, 1,... .
0
218
GENERATING FUNCTIONS
Example 6.18 Let us determine the sequence of probabilities Pk, k moments
=
b(r)
(;)pr,
r
= 0, 1, ... , with binomial
= 0, 1, ... ,
where 0 < p < 1 and n is a positive integer. The binomial moment generating function, on using Newton's binomial formula, is obtained as
Consequently, by virtue of the second of (6.43), the generating function of the sequence Pk· k = 0, 1, ... , is given by
and so
Example 6.19 Let us consider the number Nn,k. k = 0, 1, ... , n, of elements that are contained ink among n subsets A1 , A2 , •.. , An. of a finite set fl, and
where the summation is extended over all rcombinations {i 1 , i 2 , ... , ir} of the n indices {1, 2, ... , n}. Then, according to (4.14) and (4.20),
Consequently, Sn,r• r = 0, 1, ... , n, is the sequence of binomial moments of the sequence Nn,k, k = 0, 1, ... , n. Thus, if n
Nn(t)
=L
n
Nn,ktk, Sn(t)
k=O
= L Sn,rtr,
(6.45)
r=O
then, according to (6.43), Sn(t) = Nn(t
+ 1),
Nn(t) = Sn(t 1).
(6.46)
219
6.5. MULTIVARIATE GENERATING FUNCTIONS
In the particular case with D the set of permutations (il, h, ... , in) of { 1, 2, ... , n} and As the subsets of these permutations for which the point j s is a fixed point, s = 1, 2, ... , n, we have
.
Nnk = Dnk, Snr ' '
= n!fr!,
where Dn,k is the coincidence number (see Section 5.2). The generating function n
L Dn,ktk,
Dn(t) =
k=O according to (6.45) and (6.46), is written as n
Dn(t) =
L
I
n; (tIt. r. r=O
Differentiating it we get the relation
Also, we find
Dn(t) = nDndt)
+ (t l)n.
These relations imply the recurrence relations:
kDn,k = nDnI,k1 and
respectively.
6.5
0
MULTIVARIATE GENERATING FUNCTIONS
The notion of a generating function can be extended to cover the case of a doubleindex and, more generally, a multiindex sequence. Specifically, the following definitions are introduced.
DEFINITION 6.4
Let an,k> k
= 0, 1, ... , n = 0, 1, ...
be a sequence of real
numbers. The sum 00
A(t, u) =
L
00
L:::an,ktkun n=Ok=O
(6.47)
GENERATING FUNCTIONS
220
is called (ordinary) generating function, while the sum oo tk un L::an,k k! n! n=Ok=O oo
L
E(t, u) =
is called exponential generating function of the sequence an,k· k n = 0,1, ....
(6.48)
= 0, 1, ... ,
It should be noted that the absolute convergence of the series (6.47) and (6.48) is required for the existence of the corresponding generating functions. REMARK 6.2 cases of
Both generating functions A(t, u) and E(t, u) are particular 00
G(t, u) =
00
L L an,dk(t)gn(u),
(6.49)
n=Ok=O where the functions fk(t), k = 0, 1, ... , as well as the functions 9n(u), n = 0, 1, ... , are linearly independent and so (6.49) is uniquely defined. It is worth noting that, with 00
Fn(t)
= L an,kfk(t),
00
Gk(u)
= L an,k9n(u), n=O
k=O (6.49) may be written in the form 00
G(t, u)
= L Gk(u)fk(t), k=O
or in the form 00
G(t, u)
=L
Fn(t)gn(u).
n=O Note that the last two expressions are of the general form (6.13) of a generating function. With this remark, the study of generating functions of doubleindex sequences is reduced to the previous study of generating functions of singleindex sequences. In the particular case of triangular doubleindex sequences an,k· k = 0, 1, ... , n = 0, 1, ... , for which an,k = 0 for k > n, it is more convenient to use the generating function (6.50)
6.5. MULTIVARIATE GENERATING FUNCTIONS
221
In an analogous way, a multivariate generating function G (t 1 , t 2 , ... , tr) of a multiindex sequence ak,,k 2 , ... ,kr• k; = 0, 1, ... , i = 1, 2, ... , r, may be introduced and studied. I Example 6.20 Let us consider the doubleindex sequence: an,k=
(~),
k=0,1, ...
,n,
n=0,1, ....
For fixed n, using Newton's binomial formula, the ordinary generating function of this sequence is readily deduced as
Consequently, the bivariate generating function of the doubleindex sequence an,k. k = 0, 1, ... , n, n = 0, 1, ... , is obtained as n
oo
00
n=O
n=Ok=O 00
= L)(l + t)u]n = [1 (1 + t)ut 1 • n=O
Since the considered doubleindex sequence is triangular, its generating function of the form (6.50) is derived as
The exponential generating function of the same sequence is quite involved and thus not of any interest. D Example 6.21 Consider the coincidence number (see Theorem 5.3),
D
 n! nk (1)i
n,k  k!
L J..
. j=O
1
,
k
= 0, 1, ... , n,
n
= 0, 1, ...
.
222
GENERATING FUNCTIONS
For fixed n, the generating function of this sequence is (see Example 6.19), n
Dn(t)
=L
k
Dn,kt
k=O
n
= n! L
(tIt r!
r=O
Then, the bivariate generating function oo
D(t, u) =
n
n
L L Dn,ktk ~! ,
n=Dk=O is obtained as
D(t,u) =
f r=O
[u(t~ lW r.
0
funr = (1 u)1eu(t1). n=r
Example 6.22 Let us consider the multiindex sequence , C(n,k1,k2, ... ,kr1)=(k k n k )  k 1 n! 1, 2, · · ·, r1 1·k2! .. · kr1!kr.1
ki = 0, 1, ... ,n,i = 1,2, ... ,rl,kr = n(k1+k2+·· ·+kr_!),n = 0, 1, .... For fixed n, the multivariate generating function of this sequence, on using the multinomial formula (see Theorem 3.7), is obtained as An(t1,t2, ... ,trd=L(k k n k )t~ 1 t~ 2 1' 2, · · • , r1 = (1 + t1 + t2 + · · · + tr_I)n.
···t~:_J. 1
Consequently,
A(h,t2, ... ,tr1,u)=E'L:(k k n k n=O 1 , 2, · · · , r1
)t~ 1 t~2 ···t~:_11 un
00
= L[(l n=O
+ t1 + t2 + · · · + trdut
= ~ [(1
+ t1 + t2 + · · · + tr_I)u]n
and
~
n! n=O = e(l+fl+h+··+trdu. Other bivariate generating functions are discussed in the exercises.
0
6.6. BIBLIOGRAPHIC NOTES
6.6
223
BIBLIOGRAPHIC NOTES
The roots of generating functions may be traced in the work of Abraham De Moivre (1718). Also, generating functions were used by L. Euler (1746) in his study on the partitions of integers. The development of the theory of generating functions arose in conjunction with the calculus of probability. P. S. Laplace (1812) devoted half of his book to a systematic and complete treatment of generating functions and especially the moment generating functions; he is considered the inventor of the method of generating functions. P. MacMahon (1915, 1916) extensively used enumerating generating functions in his treatise on combinatorial analysis. The basic elements of the symbolic calculus included in this chapter are based on the paper of E. T. Bell (1940) and the book of J. Riordan (1958). The exponential Bell numbers were examined in E. T. Bell (1934a,b). Abel's generalization of Newton's binomial formula first appeared in N. H. Abel (1826) and Gould's generalization of Vandermonde's formula was derived in H. W. Gould (1956). An illuminating historical coverage of the use of generating functions in probability theory is given by H. L. Seal (1949). A wealth of information can be found in the survey article on generating function by R. P. Stanley (1978).
6.7
EXERCISES
1. Construct a generating function for the number ak of kcombinations of the set W5 = {W1, W2, ... , W5}, with repetition, in which the element W1 appears at least once, each of the elements w 2 , W3 and w4 appears at least twice and the element w 5 may appear without any restriction. Expanding the generating function, deduce the number ak, k = 7, 8, .... 2. Suppose that three distinguishable dice are rolled. Construct a generating function of the number ak of possible results in which the sum of the three numbers equals k. Derive the numbers a 9 and a 10 (see Exercise 2.11). 3. Consider ten urns, each containing five numbered balls. The balls of each of the first five urns bear the even integers {0, 2, 4, 6, 8}, while the balls of each of the other five urns bear the odd integers { 1, 3, 5, 7, 9}. Assume that one ball is drawn from each of the urns. Construct a generating function for the number ak of drawings in which the sum of the two numbers equals k. Derive the numbers a 7 , a 9 and a 11 •
GENERATING FUNCTIONS
224
4. Consider a collection of 52 cards of four different kinds with 13 cards like from each kind. Construct a generating function for the number ak of different subcollections of k cards. Expanding it, deduce the numbers a 13 and a26·
5. Suppose that five letters {h, l2, ... , l 5 } from an alphabet are given. Construct a generating function for the number of kletter words that can be formed (a) without repeated letters and (b) without any restriction. 6. Calculate the number of kpermutations of the set {0, 1, ... , 9} in which (a) each of the digits 0 and 5 appears an odd number of times and (b) each of the digits 0 and 5 appears zero or an even number of times, using suitable generating functions. 7. Calculate the number of kpermutations of the set {0, 1, ... , 9} (a) with zero or an even number of Os and odd number of 5s and (b) in which the total number of Os and 5s is zero or even. 8. In the quaternary number system each number is represented by an ordered sequence of the digits from the set {0, 1, 2, 3}. Using suitable generating functions, calculate the number of kdigit quaternary sequences in which (a) each of the digits 1 and 3 appears an even number of times and (b) the total number of appearance of the digits 1 and 3 is an even number. 9. Let T(n, k) be the number of kcombinations of n with repetition and the restriction that each element may appear at most two times. For fixed n, construct a generating function for the sequence T(n, k), k = 0, 1, ... , and derive the expressions
and
10. (Continuation). Show that
T(n, k) = T(n 1, k)
+ T(n
1, k 1)
+ T(n
1, k 2),
for k = 2, 3, ... , 2n and n = 1, 2, ... , with initial conditions
T(n, 0)
= 1,
T(n, 1)
= n,
= 0,
k > 2n.
+ 2nT(n
1, k 2),
T(n, k)
11. (Continuation). Show that
kT(n, k) = nT(n 1, k 1)
6. 7. EXERCISES
for k
= 2, 3, ...
225
, 2n, n
= 1, 2, ... , and
k(k 1)T(n, k) = 2n(2n 1)T(n 1, k 2) for k
= 2, 3, ...
, 2n and n
T(n, 0)
= 1,
+ 3n(n 1)T(n 2, k
2),
= 2, 3, ... , with initial conditions T(n, 1)
= n,
= 0,
T(n, k)
k > 2n.
=
12. (Continuation). Let Tn T(n, n) be the number of ncombinations of n with repetition and the restriction that each element may appear, at most, twice. Show that
nTn (2n 1)Tn1 3(n 1)Tn2 and
= 0,
n
= 2,3, ...
, To= T1
=1
00
T(t) = LTntn = (1  2t 3t 2 ) 1 f 2 . n=O
13*. (Continuation). From the generating function
2n Tn(t) = LT(n, k)tk k=O
and using Lagrange formula, 1
00
~
[dn n ] un dtn f(t)g (t) t=O · n!
deduce that
= f(t)
(
dg(t))1 u;[t ,u
t
= g(t),
00
T(u)
= LTnun = (1 2u 3u 2 ) 112 . n=O
14. (Continuation). Let A(n, r, k) be the number of kcombinations of with repetition and the restriction that each of n specified elements may appear at most twice, while each of the other r elements may appear at most once. For fixed n and r, construct a generating function for the sequence A(n, r, k), k = 0, 1, ... , and derive the expressions
n
+r
""' (n) (n + J)
[k/2]
A(n, r, k) = ~ ]=0
.
. J
k _r  . 2J
and k
A(n,r,k) =
~ (:)r(n,ki).
,
GENERATING FUNCTIONS
226
15. Let G(n, k) be the number of kcombinations of n with repetition and the restriction that each element appears zero or an even number of times. For fixed n, construct a generating function for the sequence G(n, k), k = 0, 1, ... and, expanding it, show that
1 G(n,2j)=(n+; ), j=0,1, ... , G(n,2j+1)=0, j=0,1, .... 16. Let H (n, k) be the number of kcombinations of n with repetition and the restriction that each element appears an odd number of times. For fixed n, construct a generating function for the sequence H(n, k), k = 0, 1, .... If n = 2r, conclude that
= (r ~ j 
H(2r, 2j)
Jr
1 ), j
= r, r + 1, ... ,
H(2r, 2j) = 0, j = 0, 1, ... , r 1, H(2r, 2j + 1) = 0, j = 0, 1, ... ,
while if n = 2r + 1, conclude that H(2r + 1, 2j + 1) =
(~ + j),
j = r, r + 1, ... ,
JT
H(2r+1,2j+1)=0, j=0,1, ... ,r1, H(2r+1,2j)=O, j=0,1, .... 17. Let B(n, k; s) be the number of kcombinations of the set Wn+I = {wo, w1 , ... , Wn} with repetition and the restriction that the element wo may appear at most s times and each of the other elements at most once. Using a suitable generating function, show that
B(n,k;s)
=
f::
j=O
(k: .), m = min{k,s} J
and
B(n,k;s)B(n,k1;s)=(n)(
k
with B(n, 0; s) = B(n, n
n
ks1
),
+ s; s) = 1.
18. Let E 8 (n, k) be the number of kcombinations of n with repetition and the restriction that each element may appear at most s times. Using a suitable generating function, show that
E 8 (n, k) = E 8 (n 1, k) + E 8 (n 1, k 1) + · · · + E 8 (n 1, k s), k 2::: s, Es(n, k) = (n + ~ ), k 5: s. 1
6. 7. EXERCISES
227
19. (Continuation). Showthat k
L Es(r,j)E.(n r, k j) = E (n, k), 8
j=O
and k
L
(~)EsI(nj,kn+j) =E (n,k),
m=max{O,nk}.
8
J
J=m
20. Let Pn be the (total) number of permutations of n: n
Pn = L(n)k, n = 0, 1, ... . k=O
Show that
and Pn = nPn1
+ 1,
n = 1, 2, ... , Po= 1.
21. Let A(n, k) be the number of kpermutations of n with repetition and the restriction that each element appears zero or an even number of times. Using a suitable generating function, show that
A(n, 2s)
=Tnt(~) (n 2j) 28 , j=O
and, for A1 (r,s) = A(2r
A(n, 2s
+ 1)
= 0, s = 0, 1, ...
J
+ 1,2s),
A2(r,s) = A(2r,2s), that
A 1 (r, s) = (2r + 1){(2r + 1)A 1 (r, s 1) 2r AI(r 1, s 1)}, A2(r,s) = 2r{2rA 2(r,s 1) (2r 1)A2(r 1,s 1)}. 22. Let B(n, k) be the number of kpermutations of n with repetition and the restriction that each element appears an odd number of times. Using a suitable generating function, show that
B(2r, 2s) = 2 2(rs)+l
i)
2 1)j ( :) (r j) 28 , J
j=O
B(2r, 2s + 1)
= 0,
228
GENERATING FUNCTIONS
and
B(2r
+ 1, 2s + 1) =
T
2 (rs)+l
i)
2 1)j ( r
+ 1, 2s) =
23. (Continuation). For B 1 (r,s) B(2r, 2s), show that
B 1 (r, s) = (2r
(r 2j
+ 1/2) 28 + 1 ,
J
j=O
B(2r
~ 1)
0.
= B(2r + 1,2s + 1)
and B 2 (r,s)
+ 1){(2r + 1)B1 (r, s 1) + 2rBt(r 1, s 1)},
and
B 2 (r, s) = 2r{2r B 2 (r, s 1)
+ (2r
1)Bz(r 1, s 1) }.
24. Let R(n, k) be the number of kpermutations of n with repetition and the restriction that each element appears at most twice. Using a suitable generating function, deduce the recurrence relations
R(n + 1, k) = R(n, k) R(n,k + 1)
and R(n, k with R(n,O)
+ kR(n, k
1)
+ (~) R(n, k 2),
= nR(n,k) nG)R(n 1,k 2)
+ 2) = n(2n 1)R(n 1, k) n(n 1)R(n 2, k)
= 1, R(n, 1) = 1, R(n,2) = n 2 .
25. (Continuation). Let R(n) be the (total) number of permutations of n with repetition and the restriction that each element appears at most twice: 2n
R(n)
=L
R(n, k), n
= 0, 1, ....
k=O
Show that
R(n) n(2n 1)R(n 1) + n(n 1)R(n 2) with R(O)
= n + 1,
= 1, R(1) = 3.
26. Parcelling out procedures. Let Kn be the number of ways of parcelling out a stick of n units length into n unitary parts when, at each step, all the parts of length greater than one unit are parceled out into two parts. Show that n1
Kn
=L
j=l
KjKnj, n
= 2, 3, ... '
Kt
=1
229
6. 7. EXERCISES
and conclude that 1 Kn = Cn = n
where Cn, n
= 1, 2, ... , are the
(2n2) n1
, n = 1,2, ... ,
Catalan numbers.
27*. Triangulation of convex polygons. Let Tn be the number of triangulations of a convex ngon (ways to cut up a convex polygon of n vertices into n 2 triangles by means of n 3 nonintersecting diagonals). Show that n1 Tn = LTkTnk+l, n = 3,4, ... ' T2 = 1. k=2
Deduce the generating function t
00
T(t) = LTntn =
2 (1 v'f=4t)
n=2
and conclude that
Tn = Cn1 = n _1 1
(2nn _ 24) ,
n = 2,3, ... ,
where Cn, n = 1, 2, ... , are the Catalan numbers. 28. Convolution of Catalan numbers. Let Cn, n = 1, 2, ... , be the sequence of Catalan numbers. For fixed k = 2, 3, ... , the sequence
c~k)
n1
= L CjC~~jl)'
n
= 1, 2, ...
'
j=l
with C~ l = Cn, n = 1, 2, ... , is called kfold convolution of the sequence Cn, n = 1, 2, ... (a) Derive the generating function
1
00
Ck(t)
=L
c~k)tn
= Tk(1 v'f=4t)k
n=l
and (b) deduce the recurrence
c~k)
= c~k+l) + c~~;l)'
k
= 2,3, ... '
n
= 1,2, ...
'
1
with C~ l = C~2 l = Cn. (c) Show that
C(k) _ ·'· _ 'n:._ n  'f'nl,nk 
(2nn_k  1) 1
, n
= 1, 2, ...
, k
= 2, 3, ...
,
GENERATING FUNCTIONS
230
where '1/Jn,k, k
= 0, 1, ...
, n, n
= 0, 1, ... , are the
ballot numbers.
29*. Let Sn be the number of permutations of {1, 2, ... , n} without a succession. If Sn,k is the number of permutations of { 1, 2, ... , n} with k successions, then (see Theorem 5.6)
1) Snk,
Sn,k= ( n k
k=0,1, ... ,n1, n=1,2, ....
Using this relation and the symbolic calculus, show that
30*. (Continuation). Show that oo
S(t, u)
n
=LL
n=Ok=O
:! n
Sn+l,ktk
= (1 u) 2 eu(tl).
31. Determine the sequence of probabilities Pk, k = 0, 1, ... , with factorial moments J.l(r) = _xr, r = 0, 1, ... , where .X > 0.
32. Let k
Sn,k(x,z)=L(x+rz)n, n=0,1, ... , k=0,1, ... r=O Show that
and
33. Let k
Cn,k(x,z) = L(x+rz)n, n=0,1, ... , k=0,1, .... r=O Show that 00 un (1 Ck(u, x, z) = '"""'Cn L... ' k(x, z)n.1 = n=O
+ u)"'[(1 + u)(k+l)z 1] (1 + u ) z 1
231
6. 7. EXERCISES
and
oo oo
C(t,u,x,z)
tk un
= "'"'cnk(x,z)k = 1 ~ ~ ' . n.1
n=Ok=O
34. Let an,k
(1
+ u)x((1 + uyet(I+u)• (1 + u )z  1
= min{n, k}, k = 1, 2, ... , n = 1, 2, ....
 et]
·
Show that
35*. Generating function of the cycles of binomial coefficients. The product ak,r
=
(r+ k) (k + r) k
r
, k
= 0, 1, ...
, r
= 0, 1, ...
,
is called cycle of binomial coefficients of length 2. Show that A(t,u) =
~~ c:k) e;r)tkur = ~~ (r:k)
= ((1 t
+ u) 2 
4ut 112 .
2
tkur
Chapter 7 RECURRENCE RELATIONS
7.1
INTRODUCTION
Recurrence relations were introduced in Chapter 1, on the basic counting principles, and encountered in several places in subsequent chapters. As has been already noted, in certain enumeration problems, the number of configurations satisfying specified conditions can only be expressed recursively. Also, even when the direct expression of this number in a closed form is possible, a recurrence relation is useful at least for tabulation purposes. This chapter presents the basic methods of solving linear recurrence relations. Specifically, after the introduction of the basic notions of linear recurrence relations, the iteration method is employed to derive the solutions of linear recurrence relations of the first order. This recursive, stepbystep, derivation of the solutions contributes to the understanding of the term recurrence relation. Then, the method of characteristic roots for the solution of linear recurrence relations with constant coefficients is presented. The last section is devoted to the use of generating functions in solving linear recurrence relations with constant or variable coefficients.
7.2
BASIC NOTIONS
Consider a sequence of numbers an, n = 0, 1, ... , and let bo(n)an+r
+ bl(n)an+r1 + · · · + br(n)an =
u(n), n = 0, 1, ... , (7.1)
where u(n) and the coefficients bj(n), j = 0, 1, ... , r, with b0 (n) :j:. 0 and br(n) :j:. 0, are given functions of n. Recurrence relation (7.1) is called linear recurrence relation of order r. If u(n) = 0, n = 0, 1, ... , (7.1) is called homogeneous; otherwise is called complete. If the coefficients are
RECURRENCE RELATIONS
234
constant (independent of n), bj(n) = bj, j = 0, 1, ... , r, n = 0, 1, ... , (7.1) is called linear recurrence relation of order r with constant coefficients. In the case of a doubleindex sequence an,k, n = 0, 1, ... , k = 0, 1, ... , the recurrence relation bo,o(n, k)an+r,k+s
+ bo,1 (n, k)an+r,k+s1 + · · · + br,s(n, k)an,k = u(n, k), (7.2)
=
=
n 0, 1, ... , k 0, 1, ... , where u(n, k) and the coefficients b;,j(n, k), i = 0, 1, ... , r, j = 0, 1, ... , s, with bo,o(n, k) =f. 0 and br,s(n, k) =f. 0, are given functions ofn and k, is called linear recurrence relation of order (r, s). Solution of recurrence relation (7.1) or (7.2) in a set S is called a se
quence that makes this equation an identity in S. A general solution of recurrence relation (7.1) includes r arbitrary constants. The knowledge of the r initial conditions (values) ao, a 1 , •.. , ar 1 specifies the constants and makes the solution unique. In the case of recurrence relation (7.2), the knowledge of the r + s initial conditions (sequences) ao,k, a 1,k, ... , arI,k and an,o, an, 1 , ... , an,s 1 guarantees the uniqueness of its solution. Let us now consider the homogeneous linear recurrence relation of order r corresponding to (7.1): bo(n)an+r
+ b1(n)an+rI + · · · + br(n)an =
0, n = 0, 1, ... ,
(7.3)
where the coefficients bi(n), j = 0, 1, ... ,r, with b0 (n) =f. 0 and br(n) =f. 0, are given functions of n. Note that, if a 1 (n) and a 2 (n) are any solutions of (7.3), then c 1 a 1 (n)+cza 2 (n), where c1 and cz are arbitrary constants, is also a solution of (7.3). In general, it can be shown that the set of solutions of (7.3) constitutes a linear r dimensional vector space. Note that r solutions a 1 (n ), az (n ), ... , ar( n) of (7.3) are linearly independent if and only if their Wronski determinant az(n) az(n + 1)
Wr(n) = a1
(n
+ r 1)
a 2 (n
+ r
1) .. · ar(n
+ r
1)
is different from zero for some index n = m. Consequently, if the r solutions a 1 (n), a 2 (n), ... , ar(n) of the homogeneous linear recurrence relation of order r (7.3) are linearly independent, then they constitute a base for the r dimensional linear vector space of its solutions. Further, every solution of (7.3) is of the form (7.4)
where c1 , c2 , ...
, Cr
are arbitrary constants. The solution (7.4) is called
general solution of (7.3). In the case where r values am, am+1, •.. , am+rI
7.3. RECURRENCE RELATIONS OF THE FIRST ORDER
235
are given, the system
c1a1(m) + c2a2(m) + · · · + Crar(m) =am, c1a1(m + 1) + c2a2(m + 1) + · · · + Crar(m
+ 1) = am+l,
since Wr(m) 1 0, has a unique solution with respect to c 1 ,c2 , ... ,cr. Introducing this solution into (7.4), the unique solution of (7.3) is deduced. Further, if w(n) is a particular solution of the complete linear recurrence relation of order r (7.1), then, according to the preceding analysis, it follows directly that
is the general solution of (7.1).
7.3
RECURRENCE RELATIONS OF THE FIRST ORDER
Let us first consider the complete linear recurrence relation of the first order with variable coefficient, an+l b(n)an = u(n), n
= m,m + 1, ...
,
(7.5)
where b(n) f:: 0, n = m,m + 1, ... and m is a given nonnegative integer. The solution of this recurrence relation is deduced in the following theorem by employing the iteration method. This recursive, stepbystep, derivation of the solution serves to the better understanding of the term recurrence relation.
THEOREM7.1 The solution of the linear recurrence relation (7.5 ), with am a given initial condition, is n2
an= ambnl,m
L
+
u(r)bnl,r+l
+ u(n 1),
(7.6)
r=m
for n
= m + 1, m + 2, ... , where n
bn,r =
II b(k). k=r
(7.7)
RECURRENCE RELATIONS
236
PROOF
Introducing the transformation
hn
= an/bnl,m,
n
= m + 1, m + 2, ...
, hm =am
and setting w(n) = u(n)/bnl,m• n = m,m + 1, ... , recurrence relation (7.5) is transformed to the recurrence relation
hn+l=hn+w(n), n=m,m+1, ... , where hm = am is a given initial condition. Iterating (applying repeatedly) this recurrence, we get hm+l = hm + w(m), hm+2 = hm+l + w(m + 1) = hm + w(m) + w(m + 1)
and
hn+l = hn + w(n) = {hm + w(m) + w(m + 1) + · · · + w(n 1)} + w(n). Consequently, n
hn+l=hm+Lw(r), n=m,m+1, .... r=m
Returning to the sequence an, n = m, m the inverse transformation
+ 1, ... , the last expression, upon using
and since u(n) = w(n)bnl,m• n = m, m
+ 1, ... , implies
nl an+l = ambn,m + L u(r)bn,r+l + u(n), n = m, m + 1, ... , r=m
which is the required expression with n
I
+ 1 instead of n.
The solution of the complete linear recurrence relation of the first order with constant coefficient,
an+l ban= u(n), n = m, m
+ 1, ...
,
(7.8)
where b ::/: 0, n = m, m + 1, ... and m is a given nonnegative integer, is readily deduced from Theorem 7.1 by setting b(n) = b, n = m, m + 1, ... . Also, its particular case with u(n) = u constant for all n = m, m + 1, ... , upon using the geometric progression summation formula, is concluded. These solutions of (7.8) are given in the following corollary.
7.3. RECURRENCE RELATIONS OF THE FIRST ORDER
237
COROLLARY 7.1 The solution of the linear recurrence relation (7.8), with am a given initial condition, is
n1
an
L u(r)bnr!,
= ambnm +
= m + 1, m + 2,...
n
.
(7.9)
r=m
In particular; with u(n)
an
= u constant for all n = m, m + 1, ... ,
= { ambnm + U · (1 bnm)/(1 b),
b::/: 1,
am+ u · (n m), b = 1, for n = m
(7.10)
+ 1, m + 2, ....
REMARK 7.1 If the term am is not given, (7.6) with am = c, an arbitrary constant, constitutes a family of solutions, which is the general solution of recurrence relation (7.5). Similarly, (7.9) and (7.10) with am = care the respective general I solutionsof(7.8).
Example 7.1 Tennis tournament Let 2n players participate in a singles tennis tournament. Determine, by the aid of a recurrence relation, the number an of different pairs that can be formed for the n matches of the first round. Consider the player with the number 2n in the list. This player can be paired with any of the other 2n 1 players. Since an! different pairs can be formed by the remaining 2(n  1) players for the other n  1 matches of the first round, it follows that an = (2n 1)an1, n = 2, 3, ... , a1 = 1.
This is a homogeneous linear recurrence relation of the first order with variable coefficient. Thus from (7.6), with m = 1, u(n) = 0, b(n) = 2n + 1, n = 1, 2, ... , and a 1 = 1, we deduce the expression
an
= 1 · 3 · · · (2n 1),
n
= 1, 2, ....
Mulliplying it by 2 · 4 · · · 2n = 2nn! and dividing the resulting expression by the same number, we find the following equivalent expression for an: 1· 2. 3 ... (2n 1)(2n)
2nn!
(2n)!
= 2nn! ,
n == 1, 2, .. ..
D
Example 7.2 Regions of a plane Determine, using a recurrence relation, the number an of regions into which a plane is separated by n circles, with each pair of circles intersecting in exactly two points and with no triple of circles having a common intersecting point.
RECURRENCE RELATIONS
238
Consider n circles on a plane, with each pair of circles intersecting in exactly two points and no triple of circles having a common intersecting point. Then these circles separate the plane into an regions. Now add another circle, which intersects each of the n circles at exactly two points and does not intersect any pair of circles at any of their intersecting points. Thus, this circle intersects the n circles at a total of 2n points, passing through 2n regions. Each of these regions is separated into two regions. Consequently, the addition of the (n + 1)st circle increases the number of regions by 2n and so
an+1
= an +
2n, n
= 1, 2, ... ,
a1
= 2.
This is a complete linear recurrence relation of the first order with constant coefficient. Hence, from (7.9), with m = 1, b = 1, a 1 = 2 and u(r) = 2r, we deduce the expression
n1 an= 2+2Lr, r=1
which, upon using the arithmetic progression summation formula, reduces to
an= n(n 1)
+ 2,
n
= 1, 2,...
.
0
Example 7.3 The transfer of the Hanoi tower Consider three pegs A, B and C and n cyclic discs of different diameters {d 1, d 2 , ... , dn}. with d 1 < d2 < · · · < dn. Initially, the discs are placed on peg A in decreasing order of size from the bottom to the top. Let an be the minimum number of movements of discs required for the transfer of the tower from peg A to peg B, using peg C as an auxiliary, under the restriction that, in each movement, only one disc is transferred and no disc is placed over a disc of smaller size. In order to find a recurrence relation for an, note that then 1 discs { d 1 , d2, ... , dn 1} of the tower can be transferred to peg C after an 1 movements. Then, in one movement, the last disc dn is transferred from peg C to peg B. Finally, the tower of then 1 discs is transferred from peg C to peg B, over the disc dn, after an1 movements. Consequently,
an
= 2an1 + 1,
n
= 2, 3, ...
, a1
= 1.
This is a complete recurrence relation of the first order with constant coefficient. Hence, from (7.10), with m = 1, a 1 = 1 and u = 1, we get
an
= 2n1
 (1  2n1)
= 2n 1.
According to legend, at the creation God established the tower of Hanoi at the temple of Benares with n = 64 gold discs. Since then, the tower is being transferred by priests, with the prediction that when the transfer of the tower is completed the world will vanish. Note that a2 = 3, a3 = 7, a 4 = 15, a 1s = 65,535, a32 = 4,294,967,295 and as 4 = 18,446,744,073,709,551,615. 0
7.4. THE METHOD OF CHARACTERISTIC ROOTS
7.4
239
THE METHOD OF CHARACTERISTIC ROOTS
Let us first consider the homogeneous linear recurrence relation of the second order with constant coefficients: (7.11) where b2 =f. 0 and m is a given nonnegative integer. Clearly, a sequence of the form a(n) = pn, n = m, m + 1, ... , is a solution of (7.11) if and only if p is a root of the equation
tn+ 2
+ b1tn+1 + b2tn = 0.
Dividing both sides of this equation by tn, for t equation of the second order,
=f. 0,
we obtain the following (7.12)
which is called the characteristic equation of the recurrence relation (7.11). Further, the discriminant bt 4~ of the characteristic equation determines the kind of its roots. Specifically, if bt 4b2 > 0, (7.12) has two distinct real roots, PI=
b~ 
..Jbr  4b2 2
, P2 =
bi +
..Jbr  4b2 2
,
while, if bi 4b2 = 0, it has one double root p = bi/2. Also, ifbt4b2 (7.12) has two conjugate complex roots Pl =
< 0,
bi i..J4b2 bi bl + i..J4b2 bi , P2 = , 2 2
where i = A is the imaginary unit. These complex numbers, on using the length p = b2 and argument 0, with tanO = ..J4b1  bi/b1, may be written in trigonometric form as PI =p(cosB+isinO), p 2 =p(cosOisinO). The general solution of recurrence relation (7.11) depends on the kind of roots of the characteristic equation (7.12). More precisely, we prove the following theorem. THEOREM7.2 The general solution of the homogeneous linear recurrence relation of the second order with constant coefficients, (7. 11 ), is given by (7.13)
RECURRENCE RELATIONS
240
where c 1 and c2 are arbitrary constants. Further, (7.12) has: (a) two distinct real roots Pl and p2 , then
if the characteristic equation
(7.14)
(b) one double root p, then (7.15)
(c) two conjugate complex roots Pl and P2 = and 0, respectively, then
p1 ,
with length p and argument() (7.16)
PROOF (a) Since p 1 and p2 are roots of the characteristic equation (7.12), a 1 ( n) = pf and a 2 ( n) = p~ are solutions of (7 .11 ). The Wronski determinant of these solutions is
with p 1 P2 "/: 0, since b2 "/: 0, and Pl "/: P2. Hence, W2 (n) "/: 0 and so the solutions in (7.14) constitute a base of the twodimensional linear vector space of the solutions of (7 .11 ). (b) Since pis a solution of the characteristic equation (7.12), a(n) = pn is a solution of (7 .11 ). A base of the twodimensional vector space of the solutions of (7 .11) is composed of two linearly independent solutions. In order to determine them, we set (7.17) where hn, n = m, m (7.11), we get
+ 1, ... , is a sequence to be determined. 2 p hn+2
Introducing it into
+ b1phn+l + b2hn = 0,
and, since b1 2p, b2 = p 2 , with p "/: 0, we deduce for hn, n m + 1, ... , the recurrence relation hn+2  2hn+l
+ hn = 0, n = m, m + 1, ....
Putting hn+l  hn
= 9n,
n
= m, m + 1, ...
)
we get the homogeneous linear recurrence relation of the first order 9n+l 9n
= 0,
n
= m, m + 1, ...
,
m,
7.4. THE METHOD OF CHARACTERISTIC ROOTS
241
the general solution of which, according to Remark 7 .l, is given by Yn = c1, n = m, m
+ 1, ....
Consequently,
hn+lhn=CI, n=m,m+1, .... The general solution of this recurrence relation, again according to Remark 7.1, is given by hn = c1 + c2n, n = m, m + 1, .... Introducing this general solution into (7.17), we deduce the required expression of the general solution of (7.11). Note that the Wronski detenninant of the solutions a 1 (n) = pn and a2(n) = npn is given by
and since p ;f. 0, W 2 (n) ;f. 0. (c) Since p and p 2 = ih are conjugate complex roots of the characteristic equation (7.12), an = pf and iin = pf are conjugate complex solutions of (7.11), which may be written in trigonometric fonn as
an = pn cos( nO) + ipn sin( nO), lin = pn cos( nO)  ipn sin( nO). Interested only in the real solutions of recurrence relation (7.11), it is necessary to isolate them. In this respect, note that, if an = a 1(n) + ia2 (n) is a complex solution of the characteristic equation (7 .12), then each of a 1(n) and a2 (n) is also a solution of the same equation since [a 1(n + 2) + b1a 1(n + 1) + b2a 1(n)] +i[a 2(n + 2) + b1a 2 (n + 1) + b2a2(n)]
= 0,
n = m, m + 1, ... ,
implies
and
a2(n + 2) + b1a2(n + 1) + b2a2(n)
= 0,
Consequently, a1 (n) = pn cos( nO), a2(n)
n = m, m + 1, ....
= pn sin(nO)
are two real solutions of (7 .11 ). The Wronski determinant of these solutions is given by
I
pn sin( nO) pn+I sin(nO
+ 0)
RECURRENCE RELATIONS
242
Thus, W 2 (n) = ln+ 1 [cos(n0) sin( nO+ 0) sin( nO) cos( nO+ 0)]
and since sin(nO + 0) = sin(nO) cos 0 +cos( nO) sin 0, cos( nO+ 0) =cos( nO) cos 0 sin(nO) sin 0, the Wronski determinant reduces to W2(n) = ln+I sinO[cos 2 (nO) + sin 2 (n0)] = /n+l sinO,
which, by virtue of p f: 0 and sin 0 f: 0, implies W 2 ( n) ¥ 0. Hence, the solutions in (7 .16) constitute a base of the twodimensional linear vector space of I the solutions of (7 .11) and the proof of the theorem is completed.
Example 7.4 Gambler's ruin A gambler plays a series of games of chance against a casino (or against an adversary). In any game the probability of the gambler to win $1 is p and to loose $1 is q = 1  p. Assume that initially the gambler possesses n dollars and the casino (or his adversary) possesses k n dollars. Find the probability Pn of the gambler's ruin. In the first game, the gambler may either win or loose $1, with probabilities p and q, respectively. If he wins the first game, then his fortune is increased ton+ 1 and so his ruin probability becomes Pn+l· If he looses the first game, his fortune is decreased to n  1 and so his ruin probability becomes Pnl· Hence Pn = PPn+l
+ QPn1,
n = 1, 2, ... , k 1
or PPn+2  Pn+l
+ QPn = 0,
n
= 0, 1, ...
, k 2,
with Po= 1, Pk
= 0.
This is a linear recurrence relation with constant coefficients. The roots of its characteristic equation, 2 pt  t + q = 0, are PI = 1 and P2 = qfp. If p ¥ 1/2, whence q ¥ 1/2, these two roots are distinct, while if p = 1/2, whence q = 1/2, p2 = p 1 = 1. Thus, according to Theorem 7 .2, the general solution of the recurrence relation: (a) for p ¥ 1/2 is
= C1 + C2 ( q / P) n,
Pn
n
= 0, 1, ...
, k,
while (b) for p = 1/2 is Pn = c1 +c2n, n = 0,1, ... ,k.
7.4. THE METHOD OF CHARACTERISTIC ROOTS
Using the initial conditions we get: (a) for p
243
:f. 1/2,
C1 + C2 = 1, C1 + C2(qfpl = 0, whence
Ct =
(qjp)k 1 1 (qjp)k' C2 = 1 (qjp)k'
while (b) for p = 1/2,
c1 = 1, c1 + c2 k = 0, whence
c1 = 1, c2 = 
1
k.
Consequently, the solution of the recurrence relation that satisfies the initial conditions: (a) for p :f. 1/2 is
Pn
=
(qjp)n _ (qjp)k , n _ (qjp)k 1
= 0, 1, ... , k
and (b) for p = 1/2 is
kn
Pn = k' n = 0, 1, ... , k.
D
Example 7.5 Determine the sequence an, n = 0, 1, ... , for which the general term is the arithmetic mean of its two preceding terms and the first two terms are 0 and 1. The sequence an, n = 0, 1, ... , according to its definition, satisfies the linear recurrence relation 1
an= 2(ant + an2), n = 2, 3, ... , or
2an+2  an+t an = 0, n = 0, 1, ... , with initial conditions a 0
= 0 and a 1 = 1. The roots of the characteristic equation, 2t 2

t 1 = 0,
are Pt = 1 and P2 = 1/2. Thus, according to Theorem 7.2, the general solution of the recurrence relation is
( l)n an=c1+c2, n=0,1, .... 2n The initial conditions require that 1
c1 + C2 = 0, C1  2C2 = 1.
RECURRENCE RELATIONS
244
Hence
and
Let us now consider the complete linear recurrence relation of the second order with constant coefficients:
(7 .18) where b2 i 0 and m is a given nonnegative integer. According to what was stated in Section 7.2, if c 1 a 1 (n) + c2a 2(n), n = m, m + 1, ... , is the general solution of the corresponding homogeneous linear recurrence relation (7.11) and w(n), n = m, m + 1, ... , is a particular solution of (7.18), then the general solution of (7.18) is given by
Consequently, the derivation of a particular solution of (7.18) is what remains to be done. Consider the case
u(n)=bntuj(~),
n=m,m+1, ... ,
(7.19)
J
j=O
where b and ui, j = 0, 1, ... , s, are constants. The method of arbitrary constants for the derivation of a particular solution of (7.18) is stated in the following theorem. THEOREM7.3 A particular solution of the complete linear recurrence relation of the second order with constant coefficients (7.18) in the case the function u(n) is given by (7.19), with b a root of the characteristic polynomial ¢(t) = t 2 + b1 t + b2 of multiplicity k ~ 0, is given by
w(n) = bnk
I: j=k
Wj
(~),
n = m, m
+ 1, ...
, k = 0, 1, 2,
(7.20)
J
where the coefficients Wj. j = k, k of the s + 1 equations: (a) fork= 0,
+ 1, ...
,k
+ s, are determined by the system
7.4. THE METHOD OF CHARACTERISTIC ROOTS
1/J(b)ws1 ¢(b)wj
+ b¢' (b)ws
=
+ b¢'(b)wj+ 1 + b2wj+2 = Uj,
245
(7.21)
U 8 I,
j
= 0, 1, ...
, s 2,
(b)for k = 1.
¢'(b)wj+I
+ bwj+2 = Uj,
j
= 0, 1, ...
, s 1,
(7.22)
(c)fork=2, Wj+2=Uj, j=0,1, ... ,s,
(7.23)
with ¢'(b) the derivative of the characteristic polynomial ¢(t) at the point t =b. PROOF Introducing (7.19) into (7.18) and requiring (7.20) to satisfy theresulting recurrence relation, we get
Using the recurrence relations
and
we deduce the relation
+b1b
~ Wj (~) + b1b ~ Wj ( . : ) 1 j=k J j=k J
+ b2 ~ Wj (~) j=k
J
= bk
t
Uj
j=O
(~). J
Introducing the characteristic polynomial ¢(b) = b2 + b1b + b2 and its derivative 1/J'(b) = 2b + b1 , we get
~ ¢(b)wj (~) +I:j=k b¢'(b)wj (J. : 1) J=k J
+I:
t
(~).
2 b wi ( . : ) = bk Uj j=k J 2 j=O J
Equating the coefficients of the binomials {]) of both sides of this expression and since (a) fork = 0, ¢(b) f. 0, (b) fork = 1, ¢(b) = 0, ¢'(b) f. 0 and
RECURRENCE RELATIONS
246
=
0, ¢>'(b)
2, ¢>(b) (c) fork respectively. I
= 0,
we conclude (7.21), (7.22) and (7.23),
Example 7.6 Expected time to gambler's ruin Consider the gambler's ruin problem of Example 7.4 and let dn be the expected value of the number of games until the gambler is ruined. Derive a recurrence relation for dn and from that deduce an explicit expression for the expected time to gambler's ruin. In the first game, the gambler may either win or loose $ 1, with probabilities p and q, respectively. If he wins the first game, then his fortune is increased ton+ 1 and, after that game, the expected time to his ruin becomes dn+i· If he looses the first game, his fortune is decreased to n  1 and, after that game, the expected time to his ruin becomes dni· In both cases, adding the first game we deduce the following recurrence relation
dn = p(dn+l
+ 1) + q(dni + 1),
n = 1, 2, ... , k 1,
or
dn+2 (1/p)dn1
+ (qfp)dn
= 1/p, n = 0, 1, ... , k 2,
with initial conditions
do= 0, dk = 0. This is a complete linear recurrence relation with constant coefficients. The general solution of the corresponding homogeneous recurrence relation,
dn+2  (1/p)dn+l
+ (qfp)dn
= 0, n = 0, 1, ... , k 2,
according to Example 7.4: (a) for p :j: 1/2 is
while (b) for p = 1/2 is
dn = c1
+ c2n,
n
= 0, 1, ...
, k.
The function u( n) of the complete linear recurrence relation is of the form (7 .19), with b = 1, s = 0 and u 0 = 1/p. Note that (a) if p "1 1/2, b = 1 is a simple root (k = 1), while (b) if p = 1/2, b = 1 is a double root (k = 2) of the characteristic polynomial ¢>(t) = t 2  (1/p)t + (qfp). Thus, according to Theorem 7.3, a particular solution of the complete linear recurrence relation with constant coefficients: (a) for p "1 1/2 is given by w(n) = w 1 n, where, by (7.22) and since ¢>'(1) = (2p 1)/p. we get w 1 = 1/(1 2p) and so
n w(n) = 1 2p'
7.4. THE METHOD OF CHARACTERISTIC ROOTS
247
Also, a particular solution (b) for p = 1/2 is given by w(n) = w 2 n(n 1)/2, where, by (7.23), w 2 = 2 and so
w(n)
= n(n 1).
Consequently, the general solution of the complete linear recurrence relation with constant coefficients: (a) for p ::/ 1/2 is given by
dn = c1
+ c2
(~) n + 1 .:: 2P,
n = 0, 1, ... , k,
while (b) for p = 1/2 is given by
+ c2n n(n 1),
dn = c1
n = 0, 1, ... , k.
Using the initial conditions, we get (a) for p ::/ 1/2, Ci
k + C2 = 0, Ci + C2 ( q)k +   = 0,
p
1 2p
whence
k c~·

1 2p
k 1 1 · ' 1  (qfp)k, c 2 = 1  2p 1  (qjp)k
while (b) for p = 1/2. c1=0, c2kk(k1)=0, whence C1 = 0, C2 = (k 1). Therefore, the unique solution of the complete linear recurrence relation: (a) for p ::/ 1/2 is given by
dn
= _n_ 1 2p
_ _ k_ . 1 (qfp)n '':',k n 1 2p 1 (qjp) ,
= 0, 1, ... , k
= n(k n),
, k.
and (b) for p = 1/2 is
dn
n
= 0, 1, ...
0
Let us finally consider the general case of the linear recurrence relation of the rth order with constant coefficients:
an+r
+ blan+r1
+···+bran= u(n), n
= m, m + 1, ...
,
(7.24)
RECURRENCE RELATIONS
248
where br 1 0 and m is a given nonnegative integer. Let
u(n)=bntuj(~),
n=m,m+1, ....
(7.25)
J
j=O
The corresponding homogeneous linear recurrence relation of the rth order with constant coefficients is given by (7.26)
where br 1 0 and m is a given nonnegative integer. The characteristic polynomial of this recurrence relation is (7.27)
The general solution of (7.26) and a particular solution of (7.24) in the case where u(n) is given by (7.25) are stated in the following two theorems. The proofs, similar to the proofs of Theorems 7.2 and 7.3, respectively, are omitted. THEOREM7.4 The general solution of the homogeneous linear recurrence relation of orderr with constant coefficients, (7.26), is given by Vt
an =
""'( L....., Cj,l
+ Cj,2n + ... + Cj,k;n k·1) ] Pjn
j=l V2
+
L
(cj,l + Cj,2n + · · · + Cj,k;nk; 1 )p'J cos(nOi)
j=v+l V2
1 + ""' )pnJ sin(nO·) L....., (d·1 ], + d·], 2n + .. · + d·J, k·nk;] J ' j=v+l
where Pi is a real root of the characteristic polynomial (7.27) of multiplicity ki ~ 0, j = 1,2, ... ,v 1 , withk 1 +k2 + ... +kv1 = vandpj(cosOi +isinBj) is a complex root of multiplicity ki ~ 0, j = v + 1, v + 2, ... , v 2, with 2(kv+l + kv+2 + ... + kv2) = r  v.
THEOREM7.5 A particular solution of the complete linear recurrence relation of order r with constant coefficients, (7.24 ), in the case the function u( n) is given by (7.25 ), with b a root of the characteristic polynomial (7.27) of multiplicity k ~ 0, is given by w(n) = bnk
k+s L j=k
Wj
(
~) J
,
n = m, m
+ 1, ...
, k = 0, 1, ... , r,
7.4. THE METHOD OF CHARACTERISTIC ROOTS
where the coefficients Wj, j = k, k of the s + 1 equations s
L
+ 1, ...
,k
+ s are determined by the system
bvj (v
+k
249
_ j)!(v+kj)(b)wv = Uj, j
= 0, 1, ... , s,
II=J
with (v+kj) (b) denoting the derivative of order v polynomial¢(t) at the point t = b.
+k
 j of the characteristic
Example 7.7 Determine the general solution of the recurrence relation
This relation is written as
which is a complete linear recurrence relation of the fourth order with constant coefficients. The corresponding homogeneous recurrence relation is
Its characteristic polynomial is given by ¢( t) = t 4

2t3 + 2t 2

2t + 1.
Note that ¢(1) = 0 and ¢'(t) = 2(2t3  3t2 + 2t 1), whence ¢'(1) = 0. Thus p = 1 is a double real root of the characteristic polynomial. Dividing it by (t 1 ) 2 we deduce the expression
which implies that the characteristic polynomial also has the conjugate complex roots Pl = i and P2 = i, with i = A. The complex root Pl = i has length I and argument 0 = 1r /2. The general solution of the homogeneous recurrence relation, according to Theorem 7 .4, is given by
an= c1 + c2n
+ c3 cos(nn/2) + c4 sin(nn/2),
n
= 0, 1, ....
Further, the function u(n) = (5n 2)2n of the complete recurrence relation is of the form (7.25), with b = 2, s = 1, u 0 = 2 and u 1 = 5. Thus, from Theorem 7.5 and since b = 2 is not root of the characteristic polynomial, whence k = 0, we conclude that w(n) = (wo +w1n)2n, n = 0, 1, ...
RECURRENCE RELATIONS
250
is a particular solution of the complete recurrence relation. The coefficients w 0 and w 1 are determined from the equations: ¢(2)wl Since u 0
= u1,
¢(2)w0
+ 2¢' (2)wl = uo.
= 2, u 1 = 5, ¢(2) = 5 and¢' (2) = 5wl = 5, 5wo
14, these equations become
+ 28w1
= 2
and so w 1 = 1 and w0 = 6. Therefore, w(n) = (n 6)2n and the general solution of the complete recurrence relation is given by an = c 1 + c 2 n forn=0,1, ....
7.5
+ c3 cos(mr /2) + c4 sin(mr /2) + (n 6)2n,
0
THE METHOD OF GENERATING FUNCTIONS
In Chapter 6, we have examined how generating functions may be used in coping with combinatorial problems. The direct determination of generating functions by invoking the assumptions and restrictions imposed by these problems was a crucial characteristic of many of them. However, a direct determination of a suitable generating function is not always possible. In many problems, a linear recurrence relation for the sequence of the numbers under consideration can be derived by appealing to the assumptions and restrictions imposed by them. We have already dealt with several such problems. In this way, the interest is converted to the solution of a linear recurrence relation. In this section, generating functions are used in solving a linear recurrence relation of order r with constant or variable coefficients and a bivariate linear recurrence relation of order (r, s) with constant coefficients. Let us, first, consider the linear recurrence relation with constant coefficients, (7.28) where b0 ::/: 0, br ::/: 0 and u(n), n = 0, 1, ... , is a given sequence (function) of n. The determination of~ sequence ak, k = 0, 1, ... , satisfying this recurrence relation requires the knowledge of r initial values ao, a1, ... , ar1· Since the coefficients b0 , b1, ... , brl are known, the numbers k ck
= l:aJbkj, j=O
k
= 0, 1, ...
,r 1
(7.29)
7.5. THE METHOD OF GENERATING FUNCTIONS
251
can be calculated. The derivation of the solution ak, k = 0, 1, ... , of (7.28) by using a generating function is carried out in two steps. At the first step, on using (7.28) and its initial values, the generating function of the sequence ak, k = 0, 1, ... is derived in the next theorem. THEOREM7.6 Letak, k = 0, 1, ... , be a sequence satisfying the linear recurrence relation (7.28). Then the generatingfunction 00
A(t)
= L>ktk,
(7.30)
k=O is given by A()= C(t) + trU(t) t Br(t) '
(7.31)
where r
r1
Br(t) = L>ktk, C(t) = :~:::>ktk k=O k=O
(7.32)
and 00
U(t) =
L u(n)tn.
(7.33)
n=O
PROOF
bo
Multiplying (7.28) by tn+r and summing for n = 0, 1, ... , we get
00
00
00
00
n=O
n=O
n=O
n=O
2: an+rtn+r +b1t 2: an+rltn+rl +·. ·+brtr L:antn = tr 2: u(n)tn.
Since
oo
s1
L an+stn+s = A(t)  L aktk'
s = 1, 2, 'r, n=O k=O upon introducing the generating functions (7 .30) and (7 .33), we deduce the relation
and so
0
0
0
RECURRENCE RELATIONS
252
Clearly, on using (7.29) and (7.32), we deduce expression (7.31).
I
After the derivation of expression (7.31) for the generating function A(t), the next step is the determination of the sequence ak, k = 0, 1, ... , which requires the expansion of the righthand side of (7.31) into powers oft. Note that the polynomial Br(t) is connected with the characteristic polynomial r
L bktrk
c/>r(t) =
k=O by the relation Br(t) = tr¢>r(1/t).
Further, when the r initial values
a 0,a1, ... ,arl are not given, the numbers ck, k = 0,1, ... ,r 1, de
fined by (7.29), enter in expression (7.31) of the generating function and, consequently, in the solution of (7.28) as r arbitrary constants. In order to expand the generating function (7.31), we express it as a sum of algebraic fractions of the form, c/(1 pt)k, by the aid of the method of partial fractions. Specifically, if p 1 , P2, ... , Ps are the roots of the characteristic polynomial ¢>r (t), with multiplicities m 1 , m 2 , ... , m 8 , respectively, so that m1 + m2 + · · · + m 8 = r, then
c/>r(t)
= bo(t PI)m' (t P2)m
2
• • •
(t Ps)m•
and the denominator in (7.31) is factored as follows:
Br(t)
= bo(1 p1t)m' (1 P2t)m
2
• • •
(1 p8 t)m•.
Therefore, the generating function (7.31) may be written in the form 8
( )
A t =
s
fflj
~~
""'""'
Ck,j
(1 Pit)k
II 1 t U(t) i=l (1 Pit)m;'
1 r
+ bo
where ck,j, k = 1,2, ... ,mj, j = 1,2, ... ,s, are constants to be determined from the system of r equations that follows from the equation of the coefficients of tk, for k = 0, 1, ... , r 1, in both members of the identity
C(t)  ~ ~ Ck,j Br(t)  L..t L..t (1  p ·t)k · ]=l k=l J The generating function A(t), on using Newton's general binomial formula, may be expanded into powers of t as
7.5. THE METHOD OF GENERATING FUNCTIONS
253
Thus
an=
t,~Ck,je::~ 1)pj +b0
1
~ u(n k r) k=O
LIT (mj:. ~; j=l
1
)P7
3
,
(7.34)
J
where, in the inner sum of the second summand, the summation is extended over all n 3 = 0, 1, ... , k, j = 1, 2, ... , s, with n 1 + n 2 + · · · + n 8 = k. The use of generating functions in solving linear recurrence relations is illustrated in the following examples.
Example 7.8 Fibonacci numbers revisited Consider n points, each being either 0 or 1, and let Qn be the number of arrangements of them in a row so that no two zeros are consecutive. The numbers Ia = 1, II = 1, In = Qn1• n = 2, 3, ... , are the Fibonacci numbers (see Example 2.28). Derive a recurrence relation for the numbers ln. n = 0, 1, ... , and solve it by using a generating function. In order to derive a recurrence relation for the numbers ln. n = 0, 1, ... , note that, in an arrangement of n points in a row, each being either 0 or 1 so that no two zeros are consecutive, the first position is occupied either by 1 or by 0. If it is occupied by 1, then there are In arrangements of the remaining n  1 points so that no two zeros are consecutive. If the first position is occupied by 0, the second position is necessarily occupied by 1 and then there are lnl arrangements of the remaining n  2 points so that no two zeros are consecutive. Therefore, according to the addition principle, we deduce the following recurrence relation
In+ 1 = In + Inl, n
= 3, 4, ....
In particular, for n = 1, 2 we have the initial conditions h = 2, h = 3. Note that these initial conditions can be replaced by the values of Ia and II, which it should be noted that they do not have any combinatorial meaning. The values Ia and II are deduced from the extension of the recurrence relation at the point n = 1: h  h  II = 0, whence II = h  h = 3  2 = 1 and at the point n = 2: h  II  Ia = 0, whence Ia = h  II = 2  1 = 1. Thus, we may have the second order recurrence relation with constant coefficients,
ln+2
ln+l In= o, n = o, 1, ...
(7.35)
and initial conditions Ia = 1 and !I = 1. Using recurrence relation (7.35) and its initial conditions, Table 7.1 of Fibonacci numbers ln. n = 0, 1, ... , 14, is constructed.
RECURRENCE RELATIONS
254
Table 7.1
Fibonacci Numbers
0
2 2
3 3
4 5
5 8
6 13
7 21
8 34
9 55
10 89
II 144
12 233
13 377
14 610
The generating function
n=O is determined by multiplying the recurrence relation (7.35) by tn+ 2 and summing forn = 0, 1, .... Then 00
L
fn+2tn+ 2  t
n=O and
00
00
n=O
n=O
L fn+ltn+I t2 L
fntn = 0
[F(t) 1 t]  t[F(t)  1] t 2F(t)
= 0,
whence
F(t) =
1 • 1  t  t2
(7.36)
In order to expand it into powers of t, we calculate the roots of the characteristic polynomial (h(t) = t 2  t 1. These roots are
PI
1+VS
= 2,
P2
1vs
= 2
and so
Then, the generating function (7.36) is written in the form 1
F(t) where
CI
C1
C2
= (1 Pit)(1 p2t) = 1 Pit+
and c2 are constants to be determined. Since
these constants satisfy the following system of equations: C1
+ C2
= 1,
C1{J2
+ C2Pl
= 0.
1 P2t'
7.5. THE METHOD OF GENERATING FUNCTIONS
Therefore, C1
P1
P1
= 'P1 Pz
v's'
cz
Pz
255
Pz
=  Pl.....:....:_'P2 =  v'5
and
whence
!.=
v'5r(' v'5rl
~{ (1+2
2
n=D,i, ... ·
0
Example 7.9 Partial sums Consider a sequence ak, k = 0, 1, ... , and let 00
A(t) = L aktk. k=O
The sequence of partial sums n
sn=Lak, n=0,1, ... , k=O
is of interest, especially in probability theory. Clearly, this sequence satisfies the relation Sn Sn1 =an, n = 1,2, ... , which is a first order linear recurrence relation with constant coefficients. Multiplying it by tn and summing for n = 1, 2, ... , we get 00
00
L Sntn t L n=1
00
Sn1tn
1
= L
n=1
antn.
n=1
Thus, the generating function 00
S(t) = L Sntn n=O
satisfies the relation S(t) s0

tS(t) = A(t) a0 and since s 0 = a0 ,
S(t) = (1 t) 1 A(t). Let us apply this result to the sequence
ak=(x+~ 1 ),
k=O,l, ... ,
(7.37)
RECURRENCE RELATIONS
256
where xis a real number. Then, according to Newton's general binomial formula,
the generating function of the partial sum
on using (7.37), is obtained as 00
S(t) =
L Sntn = (1 t)x
1
.
n==O
This generating function is expanded into powers oft as
and so
Consider now the linear recurrence relation of order r with variable coefficients:
bo(n)an+r
+ b1(n)an+r1 + · · · + br(n)an = u(n),
n
= 0, 1, ... , (7.38)
where u(n) and the coefficients bj(n), j = 0,1, ... ,r, with b0 (n) :f. 0 and br(n) :f. 0, are given functions of n. The derivation of the solution ak, k = 0, 1, ... , of this recurrence, by using a generating function, when the coefficients bj(n) are polynomials in n of order sj, j = 0, 1, ... , r, may be carried out as follows. Multiplying (7.38) by tn+r and summing for n = 0, 1, ... , we get for the generating function 00
A(t) =
L aktk, k==O
a linear differential equation of order at most s = max{ s 0 , s1, ... , Sr}. The solution of this equation gives A(t) and its expansion into powers oft yields ak, j = 0, 1, .... This method is illustrated in the following examples.
7.5. THE METHOD OF GENERATING FUNCTIONS
257
Example 7.10 Let ak, k = 0, I, ... , n, be the number of kcombinations of the set Wn { w 1 , w2, ... , Wn}. A recurrence relation for the sequence ak. k = 0, I, ... , n may be obtained as follows. Attaching in every kcombination { Wi 1 , Wi 2 , ••• , wik} of Wn any one of then k elements of Wn {Wi 1 , Wi 2 , ••• , wik }, we get all the (k +I)combinations { Wi 1 , Wi 2 , ••• , wik+l} ofWn, k +I times each. Specifically, the (k +I)combination { wii, Wi 2 , ••• , Wik+'} of Wn, is obtained from each of the kcombinations
by attaching the elements integer n, we get
wi,, Wi 2
, ••• ,
respectively. Thus, for a fixed
Wik+l,
(k + I)ak+l  (n k)ak = 0, k = 0, I, ... , n I, ao = I,
which is a first order linear recurrence relation with variable coefficients. Its solution may be obtained by using a generating function n
A(t) =
L>ktk, k=O
as follows.
Multiplying the recurrence relation by
tk
and summing for k
0, I, ... , n I, we get n1
n
n
L
L
L(k + I)ak+1tk n aktk + t kaktkl = 0. k=O k=O k=l
Further, :t A(t) =
n
n1
k=l
r=O
L kaktkl =
and so
d
(I+ t) dt A(t) Therefore
1
1 0
dA(s) n A(s) 
L(r + I)ar+lt''
= nA(t).
1 1
0
dsI+ s
and log A(t) log A(O) = n log(I + t)  n log 1. Since A(O) = a 0 = I and log I = 0, it follows that
RECURRENCE RELATIONS
258
and so
ak =
(~),
Example 7.11 Consider the sequence ao = 1, an =
0
k=0,1, ... ,n.
e;), n = 1, 2, ... , and let
In order to determine this generating function, note that
an 1 = (2n + 2) = 2(2n + 1) (2n) = 2(2n + 1) an. + n+1 n+1 n n+1 Thus, the sequence an, n = 0, 1, ... , satisfies the first order linear recurrence relation with variable coefficients:
(n + 1)an+1 2(2n + 1)an = 0, n = 0, 1, ... , ao = 1. Multiplying it by tn and summing for n = 0, 1, ... , since d
dtA(t)
00
00
n=1
k=O
= L nantn 1 = L(k + 1)ak+1tk,
we get the first order differential equation d (1 4t) dt A(t) = 2A(t).
Hence
t }0
t
dA(s) 2ds A(s) = } 0 1 4s
and so log A(t) log A(O) = 
1
{log(1  4t) log 1}.
2
Since A(O) = a0 = 1 and log 1 = 0, it follows that
A(t)
= (1 4t) 112 .
0
Let us finally consider the bivariate linear recurrence relation of order (r, s) with constant coefficients:
bo,oan+r,k+s + bo,!an+r,k+s1 + b1,oan+r1,k+s +
0
0
0
+ br,san,k = w(n, k), (7.39)
7.5. THE METHOD OF GENERATING FUNCTIONS
259
where w(n,k) is a given function and the coefficients bi,j, i = 0,1, ... ,n, The determination of the doubleindex sequence an,k, n = 0, 1, ... , k = 0, 1, ... , that satisfies the recurrence relation (7.39) requires the knowledge of r + s independent sequences,
j = 0, 1, ... , k, are given constants.
ai,k, k=0,1, ... , i=0,1, ... ,r1,
(7.40)
an,j, n = 0, 1, ... , j = 0, 1, ... , s 1,
(7 .41)
the initial conditions. The derivation of the solution of (7.39) by using a generating function is carried out in two steps. In the first step, we determine the bivariate generating function 00
00
A(t,u) = L Lan,ktkun, n=Dk=O which, on using the sequence of generating functions 00
An(t) = L an,ktk, n = 0, 1, ... , k=O is written in the form 00
A(t, u) = L An(t)un. n=O For the derivation of the sequence of generating functions An(t), n 0, 1, ... , we multiply (7.39) by tk+ 8 and sum fork= 0, 1, .... Then r
L i=O
s
oo
oo
L bi,jti L an+ri,k+8jtk+oj = t 8 L w(n, k)tk, n j=O k=O k=O
= 0, 1, ...
and, since oo
81
""a ·k ·tk+ 8j A · ·tmj , ~ n+r~, +SJ  n+r~·(t) ""a ~ n+r~,mJ k=O m=j for j = 0, 1, ... , s 1, we deduce the relation r
L i=O
r
8
oo
81 81
L bi,jtj An+ri(t)  L j=O i=O
L L an+ri,mjtm = t 8 L w(n, k)tk' j=O m=j k=O
for n = 0, 1, .... Introducing 8
bi(t) =
'2.:: bi,jti, j=O
i
= 0, 1, ...
, r,
RECURRENCE RELATIONS
260
and
00
Wn(t) = L w(n, k)tk, k=O this relation may be written in the form bo (t)An+r(t)
+ b1 (t)An+r1 (t) + 000+ br(t)An(t) = Cn (t) + t"Wn(t), (7.42)
for n = 0, 1, 0000 The last relation is a linear recurrence relation of order r for the sequence of generating functions An(t), n = 0, 1, 000 with constant, with respect ton, coefficients bi(t), i = 0, 1, 000 , ro Note that the generating functions 00
ai,kl, i = 0, 1, 000 , r  1,
Ai(t) = L
(7.43)
k=O which, according to the initial conditions (7.40), are known, constitute the initial conditions of (7.42)0 Considering t as a parameter, the recurrence relation (7.42) is solved in exactly the same way as the recurrence relation (7028)0 So, multiplying (7.42) by un+r and summing for n = 0, 1, 000, we deduce the relation
A( t, u )
= C(t,u) +urt"W(t,u) Br,s(t, U)
where
r
r
i=O
=L
L bi,jtiui,
i=O
00
W(t, u)
s
= L bi(t)ui = L
Br,s(t, u)
00
Wn(t)un
=L
'
j=O
00
L w(n, k)tkun
n=O
n=Ok=O
Ani(t)bi(t)un
+L
and r1
C(t, u) =
s1 m
n
LL
L A:n_j(u)bj(u)tm
m=Oj=O
n=Oi=O r1 s1
n
L L
n
LLani,mjbi,jtmun, n=O m=O i=O j=O
with
00
Aj(u)=Lan,jUn, j=0,1,00o,s1, n=O
(7044)
7.5. THE METHOD OF GENERATING FUNCTIONS
261
known generating functions, according to the initial conditions (7.41), and r
bj(u)
= L)i,ju,
j
= 0, 1, ...
, s.
i=O
After the derivation of the generating function A(t, u), the next step is the determination of the sequence an,k, k = 0, 1, ... , n = 0, 1, ... , which requires the expansion of the righthand member of (7.44) into powers of t and u. For this purpose, considering t as a parameter, we expand this function into powers of u, and then we expand the resulting expression into powers of t. This procedure is further clarified in the following examples. Example 7.12 Sums of sums Let us consider a sequence ak, k
= 0, 1, ... , with generating function 00
A(t)
=L
aktk.
k=O
Further, consider the sum j s1,j
= ~ ai, j = 0, 1, ... , i=O
the double sum r
r
j
B2,r=~~ai=Lsl,j, r=0,1, ... j=O i=O
j=O
and, generally, the ntuple sum k
Sn,k = ~ Snl,r, k = 0, 1, ... , n = 0, 1, .... r=O
Clearly, Sn,k  Sn,k1 = Snl,k, k = 1, 2, ... , n = 1, 2, ....
Multiplying this relation by tk and summing fork = 1, 2, ... , we get 00
00
00
~ Sn,ktk t ~ Sn,kltkl k=l
and, since sn,o
k=l
= ~ Snl,ktk k=l
= Bn 1,o, we deduce for the generating function 00
Sn(t)
=L k=O
Sn,ktk ', n
= 0, 1, ...
,
RECURRENCE RELATIONS
262
the first order linear recurrence relation with constant, with respect ton, coefficients
(1 t)Sn(t)
= Bn1 (t),
n
= 1, 2, ... ,
and initial condition 00
So(t)
00
= "2:: so,ktk = "2:: aktk = A(t). k=O
k=O
Multiplying it by un and summing for n 00
(1  t)
00
L
Sn(t)un
= U"2:: Sn1 (t)un 1.
n=l
n=l
Further, using the initial condition S 0 (t) variate) generating function 00
S(t, u)
= 1, 2, ... , we get
= A(t), we deduce for the double (bi
00
00
= L "2:: Bn,ktkun = L
Sn(t)un,
n=O
n=Ok=O the expression
S(t, u) = (1 t)(1 t u) 1 A(t). Expanding the righthand side of this expression into powers of u, we find
Sn(t) = (1 t)n A(t) and since
we conclude that
Bn,k =
2:: j=O k
0
(
n+J
.
1) akj·
D
J
Example 7.13 Distribution of shares Let us consider two players K and R playing in a series of games in which winner is declared the one who first wins n games. Assume that the probability of K to win in a game is p and so that of R is q = 1  p. Further, suppose that, for some reason, the series of games is interrupted when K has won n  k games and n n  r games, with k, r < n. In this case, what should be the shares of the two players from a total stake of s dollars? The total stake should be distributed to the two players in shares proportional to the probability that each one has to win the series of games if it is continued.
7.5. THE METHOD OF GENERATING FUNCTIONS
263
So, let Pk,r be the probability of K to win the series of games, when k wins of K are required before r wins of R. Note that K may win or lose the next game with probability p or q = 1  p, respectively. If K wins the next game, then the probability of winning the series is Pk 1 ,r. while if K loses (whence R wins) the next game, then the probability of winning the series is Pk,r 1 • Consequently, the probability Pk,r• r = 1, 2, ... , k = 1, 2, ... , satisfies the recurrence relation
Pk,r
= PPk1,r + QPk,r1,
r
= 1, 2, ... ,
k
= 1, 2, ...
,
with initial conditions
Po,r
= 1,
= 1, 2, ...
r
= 0,
, Pk,O
k
= 1, 2, ....
Multiplying this recurrence relation by tr and summing for r 00
L
00
Pk,rtr
and, since Pk,o
00
=PL
r=1
= 1, 2, ... , we get
Pk1,rtr
+ qt LPk,r1tr 1 ,
r=1
k
= 1, 2, ...
r=1
= 0, we deduce for the sequence of generating functions 00
Pk(t) = LPk,rtr, k = 1,2, ...
1
r=1
the first order recurrence relation with constant, with respect to k, coefficients:
(1 qt)Pk(t) = pPk 1(t), k = 1, 2, ... and initial condition 00
Po(t)
00
= LPo,rtr = Ltr = t(1 t) 1 . r=1
r=1
Multiplying this recurrence relation by uk and summing fork = 1, 2, ... , we get 00
(1  qt) L
00
Pk (t)uk
= pu L
k=1
Pk 1(t)uk 1
k=1
and using the initial condition P0 (t) generating function 00
=
t(1  t) 1 , we deduce for the double 00
00
P(t,u) = LPk(t)uk = LLPk,rtruk, k=1
the expression
k=lr=1
P(t, u) = t(1 t) 1 (1 qt pu) 1pu.
RECURRENCE RELATIONS
264
Expanding the righthand side of this expression into powers of u, we get
P(t,u) = t(1 t) 1 (1 p(1 qt) 1 u) 1 p(1 qt) 1 u 00
= t(1 t)1 LPk(l qt)kuk k=l
and so
Pk(t) = t(1 t) 1pk(1 qt)k. Finally, expanding this generating function into powers oft, by using Newton's negative binomial formula, we deduce for the probability Pk,r the expression
Pk,r
=p
k r1
L
(k + 1) ql.. j k  1
J=O
Consequently, from the total stake of s dollars, player K might get a share of SPk,r and player R a share of s(1  Pk,r) dollars. Note that, for p = q = 1/2, the probability Pk,r reduces to the probability derived in Section 2. 7 .I. 0
7.6 BIBLIOGRAPHIC NOTES The first recurrence relation was given in 1202 by Leonardo Fibonacci in his book on abacus (Liber Abaci). This recurrence relation of the Fibonacci numbers is deduced and solved in Example 7.8. The derivation of the solution by using a generating function is due to Abraham De Moivre (1718). E. Lucas (1891) named this sequence of numbers after Fibonacci and also examined the related sequence of Lucas numbers. An extensive coverage of the Fibonacci and Lucas numbers and some of their extensions and generalizations is provided by V. E. Hoggatt (1969). The interest of E. Lucas (1891) in the problem of the Hanoi tower contributed the most in its popularization. The gambler's ruin is an old problem; in 1657, Christian Huyghens' treatise, which was included in the book Ars Conjectandi of J. Bernoulli, discussed the particular case of n = k n = 12 tokens and pfq = 5/7. Bernoulli considered and solved the general case.
7.7
EXERCISES
1. An amount of a0 = $10,000 is deposited in a bank at the beginning of an interest period at interest rate r. If the interest is compounded each
7. 7. EXERCISES
265
period, show that the amount an on deposit after n periods satisfies the first order linear recurrence relation an+l = (1
+ r)an,
n = 0, 1, ....
Iterate it to derive an. 2. Let an be the number of ndigit nonnegative integers in which no two same digits are consecutive. Show that an+ I
with a 1
= 10.
= 9an,
n
= 1, 2, ...
,
Iterate this recurrence relation to derive an.
3. A player decides to play a series of games against a casino (or against an adversary) until he wins. In any game, if he wins he gets b + 1 times the amount of his stake, while if he looses he stakes a new amount. Let an be the player's stake on the nth game, n = 1, 2, .... Provided that, if the player wins the nth game, he gets back not only the stakes he lost in the n 1 previous games but also an amount a, which is fixed in advance, show that ban= (b+ 1)anI, n = 2,3, ... , with a 1
= a/b.
Iterate this recurrence to find an.
4. Consider the set of sequences of flips of a coin that are terminated when heads appear for the second time. Let an be the number of such sequences that are terminated before or at the nth flip. Show that an+l
with a2
= 1.
= an + n,
n
= 2, 3, ...
,
Iterate this recurrence relation to derive an.
5. Let Rn be the number of regions in which a plane can be divided by n lines, each two of them having a point in common but no three of them having a point in common. Show that
Rn = Rn1
+ n,
n = 2, 3, ...
with R 1 = 2. Further, show that the unique solution of this recurrence relation is given by
6. Let an be the number of npermutations of the set {0, 1, 2} with repetition that include an even number of zeros. Show that an+l=an+3n, n=1,2, ... ,
RECURRENCE RELATIONS
266
with a 1 = 2. Iterate this recurrence relation to derive an. 7. Assume that the sequence an, n = 0, 1, ... , satisfies the recurrence relation 2(n + 1)an+l = (2n + 1)an, n = 0, 1, ... , with initial condition a0 = 1. Show that its unique solution is given by
= 2!n
an
c:),
n
= 0, 1, ....
8. Let an and bn be the numbers of npermutations of the set {0, 1, ... , 9} with repetition that include an odd and even number of 5s, respectively. Show that with a 1 = 1, a 2 = 18 and
with b1 = 9, b2 = 82. Note that both sequences an, n = 1, 2, ... , and bn, n = 1, 2, ... , satisfy the same recurrence relation but with different initial conditions. Further, derive the unique solutions a n = ~2 ( 10n  8n) , n
= 1, 2, ...
and
9. Let an be the number of ndigit nonnegative integers in which no three same digits are consecutive. Show that
with a 1 = 10 and a 2 = 100. Further, show that the unique solution of this recurrence relation is given by an =
9
5 (3)n v'l3 2 { ( 3 + v'l3)n+l 
(3 
v'l3)n+l} .
10. Let an be the number of npermutations of the set {0, 1, 2} with repetition and the restriction that no two zeros are consecutive. Show that an+2
= 2an+1 + 2an,
n
= 0, 1, ... ,
7. 7. EXERCISES
267
with ao = 1 and a 1 = 3. Further, show that the unique solution of this recurrence relation is given by an =
4
1 { ( 1 + J3 )n+2 J3
(1
J3)n+2} .
11. Let an be the number of npermutations of the set {0, 1, 2} with repetition and the restriction that no two zeros and no two ones are consecutive. Show that an+2 = 2an+l +an, n = 0, 1, ... ,
with a0 = 1 and a 1 = 3. Further, show that the unique solution of this recurrence relation is given by
12. Let
Show that
and thus conclude the recurrence relations: an= 2anl an2, n
= 2,3, ... ,
with a0 = 1, a 1 = 2, and bn with bo
= 1, b1 = 3.
= 2bnl 
bn2, n
= 2, 3, ...
'
Solve these recurrence relations to get
an = n
+ 1,
bn = 2n + 1, n = 0, 1, ....
13. Find the unique solution of the recurrence relation
an+l
1
= 4(an+2 an),
with initial conditions a0
n
= 0, 1, ...
,
= 1 and a 1 = 4.
14. Find the general solution of the recurrence relation
RECURRENCE RELATIONS
268
15. Find the general solution of the recurrence relation
16*. Let Qr(n, k) be the number of kpermutations of n with repetition in which no r like elements are consecutive r = 2, 3, .... Show that r1
Qr(n, k)
= (n 1) L
Qr(n, k j), k?. r, Qr(n, k)
= nk,
k
j=1
and ~
Qn,r(t)
nt(1
+ t + ... + tr2)
= ~ Qr(n, k)tn = 1 (n 1)t(1 + t + ... + tr2).
17*. Let fn+ 1(s) be the number of npermutations of the set {0, 1, ... , s} with repetition and the restriction that no two zeros are consecutive. (a) Derive the recurrence relation fn+1(s)
with initial conditions
hk+1(s) and hk+2(s)
= s{fn(s) + fn1(s)}, J0 (s)
n
= 1, 2, ... ,
= 1/s, /I(s) = 1. (b) Show that
skt (~)!ki+1(s), = t (~) =
j=O
J
j=O
J
sk
fki+2(s),
k
= 0, 1, ... ,
k
= 0, 1, ....
(c) Derive the generating function 1
00
F(t; s)
= """'fn(s)t = s (1st st 2 )" ~ 11
n=O
18*. (Continuation). Showthat [n/2]
fn(s) =
L
(n ~
k)snk+ 1, n = 0, 1, ....
k=O
19*. (Continuation). Show that
Jn (8 ) =
1
1
1 sJs(s+4)
{ (
S
+J
S (S
2
+ 4) )
n+
_
( S 
J
S (S
2
+ 4)) n+
}
269
7. 7. EXERCISES
and conclude that
Note that fn(1) = fn, n = 0, 1, ... , are the Fibonacci numbers. 20*. Let fn+I (r, s) be the number of npermutations of the set W = W 0 + W1 , where N(Wo) = r + 1, N(WI) = s, with repetition and the restriction that no two like elements of W 0 are consecutive. (a) Derive the recurrence relation fn+l(r, s) = (r
+ s)fn(r, s) + sfnI(s),
with initial conditions fo (r, s) = 1/ s, erating function
h (r, s)
oo
F(t;r,s) =
L
fn(r,s)tn
n=O
= s (1 
n = 1, 2, ... ,
= 1 and (b) deduce the gen
(
1 rt )
r
+s
t st 2
]·
Note that fn(O, 1) = fn, n = 0, 1, ... , are the Fibonacci numbers. 21 *. Fibonacci numbers of order s. Let Fn+l,s be the number of npermutations of the set {0, 1} with repetition and the restriction that no s zeros are consecutive. Show that (a) m
Fn,s = LFnj,s, m = min{n,s}, n
= 1,2, ...
, Fo,s
= F1,s =
1
j=l
and (b)
= 1,2, ... ,s, Fnsl,s, n = s + 1, s + 2, ....
Fn,s = 2n 1 , n Fn,s = 2Fnl,s 
(c) Derive the generating function 1
00
Fs (t) =
L Fn,s tn = 1tt.,2_,.._ts. n=O
22. (Continuation). Show that [nnfk]
Fn,s=
L
Esl(nk,k), n=0,1, ... ,
k=O
where Esl (r, k) is the number of kcombinations of r with repetition and the restriction that each element appears at most s  1 times.
RECURRENCE RELATIONS
270
23. Lucas numbers revisited. Let 9o = 2 and 9n, n = 1, 2, ... , be the number of combinations of then numbers {1,2, ... ,n}, displayed on a circle (whence (n, 1) is a pair of consecutive elements), which include no pair of consecutive elements. These numbers satisfy the recurrence relation 9n+2 9n+1  9n
= 0,
n
= 0, 1, ...
with initial conditions 9o = 2, 91 = 1. Show that 00
2t
G(t) = "'9ntn = :::2 ~ 1 t t n=O
and conclude that
24*. Lucas numbers of order s. Let Gn,s be the number of cyclic npermutations of {0, 1} with repetition and with one element marked by a star and the restriction that no s zeros are consecutive. Note that, for s = 2, Gn,2 = 9n, n = 0, 1, 2, ... , are the Lucas numbers. Show that (a) s
Gn,s =
L Gnj,s,
n
= s + 1, s + 2, ... '
Gn,s = 2n  1, n = 1, 2, ...
'8
j=l
and (b) Gn,s = 2Gn2,s GnsI,s, n
= s + 2, s + 3, ....
(c) Derive the generating function
25. Generalized Fibonacci numbers. Let fn+s,s be the number of combinations of then numbers {1, 2, ... , n} possessing the property: between any two elements belonging to such a combination there exist at least s elements of n that do not belong to it. Show that the numbers fn,s, n = s, s + 1, ... satisfy the recurrence relation fn+s,s = fn+sl,s
+ fnI,s,
n =
S
+ 1, S + 2, · · ·
with initial conditions fn+s,s = n+ 1, n = 1,2, ... ,s.
7. 7. EXERCISES
Putting fn,s
271
= 1, n = 0, 1, 2, ... , s, show that 1
00
=L
Jn,stn
= 1 t tsH.
n=O
The numbers fn,s, n = 0, 1, ... , which for s = 1 reduce to the Fibonacci numbers, are called generalized Fibonacci numbers.
26. Generalized Lucas numbers. Let gn,s be the number of combinations of then numbers {1, 2, ... , n }, displayed on a circle, possessing the following property: between any two elements belonging to such a combination there exist at least s elements of n that do not belong to it. Show that the numbers gn,s, n = 1, 2, ... , g0 ,s = s + 1, satisfy the recurrence relation gn,s = gnl,s + gns1,8> n =
S
+ 1, S + 2, ...
with initial conditions go,s = s + 1, gn,s = 1, n = 1, 2, ... 's
and
00
G 8 (t) = ~ gn,stn = L....t
s +1st . 1  t  ts+l
The numbers gn,s, n = 0, 1, ... , which for s = 1 reduce to the Lucas numbers, are called generalized Lucas numbers and are connected with the generalized Fibonacci numbers by the relation gn,s = fn,s + sfnsl,s, n = s + 1, s + 2, ....
27. Convolution of Fibonacci numbers. Let fn, n = 0, 1, ... , be the sequence of Fibonacci numbers with fo = h = 1. The sequence n 2
fA )=LiJfnj, n=0,1, ... j=O
is called convolution of the sequence fn, n sequence n
= 0, 1,....
In general, the
1
fAk) = LiJf~~j ), n=0,1, ... , k=2,3, ... , j=O
with JA ) = fn, n = 0, 1, ... , is called kfold convolution of the sequence fn, n = 0, 1, .... (a) Derive the generating function 1
00
Fk(t) =
L: !Ak)el = (1 t e)k n=O
RECURRENCE RELATIONS
272
and deduce the expression
(b) Show that
28. Let Rn,k be the number of regions in which a plane can be divided by n lines, k of them being parallel to each other but no three of them having a point in common. Derive the recurrence relation
Rn,k=Rn1,k1+nk+1, k=1,2, ... ,n, n=2,3, ... , with R 1,0 = R 1,1 = 2 and show that its solution is given by
Rn,k = 1 + (n k
+ 1)k + ( n 2k +
1)
·
29. Let Sn be the number of permutations of {1, 2, ... , n} without a succession. Using the recurrence relation
Sn = (n 1)Sn1
+ (n 2)Sn2, n
= 3, 4, ... , S1 = S2 = 1,
show that
30*. Consider the sum of inverse binomial coefficients:
Sn =
L n
(
~
) 1
, n = 0, 1, ....
k=O
(a) Derive the recurrence relation
2nSn (n with initial condition S 0
=
+ 1)Sn1
(
= 1,2, ...
1. (b) Show that
00
S(t) = ~ Sntn =
= 2n, n
1
t)1
2
1 ( 1(1 t) 1  2
,
t) 2
2
log(1 t),
7. 7. EXERCISES
273
and (c) conclude that
Sn
_ n
+1

2n
2k
""'n
L....t k + 1 . k=O
31. Consider the sequences
and let 00
00
n=O
n=l
Show that
A(t) 1 = tA(t)
+ B(t), B(t)
=
tA(t)
+ tB(t)
and thus, conclude that
A(t) 1 t B( ) t  (1  t) 2  t' t = (1  t) 2

t.
32*. Let En be the number of partitions of a finite set of n elements. Show that
tn
L Bn In.= exp(et 1) 00
B(t) =
n=O and
oo
Bn
n
r ' = e 1 ""' L....t f r.
n
= 0, 1, ....
r=O
33*. Let An be the number of partitions of a finite set of n elements into subsets with an even number of elements. Show that [n/2]
An=
n 1
L (2 k _ 1) An2k,
n = 2, 3, ... , Ao = 1, A1 = 0
k=l
and
tn L An I = exp(cosh t 1), n=O n. 00
A(t) = where cosh t
= (et + et)/2 is the hyperbolic cosine oft.
RECURRENCE RELATIONS
274
34. (Continuation). Show that
 1 An e
Loo G(r,n) r!
r=O
' n
= 0, 1, ... ,
where G(r, n) is the number of npermutations of r with repetition and the restriction that each element appears zero or an even number of times. 35*. (Continuation). Let En be the number of partitions of a finite set of n elements into subsets with an odd number of elements. Show that
n _
[(n1)/2]
En =
~
(
2
1 k ) En2k1, n = 1, 2, ... , Eo = 1
and 00
tn E(t) = LEn I = exp(sinh t), n. n=O where sinh t
= (et et)/2 is the hyperbolic sine oft.
36. (Continuation). Show that
En
=~ H(r,n) ~ , r=O
r.1
n
= 0, 1, ... ,
where H(r, n) is the number of npermutations of r with repetition and the restriction that each element appears an odd number of times. 37*. Ballot numbers revisited. In a ballot between two candidates N and K, N receives n votes and K receives k votes, with n 2: k. Let Xr and Yr be the numbers of votes for N and K, respectively, after the counting of r votes, r = 1, 2, ... , n + k and 1/Jn,k be the number of ways of counting the votes in which Xr 2: Yr, r = 1, 2, ... , n + k. Show that
1/Jn,k = 1/Jn,k1 + 1/JnI,k, k = 1, 2, ... ,n 1, n = 1, 2, ... , With 1/Jn,n
= 1/Jn,n1, 1/Jn,O = 1, and 00
'ljJ(t, u)
=L
n
k
L 1/Jn,kt u n=Ok=O
n
=
(14tu) 1 12 +2u1 2(1 t u)u ·
38*. Delannoy numbers. Let us consider an electoral district in which each voter may vote for, at most, two candidates of the same party and assume that two candidates of the same party N and K receive n and k
7. 7. EXERCISES
275
votes, respectively. The total number D(n, k) of ways of counting the votes is called Delannoy number. (a) Show that D(n, k)
= D(n, k 1) + D(n 1, k 1) + D(n 1, k),
for k = 1, 2, ... , n = 1, 2, ... , with D(n, 0) generating function 00
G(t, u)
=L
= D(O, k) = 1.
(b) Derive the
00
L D(n, k)tkun = (1 t u tu)
1
n=Ok=O
and conclude that
39*. (Continuation). Let D(n) generating function
=D(n, n), n
= 0, 1, ....
From the
00
Gn(t) =
L D(n, k)tk k=O
and using Lagrange formula, 00
d(t))L:: [dn n J(t)gn(t) ] . ; n = J(t) ( 1 u_j!_d n=O dt t=O n. t deduce that
1
t , u = g(t),
00
n=O
and conclude that nD(n) (6n 4)D(n 1)
+ (n
5)D(n 2) = 0, n = 2,3, ... ,
with D(O) = 1 and D(1) = 3. 40*. Leibnitz numbers. The numbers
k! (n
k
+ 1)k+I'
= 0, 1, ... ,n,
n = 0,1, ...
are called Leibnitz numbers (see Exercise 3.6). Show that L(n, k)
+ L(n, k 1) = L(n 1, k 1),
k
= 1, 2, ...
, n, n
= 1, 2, ...
and G( t, u
k)k n= ) =~~L( L.... L.... n, t u n=Ok=O
log{(1u)(1tu)} u{1 + t(1  u)} .
ChapterS STIRLING NUMBERS
8.1
INTRODUCTION
The Stirling numbers of the first and second kind, which are the coefficients of the expansions of the factorials into powers and of the powers into factorials, respectively, were introduced by James Stirling in 1730. Since the factorials occupy the same central position in the calculus of finite differences as the powers in the infinitesimal calculus, the Stirling numbers constitute a part of the bridge connecting these two calculi. In the classical occupancy problem, the number of ways of distributing n distinguishable balls into k distinguishable urns so that no urn remains empty was expressed by Abraham De Moivre, in 1718, in the form of a simple sum of elementary terms with alternating sign; it is essentially the Stirling number of the second kind multiplied by k!. The number of different results of a tossing of n distinguishable dice in which each of the k = 6 faces appears (at least once), derived by P. R. Montmort in 1708, probably inspired De Moivre's more general result. The Stirling numbers under different names attracted the attention of several other wellknown mathematicians of the 18th and 19th centuries. The classical book of Ch. Jordan (1939a) on the calculus of finite differences revived the interest in these numbers. A variety of applications of the Stirling numbers in combinatorics and in probability theory was provided. The coefficients of the expansion of the generalized factorials into factorials are connected with the Stirling numbers and have applications in combinatorics, in occupancy problems, and probability theory. In this chapter, the Stirling numbers and the generalized factorial coefficients are thoroughly examined. Basic properties, generating functions, explicit expressions and recurrence relations are presented. In the examples many of their applications are discussed. The Stirling numbers have been generalized to several directions. The noncentral Stirling numbers are briefly examined in the last section of this chapter.
278
STIRLING NUMBERS
8.2
STIRLING NUMBERS OF THE FIRST AND SECOND KIND
Consider the factorial of t of order n: (t)n = t(t 1) · · · (t n
+ 1),
n = 1, 2 ... , (t)o = 1.
(8.1)
Clearly, this is a polynomial oft of degree n. Executing the multiplications and arranging the terms in ascending order of powers oft, we get n
(t)n = Ls(n,k)tk, n=0,1, .... k=O
(8.2)
Inversely, the nth power oft may be expressed in the form of a polynomial of factorials oft of degree n. Specifically, using (8.1), we get successively the expressions t 0 = (t) 0 = 1, t 1 = (t)l, t 2 = t[1
+ (t
1)] = (th
+ (t)2,
t 3 = (t)lt + (tht = (t)1[1 + (t 1)] + (t)2[2 + (t 2)] = (th + 3(t)2 + (t)J
and, generally, n
tn
=L
S(n, k)(t)k, n
= 0, 1, ....
(8.3)
k=O
Then, the following definition is introduced. DEFINITION 8.1 The coefficients s(n, k) and S(n, k) of the expansions (8.2) and (8.3) of the factorials into powers and of the powers into factorials are called Stirling numbers of the first and second kind, respectively.
Clearly, this definition implies s(n, k) = S(n, k) = 0, k
> n, s(O, 0) = S(O, 0) = 1.
Further, replacing in (8.2) t by t and, since (t we deduce the expression
+ n 1)n
= ( 1)n( t)n,
n
(t
+ n 1)n =
L is(n, k)it\ n = 0, 1, ... , k=O
(8.4)
where is(n, k)l
= (l)nks(n, k).
(8.5)
8.2. STIRLING NUMBERS OF THE FIRST AND SECOND KIND 279
Note that ls(n, k)l, according to (8.4), as sum of products of n k positive integers from the set {1, 2, ... , n 1} is a positive integer (see Theorem 8.1 that follows). The coefficient ls(n, k)l, in expansion (8.4), of the rising factorials into powers, is called signless or absolute Stirling number of the
first kind. Expansions (8.2) and (8.3) readily imply that the Stirling numbers of the first kind are derivatives of factorials and the Stirling numbers of the second kind are finite differences of powers. Specifically, a function f(t) for which the derivatives at zero, [Dk j(t)]t=O, k = 0, 1, ... , exist may be expanded, according to Maclaurin formula, into powers of t as
j(t)
=
t
k=O
[~!Dkj(t)]
_ ·tk. t0
In the case of the nth order factorial oft, we have Dk(t)n so
=
(t)n
t [~!Dk(t)n] k=O
_ · tk, n t0
= 0, k > n, and
= 0, 1, ....
(8.6)
From (8.2) and (8.6) it follows that
[~ Dk(t)n]
s(n, k) =
k.
t=O
, k = 0, 1, ... , n, n = 0, 1,... .
(8.7)
Similarly
1
ls(n,k)l = [k 1 Dk(t+n1)n]
.
t=O
, k=0,1, ... ,n, n=0,1, .... (8.8)
Further, a function f(t) for which the differences at zero, [.::1k j(t)]t=O, k = 0, 1, ... , exist may be expanded, according to Newton's formula, into factorials of t as
In the case of the nth power t, we have dktn = 0, k
> n, and so (8.9)
From (8.3) and (8.9) it follows that
S(n, k)
=
[:!dktn
L=O, k = 0, 1, ... , n, n = 0, 1,....
(8.10)
280
STIRLING NUMBERS
THEOREM8.1 The signless Stirling number of the first kind ls(n, k)l. k 2, 3, ... , is given by the sum
I,2, ... ,n, n
(8.11)
where the summation is extended over all (n k)combinations {i 1, i2, ... , ink} of then I positive integers {I, 2, ... , n I}. PROOF
According to definition (8.4) of is(n, k)l, we have for n = 2, 3, ... , n
(t
+ I)(t + 2) · · · (t + n I) = L
is(n, k)itk 1 .
k=1
Note that the ith factor of the product of the lefthand side,
Pi(t) = t + i, i = 1, 2, ... , n 1, is a monomial with constant term i. Executing the multiplications, the (k .:... 1)th order power oft is formed by multiplying the constant terms of any n  k factors {i 1, i2, ... , ink}, out of then I factors {1, 2, ... , n I}, together with the first order terms t of the remaining k  I factors. Since the coefficient of the first order term t, in any factor, equals one, by the multiplication principle, (8.11) is I deduced. REMARK 8.1 Expression (8.11) of the signless Stirling number of the first kind ls(n, k)l may be transformed as
is(n, k)l = (n I)!'"' . .
I
.
L..., ]1]2 ... ]k1
(8.12)
,
where the summation is extended over all (k I)combinations {ii, h, ... ,ikd ofthenI positive integers {I, 2, ... , n1 }. Indeed, to each (nk)combination {i 1,i2, ... ,ini.} of {I,2, ... ,n I}, there uniquely corresponds the (k I)combination {j 1, h, ... ,ikd = {I, 2, ... , nI} {i 1, i 2, ... , ink} and vice versa. Therefore . . ... . _ ( _ I) I Z1 Z2 ···ink _ ( _ I)! I Znk  n ·I · 2 · · · (n I)  n ·· · · · J1]2 · · · Jk1
2122
Introducing this expression into (8.11 ), (8.12) is deduced.
I
The Stirling numbers of the first and second kind constitute a pair of orthogonal bivariate sequences. This is shown in the next theorem.
8.2. STIRLING NUMBERS OF THE FIRST AND SECOND KIND 281
THEOREM8.2 The Stirling numbers of the first and second kind satisfy the following orthogonality relations: n
n
= 15n,k,
L s(n, r)S(r, k) r=k
L S(n, r)s(r, k) r=k
= 15n,k,
(8.13)
where 15n,k = 1, if k = n and 15n,k = 0, if k ,P n is the Kronecker delta. PROOF Expanding the nth order factorial oft into powers oft, by using (8.2), and in the resulting expression expanding the powers of t into factorials of t, by using (8.3), we get the relation n
(t)n
= =
r
n
L s(n, rW r=O
=L
s(n, r) L S(r, k)(t)k r=O k=O
~ {~ s(n, r)S(r, k)} (t)k,
which implies the first of (8.13). Similarly, expanding the nth power of t into factorials of t and, in the resulting expression, expanding the factorials of t into I powers oft, we deduce the second of (8.13). Note that, for fixed n, the (usual) generating function of the sequence of Stirling numbers of the first kind s(n, k), k = 0, 1, ... , n, according to (8.2), is given by n
sn(t)
=L
s(n, k)tk
= (t)n,
n
= 0, 1, ...
,
k=O
while, the factorial generating function of the sequence of Stirling numbers of the second kind S(n, k), k = 0, 1, ... , n, according to (8.3), is given by n
Sn(t)
=L
S(n, k)(t)k
= tn,
n
= 0, 1, ....
k=O
Further, from (8.2), using Newton's general binomial theorem, we deduce the double generating function oo
n
n
g(t,u) = L L s(n, k)tk; = (1 n=O k=O n.
+ u)t,
(8.14)
while from (8.3) we get oo
f(t,u)
n
n
= LLS(n,k)(t)k; = et". n. n=Dk=O
(8.15)
STIRLING NUMBERS
282
The exponential generating functions of the sequences s(n, k), n = k, k + 1, ... , and S(n, k), n = k, k + 1, ... , for fixed k, are derived in the next theorem.
THEOREM8.3 (a) The exponential generating function of the Stirling numbers of the first kind s(n, k), n = k, k + 1, ... , for fixed k, is given by ~ un [log(1 + u)jk gk(u)=Ls(n,k) 1 = k' ,k=0,1,., .. n. . n=k
(8.16)
(b) The exponential generating function of the Stirling numbers of the second kind S(n, k), n = k, k + 1, ... .for fixed k, is given by
un (eu 1)k S(n, k)! = , k = 0, 1,. n. k!
L 00
!k(u) =
n=k PROOF
o o
•
(8.17)
(a) Interchanging the order of summation in (8.14), we get
and since
g(t,u) = (1
+ u)t
= exp{tlog(1
+ u)}
~ [log(1 + u)Jk = L k! tk,
k=O we deduce (8.16). (b) Similarly, interchanging the order of summation in (8.15), we get oo
f(t, u)
=L
n
oo
L
k=On=k
S(n, k);(t)k n.
oo
=L
fk(u)(t)k
k=O
and since
f(t,u) = [1
+ (eu
1W =
L 00
k=O (8.17) is established.
(
!
)
00
(eu 1)k =
L
(
e
u
~! 1)k (t)k,
k=O
I
The generating functions of the signless Stirling numbers of the first kind may be obtained from the corresponding generating functions of the
8.2. STIRLING NUMBERS OF THE FIRST AND SECOND KIND 283
numbers s(n, k) by using (8.5). Thus, from the first part of Theorem 8.3, we deduce the following corollary. COROLLARY 8.1 The exponential generating function of the signless Stirling numbers of the first kind is(n, k)l. n = k, k + 1, ... ,for fixed k, is given by ~
hk(u) = L
un
[log(1 uW k! , k = 0, 1, ....
is(n, k)l n! =
(8.18)
n=k
In the following example, a probabilistic application of the signless Stirling numbers of the first kind is given. Example 8.1 Bernoulli trials with varying success probability Suppose that balls are successively drawn one after the other from an urn initially containing m white balls, according to the following scheme. After each trial the drawn ball is placed back in the urn along with s black balls. Determine (a) the probability p( k; n) of drawing k white balls in n trials and (b) the probability q( n; k) that n trials are required until the kth white ball is drawn. (a) Let Aj be the event of drawing a white ball at the j th trial, j = 1, 2, ... , n. Then, setting() = m/ s, we get ()
Pi
= P(Aj) = () + j
_ 1,
qi = P(Aj) = 1 P(Aj) = ()
j 1 .
+J 1
, j = 1,2, ... ,n
and P(A )1· A)2 · · · A)k A')k+l · · · A')n ) = P(A)1 )P(A)2 ) · · · P(A)k )P(A')k+l ) · · · P(A')n ) ()k
(0+n1)n
Uk+l 1)(jk+2 1) · · · Un
1).
Therefore, the probability p( k; n) of drawing k white balls in n trials is given by the sum ()k
p(k; n) = (()
+ n _ 1)n LUk+l 
1)(jk+2  1) · · · Un 1),
where, since in the first trial the probability of drawing a black ball is q1 = 0, the summation is extended over all (n k)combinations {jk+l, Jk+ 2, ... , Jn} of the
STIRLING NUMBERS
284 n1 positive integers {2, 3, ... , n }. Puttingim and using (8.11 ), we get
. )=
p (k , n
ls(n, k)IBk (B + n _ 1)n, k
= ik+m 1, m = 1, 2, ... , nk,
= 1, 2, ... , n.
Note that, according to (8.4), these probabilities sum to unity. (b) The probability q( n; k) that n trials are required until the kth white ball is drawn equals the probability p( k  1; n  1) of drawing k  1 white balls in n  1 trials multiplied by the probability Pn = P(An) = () /(0 + n 1) of drawing a white ball at the nth trial. Therefore
·k)= ls(n1,k1)10k ( qn, (O+n1)n 'n=k,k+1, ....
0
An interesting application of the orthogonality property of the Stirling numbers is presented in the following example.
Example 8.2 Inverse relations Consider a sequence of real numbers Xk, k = 0, 1, ... , and let r
Yr=Lar,kXk, r=0,1, ... , k=O
where ar,k. k = 0, 1, ... , r, r = 0, 1, ... , are given coefficients. If there exist coefficients bn,r• r = 0, 1, ... , n, n = 0, 1, ... , orthogonal to the coefficients ar,k. k = 0, 1, ... , r, r = 0, 1, ... , n
L
bn,rar,k = 8n,k,
r=k
then
and so
n
Xn
=L
bn,rYr, n
= 0, 1,
0
0
0
°
r=O
The expressions of Yr. r = 0, 1, ... , in terms of Xk, k = 0, 1, ... , r, and inversely of Xn, n = 0, 1, ... , in terms of Yr· r = 0, 1, ... , n constitute a pair of inverse relations. Similarly if 00
Ur = Lan,rWn, r = 0,1, ... ' n=r
8.2. STIRLING NUMBERS OF THE FIRST AND SECOND KIND 285
and
n
L an,rbr,k = 8n,k, r=k then
00
Wk = Lbr,kUr, k=8,1, .... r=k In conclusion, every pair of orthogonal sequences entails a pair of inverse relations. According to Theorem 8.2, the Stirling numbers of the first and second kind constitute a pair of orthogonal sequences. Consequently, if r Yr = L s(r, k)xk, r = 0, 1, ... , k=O then
n Xn = L S(n,r}yr, n = 0, 1, ... r=O
and inversely. Also, if 00
s(n, r)wn, r = 0, 1, ... ,
Ur = L n=r
then 00
Wk = L S(r, k)ur, k = 0,1, ... r=k and inversely.
0
Example 8.3 Connection of ordinary and factorial moments Consider a sequence aj, j = 0, 1, ... (of numbers with a combinatorial interpretation or probabilities or masses). The rth order (power) moment J.L~ and the rth order factorial moment J.L(r) of this sequence are defined by 00
00
J.L~ = L raj, J.L(r) = 2:U)raj, r = 0, 1, .... j=O j=O Using (8.2), the rth order factorial moment Jl(r) may be expressed as ~
J.L(r) and so
r
r
=L
L s(r, k)/aj j=Ok=O
=L k=O
~
s(r, k) L jka1 j=O
r
Jl(rJ=Ls(r,k)J.L~, r=0,1, .... k=O
STIRLING NUMBERS
286
In the same way, using (8.3), we conclude that n
J.l~ = LS(n,r)J.L(r)• n = 0, 1, .... r=O
Note that these two relations constitute a pair of inverse relations.
0
Example 8.4 The operator e = tD In the infinitesimal calculus, besides the usual derivative operator D, the operator e = tD is frequently used. Express the operator e in terms of the operator D. Clearly, e J(t) = tD J(t),
e
2
f(t) = e(ej(t)) = tD(tDf(t)) = tDf(t)
+ t 2 D 2 f(t),
e 3 f(t) = e(e 2 f(t)) = tD(tDf(t) + t 2 D 2 f(t)) = tDf(t)
+ 3t 2 D 2 f(t) + t 3 D 3 f(t)
and, generally, n
en f(t) =
L Cn,rtr Dr f(t), r=l
where the coefficients Cn,r• r = 1, 2, ... , n are independent of the function f(t). Thus, for their determination the most convenient function can be chosen. Let f(t) = tu, whence
Consequently,
r=l
and,accordingto(8.3),Cn,r
= S(n,r),r = 1,2, ... ,n. Thus n
en f(t) =
L S(n, rW Dr f(t). r=l
The inverse of this relation (see Example 8.2) is r
Dr f(t) = cr
L s(r, k)ek f(t).
D
k=l
Example 8.5 The operator .P = t..:l In the calculus of finite differences, besides the usual difference operator Ll, the operator lJt = tLl is frequently used. Express the operator lJt in terms of the operator Ll.
8.2. STIRLING NUMBERS OF THE FIRST AND SECOND KIND 287
Clearly,
t/1 f(t)
= t11f(t), + (t + 1)211 2 f(t),
2
t/1 f(t) = t/l(t/1 f(t)) = t11(t11f(t)) = t11f(t)
+ (t + 1)2112 f(t)) = t11f(t) + (t + 1)211 2 f(t) + (t + 2)311 3 f(t)
t/1 3 f(t) = t/l(t/1 2 f(t) = t11(t11f(t)
and, generally, n
t/ln f(t) =
2:: Bn,r(t + r 1)r11r f(t), r=1
where the coefficients Bn,r• r = 1, 2, ... , n are independent of the function f(t). Thus, for their determination the most convenient function can be chosen. Let f(t) = (t + u 1)u, whence
t/ln(t
+ U  1)u = un(t + U  1)u, 11r(t + U  1)u = (u)r(t + U  1)ur·
Consequently,
r=1 and, according to (8.3), Bn,r = S(n, r), r
= 1, 2, ...
, n. Thus
n
t/lnf(t) = l:S(n,r)(t+r1)r11rf(t). r=1 The inverse of this relation (see Example 8.2) is 1
11r f(t) = (
r
) "'s(r, k)t/lk f(t).
t+r1rL k=1
0
Example 8.6 Convolution of a logarithmic distribution The probability function of a logarithmic distribution is
on
Pn=[log(1B)] 1  , n=1,2, ... , 0<8<1. n The generating function ofthe sequence of probabilities Pn• n = 1, 2, ... , is readily obtained as oo
h(u) = l:Pntn = [log(1 8)] 1 =1
(8 )n 2:: _u_ oo
=1 1
= [log(1 B)] [log(1 Bu)].
n
STIRLING NUMBERS
288 Therefore, if Pn(k), n = k, k 1, 2, ... , then
+ 1, ... ,
is the kfold convolution of Pn. n =
00
2: Pn(k)tn = [h(u)]k = [log(1 O)]k[log(1 Ou)]k.
hk(u) =
n=k
Using (8.14), we deduce the expression of the probability function
Pn(k)
on = [log(1 O)r 1 k!Js(n, k)J,, n.
n
= k, k + 1, ....
The exponential generating function
of the factorial moments 00
J.L(r)(k) = l:(n)rPn(k), r = 0, 1, ... , n=k
of the kfold convolutionpn(k), n = k, k + 1, ... , of the logarithmic distribution, is connected with the generating function hk (u) by the relation bk (u) = hk (u + 1). Therefore, using (8.14 ), we successively get
bk(u) = (1 =
+ [log(1 O)r 1 [log(1 0(1 0) 1 u)l)k
f
(~) [Iog(1 O)rj[log(1 0(1 o)
j=O
1
u)]j
J
00
00
or
r
=
~(k)j[log(1 0)]j ~s(r,j) (1 oy ~!
=
~
or
oo {
min{r,k}
}
( _ O)r
~
(k)jis(r,j)l[log(1
or
min{r,k}
f;
(k)jis(r,j)l[log(1 0)]J,
1
o)rj
and so J.L(rJ(k) = ( _ O)r 1 for r = 0, 1,. . . .
D
.
r
~!
8.3.
EXPLICIT EXPRESSIONS AND RECURRENCES
8.3
EXPLICIT EXPRESSIONS AND RECURRENCE RELATIONS
289
An explicit expression of the Stirling numbers of the second kind is deduced in the following theorem. THEOREM8.4 The Stirling number of the second kind S(n, k), k = 0, 1, ... , n, n = 0, 1, ... , is given by the sum
;! ~( k
S(n, k) =
1)kr (
~) rn.
(8.19)
PROOF Expanding the generating function (8.17) into powers of u, we get the expression
= 2_ ~( 1 )kr k! L
r=O
=
f {;!
n=O
which implies (8.19).
(k)r ~rnun L n! n=O
t(1)kr r=O
(~) rn} :~,
I
REMARK 8.2 The explicit expression (8.19) of the Stirling numbers of the second kind may also be deduced from (8.1 0), by using the expression of the kth power of the difference operator in terms of the shift operator,
Thus
and, since [Ertn]t=O = rn, (8.19) is deduced.
I
STIRLING NUMBERS
290
The Stirling numbers of the second kind, according to Theorem 8.4, are expressed in the form of a single summation of elementary terms, which are products and quotients of factorials and powers. There does not exist an analogous expression for the Stirling numbers of the first kind. An expression of the Stirling numbers of the first kind in the form of a single summation of terms that are products of binomial coefficients and Stirling numbers of the second kind is derived in the next theorem due to Schlomilch (1852). For the Stirling numbers of the first kind, it leads to an expression in the form of a double summation of elementary terms. THEOREM 8.5 Schlomilch 's formula The Stirling numbers of the first kind are expressed in terms of the Stirling numbers of the second kind by
nk ( n + r 1 s(n,k)=~(1Y k 1 )
( 2n k nkr)s(nk+r,r). (8.20)
PROOF The power series t = ¢ 1 (u) = log(1 + u) is the inverse ofthe power series u = ¢(t) = et 1, with ¢(0) = 0. Thus, by (8.16) and applying Lagrange inversion formula (see Theorem 11 .11 ),
1 [ dn _ 1 k] dun(¢ (u)) u=O =
k!
(n 1) [ dtnk dnk (¢(t)) nl tk 1
t=O'
we get
k] (n1) [dnk (et1)n] 1 [dn s(n, k) = k! dun (log(1 + u)) u=O = k 1 dtnk  t t=O. Since (see Remark 11.5)
for sa real number and h(t) a function with h(O) = 1, for which the derivatives at zero exist, we deduce the expression
s(n,k)=
( ) ( 2 (~=11) nk ~ ~n n:::;_r
T 1) r]
k ) [ dnk ( t _
dtnk
Further, by (8.17), we have
r  L
(et  1
t
r
' tnr r.S(n, r ) n.1 n=r 00
L r.S(J + r, r) ("J +tJ r )',. 00
j=O
'
•
t=O·
8.3.
EXPLICIT EXPRESSIONS AND RECURRENCES
whence
dnk [
dtnk
(et1)r] t
t=O
291
=S(nk+r,r) ( ~+
n r)
and so, using the relation
we conclude (8.20). REMARK 8.3
I
Schlomilch's formula (8.20), by replacing k by n k and since
(1)ks(n,n k) = ls(n,n k)l, n+r1) (k+n) = ( 1)k+r (kn) (k+n), ( nk1 kr k+r kr may be rewritten in the following symmetric form
is(n,nk)l=~ k
(
kn k+n k+r) (kr ) S(k+r,r).
The inverse of this relation is given in Exercise 2.
I
Introducing expression (8.19) into Schlomilch's formula (8.20), we conclude the following corollary.
COROLLARY 8.2 The Stirling number of the first kind s(n, k), k is given by the double sum
s(n, k)
= ~ ~( 1)j (~) L.... L....
r=Oj=O
J
(n + r
k  1
= 0, 1, ... , n,
1) (n 2nk ) r k 
n
= 0, 1, ... ,
jnk+r. (8.21) r!
Expressions of the Stirling numbers s(n, k) and S(n, k) as multiple sums over all compositions, as well as over all partitions of n into k parts, are derived in the following theorem.
THEOREM8.6 The Stirling numbers of the first and second kind s(n, k) and S(n, k) are given by
n!L
ls(n, k)l = k!
1 T1 • T2 · · · Tk
(8.22)
STIRLING NUMBERS
292
and
1 , S(n,k) = n!""' k! ~ r1 lr2! · · · rk!
(8.23)
respectively, where the summation in both sums is extended over all nonnegative integer solutions of the equation r1 + r2 + · · · + Tk = n. Alternatively, (8.24)
and (8.25)
respectively, where the summation in both sums is extended over all nonnegative integer solutions of the equations k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. PROOF The exponential generating function of the singless Stirling numbers of the first kind (8.18), upon using the expansion log(1  u) = 1 ur /rand then the Cauchy rule of products of series, may be written as
I:::
00
~
nk
00
un 1 ( ur) k 1 is(n, k)i n! = k! ~;: = k! T1
00
k
!_J
t1
(
~
T;1
ur•) ;::
yielding (8.22). Similarly, the exponential generating function ofthe Stirling numbers of the second kind (8.17) may be expressed as
yielding (8.23). Using the multinomial theorem instead of the Cauchy rule, the generating functions (8.18) and (8.17) may be expressed as
L oo
hk(u) =
un 1 ( is(n, k)i n! = k! u +
u2
un
2 + · ·· + ~ + · ··
)
k
n=k
=
~ { L: k1lk2~·'· ·kn! (~rl (~r, ···(~) kn} :~,
8.3.
293
EXPLICIT EXPRESSIONS AND RECURRENCES
and
un
oo
fk(u)
u2
1 (
= n=k LS(n,k)f = k' n. .
un
21. + ·.·· + 1n. + ···
u+
) k
00
=~
{
L
n! ( 1 ) kt ( 1 ) k2 ( 1 ) kn } un k1!k2! ···kn! 1! 2! ... n! n!'
I
yielding (8.24) and (8.25), respectively.
In the next theorem, triangular recurrence relations of the Stirling numbers of the first kind and second kind are deduced.
THEOREM8.7 (a) The Stirling numbers of the first kind s(n, k), k satisfy the triangular recurrence relation s(n fork = 1, 2, ... , n
+ 1, k)
+ 1, n
= 0, 1, ...
, n, n
= 0, 1, ... , (8.26)
= s(n, k 1) ns(n, k),
= 0, 1, ... , with initial conditions
s(0,0)=1, s(n,O)=O, n>O, s(n,k)=O, k>n.
(b) The Stirling numbers of the second kind S(n, k), k 0, 1, ... , satisfy the triangular recurrence relation
S(n fork = 1, 2, ... , n
+ 1, k)
+ 1, n
= S(n, k 1)
0, 1, ... ,n, n
+ kS(n, k),
(8.27)
= 0, 1, ... , with initial conditions
S(O, 0) = 1, S(n, 0) = 0, n
> 0, S(n, k)
= 0, k
> n.
PROOF (a) Expanding both members of the recurrence relation (t)n+I (t n)(t)n into powers oft, according to (8.2), we get the relation
n+l L s(n k=O
n
+ 1, k)tk = (t n) L
s(n, rW
r=O n+l n = Ls(n,k 1)tk L:ns(n,k)tk,
k=O which implies (8.26). The initial conditions follow directly from (8.2). (b) Expanding both members of the recurrence relation tn+I = t · tn into factorials oft, according to (8.3 ), we have k=l
n+l L S(n k=O
n
+ 1, k)(t)k = t L
S(n, r)(t)r·
STIRLING NUMBERS
294 Since (t)r+l
= (t r)(t)r. whence t(t)r =
(t)r+l + r(t)r, we deduce the relation
n+l
n
n
k=O
r=O
r=O
L S(n + 1, k)(t)k = L S(n,r)(t)r+l + L rS(n,r)(t)r n+l
=L
n
S(n, k 1)(t)k
+L
kS(n, k)(t)k,
k=O
k=l
I
which implies (8.27). The initial conditions follow directly from (8.3).
Multiplying the triangular recurrence relation (8.26) by ( I)nk+I and using (8.5), we deduce the following corollary.
COROLLARY 8.3 The signless Stirling numbers of the first kind ls(n, k)l, k 0, 1, ... , satisfy the triangular recurrence relation ls(n fork
+ 1, k)l = ls(n, k
1)1
= 0, 1, ...
, n, n
+ nls(n, k)l,
=
(8.28)
= 1, 2, ... , n + 1, n = 0, 1, ... , with initial conditions ls(O,O)I
= 1,
ls(n,O)I = 0, n > 0, ls(n,k)l
= 0, k > n.
The signless Stirling numbers of the first kind ls(n, k)l can be tabulated by using the triangular recurrence relation (8.28) and its initial conditions. Table 8.1 gives the numbers ls(n, k)l, k = 1, 2, ... , n, n = 1, 2, ... , 9.
Table 8.1
k
Signless Stirling Numbers of the First Kind ls(n,k)l
1
2
3
4
5
6
7
8 9
n I
2 3 4 5
6 7 8 9
1 1 2 6 24 120 720 5040 40320
I
3
I
11
6 35 225 1624 13132 188124
50 274 1764 13068 109584
I 10
I
85 735 6769 67284
15 175 1960 22449
1 21 I 322 28 4536 546
I
36
1
8.3.
EXPLICIT EXPRESSIONS AND RECURRENCES
295
The Stirling numbers of the second kind S(n, k) can also be tabulated by using the triangular recurrence relation (8.27) and its initial conditions. Table 8.2 gives the numbers S(n, k), k = 1, 2, ... , n, n = 1, 2, ... , 10.
Table 8.2 Stirling Numbers of the Second Kind S(n, k) k
n 1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
1 1 1 1 3 1 1 6 1 7 25 1 15 1 10 1 15 1 31 90 65 21 301 140 1 1 63 350 966 266 28 1 1 127 1701 1050 2646 462 1 7770 6951 36 1 255 3025 1 511 9330 34105 42525 22827 5880 750 45
1
Vertical and horizontal recurrence relations of the Stirling numbers of the first and second kind are derived in the following theorems. THEOREM8.8 The Stirling numbers of the first kind s(n, k), k = 0, 1, ... , n, n = 0, 1, ... , with s(O, 0) = 1, satisfy (a) the vertical recurrence relation n
s(n + 1, k
+ 1) = L( 1)nr(n)nrs(r, k)
(8.29)
r=k
and (b) the horizontal recurrence relation
s(n+1,k+1)
= ~(1rk (~) s(n,r).
(a) According to (8.2), for n = 0, 1, ... and since s(n
PROOF
(8.30)
+ 1, 0)
we have n
L k=O
s(n + 1, k + 1)tk
= (t 1)(t 2) · · · (t n) = (t 1)n·
= 0,
STIRLING NUMBERS
296
Expanding (t 1)n into factorials oft by Vandennonde's formula,
(t 1)n we find
= ~ ( ~) (1)nr(t)r = ~( 1)nr(n)nr(t)r,
n
L
n
= L( l)nr(n)nr(t)r·
s(n + 1, k + 1)tk
r=O
k=O
Further, expanding the factorials in the righthand side into powers oft, by using (8.2), we get the relation n
n
r
k=O
r=O
k=O
L s(n + 1, k + 1)tk = L( 1)nr(n)nr L
s(r, k)tk
from which (8.29) is deduced. (b) Differentiating the generating function (8.16),
un+l
oo
[log(1 + u)]k+l (k+ 1)! ,
~s(n+1,k+1)(n+ 1 )! = we get
~ (
L.....t s n + 1' n=k
k
)un = (1+u) 1 [log(1+u)Jk k!
+ 1 n!
and so
un
oo
~s(n+1,k+1)n! = =
elog(l+u)[log(1 + u)]k k!
~( _ 1r_dlog(1 +
~
uW
(r k)!k!
Expanding the terms of the last member of this relation into powers of u, by using (8.16), and interchanging the order of summation, we deduce the relation n
n
~s(n+1,k+l):! =~(1rk ~ ~s(n,r):! oo
oo
= which implies (8.30).
I
(
)
oo
~ {~(1rk (~) s(n,r)} :~,
8.3.
EXPLICIT EXPRESSIONS AND RECURRENCES
297
THEOREM8.9 The Stirling numbers of the second kind S(n, k), k = 0, 1, ... , n, n with S(O, 0) = 1, satisfy (a) the vertical recurrence relation
S(n
't (~)
+ 1, k + 1) =
= 0, 1, ... , (8.31)
S(r, k)
r=k and (b) the horizontal recurrence relation n
S(n, k) = 2:)1tk(r)rkS(n + 1,r + 1). r=k
PROOF
(8.32)
(a) According to (8.3), for n = 0, 1, ... and since S(n
we have
+ 1, 0)
= 0,
n
L
S(n
+ 1, k + 1)(t + 1)k+l = (t + 1)n+l.
k=O Further, since (t
+ 1)k+I = (t + 1)(t)k, we conclude that n
L S(n k=O
+ 1, k + 1)(t)k =
(t
+ 1)n.
Expanding the righthand side into powers oft, by Newton's binomial formula, and in the resulting expression, expanding the powers into factorials, by using (8.3), we successively get
~ S(n + 1, k + 1)(t)k = ~ ( ~) ~ S(r, k)(t)k =
~ {~ ( ~) S(r,k)} (t)k,
yielding (8.31). (b) Multiplying the recurrence relation
S(n
+ 1, r + 1) =
+ (r + 1)S(n, r + 1) k, k + 1, ... , n, since S(n, n + 1)
S(n, r)
by ( 1Yk(r)rk and summing for r = we get
= 0,
n
L( 1tk(r)rkS(n + 1, r r=k
+ 1)
n
n1
= L( 1tk(r)rkS(n,r) + L( 1tk(r + 1)r+IkS(n,r + 1) r=k n
= L(1tk(r)rkS(n,r)r=k
r=k n
L (1)jk(j)jkS(n,j) j=k+l
= S(n,k),
STIRLING NUMBERS
298
I
establishing (8.32).
The triangular recurrence relation (8.27) can be used for the determination of the (power) generating function of the sequence of the Stirling numbers of the second kind S(n, k), n = k, k + 1, ... , for fixed k. This generating function (a) suitably expanded leads to an interesting expression of the numbers S(n, k) and (b) transformed yields the expansion of the inverse factorials into inverse powers. Specifically, we have the following theorems.
THEOREM 8.10 The (power) generatingfunction ofthe Stirling numbers ofthe second kindS (n, k ), n = k, k + 1, ... ,for fixed k, is given by k
00
IT
(8.33)
= k 1, k, ... , since
00
00
L S(n+1,k)un+ 1 =u L S(n,k1)un+kuLS(n,k)un, n=k1 n=k1 n=k fork= 1, 2, .... Consequently, ¢k(u)
= u
k
= 1, 2, ...
and
¢k(u) = u(1 ku) 1
¢k(u) = u 2(1 ku) 1 (1 (k 1)u) 1 ¢k2(u) = · · · = uk(1 ku) 1 (1 (k 1)u) 1 · · · (1 u) 1 ¢o(u) and since, according to the initial conditions of (8.27) ¢ 0 (u) = 1, we deduce (8.33). I
THEOREM 8.ll The Stirling number of the second kind S(n, k), k given by the sum
= 0, 1, ...
, n, n
= 0, 1, ... , is (8.34)
8.3.
EXPLICIT EXPRESSIONS AND RECURRENCES
299
where the summation is extended over all integers Tj 2: 0, j = 1, 2, ... , k, with r1 + r2 + · · · + Tk = n  k, or equivalently by the sum (8.35)
where the summation is extended over all (n k )combinations {i 1 , i 2 , ••. , ink} with repetition of the k positive integers {1, 2, ... , k }. PROOF
Expanding each factor in (8.33) by using the geometric series, we find
=
f {:~::::>r12r2 ...
krk} un,
n=k
where, in the inner sum, the summation is extended over all integers Tj 2: 0, j = 1, 2, ... , k, with r 1 + r 2 + · · · + Tk = n k. This relation implies (8.34), as well as its equivalent expression (8.35). I THEOREM 8.12 (a) The inverse factorial (t)k. k = 1, 2, ... , is expanded into a series of inverse powers tn, n = k, k + 1, ... , as 00
(t)k = z)1)nks(n,k)cn, k = 1,2, ... , t > k. n=k
(8.36)
(b) The inverse power ck, k = 1, 2, ... , is expanded into a series of inverse factorials (t)n. n = k, k + 1, ... , as 00
ck = z)1)nks(n,k)(t)n, k = 1,2, .... n=k PROOF
(a) Setting u = 1/t in (8.33), we get
f(1)nS(n,k)Cn n=k
=
(1)k + k}k
(t
and, since (t)k = 1/(t + k)k, k = 1, 2, ... , we conclude (8.36). (b) Multiplying both members of the expansion 00
(t)n =
L( 1rns(r, n)cr r=n
(8.37)
STIRLING NUMBERS
300 by ( l)nk s(n, k) and summing for n 00
= k, k + 1, ... , we find
00
2:(1)nks(n,k)(t)n n=k
00
= 2: 2:(1)'"nS(r,n)s(r,k)t_,. n=k r=n =
~( 1)'"k {~ S(r, n)s(n, k)} t'".
By the second of the orthogonality relations (8.13), it holds
,.
2: S(r, n)s(n, k) = o,.,k n=k and so
00
00
n=k
r=k
I
which implies (8.37).
REMARK8.4 Expansion (8.36), by replacingt by t and since ( 1)k( t)k 1/(t 1)k, may be transformed to the expansion 1
00
t k+l
n=k
1
() = l:S(n,k)1 , tn+
k= 1,2, ... , t
=
> k.
Similarly, (8.37) may be rewritten as
1 k1 t +
00
1
n=k
(t n+l
= 2: s(n, k)),
k
= 1, 2, ...
or
1 k1 t +
00
= 2: is(n, k)l ( n=k
1 t
+n
)
n+l
,k
= 1, 2, ...
Note that, according to the last expansion, the probabilities q(n; k), n = k, k ... , derived in the second part of Example 8.1, also sum to unity. I
+ 1,
Example 8.7 A Markov chain Suppose that balls are successively drawn one after the other from an urn initially containing m white balls, according to the following scheme. After each trial, if the drawn ball is white, a black ball is placed in the urn, while if the drawn ball is black, it is placed back in the urn. Determine (a) the probability p(k; n) of drawing k white balls inn trials, with n :::; m, and (b) the probability q(n; k) that n trials are required until the kth white balls is drawn, with k :::; m.
8.4. GENERALIZED FACTORIAL COEFFICIENTS
301
(a) The probability of drawing a white ball at any trial, given that j white balls are drawn in the previous trials, is given by PJ = (m j) /m j = 0, 1, ... , n 1. Then the probability p(k; n) of drawing k white balls inn trials satisfies the triangular recurrence relation mk+1 k p(k;n) = p(k;n 1) + p(k 1;n 1), m m fork = 1, 2, ... , n, n = 1, 2, ... , with initial conditions p(O;O) = 1, p(O;n) = 0, n
> 0,
p(k;n) = 0, k
> n.
Further, since p(1; n) = m/mn and p(n; n) = (m)n/mn, we may set (m)k p(k;n) = Cn,k mn , k = 0, 1, ... ,n.
Then the coefficient Cn,k satisfies the triangular recurrence relation Cn,k = kCn1,k +
Cn1,k1
1
k = 1, 2, ... , n, n = 1, 2, ... ,
with initial conditions Co,o = 1, Cn,o = 0, n
> 0,
Cn,k = 0, k
> n.
Comparing this recurrence with the triangular recurrence relation (8.27) of the Stirling numbers of the second kind, we get Cn,k = S(n, k) and so p(k;n)=
S(n, k)(m)k ,k=1,2, ... ,n. mn
Note that, according to (8.3), these probabilities sum to unity. (b) The probability q(n; k) that n trials are required until the kth white ball is drawn equals the probability p(k 1; n 1) of drawing k 1 white balls inn 1 trials multiplied by the probability Pk 1 = (m k + 1)/m of drawing a white ball at the ntrial, given that the k  1 balls are drawn in the previous trials. Thus q(n; k) =
S(n 1, k 1)(m)k , n = k, k mn
+ 1, ....
Note that these probabilities, according to the first expansion in Remark 8.4, also sum to unity. D
8.4
GENERALIZED FACTORIAL COEFFICIENTS
Consider the generalized factorial oft of order n and scale parameters, (st)n = st(st 1) · · · (st n + 1), n = 1, 2, ... , (st)o = 1,
(8.38)
STIRLING NUMBERS
302
with s a real number. This can be expressed as a polynomial of factorials oft of degree n. Specifically, we successively get the expressions
= (t)o = 1,
(st)o
= s(t)l,
(st)l
(st)2 = st[s(t 1) + (s 1)] = s 2 (t)2
+ (s)2(t)l,
(st)J = s 2 (t)2[s(t 2) + 2(s 1)] + (s)2t[s(t 1) + (s 2)] = s 3 (t)3 + 3s(s)2(t)2 + (s)J(t)l and, generally, n
L C(n, k; s)(t)k,
(st)n =
n = 0, 1, ....
(8.39)
k=O In particular, for s = 1 and introducing the coefficient L( n, k) C(n, k; 1), we deduce the expression
=
n
( t)n
=L
L(n, k)(t)k, n
= 0, 1, ....
(8.40)
k=O Further, since ( t)n (1)nL(n,k), we get
(l)n(t
+
n 1)n and setting IL(n,k)l
=
n
(t
+ n 1)n =
L IL(n, k)l(t)k,
n = 0, 1,... .
(8.41)
k=O Then, the following definition is introduced.
DEFINITION 8.2 The coefficient C(n, k; s) of the kth order factorial oft of the expansion (8.39) of the nth order generalized factorial oft, with scale parameter s, is called generalized factorial coefficient. In particular, the coefficients L(n, k) and IL(n, k)l of expansions (8.40) and (8.41) are called Lah and signless or absolute Lah numbers, respectively. Clearly,
C(n,k;s) = 0, k > n, C(O,O;s) = 1. Further, expansion (8.39) implies that the generalized factorial coefficients are differences of generalized factorials. Specifically, expanding the function f(t) = (st)n into factorials oft, by using Newton's formula and since _dk(st)n = 0, k > n, we get
(st)n =
t
k=O
[:!Llk(st)n] _ · (t)k, n = 0, 1, ... t0
8.4. GENERALIZED FACTORIAL COEFFICIENTS
303
and so, by (8.39), we deduce that
C(n,k;s)
= [;!Llk(st)nL=O, k = 0,1, ...
,n, n
= 0,1,....
(8.42)
Also, putting t = bu and s = afb in (8.39), it follows that n
(au)n
=L
C(n, k; s)(bu)k, n
= 0, 1, ... , s = afb.
(8.43)
k=O
Finally, note that the rising factorial oft of order n, (t + n 1)n, using Vandermonde's formula, may be expanded into factorials oft as (t
n + n 1)n = [;
(n) k (n 1)nk(t)k
n n! (n = (; k! k _
1) 1 (t)k
and so, by (8.41) we get for the Lah numbers the expression IL(n,k)l = (l)nL(n,k) = n! k! (nk _ 1) , 1 for k
= 1, 2, ...
, n, n
(8.44)
= 1, 2, ....
REMARK 8.5 (a) The generalized factorial oft of order nand increment a, denoted by (t)n,a. is defined fort and a real numbers and n integer by (t)n,a
= t(t a)··· (t na +a), n = 1, 2, ... , (t)o,a = 1
and fort :j:. ra, r = 1, 2, ... , n, by 1
(t)n,a
= (t + na)(t + na a)··· (t +a)' n = 1 ' 2' · · · ·
Clearly, bn(t)n,b = (tfb)n, ak(t)k,a = (tja)k and so, replacing t by tfa in (8.39), the generalized factorial oft of order n and increment b is expressed as a polynomial of generalized factorials oft with increment a as n
(t)n,b
=L
bnakC(n, k; s)(t)k,a, s
= ajb, n = 0, 1, ....
k=O
(b) A function f(t) for which the differences with increment a at zero, [Ll~f(t)]t=o, k = 0, 1, ... , exist may be expanded, according to Newton's formula, into generalized factorials oft as f(t) =
~ [ a;!k Ll~f(t)] t=O · (th.a·
STIRLING NUMBERS
304
Note that this formula may be deduced as follows:
f(t) = [E~fa J(u)]u=O = [(Lla 00
= [;
[
+ 1)tfa J(u)]u=O
k! Ll~f(u) u=O · (tja)k
In the case of f(t)
1
]
00
= [;
[
k ak! Ll~f(u) ] u=O · (t)k,a.
= (t)n,b and since Ll~(t)n,b = 0, k > n, we get (t)n,b
n = {;
[
k
ak!
Ll~(t)n,b
]
t=O · (t)k,a
and so
bn ] C(n, k; s) = [k!Ll~(t)n,b t=O, k = 0, 1, ... , n, n = 0, 1, ....
I
where s = ajb.
The generalized factorial coefficient C(n, k; s) is a polynomial in s of degree n. Specifically, the following theorem is derived. THEOREM 8.13
The generalized factorial coefficient C(n, k; s) is a polynomial ins of degree n, the coefficient of the general term of which is a product of the Stirling numbers of the first and second kind: n
C(n,k;s) = Ls(n,r)S(r,k)sr. r=k
(8.45)
PROOF Expanding the generalized factorial oft of order nand scale parameter s into powers of u = st by using (8.2) and, in the resulting expression, expanding the powers oft into factorials oft by using (8.3), we deduce the expression n
(st)n
=L r=O =
n
s(n, r)srtr
r
=L
s(n, r)sr L S(r, k)(t)k r=O k=O
~ {~ s(n, r)S(r, k)sr} (t)k,
which, compared with (8.39), implies (8.45). REMARK 8.6
In the particular cases
I
= 1 and, since by (8.39) C(n, k; 1) =
on,k, it follows as a corollary of (8.45) the first of (8.13). Also, in the cases = 1,
8.4. GENERALIZED FACTORIAL COEFFICIENTS
305
we have C(n, k; 1) = L(n, k) and so from (8.45), using (8.5) and (8.44), we deduce the relation n!
~ ls(n, r)IS(r, k) = k! n
fork = 1, 2, ... , n, n = 1, 2,....
(n 1) k_ 1
.
I
Note that, for fixed n, the factorial generating function of the sequence of the generalized factorial coefficients C(n, k; s), k = 0, 1, ... , n, according to (8.39), is given by n
Cn(t; s)
=L
C(n, k; s)(t)k
= (st)n,
n
= 0, 1, ...
.
k=O Further, from (8.39), using Newton's general binomial theorem, we deduce the double generating function oo
g(t, u; s) =
n
n
L L C(n, k; s)(t); = (1 + uyt.
n=Ok=O
(8.46)
n.
The exponential generating function of the sequence of the generalized factorial coefficients C(n, k; s), n = k, k + 1, ... , for fixed k, is derived in the next theorem. THEOREM 8.14 The exponential generating function of the generalized factorial coefficients C(n, k; s), n = k, k + 1, ... , for fixed k, is given by
fk(u; s)
=L oo
n=k
PROOF
un C(n, k; s) 1 n.
=
[(1
+ u)sk'
1Jk
.
, k
= 0, 1,....
Interchanging the order of summation in (8.46) we get oooo
g(t, u; s) =
LL
n
oo
C(n, k; s);(t)k k=O n=k n.
=L
fk(u; s)(t)k
k=O
and, since
g(t,u;s) = (1 + [(1 +u) 8 1])t = =
~ (!) [(1 +u) ~ [(1 + u)s ~
k=O
k!
8
1]k
lJk (t) k,
(8.47)
STIRLING NUMBERS
306
I
we deduce (8.47).
An explicit expression of the generalized factorial coefficient C(n, k; s) is derived in the following theorem. THEOREM 8.15 The generalized factorial coefficient C(n, k; s ), k = 0, 1, ... , n, n = 0, 1, ... , is given by the sum k
C(n,k;s) =
~! ~(1)kr (~) (sr)n.
(8.48)
PROOF Expanding the generating function (8.47) into powers of u, we get the expression
E
C(n, k; s)
which implies (8.48).
:~ =
:!
t.( 1)kr (
~) (1 + u)sr
=
~! t.(1)kr (~) ~ ( : ) un
=
~ { ~! t.(1)kr (~) (sr)n} :~,
I
REMARK 8.7 (a) Expression (8.48), of the generalized factorial coefficient, may also be deduced from (8.42), by using the expression of the kth power of the difference operator in terms of shift operator, k
.jk
= ~(1)kr (~) er.
Thus C(n,k;s)
= [~!.dk(st)nL=O =
:, t.(1)kr
and, since [Er(st)n]t=O = (sr)n. (8.48) is deduced.
(~) [Er(st)n]t=O I
Expressions of the generalized factorial coefficient C(n, k; s) as multiple sums over all compositions, as well as over all partitions of n into k parts, are given in the following theorem.
8.4. GENERALIZED FACTORIAL COEFFICIENTS
307
THEOREM 8.16 The generalized factorial coefficient C(n, k; s ), k = 0, 1, ... , n, n = 0, 1, ... , is given by C(n, k; s) =
~: L (:. ) ( :2) ··· (:k),
(8.49)
where the summation is extended over all nonnegative integer solutions of the equation r 1 + r2 + · · · + Tk = n. Alternatively, (8.50)
where the summation is extended over all nonnegative integer solutions of the equations k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. Expanding (1 + u) 8  1 into powers of u and then using the Cauchy rule of multiplication of series, the exponential generating function (8.47) may be expressed as
PROOF
f.c(n,k;'):: = ~! [~
(;) •']'
= ~!
li [; (:,) •"]
E{~: L (:. ) (:2 ) ... (:k )}:~'
=
yielding (8.49). Using the multinomial theorem instead of the Cauchy rule, we get the expansion
E
C(n, k; s)
=
:~ = ~! [ ( ~) u + (;) u
E{L
which implies (8.50).
kl !k2
2
+ ... + ( ~) un + ...
~! .. kn! ( ~ ) kt (; )
k2 • . • (
r
~) kn } :~ '
I
Limiting expressions as s + 0 and s + oo and an orthogonality relation for the generalized factorial coefficient are deduced in the following theorems.
THEOREM 8.17 Let C(n, k; s) be the generalized factorial coefficient. Then lim skc(n, k; s)
s+0
= s(n, k),
lim snc(n, k; s)
s+oo
= S(n, k),
(8.51)
STIRLING NUMBERS
308
where 8(n, k) and S(n, k) are the Stirling numbers of the first and second kind, respectively.
=
PROOF By using lims4 0 8 1 [(1 + u) 5 1] log(1 generating function (8.47) the limiting expression oo
+ u), we deduce for the
n
lim 8k fk(u; 8) = "'[lim 8kc(n, k; 8)]~, ~ S40 n=k
S40
n.
!~ 8k[(1 + uY 1]k ~~~~
k!
=
[Iog(1
+ u)]k
~~~~
k!
which, compared with the generating function (8.16) of the Stirling numbers of the first kind, implies the first of (8.51). Also, since lims4oo (1 + uj 8 ) 5 = eu, we deduce the limiting expression oo lim /k(u/8; 8) =""[lim 8nqn, k; S400 ~ S400 n=k
s~~[(1
+ u/8)
5
1]k

n
8)]~, n. (eu _ 1)k
~~~~= ~~~
k!
k!
which, compared with the generating function (8.17) of the Stirling numbers of the second kind, implies the second of (8.51 ). I
THEOREM 8.18 The generalized factorial coefficients C(n, k; 8), k satisfy the relation
= 0, 1, ...
, n, n
= 0, 1, ... ,
n
L C(n, r; 81 )C(r, k; 82) = C(n, k; 8182).
(8.52)
r=k In panicular, they satisfy the orthogonality relation n
L C(n, r; 8)C(r, k;
1 8 )
= 6n,k,
(8.53)
r=k where 6n,k = 1,
PROOF
if k
= nand 6n,k = 0,
if k 1 n
is the Kronecker delta.
Expanding the nth order factorial of 8 1 82t into factorials of 82t, using (8.39), and in the resulting expression expanding the factorials of 82t into factorials
8.4. GENERALIZED FACTORIAL COEFFICIENTS
309
oft, we get the relation n
(stSzt)n =
L C(n, r; st)(szt)r r=O
= =
n
r
r=O
k=O
L C(n, r; St) L C(r, k; s 2)(t)k
~ {~C(n,r;st)C(r,k;sz)} (t)k.
Further, according to (8.39), we have n
(s1s2t)n =
L C(n, k; StSz)(t)k k=O
and so
The last relation implies (8.52). In the particular case of s 1 since C(n, k; 1) = 8n,k. (8.53) is deduced. I
= sand s 2 = s 1,
Recurrence relations for the generalized factorial coefficients are given in the following theorems. THEOREM 8.19 The generalized factorial coefficients C (n, k; s ), k satisfy the triangular recurrence relation
C(n
+ 1, k; s)
for k = 1, 2, ... , n
C(O,O;s)
+ 1, n = 1,
= (sk n)C(n, k; s)
= 0, 1, ... , n, n = 0, 1, ... ,
+ sC(n, k
1; s),
(8.54)
= 0, 1, ... , with initial conditions
C(n,O;s)
= 0,
n > 0, C(n,k;s)
= 0,
k > n.
PROOF Expanding both members of the recurrence relation (st)n+l (st  n)(st)n into factorials oft, according to (8.39), we have n+l
n
n
k=O
r=O
k=O
L C(n + 1, k; s)(t)k = st L C(n, r; s)(t)r  n L C(n, k; s)(t)k.
STIRLING NUMBERS
310
Since (t)r+l = (t r)(t)r, whence t(t)r = (t)r+l + r(t)n the righthand side of this relation may be written as n n n s L C(n, r; s)(t)r+I + s L rC(n, r; s)(t)r n L C(n, k; s)(t)k r=O
r=O
and so n+l L C(n
k=O
n
n+l
+ 1, k; s)(t)k = L(sk n)C(n, k; s)(t)k + L
k=O
sC(n, k 1; s)(t)k.
k=l
k=O
Equating the coefficients of (t)k in this relation, we conclude (8.54). The initial conditions follow directly from (8.39). I REMARK 8.8 (a) If the parameters is a positive integer, then the numbers C(n, k; s), n = k, k + 1, ... , sk, k = 0, 1, ... , are positive integers. Indeed, from the triangular recurrence relation (8.54) and its initial conditions, it follows that these numbers result from successive summation of positive integers. Further, in addition to C(n, k; s) = 0, k > n, it follows from (8.48) that C(n, k; s) = 0, sk < n. (b) Also, if sis a positive integer, then the numbers
!C(n, k; s)l = ( 1)nC(n, k; s), n = k, k
+ 1, ...
, k = 0, 1, ... ,
are positive integers since, by (8.54 ), we have
+ 1, k; s)l = (sk + n)!C(n, k; s)l + s!C(n, k1, 2, ... , n + 1, n = 0, 1, ... , with
!C(n fork=
= 1,
IC(O, 0; s)l
IC(n, 0; s)l
= 0,
n > 0, !C(n, k; s)l
1; s)!,
= 0,
(8.55)
k > n;
this means that these numbers result from successive summation of positive integers. I THEOREM 8.20 The generalized factorial coefficients C(n, k; s), k =· 0, 1, ... , n, n with C (0, 0; s) = 1, satisfy the vertical recurrence relation C(n
+ 1, k + 1; s)
=
t (~)
(s)nr+l C(r, k; s).
= 0, 1, ... , (8.56)
r=k
PROOF we have
(a) According to (8.39) and since C(n n
+ 1, 0; s) = 0, n = 0, 1, ... ,
2: C(n + 1, k + 1; s)(t + 1)k+I = (st + s)n+l· k=O
8.4. GENERALIZED FACTORIAL COEFFICIENTS
Further, since (t
+ 1)k+ 1
= (t
311
+ 1)(t)k, we conclude that
n
L C(n + 1, k + 1; 8)(t)k = 8(8t + 8 1)n· k=O
Expanding the righthand side into factorials of 8t, by Vandennonde's formula, and in the resulting expression expanding the factorials of 8t into factorials oft, by using (8.39), we successively get
~ C(n + 1, k + 1; 8)(t)k n
=
~ n
(
;
)
(8)nr+l
= ~ {~ (;) yielding (8.56).
~ C(r, k; 8)(t)k r
(8)nr+IC(r, k; 8)} (t)k,
I
The generalized factorial coefficients C(n, k; 8) can be tabulated by using the triangular recurrence relation (8.54) and its initial conditions. Table 8.3 gives the numbers C(n, k; 8), k = 1, 2, ... , n, n = 1, 2, ... , 5.
Table 8.3 Generalized Factorial Coefficients C(n, k; s)
k
2
1
3
4
5
n I
(8)1
2
(8)2
82
3
(8)3
3(8)28
4
(8)4
5
(8)5
7(8)38
+ 3(8)28
15(8)48 + 20(8)38
83 6(8)28 2
84
25(8)38 2 + 15(8)28 2
10(8)283
85
The numbers C(n, k; 8), k = 0, 1, ... , n, n = 0, 1, ... , are defined as the coefficients of the factorials in the expansion of the generalized factorial, with scale parameter 8. These numbers emerge also in the expansion of the
STIRLING NUMBERS
312
inverse generalized factorials into inverse factorials. Specifically, we have the following theorem.
THEOREM 8.21 The inverse generalizedfactorial oft, with scale parameters, (st) k· k = 1, 2, ... , is expanded into a series of inverse factorials (t)n, n = k, k + 1, ... , as 00
(st)k = L(1tkC(n,k;s 1)(t)n, k = 1,2,....
(8.57)
n=k
PROOF
Multiplying both members of the recurrence relation
by ( 1)nk+ 1s(t)n and, on the lefthand side, using the recurrence relation (t)n = (t + n + 1)(t)nt. we get the relation
( 1)nk+ 1sC(n + 1, k; s 1)t(t)n1 +( 1)nk+ 1s(n + 1)C(n + 1, k; s 1)(t)n1 = (1)nksnC(n,k;s 1)(t)n (1tkkC(n,k;s 1)(t)n +(1)nk 1 C(n,k 1;s 1)(t)n, Summing the last relation for n = k 1, k, ... , we derive for the sum 00
= L(1tkC(n, k; s 1)(t)n, k = 0, 1, ... ,
Ck(t; s)
n=k
the recurrence relation ck(t; s)
= (st + k) 1ckl (t; s), k = 1, 2, ...
, eo(t; s)
= 1.
Iterating it, we conclude that
1
ck(t; s) = [(st + k)(st + k 1) · · · (st + 1)r co(t; s)
and so (8.57) is established. REMARK 8.9
= (st)k,
I
(a) Expansion (8.57) with t
= bu and s = a/b becomes
00
(au)k = L(1)nkC(n,k;s 1 )(bu)n, s = ajb, k n=k
= 1,2, ....
8.4. GENERALIZED FACTORIAL COEFFICIENTS
(b) Using the relations (t)n = an(u)n,a• (st)k and u = at, (8.57) may be rewritten as
313
= bk(u)k,b· with s = afb
00
(u)k,b =
L (1)nkbkanC(n, k; s 1)(u)n,a, s = afb, k = 1, 2, .... n=k
(c) Setting u = tin (8.57) and since (1)k(su)k (l)n(u)n = 1f(u1)n,weget 00
1
1 (SU  1)k = '"'c(n,k;s)( L..,.. U 
n=k
1
=
1/(su 1)k,
) , k= 1,2, .... 1n
Dividing both members by su, we deduce the expansion 00
(
1) '"' 1 C(n,k;s 1 )() 1 ,k=0,1, ... , =L....s SU k+1 n=k U n+l
which, equivalently, may be written as
or as 00
1 1 ()  = sC(n, k; s) ( ) , k t k+1 n=k st n+1
L
= 0, 1, ... ,
an expression useful in waitingtime probability problems.
I
Example 8.8 Coupon collector's problem Consider an urn containing s identical series of coupons, each consisting of m coupons bearing the numbers 1, 2, ... , m. Suppose that coupons are randomly drawn, one after the other, without replacement. Calculate (a) the probability p(k; n) of drawing exactly k of the numbers {1, 2, ... , m} inn drawings, with n :S: m, and (b) the probability q( n; k) that n drawings are required until the kth different number is drawn, with k ::; m. (a) Let Aj be the event that number j is not drawn in the n drawings, j = 1, 2, ... , m. Then p(k; n) is the probability that exactly m k among them events A 1 , A 2 , ... , Am occur. Further, for any r indices {i 1 , i 2 , ... , ir} out of them indices { 1, 2, ... , m },
STIRLING NUMBERS
314
Therefore, the events A 1 , A 2 , ... , Am are exchangeable and according to Corollary 4.4, the probability p(k; n) is given by
=
p(k;n)
(m) "(k
LJ
n
r=O
1)kr (k) (sr)n r (sm)n
and so, by virtue of (8.48), p(k;n)
k; s)(m)k = C(n,(sm )n , k = 1,2, ...
,n.
(b) The probability q(n; k) that n drawings are required until the kth different number is drawn equals the probability p( k 1; n 1) of drawing k 1 of them numbers inn 1 drawings multiplied by the probability s(m k+ 1)/ ( smn+ 1) that one of them  k + 1 not already drawn numbers is drawn at the nth drawing. Hence q(n;k)=
C(n 1, k 1; s)s(m)k () ,n=k,k+1, ... ,sm. sm n
Note that these probabilities, according to Remark 8.9(c), sum to unity.
0
8.5 NONCENTRAL STIRLING AND RELATED NUMBERS The Stirling numbers of the first and second kind and the generalized factorial coefficients have been generalized in several directions. In this section, the corresponding noncentral numbers are briefly presented. Additional properties and applications of these numbers are given in the exercises. Consider the noncentral factorial oft of order n, (t  r)n, where the noncentrality parameter r is a real number and let n
(t r)n =
L s(n, k; r)tk, n = 0, 1, ....
(8.58)
k=O
Also, let n
(t
+ r)n = L
S(n, k; r)(t)k, n
= 0, 1, ....
(8.59)
k=O
The coefficients s(n, k; r) and S(n, k; r) are called noncentral Stirling numbers of the first and second kind, respectively. Clearly, s(n, k; r)
= S(n, k; r) = 0, k > n,
s(O, 0; r)
= S(O, 0; r) = 1.
8.5.
NONCENTRAL STIRLING AND RELATED NUMBERS
Further, replacing t by t in expansion (8.58) and since (t (1)n(t r)n, we deduce the expression
315
+ r + n 1)n =
n
(t
+ r + n 1)n =
L
(8.60)
ls(n, k; r)ltk, n = 0, 1, ... ,
k=O
where the coefficient
= (1)nks(n, k; r), k = 0, 1, ... , n,
ls(n, k; r)l for r
n
= 0, 1, ...
, (8.61)
> 0, as sum of products of positive numbers, is positive. Specifically, ls(n, k; s)l =
L(r + i1)(r + i2) · · · (r +ink),
where the summation is extended over all (n k)combinations {i 1, i 2, ... , ind of the n integers {0, 1, ... , n 1}. The coefficient ls(n, k; r)l, for r > 0, of expansion (8.60), of the rising noncentral factorials into powers, is called noncentro/ signless or absolute Stirling number of the first kind. Expansions (8.58) and (8.60) imply that the noncentral Stirling numbers of the first kind are derivatives of factorials, while expansion (8.59) entails that the noncentral Stirling numbers of the second kind are differences of powers. Specifically,
s(n, k;r)
ls(n, k; r)l
=
[:,Dk(t)n L=r, k = 0, 1, ... , n, n = 0, 1, ... ,
= (:, Dk(t + n 1)n
L=r, k = 0, 1, ... , n, n = 0, 1, ...
and
S(n, k; r)
= [:, Llktn]
, k
= 0, 1, ... , n, n = 0, 1, ....
t=r
Note that, for r = 0, these numbers reduce to the corresponding usual (central) Stirling numbers. For r f= 0 the noncentral Stirling numbers may be expressed in terms of the corresponding central Stirling numbers. Specifically, expanding the rising noncentral factorial of t of order n, (t + r + n 1)n, into powers of u = t + r, using (8.4), and then expanding the power of u = t + r into powers oft, using Newton's binomial formula, we deduce the expression ls(n, k; r)l
=
t (~)
r 1kls(n,j)l.
J=k
Also, expanding the noncentral factorial oft of order n, (t + r + n 1)n, into factorials oft, using Vandermonde's formula, and then expanding the
STIRLING NUMBERS
316
factorials oft into powers oft, using (8.2), we conclude the expression
ls(n, k; r)l =
t (~) j=k
(r
+ n j 
1)njls(j, k)l.
J
Similarly
t. (t)
S(n, k; r)
=
S(n,k;r)
=
and
t (~) j=k
(r)jkS(n,j)
rnis(j,k).
J
The noncentral Stirling numbers retain the orthogonality relation of the central Stirling numbers. Specifically, for any real parameter r, n
n
= 8n,k, "'L,S(n,j;r)s(j,k;r) = 8n,k,
"'L,s(n,j;r)S(j,k;r) j=k
j=k
where 8n,k = 1, if k = n and 8n,k = 0, if k =f. n is the Kronecker delta. The exponential generating functions of Is( n, k; r) I, n = k, k + 1, ... and S(n, k; r), n = k, k + 1, ... , for fixed k and r, may be obtained as ~ un 9k(u; r) = L... ls(n, k; r)l n! n=k
= (1 u) rllog(1u)jk k! , k=
0, 1, ...
and .

fk(u,r)
L S(n,k,r) unn! oo
•
ru (eu 1)k k! , k 0, 1, ... ,
e
n=k respectively. The noncentral Stirling number of the second kind, on using the expression of the kth power of the difference operator in terms of the shift operator, is expressed in the form of a single sum of elementary terms as k
S(n,k;r) =
~! "'L,(1)kj (~)
(r+j)n.
(8.62)
J
j=O
Triangular recurrence relations for the noncentral Stirling numbers, analogous to those for the central Stirling numbers, can be similarly deduced:
s(n + 1, k; r) = s(n, kfork= 1, 2, ... , n
+ 1, n
1; r)
(n
+ r)s(n, k; r),
= 0, 1, ... , with
s(O,O;r) = 1, s(n,O;r) = (r)n, n > 0, s(n,k;r) = 0, k > n.
8.5.
NONCENTRAL STIRLING AND RELATED NUMBERS
317
and
+ 1, k; r) = S(n, k 1; r) + (k + r)S(n, k; r), , n + 1, n = 0, 1, ... , with
S(n
for k = 1, 2, ...
S(O,O;r) = 1, S(n,O;r) = rn, n > 0, S(n,k;r) = 0, k > n.
Using (8.61), the first recurrence relation yields for the noncentral signless Stirling numbers of the first kind the triangular recurrence relation ls(n + 1, k; r)l = ls(n, k 1; r)l for k = 1, 2, ... , n
+ 1, n
+ (n + r)ls(n, k; r)l,
= 0, 1, ... , with
is(O, 0; r)l = 1, is(n, 0; r)l = (r
+ n 1)n,
n > 0, is(n, k; r)l = 0, k > n.
The triangular recurrence relation for the noncentral Stirling numbers of the second kind can be used for the derivation of the (power) generating function k
00
IT
¢k(u; r) = L S(n, k; r)un = uk (1 ru ju) 1 , n=k j=O
fork= 1, 2, .... Setting u = 1/t, we get 00
1
1
LS(n,k;r)=i=T=( ) ,k=1,2, .... tn t r k+l n=k
Using the orthogonality relation of the noncentral Stirling numbers, this expansion can be inverted as 1
= tk+l' k = 1, 2, ... . Equivalently, this expression may be written as 00
Lls(n,k;r)l( n=k t
1
+r +n
)
n+ 1
=
1
t
k+I'
k= 1,2, ....
Consider the noncentral generalized factorial of t of order n, scale parameters and noncentrality parameter r, (st + r)n, and let n
(st
+ r)n = L k=O
C(n, k; s, r)(t)k, n
= 0, 1, ... .
(8.63)
STIRLING NUMBERS
318
The coefficients C(n, k; s, r) are called noncentral generalized factorial coefficients or GouldHopper numbers. Clearly, C(n,k;s,r)
= 0, k > n, C(O,O;s,r) = 1.
Further, expansion (8.63) implies that the noncentral generalized factorial coefficients are differences of noncentral generalized factorials. Specifically, , k=0,1, ... ,n, n=0,1, ....
C(n,k;s,r) = [k\Llk(st+r)n] ·
t=O
Note that, for r = 0, these numbers reduce to the corresponding central numbers. For r =f. 0, the noncentral generalized factorial coefficients may be expressed in terms of the corresponding central generalized factorial coefficients. Specifically,
=
C(n, k; s, r)
:t (~)
(rfs)jkC(n,j; s)
]=k
and C(n, k; s, r)
=
:t (~) j=k
(r)njC(j, k; s).
J
The noncentral generalized factorial coefficient C(n, k; s, psr) is a polynomial ins of degree n, the coefficient of general term of which is a product of the noncentral Stirling numbers of the first and second kind, n
C(n,k;s,psr)
= "L,s(n,j;r)S(j,k;p)si. j=k
k
The exponential generating function of the sequence C(n, k; s, r), n k and r, may be obtained as
+ 1, ... , for fixed
oo
= k,
n
!k(u;s,r) = "L,c(n,k;s,r); n. n=k = (1
+ u)
r[(1+u)s1]k , k = 0, 1, ... . k.1
The noncentral generalized factorial coefficients, on using the expression of the kth power of the difference operator in terms of the shift operator, is expressed in the form of a single sum of elementary terms as k
C(n,k;s,r) =
~! ~)1)kj (~) j=O
J
(sj+r)n·
8.6. BIBLIOGRAPHIC NOTES
319
A triangular recurrence relation for the noncentral generalized factorial coefficients, analogous to that for the central generalized factorial coefficients, can be similarly deduced:
C(n + 1, k; s, r) = (sk for k
+ r n)C(n, k; s, r) + sC(n, k 1; s, r),
= 1, 2, ... , n + 1, n = 0, 1, ... , with
C(O, 0; s, r) = 1, C(n, 0; s, r) = (r)n, n
> 0, C(n, k; s, r)
= 0, k
> n.
This triangular recurrence relation can be used to show that 00
1 ()  = L:sC(n,k;s,r)(
t
8.6
k+l
n=k
1
st + r
)
n+l
.
BffiLIOGRAPHIC NOTES
The Stirling numbers were so named by N. Nielsen (1906) in honor of James Stirling, who introduced them in his Methodus Differentialis (1730) without using any notation for them. The notation adopted in this book is due to J. Riordan (1958). Recurrence relations and certain number theoretic properties of the Stirling numbers of the first kind were derived by L. Lagrange (1770). P. S. Laplace (1812) and A. Cayley (1887) provided several approximations of the Stirling numbers of the second kind. J. A. Grunert (1822, 1843), A. Cauchy (1833), 0. Schlomilch (1852, 1895), G. Boole (1860), L. Schlafli (1867), J. Blissard (1867) and J. Worpitzky (1883) explored further the Stirling numbers of both kinds. The work of N. E. Norlund (1924) inspired several publications on certain generalizations of the Stirling numbers and their connection with the Bernoulli and the generalized Bernoulli numbers. Also, the books by N. Nielsen (1906, 1923), E. Netto (1927), J. F. Steffensen (1927) and L. M. MilneThomson (1933) are worth mentioning. A thorough presentation of the Stirling numbers and their most important properties provided by Ch. Jordan (1933) in an excellent paper, which was included as Chapter 4 in his classical book on the calculus of finite differences (1939a), revived the interest in the Stirling numbers. Since then, a large number of publications on these numbers appeared in the literature. The more recent book of L. Comtet (1974) devotes a chapter to these numbers and provides a very rich bibliography. The expression of the nth power of the operator e = tD in terms of powers of the operator D was obtained by J. A. Grunert (1843). The derivation of this expression in Example 8.4 is due to Ch. Jordan (1933, 1939a), who also derived the
STIRLING NUMBERS
320
expression of the nth power of the operator tJi = tL.l in terms of powers of the operator L.l given in Example 8.5. H. W. Gould (1964) expressed the operator (a 1D)n in terms of powers of the operator D with the Stirling numbers of the first kind as coefficients. L. Comtet (1973) expressed the more general operator (>.(t)D)n in terms of powers of the operator D with coefficients Bell partition polynomials. A generalization to another direction was discussed by L. Carlitz (1932). 0. Schlomilch (1852) used the same symbol, Cf, to denote the Stirling numbers of both kinds calling them factorial coefficients. In this unified notation, Cf = ls(n, k)l and CJ:n = S(k, n). N. E. Norlund (1924) introduced the generalized Bernoulli numbers Br;), n = 0, 1, ... , with generating function the rth power of the generating function of usual Bernoulli numbers.
=
(n1)
(n)
(n) (k) . Bnk and S(n, k) = k Bnk. (see Exerctse 1 19). Properties of the generalized Bernoulli numbers were studied by L. Carlitz (1960). F. N. David and D. E. Barton (1962) devoted a chapter to these numbers. The first short table of the Stirling numbers of the second kind, up to n = 9, was published by James Stirling (1730). Extensive tables of the Stirling numbers of both kinds were constructed by H. Gupta (1950), R. A. Fisher and F. Yates (1953), F. N. David, M. G. Kendall and D. E. Barton (1966) and M. Abramowitz and I. A. Stegun (1965). A variety of asymptotic expressions for the Stirling numbers exist in the literature. The approximate expressions given in Exercise 7 are due to Ch. Jordan (1933, 1939a). Several references on other approximations are given in the review paper by Ch. A. Charalambides and J. Singh (1988). The coefficients L(n, k), k = 0, 1, ... , n, n = 0, 1, ... of the expansion of the rising factorials into falling factorials, introduced by I. Lah (1955), were called Lah numbers by J. Riordan (1958). Extending these numbers, Ch. A. Charalambides (1976, 1977a, 1979a) systematically studied the coefficients C (n, k; s) of the expansion of the generalized factorials into falling factorials. These numbers were noted before by Ch. Jordan (1933) and appeared as coefficients of a generalized Hermite polynomial in E. T. Bell (1934b), and in several other forms in H. W. Gould (1958), R. Shumway and J. Gurland (1960), L. Bernstein (1965), L. Carlitz (1965), W. Feller (1968), L. Comtet (1973). Later, L. Carlitz (1979) studied these numbers under the name degenerate Stirling numbers. The noncentral Stirling numbers of the second kind appeared in N. Nielsen (1906) as differences of the powers at an arbitrary point. J. Riordan (1937) used them as connection constants of power moments about an arbitrary point and factorial moments (see Example 8.3). Also, they appeared as coefficients in a modification of the classical occupancy problem discussed by D. E. Barton and F. N. David (1959). Recently, these numbers were studied by L. Carlitz (1980a,b) as weighted Stirling numbers, by M.
Then Js(n, k)l
k_
8. 7. EXERCISES
321
Koutras (1982) as noncentral Stirling numbers, by A. Z. Broder (1984) as rStirling numbers and also by R. Shanmugan (1984). The differences of the generalized factorials at an arbitrary point (GouldHopper numbers) were studied by Ch. A. Charalambides and M. Koutras (1983). The associated Stirling numbers, introduced by J. Riordan (1958), are closely related to the numbers of Ch. Jordan (1933, 1939a) and M. Ward (1934), which are the coefficients of the representation of s(n, n k) and S(n, n k) as sums of binomials of n. These numbers were further discussed by L. Carlitz (1971). Recurrence relations and other properties of the rassociated Stirling numbers and generalized factorial coefficients were discussed by J. Riordan (1958), L. Comtet (1974) and Ch. A. Charalambides (1974).
8. 7 EXERCISES 1.
Additional vertical recurrence relations for the Stirling numbers.
Show that (a) the Stirling numbers of the first kind s(n, k), k = 0, 1, ... , n, n = 0, 1, ... , with s(O,O) = 1, satisfy the recurrence relation n
s(n, k)
=L
nrk s(n + 1, r + 1), k
= 0, 1, ...
, n, n
= 0, 1, ...
,
r=k and (b) the Stirling numbers of the second kind S(n, k), k = 0, 1, ... , n, n = 0, 1, ... , with S(O, 0) = 1, satisfy the vertical recurrence relation n
S(n,k)
= Lknrs(r 1,k 1), k = 1,2, ...
,n, n
= 1,2, ....
r=k 2. Show that
t S(n, r)s(r r=k
+ 1, k + 1) = (1)nk (~)
and
ts(n+1,r+1)s(r,k)= r=k
(~).
3". Show that
k (kn) S(n,nk)=?; k+r (k+n) kr !s(k+r,r)!.
STIRLING NUMBERS
322
4. Show that
( k; r) s(n, k + r)
=~ (
J) s(j, k)s(n
j, r)
and conclude its inverse relation
(7) s(ni,r)
=
~ (k;r) S(k,i)s(n,k+r).
5. Show that
and conclude its inverse relation
(7) S(ni,r)
=
~ (k;r) s(k,i)S(n,k+r).
6*. Let C(n, k; s) be the generalized factorial coefficient. Show that
(1 ) k C(n,nk;s 1 ) = ~ ~ (kn) k+r (k+n) kr C(k+r,r;s). 7. (Continuation). Show that
( k;
r) C(n, k + r; s) = ~ ( J) C(j, k; s)C(n
j, r; s)
and
8*. Show that the signless Stirling number of the first kind is given by
L;=
8 where (n(s) = 1 lfj , m = [n/2] and the summation is extended over all nonnegative integer solutions of the equation r 1 + 2r2 + · · · + krk = k.
8. 7. EXERCISES
323
9*. Asymptotic expressions for the Stirling numbers. For fixed k and n too, show that
ls(n
+ 1, k + 1)1
~ n![log(n
+ 1) + C]k/k!
and
S(n,k)
~
knfk!,
where C = 0.57721 is the Euler's constant. Also, for fixed k, s and n t oo derive for the generalized factorial coefficients C(n, k; s) the asymptotic expression C(n, k; s) ~ (sk)nfk!. 10. Associated Stirling numbers of the first kind. Using the triangular recurrence relation
s(n
+ 1, k) = s(n, k 1) s(O,O)
= 1,
 ns(n, k), k
s(n,O)
= 0,
= 1, 2, ...
n > 0, s(n,k)
,n
+ 1,
n
= 0, 1, ...
= 0, k > n,
show that
s(n, n)
= 1,
s(n, n 1)
=  ( ~) , s(n, n 
2)
= 2 ( ~) + 3 ( ~)
and, generally, setting k
s(n,n
k) = Ls2(k + (k: ·), j,j)
J
j=O
derive the bivariate generating function
and deduce the generating function /k.2(u)
=
~
un L... s2(n, k) 1 n. n=2k
=
[log(1
+ u) u]k k' .
, k
= 0, 1, ...
Further, show that
s 2(n for n
+ 1, k) =
ns 2(n, k) ns 2(n 1, k 1),
= 2k, 2k + 1, ... , k = 1, 2, ... , with s2(0, 0)
= 1,
s2(n, 0)
= 0,
n > 0, s2(n, k)
= 0,
2k > n,
.
,
324
STIRLING NUMBERS
and
k
s 2(n,k)
= ~)1)i (~)
for n
s(nj,kj),
J
j=O
= 2k, 2k + 1, ... , k = 0, 1, ....
11. (Continuation). Consider, more generally, the numbers Sr (n, k), n = rk, rk + 1, ... , k = 0, 1, ... , r = 2, 3, ... , defined by their exponential generating function n
OC>
!k,r(u) =
L
1
[
sr(n, k); = kl log(1 n. ·
n=rk
k
rl
j
j=l
J
+ u) L( 1)jl ~
Derive the recurrence relation sr(n + 1, k) for n
l
= (1r 1 (n)r 1 sr(n r + 1, k 1) nsr(n, k),
= rk, rk + 1, ... , k = 1, 2, ... , r = 2, 3, ... , with = 1,
sr(O, 0)
sr(n, 0)
= 0,
n
> 0, sr(n, k) = 0, rk > n.
Also, show that k
(n)rj Sr+l (n, k)  "'(1)rj L .1 j Sr (n  rJ,. k J") , r  1, 2, ... i=O J .r and k
(n)rj Sr (n, k)  "'(1)rj+l L .1 j Sr+l (n  rJ,. k J") , r  1, 2, .... j=O J.r 12. Associated Stirling numbers of the second kind. Using the triangular recurrence relation S(n
+ 1, k) = S(n, k 1) + kS(n, k), S(O,O)
= 1,
S(n,O)
= 0,
n
k
= 1, 2, ...
,n
+ 1,
n
= 0, 1, ...
> 0, S(n,k) = 0, k > n,
show that S(n, n)
= 1,
S(n, n 1)
= ( ~) , S(n, n 2) = ( ~) + 3 ( ~)
and, generally, setting S(n,nk) = ts2(k+j,j) (k n .) , j=O +J
,
8. 7. EXERCISES
325
derive the bivariate generating function
g2(t,u)
oo [n/2]
n
n=O k=O
S2(n,k)tk; n.
=L
L
= exp[t(eu 1 u)]
and deduce the generating function
Also, show that
S2(n for n
+ 1, k)
= kS2(n, k)
+ nS2(n 1, k
1),
= 2k, 2k + 1, ... , k = 1, 2, ... , with S2(0, 0) = 1, S2(n, 0) = 0, n
> 0, S 2(n, k)
= 0, 2k
> n,
and S2(n, k) for n
k
(~)
j=O
J
= L( 1)i
S(n j, k j),
= 2k, 2k + 1, ... , k = 0, 1, ....
13. (Continuation). Consider, more generally, the numbers Sr (n, k), n = rk, rk + 1, ... , k = 0, 1, ... , r = 1, 2, ... , defined by their exponential generating function n
oo
!k,r(u)
=
L
Sr(n,k);
n=rk
n.
r1
1
= kl
·
eu(
k j
L ~I j=O ).
)
Derive the recurrence relation
for n = rk, rk
+ 1, ... , k
= 1, 2, ... , r = 1, 2, ... , with
Sr(O, 0) = 1, Sr(n, 0) = 0, n > 0, Sr(n, k) = 0, rk > n. Further, show that k
Sr+t(n,k)
= L(1)j J..~n()~)iSr(nrj,kj), r. J j=O
r= 1,2, ...
326
STIRLING NUMBERS
and
k
(n )rj =" ~' :r( !)J Sr+I (n TJ,. k  J') , r j=O J. r.
Sr (n, k )
= 1, 2, ....
14. Associated generalized factorial coefficients. Using the triangular recurrence relation C(n
+ 1, k; s) + 1, n
for k = 1, 2, ... , n
= (sk n)C(n, k; s)
+ sC(n, k
1; s),
= 0, 1, ... , with initial conditions
> 0, C(n,k;s)
C(O,O;s) = 1, C(n,O;s) = 0, n
= 0, k
> n,
show that C(n,n;s)
= sn,
C(n,n 1;s)
C(n, n 2; s) = sn 3 (s)3 (
= sn 2(s)2 (~),
~) + 3sn 4 [(shf ( ~)
and, generally, setting k
C(n,nk;s)=LC2(k+j,j;s)snkj j=O
(k: .), J
derive the bivariate generating function oo [n/2]
n
92(t,u;s) = L L C2(n,k;s)tk; = exp{t[(1 +uV 1 su]}, n=O k=O n.
and conclude the generating function 00
un [(1 !k,2(u; s) = L C2(n, k; s) n! = n=2k
+ u)
8

k!
1  su]k
, k = 0, 1, ....
Also, show that C2(n
for n
+ 1, k; s)
= (sk n)C2 (n, k; s)
+ n(s)2C2(n
1, k 1; s),
= 2k, 2k + 1, ... , k = 1, 2, ... , with C2(0,0;s) = 1, C 2 (n,O;s) = 0, n
and
> 0, C 2(n,k;s)
= 0, 2k
k
C 2(n, k; s)
= L( 1)1 (~) j=O
J
siC(n j, k j; s),
> n,
8. 7. EXERCISES
327
for n = 2k, 2k + 1, ... , k = 0, 1, .... 15. (Continuation). Consider, more generally, the numbers Cr(n, k; s), n = rk, rk + 1, ... , k = 0, 1, ... , r = 1, 2, ... , defined by their exponential generating function
Derive the recurrence relation Cr(n
+ 1, k; s) =
(sk n)Cr(n, k; s)
+ ( r ~ 1) for n = rk, rk
+ 1, ... , k =
(s)rCr(n r
+ 1, k
1; s),
1, 2, ... , r = 1, 2, ... , with
> 0, Cr(n, k; s)
Cr(O, 0; s) = 1, Cr(n, 0; s) = 0, n
= 0, rk
> n.
Also, show that Cr+l (n, k; s)
)j r
k = ~( 1)1 (n!,rj J.
( 8
j=O
Cr(n rj, k j; s), r
= 1, 2, ...
and k
Cr(n, k; s)
j
= ~ (n!!rj j=O
(;) Cr+l (n rj, k j; s), r
= 1, 2, ....
J
16. Cauchy numbers. The sequence of the Cauchy numbers Cn, n = 0, 1, ... , has generating function oo
f(t)
tn
t
= ~ Cnn! = log(1 + t) ·
Show that (n)kl C nk, n = 1, 2, ... , vo ,, = 1 Cn = ~( L.J 1 )k1 k
+
k=l
and n
Cn =
1
r
1
L k + 1 s(n, k), n=O L S(r, n)Cn = r + 1 , k=O
STIRLING NUMBERS
328
where s(n, k) and S(r, n) are the Stirling numbers of the first and second kind, respectively. 17. Bernoulli numbers. The sequence of the Bernoulli numbers Bn,
n
= 0, 1, ... , has generating function oo
g(t)
=L
tn Bn n!
n=O
Show that Bn
=
t (~)
Bk, n
t
= et 
1.
= 1, 2, ... , B 0 = 1
k=O
and n ( 1)kk! r ( 1Yr! Bn=L k+ S(n,k),Ls(r,n)Bn= r+ , 1 1 k=O n=O
where s(n,k) and S(r,n) are the Stirling numbers of the first and second kind, respectively. 18. (Continuation). (a) Show that B 2r+l = 0, r = 1, 2, ... , and conclude that oo t2r 2 1+ 1Y2 r B2r ( r)! = t cot t. 2
?;(
(b) Using the expansion 00
t2
1 + 2""' 2 ~ t  n 21r 2 n=O
= t cot t,
show that 00
B2r
= (1Y+l2(2r)!(27r) 2r((2r),
((s) =
L ns n=l
and conclude that IB2rl = ( 1Y+l B2r, r
= 1, 2, ... , and
(27ryr ((2r) =  ( 1 IB2rl, r = 1, 2, .... 2 2r). 19. NorlundBernoulli numbers. The sequence of numbers Btl, n = 0, 1, ... , where r is a real number, with generating function
8. 7. EXERCISES
329
has been defined by Norlund and, in the particular case of r = 1, reduces to the sequence of Bernoulli numbers. Show that
B~+t)
= ( 1  ~) B~)  nB~~ 1 , n = 1, 2, ... , B~r) = 1
and conclude that
B~) = s(r, r n)/ ( r ~ 1 )
,n
B~r) = S(n + r, r)/ ( n: r)
= 0, 1, ... ,n
, r 1, r
= 1, 2, ... ,
= 0, 1, ... , r = 1, 2, ...
,
where s(n, k) and S(n, k) are the Stirling numbers of the first and second kind, respectively. 20. Consider the sequence B(n; s), n = 0, 1, ... , where sis a real number, with generating function tn
oo
g(t;s) =
~B(n;s)n!
t = ( 1 + t)B
1'
Show that B(n; s)
=
t (~)
(s)nkB(k; s), n
= 1, 2, ... , B(O; s) = 1/s
k=O
and
~ ~
B = n
k=O
(1/sh+t ) k+ 1 C( n, k·,s,
where C(n, k; s) is the generalized factorial coefficient. Further, show that lim sB(n; s) = Cn, lim sn+I B(n; s) = Bn,
s+ 0
stoo
where Cn and Bn are the Cauchy and Bernoulli numbers, respectively. 21 *. GarlitzRiordan numbers of the first kind. Consider the expansion of central factorial of t of order n,
t[n] = t (t
+~
= t (t + ~2 with t[OJ
1) (t
+ ~ 2) ··· (t ~ + 2)
1) , n1
n
= 1, 2, ...
,
= 1, into powers oft: n
t[n] =
L r(n, k)tk, k=O
n = 0, 1, ... .
(t
~ + 1)
STIRLING NUMBERS
330
Show that oo
n
U
n
g(t,u)= LLr(n,k)tk:! n=Ok=O
= ((u/2)+V1+(uf2) 2 )
,
and conclude that oo un [2log((u/2)+J1+(u/2F)r fk(u)=Lr(n,k)n! = k! , k=0,1, .... n=k
Derive the recurrence relation r(n
fork= 2, 3, ... , n
+ 2, k) = r(n, k 2) (n/2) 2 r(n, k),
+ 2, n = 0, 1, ... , with
r(O, 0) = 1, r(n, k) = 0, k
> n, r(2n, 2k + 1) = 0, r(2n + 1, 2k) = 0.
22*. GarlitzRiordan numbers of the second kind. Consider the expansion of the nth order power of t into central factorials of it: n
t 0 =LR(n,k)t[kJ, n=0,1, .... k=O
Show that n
n
Lr(n,j)R(j,k)
= c5n,k,
j=k
LR(n,j)r(j,k)
= c5n,k
j=k
and
Derive the explicit expression
(k) (k2  )n
1 "'(1)3 k . R(n k) = ' k! L.....t j j=O
j
23*. (Continuation). Show that R(n + 2, k) = R(n, k 2) (k/2) 2 R(n, k),
fork= 2, 3, ... , n
+ 2,
n = 0, 1, ... , with
R(O, 0) = 1, R(n, k) = 0, k > n, R(2n, 2k + 1) = 0, R(2n + 1, 2k) = 0
8. 7. EXERCISES
331
and deduce the generating function
k = 2s + 1.
24. Consider the sequence of noncentral signless Stirling numbers of the first kind ls(n, k; r)l, k = 0, 1, ... , n, n = 0, 1, ... , for fixed r, which are defined by (see Section 8.5) n
(t
+ r + n 1)n =
L
ls(n, k; r)itk, n
= 0, 1, ....
k=O
Show that oo
g(t, u; r)
=L
n
L
n=Dk=O
n
ls(n, k;r)ltk; n.
= (1 u)tr
and conclude that
. ~ . un_ rflog(1u)]k _ fk(u, r)  L....is(n, k, r)l n!  (1 u) k! , k 0, 1, .... n=k
25. (Continuation). Show that the noncentral signless Stirling numbers of the first kind ls(n, k; r)l, k = 0, 1, ... , n, n = 0, 1, ... , satisfy the triangular recurrence relation ls(n for k
+ 1, k; r)l
= ls(n, k 1; r)l
= 1, 2, ... , n + 1, n =
+ (n + r)ls(n, k; r)l,
0, 1, ... , with initial conditions
ls(O, 0; r)l = 1, ls(n, 0; r)l = (r
+ n 1)n,
n > 0, ls(n, k; r)l = 0, k > n.
Also, show that ls(n, k; r)l =
t (~) j=k
(m
+ n j 
1)njls(j, k; r m)l
J
and conclude that n
ls(n, k; r )I = L(n)nj ls(j, k; r 1)1 j=k
STIRLING NUMBERS
332
and
ls(n, k; r)l =
t (~) j=k
+ n j 
(r
1)njls(j, k)l·
J
26. Suppose that, balls are successively drawn one after the other from an urn initially containing w white and b black balls, according to the following scheme. After each trial the drawn ball is placed back in the urn along with s black balls. Show that (a) the probability p(k; n, r) of drawing k white balls in n trials is given by
is(n,k;r)IOk . )_ p (k ,n,r (O+r+n 1)n' k=0,1, ... ,n and (b) the probability q( n; k, r) that n trials are required until the kth white ball is drawn is given by
q(n;k,r) =
ls(n 1, k 1; r)IOk (O ) , n = k,k +r+n1n
+ 1, ...
,
where 0 = w/s and r = bfs.
27. Consider the sequence of noncentral Stirling numbers of the second kind S(n, k; r), k = 0, 1, ... , n, n = 0, 1, ... , for fixed r, which are defined by (see Section 8.5) n
(t
+ r)n = L
S(n, k; r)(t)k, n
= 0, 1, ....
k=O Show that
n
oo
f(t, u; r)
n
=L
L S(n, k; r)(t)k; n=O k=O n.
= e(t+r)u
and conclude that .

oo
•
Un 
fk(u,r) LS(n,k,r)n! e n=k
ru ( eu 
k!
1) k

, k0,1, ....
Further, derive the explicit expression
~(S( n, , r  !._ k! L....t j=O
k. )_
1 )kj (k). (r + J·)n . J
28. (Continuation). Show that the noncentral Stirling numbers of the second kind S(n,k;r), k = 0,1, ... ,n, n = 0,1, ... , satisfy the triangular recurrence relation
S(n
+ 1, k; r)
= S(n, k 1; r)
+ (k + r)S(n, k; r),
8. 7. EXERCISES
for k = 1, 2, ... , n
333
+ 1, n
= 1,
S(O,O;r)
= 0, 1, ... , with initial conditions
S(n,O;r) =rn, n > 0, S(n,k;r)
= 0,
k > n.
Also, show that S(n,k;r)
=
t
(~) mnjS(j,k;rm)
j=k
J
and conclude that S(n, k; r)
=
and S(n,k;r) =
t (~) t (~) j=k
J
j=k
J
S(j, k; r 1)
rnjS(j,k).
29. (Continuation). Show that
~s(n,j;r1)S(j,k;r2) = (~) (r2 rl)nk,
t
S(n,j; ri)s(j, k; r2) = (
~) (r1 
r 2 )nk,
J=k and conclude that n
n
Ls(n,j;r)S(j,k;r) j=k
= On,k.
LS(n,j;r)s(j,k;r) j=k
= On,k·
30. (Continuation). Show that ( k
t m ) s (n, k + m; r 1 + r 2) = %' (;) s (j, k; r1) s( n  j, m; r2)
and conclude its inverse relation
31. (Continuation). Show that ( k
t
m) S(n, k
+ m;r 1 + r 2) =
~ ( ; ) S(j, k;r1)S(n j,m;r2)
STIRLING NUMBERS
334
and conclude its inverse relation
(7)
~ ( k ~ m) s(k, i; r1)S(n, k + m; r 1 + r2).
S(n i, m; r 2) =
32. (Continuation). Show that k
00
¢k(u; r) =
2: S(n, k; r)un = uk II (1 ru ju)n=k
1
,
k = 1, 2, ...
j=O
and conclude the inverse relations 00
(t+r)k
= '2:(1)nkS(n1,k1;r+ 1)tn, k = 1,2, ... , n=k 00
(t r)k
= 2:( 1tks(n 1, k 1; r + 1)(t)n, k = 1, 2, .... n=k
33. Suppose that, balls are successively drawn one after the other from an urn initially containing m white and r black balls, according to the following scheme. After each trial, if the drawn ball is white, a black ball is placed in the urn, while if the drawn ball is black, it is placed .back in the urn. Show that (a) the probability p(k; n, r) of drawing k white balls in n trials, with n ~ m, is given by
. _S(n,k;r)(m)k _ p(k,n,r)( ) ,k0,1, ... ,n m+r n and (b) the probability q(n; k, r) that n trials are required until the kth white ball is drawn, with k ::::; m, is given by
q(n; k, r) =
S(n 1, k 1; r)(m)k ( ) , n = k, k m+r n
+ 1, ....
34. Consider the sequence of the noncentral generalized factorial coefficients C(n, k; s, r), k = 0, 1, ... , n, n = 0, 1, ... , for fixed s and r, which are defined by (see Section 8.5) n
(st
+ r)n
=
'2: C(n, k; s, r)(t)k> n = 0, 1, .... k=O
Show that oo
n
n
f(t,u;s,r) = LLC(n,k;s,r)(t)k; = (1 +u)st+r n=O k=O n.
8. 7. EXERCISES
335
and conclude that oo un [(1 fk(u;s,r)=l:C(n,k;s,r),=(1+uY
+ u)s
n.
n= k
k'
1Jk
.
,k=0,1, ....
Using a suitable generating function, show that
C(n,k;s,r)
=
t (t)
(r/s)JkC(n,j;s),
J=k
and
C(n,k;s,r)
=
t (~) j=k
(r)njC(j,k;s).
J
35. (Continuation). Show that the noncentral generalized factorial coefficient C(n, k; s, r), k = 0, 1, ... , n, n = 0, 1, ... , is given by the sum
C(n,k;s,r)= :,t(1)kJ (;) (sj+r)n. J=O
36. (Continuation). Show that the noncentral generalized factorial coefficients C(n, k; s, r), k = 0, 1, ... , n, n = 0, 1, ... , satisfy the triangular recurrence relation
C(n for k
+ 1, k; s, r) = (sk + r n)C(n, k; s, r) + sC(n, k
= 1, 2, ...
,n
+ 1, n = 0, 1, ... , with
1; s, r),
initial conditions·
C(O,O;s,r) = 1, C(n,O;s,r) = (r)n, n > 0, C(n,k;s,r) = 0, k > n. 37. (Continuation). Show that lim skc(n, k; s, r)
s+0
= s(n, k; r)
and, for lim r / s = p, that s+oo
lim snc(n, k; s, r)
s+oo
= S(n, k; p).
38. (Continuation). Noncentral Lah numbers. Consider the expansion n
((t r))n =
2::: L(n, k; r)(t)k, n = 0, 1, ... . k=O
STIRLING NUMBERS
336
The coefficient L(n, k; r) is called noncentral Lah number; it is a particular case, s = 1, of the noncentral generalized factorial coefficient. Show that n
n! (n r 1)
L(n,k;r)=(1) k!
kr 1
.
39. (Continuation). Show that n
C(n, k; s, ps r)
=L
s(n,j; r)S(j, k; p)sl
j=k
and conclude that ts(n,j;r)S(j,k;p) )=k t(1)njs(n,j;r)S(j,k;p) )=k
= (~) (pr)nk, = (~)
(pr+n1)nk·
40. (Continuation). Show that n
LC(n,j;sl,ri)C(j,k;s2,r2) = C(n,k;s1s2,r1 +r2s1) j=k and conclude that 1 tc(n,j;s,ri)C(j,k;s ,r2) = J=k
and
(~) (r1 +r2s)nk,
n
L C(n, j; s, r)C(j, k; s 1, rs 1) j=k
= 8n,k·
41. (Continuation). Let 1
00
cPk(u;s,r) = LC(n1,k1;s,r)(u)n' k = 1,2, .... n=k
Show that cPk(u; s, r)
= (u
1
_ r)js _ (k _ 1) c/Jkdu; s, r), k
= 2, 3, ...
,
8. 7. EXERCISES
with ¢ 1 (u;s,r)
337
= 1/(u r), and conclude that 1
¢k(u; s, r)
= s (( ur )/ s )k
and 00
1
sC(n 1, k 1; s, r) (
)( = L
t k
n=k
1
st
+r
) . n
42. (Continuation). Show that
(
k+m) k C(n,k+m;s,r 1 +r2 )
=
~ (~) C(j, k; s, rt)C(n j, m; s, r2) j=k
J
and conclude its inverse relation
(7) C(ni,m;s,r2)
'E' (k~m)
=
C(k,i;s 1 ,r1 s 1 )C(n,k+m;s,r, +r2).
43. (Continuation). Show that n
(tb+! D)n f(t) = L( 1)nkbnakC(n, k; s, r)tbnak(Dta+l )k f(t), k=O where s
= ajb, r =
(a+ 1)/b, and conclude that n
(tb+l D)n J(t)
= L( 1)nbnC(n, k; s)tbn+k Dk f(t), s = 1/b k=O
and
n
(tDt f(t) = LS(n, k)tkDk f(t). k=O 44. Consider an urn containing s identical series of coupons, each consisting of m coupons bearing the numbers 1, 2, ... , m and r additional coupons, all bearing the number m + 1. Suppose that coupons are drawn one after the other, at random, without replacement. Show that (a) the probability
STIRLING NUMBERS
338
p(k; n) of drawing exactly k of them numbers {1, 2, ... , m} in n drawings, with n ::=; m, is given by . ) = C(n, k; s, r)(m)k k = 1 0 , , ... ,n (k p,n ( ) , sm +r n and (b) the probability q(n; k) that n drawings are required until the kth different number, among the m numbers { 1, 2, ... , m}, is drawn, with k ::=; m, is given by
q(n; k)
=
C(n 1,k 1;s,r)s(m)k (sm +r )n ,n
= k, k + 1, ...
, sm
+ r.
45. (Continuation). Suppose that from the urn coupons are drawn one after the other, at random, by returning in the urn, after each drawing, the chosen coupon together with another coupon bearing the same number. Show that (a) the probability p(k; n) of drawing exactly k of the m numbers {1, 2, ... , m} in n drawings is given by p (k ,n)
= JC(n,k;s,r)J(m+k1)k k=O ( ) , , 1, ... ,n sm + r + n 1 n
where JC(n, k; s, r)J = ( 1)nC(n, k; s, r). Further show that (b) the probability q(n; k) that n drawings are required until the kth different number, among the m numbers { 1, 2, ... , m}, is drawn is given by
. k) _ JC(n 1, k 1; s, r)Js(m q (n, ( ) sm + r + n 1 n
+ k 1)r
_ k k
,n
, + 1, ....
Chapter 9 DISTRIBUTIONS AND OCCUPANCY
9.1
INTRODUCTION
A considerable number of combinatorial configurations, such as permutations, combinations and partitions of a finite set under various conditions, can be described by the model of distribution (allocation) of balls (objects) into urns (cells, boxes), introduced by P. MacMahon. It is an advantageous approach, since urn models can be easily visualized and are very flexible. Further, these models admit many equivalent interpretations. In general, the balls are of the type (r 1 , r2, ... , rm) with r1 + r2 + · · · + rm = n, that is, ri of the n balls are of the ith kind, i = 1, 2, ... , m. The urns are of the type (si,s2,··· ,sv) with s1 + s2 + ··· + Sv = k, that is, Sj of the k urns are of the jth kind, j = 1, 2, ... , v. Moreover, the urns may be of limited or unlimited capacity and the balls in each urn may be ordered or unordered. The enumeration of the assignments of the balls to the urns is a distribution problem, while the enumeration of the balls in specified or arbitrary urns is an occupancy problem. This general consideration reveals a host of enumeration problems. Particular cases of such problems have been examined in Chapter 2 as applications of enumeration of certain permutations and combinations. In the present chapter, more general distribution and occupancy problems are studied by using the inclusion and exclusion principle and generating functions. Specifically, the classical occupancy problem and some of its modifications are examined in length. Then the distributions of balls into urns, when the ordering of the balls in the urns counts, are enumerated. Further, the problem of enumeration of the distributions of balls of a general specification into distinguishable urns is treated by using the inclusion and exclusion principle. As applications, a variety of committee problems is discussed. Finally, the use of generating functions in coping with the same general problem is demonstrated.
DISTRIBUTIONS AND OCCUPANCY
340
9.2
CLASSICAL OCCUPANCY AND MODIFICATIONS
Let us first consider n distinguishable balls and k distinguishable urns and assume that the capacity of each urn is unlimited. The urns as distinguishable, without any loss of generality, may be numbered from 1 to k. Then, the number of distributions of n distinguishable balls into k distinguishable urns equals
the number of npermutations of the set {1, 2, ... , k }, with repetition (see Theorem 2.3). Further, the number of distributions of n distinguishable balls into k distinguishable urns with rj balls in the jth urn, j = 1, 2, ... , k, r1 + r2 + · · · + rk = n, equals
n!
the number of divisions of the set of n balls {1, 2, ... , n} into k subsets containing r 1, r2, ... , rk balls, respectively (see Theorem 2.8). In the classical occupancy problem, the number of distributions with a given number of occupied urns is of interest. The next theorem is concerned with this number.
THEOREM9.1 The number of distributions of n distinguishable balls into k distinguishable urns so that r urns are occupied is given by N(n, k, r) = (k)rS(n, r),
where S(n, r) =
~ r.
i)
1)j
j=O
(9.1)
(~) (r j)n, J
is the Stirling number of the second kind. PROOF Consider the set [l of distributions of n distinguishable balls into k distinguishable urns and let Ai be the subset of these distributions in which the ith urn remains empty, i = 1, 2, ... , k. Then, for any selection of s indices {i 1 ,i2, ... ,i.},outofthekindices{l,2, ... ,k},
9.2.
CLASSICAL OCCUPANCY AND MODIFICATIONS
341
which is the number of distributions of n distinguishable balls into the remaining (after excluding the s specified urns) k  s distinguishable urns. This expression implies the exchangeability of the sets A 1 , A2 , ... , Ak. Further, the number N(n, k, r) of distributions of n distinguishable balls into k distinguishable urns, which leave k  r urns empty (and consequently r urns occupied), equals the number Nk,kr of elements of [l that are contained in k  r among the k sets A 1 , A2, ... , Ak. Thus, according to Corollary 4.4, we deduce the expression
which, upon using the explicit expression of the Stirling numbers of the second I kind (see Section 8.3), implies (9.1). COROLLARY 9.1 The number of distributions of n distinguishable balls into k distinguishable urns so that, out of s specified urns, r urns are occupied is given by
N(n,k,s,r) where S (n, r; k  s) =
~
= (s)rS(n,r;ks)
t(
1 )i
j=O
(~) ( k 
s
+r
(9.2)
 j) n,
J
is the noncentral Stirling number of the second kind. PROOF Note that i balls can be selected from n distinguishable balls in (~) ways, i = 0, 1, ... , n. Further, for each selection, thei balls, according to Theorem 9.1, can be distributed into the s specified urns, so that r among them are occupied in
ways, while the remaining n  i balls can be distributed into the remaining k  s urns, without any restriction, in (k  s )ni ways, i = 0, 1, ... , n. So, according to the multiplication principle, there are
different such distributions. Thus, summing fori = 0, 1, ... , n, according to the addition principle, the number of distributions of n distinguishable balls into k
DISTRIBUTIONS AND OCCUPANCY
342
distinguishable urns so that r among s specified urns are occupied is obtained as
N(n, k, s, r) =
The last expression, using the explicit expression of the noncentral Stirling numI bers of the second kind (see Section 8.5), implies (9.2). COROLLARY 9.2 The number of distributions of n distinguishable balls into k indistinguishable urns so that r urns are occupied is given by
·(r)
1 ~ S(n,r)=;:yL...(1)1
. (rj)n,
(9.3)
J
j=O
the Stirling number of the second kind, while, without any restriction, it is given by k
B(n,k)
= LS(n,r),
(9.4)
r=l
the Bell number. PROOF Note first that r urns can be selected from k indistinguishable urns in only one way. Further, let R(n, r) and N(n, r) be the number of distributions of n distinguishable balls into r indistinguishable urns and into r distinguishable urns, respectively, so that no urn remains empty. Consider a distribution of n distinguishable balls into r indistinguishable urns so that no urn remains empty. If the r indistinguishable urns are transformed to distinguishable, by numbering them, and permuted in all possible ways, r! distributions of n distinguishable balls into r distinguishable urns, so that no urn remains empty, are constructed. Consequently r!R(n, r) = N(n, r) and since, by Theorem 9.1, N(n, r) = r!S(n, r), it follows that R(n, r) = S(n, r). Thus, the number R(n, k, r) of distributions of n distinguishable balls into k indistinguishable urns so that r urns are occupied is given by (9.3). As regards the evaluation of the number B(n, k) of distributions of n distinguishable balls into k indistinguishable urns, without any restriction, note that, in such a distribution, 1 or 2 or , ... , or k urns are occupied. Since the number of distributions of n distinguishable balls into k indistinguishable urns with r occupied urns equals S(n, r), summing for all values r = 1, 2, ... , k, according to the addition principle, we deduce (9.4). I
9.2.
CLASSICAL OCCUPANCY AND MODIFICATIONS
343
Consider now n indistinguishable (like) balls and k distinguishable urns. Again, the urns as distinguishable may be numbered from 1 to k. The number of distributions of n indistinguishable balls into k distinguishable urns equals the number of nonnegative integer solutions of the linear equation r 1 + r2 + · · · + rk = n, where Tj is the number of balls in the jth urn, j = 1, 2, ... , k. Thus, according to Theorem 2.12, the number of distributions of n indistinguishable balls into k distinguishable urns equals
the number of ncombinations of the k urns {1, 2, ... , k} with repetition. Further, the number of distributions of n indistinguishable balls into k distinguishable urns so that no urn remains empty equals the number of positive integer solutions of the linear equation r 1 +r 2 +· · · +rk = n, which, according to Corollary 2.5, equals
(n1)
k 1 ,
the number of ncombinations of the k urns { 1, 2, ... , k}, with repetition and the restriction that each urn is included at least once. As a consequence of these remarks, the following corollary, analogous to Theorem 9.1, is deduced. COROLLARY 9.3 The number of distributions of n indistinguishable balls into k distinguishable urns so that r urns are occupied is given by (9.5)
PROOF Note that r urns can be selected from k distinguishable urns in (~) ways and n indistinguishable balls can be distributed into these r urns, so that no urn remains empty, in (~::::D ways. Thus, according to the multiplication principle,
the required number of distributions is deduced as (9.5).
I
Let us, more generally, consider n indistinguishable (like) balls and k distinguishable urns, each divided into s distinguishable cells (compartments). Further, assume that each cell is of capacity limited to one ball. Then, the number of distributions of n indistinguishable balls into the k distinguishable urns (sk distinguishable cells) equals
DISTRIBUTIONS AND OCCUPANCY
344
while the number of distributions of n indistinguishable balls into the k distinguishable urns with Tj balls in the jth urn, j = 1, 2, ... , k, Tt + r2 + · · · + Tk = n, equals
(~) (;J ... (~).
The next theorem is analogous to Theorem 9.1.
THEOREM9.2 The number of distributions of n indistinguishable balls into k distinguishable urns, each with s distinguishable cells ofcapacity limited to one ball, so that r urns are occupied, is given by
(k)r R(n,k,s,r) =  1C(n,r;s) n.
where
C(n,r;s)
(9.6)
= ~ t(1ri(~)(sj)n J
j=O
is the generalized factorial coefficient. PROOF Consider the set[} of distributions of n indistinguishable (like) balls into k distinguishable urns, each with s distinguishable cells of capacity limited to one ball. Let Ai be the subset of these distributions in which the ith urn remains empty, i = 1, 2, ... , k. Then, for any selection of j indices {i~, i 2 , .•. , ij }, out of the k indices {1, 2, ... , k },
vi= N(Ai,Ai 2 ···AiJ = (
s(kn
j)) .
, J = 1,2, ... ,k,
which is the number of distributions of n indistinguishable balls into the remaining (after excluding the j specified urns) k  j distinguishable urns (s(k  j) distinguishable cells each of capacity limited to one ball). This expression implies the exchangeability of the sets At, A2 , ••• , Ak. Further, the number R( n, k, s, r) of distributions of n indistinguishable balls into k distinguishable urns that leave k r urns empty (and consequently r urns occupied) equals the number Nk,kr of elements of [} that are contained in k  r among the k sets At, A2, ... , Ak. Thus, according to Corollary 4.4, we get the expression
which, using the explicit expression of the generalized factorial coefficient (see Section 8.4), implies (9.6). I
9.2.
CLASSICAL OCCUPANCY AND MODIFICATIONS
345
REMARK 9.1 The number of distributions of n distinguishable balls into k distinguishable urns, each with s distinguishable cells of capacity limited to one ball, equals (sk)n,
while, with the restriction that r urns are occupied, is given by n!R(n, k, s, r) = (k)rC(n, r; s),
where C (n, r; s) is generalized factorial coefficient.
(9.7)
I
Assume now that, in the preceding model, the cells are of unlimited capacity. Then, the number of distributions of n indistinguishable balls into the k distinguishable urns (sk distinguishable cells), equals
while the number of distributions of n indistinguishable balls into the k distinguishable urns, each with s distinguishable cells, with r1 balls in the jth urn, j = 1, 2, ... , k, r 1 + r2 + · · · + Tk = n, is given by
The next theorem is analogous to Theorem 9.1 and 9.2. THEOREM9.3 The number of distributions of n indistinguishable balls into k distinguishable urns, each with s distinguishable cells of unlimited capacity, so that r urns are occupied, is given by
T(n, k, s, r)
= (k)r IC(n, r; s)l, n.1
(9.8)
where IC(n, r; s)l = ( 1)nC(n, r; s) =
~ r.
i)
J=O
1rj (~) (sj J
+ n 1)n
is the generalized factorial coefficient. PROOF Consider the set fl of distributions of n indistinguishable (like) balls into k distinguishable urns, each with s distinguishable cells of unlimited capacity. Let Ai be the subset of these distributions, in which the ith urn remains empty,
DISTRIBUTIONS AND OCCUPANCY
346
i = 1, 2, ... , k. Then, for any selection of j indices {it, i 2 , •.. , ij }, out of the k indices {1, 2, ... , k },
v1 = N(A;,A; 2
...
A;,)= (
s(k j) + n 1) . n , J = 1,2, ... ,k,
which is the number of distributions of n indistinguishable balls into the remaining (afterexcluding the j specified urns) k  j distinguishable urns (s(k  j) distinguishable cells, each of unlim1ted capacity). This expression implies the exchangeability of the sets At, A2, ... , Ak. Further, the number T(n, k, s, r) of distributions of n indistinguishable balls into k distinguishable urns, which leave k r urns empty (and consequently r urns occupied), equals the number Nk,kr of elements of fl that are contained in k  r among the k sets A 1 , A 2 , . . . , Ak. Thus, according to Corollary 4.4, we deduce the expression
which, upon introducing the explicit expression of the generalized factorial coefficient, implies (9.8). I REMARK 9.2 The enumeration of distributions of indistinguishable balls into indistinguishable urns with or without restrictions cannot be handled by using the inclusion and exclusion principle. Further, it cannot be connected with the enumeration of distributions of indistinguishable balls into distinguishable urns, as the enumeration of distinguishable balls into indistinguishable urns is connected with the enumeration of distributions of distinguishable balls into distinguishable I urns. This problem will be separately examined in the next chapter.
Example 9.1 The Morse code A letter in the Morse code is formed by a succession of dashes and dots with repetitions allowed. The two symbols, dash and dot, may be considered as two distinguishable urns and then different positions (first, second, ... , nth) of a letter as n distinguishable balls. Thus, a formation of a letter of n symbols corresponds to a distribution of n distinguishable balls into k = 2 distinguishable urns. Consequently, there are 2n different letters of n symbols. Further, the number of letters of n symbols that include both the dash and dot symbols, according to Theorem 9.1, equals
t(1)j j=O
(~) (2 j)n = 2n 2. J
In the more general ksymbol code, there are kn different letters of n symbols. Further, the number of letters of n symbols that include rout of s specified symbols,
9.2.
CLASSICAL OCCUPANCY AND MODIFICATIONS
347
according to Corollary 9 .I , is given by N(n,k,s,r) = (8)rS(n,r;k 8), where S (n, r; k  8) is the noncentral Stirling number of the second kind.
0
Example 9.2 Decomposition of a product of primes into factors Consider a product Pn = a 1 a2 ···an. where ai, i = 1, 2, ... , n, are different prime numbers. Find the number of decompositions of the product Pn· of n different prime numbers, into k factors. The n different prime numbers {a 1 , a 2 , •.• , an} may be considered as n distinguishable balls and the k factors, which are not ordered, as k indistinguishable urns. Thus, a decomposition of a product of n different prime numbers into k factors corresponds to a distribution of n distinguishable balls into k indistinguishable urns with no urn empty and vice versa. For instance, the different decompositions of the product p3 = a 1 a 2 a 3 of n = 3 different prime numbers into k = 2 factors, and their corresponding distributions of three distinguishable balls into two indistinguishable urns with no urn empty, are the following:
Hence, the number of decompositions of a product of n different prime numbers into k factors equals the number of distributions of n distinguishable balls into k indistinguishable urns with no urn empty. According to Corollary 9.2, this is given by S (n, k), the Stirling number of the second kind. 0
Example 9.3 Selection of a central committee For the selection of the n members of the central committee of a federation of k associations, each association submits a list of 8 candidates. Find the number of different selections of the n members of the central committee in which r associations are represented in the committee. The n members of the central committee may be considered as n indistinguishable balls and the k associations ask distinguishable urns, each with 8 distinguishable cells of capacity limited to one ball. Thus, a selection of the n members of the central committee from the k lists, each with 8 candidates, corresponds to a distribution of n indistinguishable balls into k distinguishable urns, each with 8 distinguishable cells of capacity limited to one ball. Consequently, the number of selections of the n members of the central committee from the k lists, each with 8 candidates, so that r associations are represented in the committee, according to Theorem 9.2, is given by (k)r R (n,k,8,r ) =  1C(n,r;8),
n.
where C(n, r; s) is the generalized factorial coefficient.
0
348
9.3
DISTRIBUTIONS AND OCCUPANCY
ORDERED DISTRIBUTIONS AND OCCUPANCY
The case with ordered balls in each urn is of particular interest. The urns with ordered balls are simply called ordered urns. The enumeration of the distributions of balls of any specification into distinguishable ordered urns, with or without restrictions, is closely related to the enumeration of the distributions of indistinguishable (like) balls into distinguishable (nonordered) urns. Specifically, we have the next theorem.
THEOREM9.4 The number ofdistributions ofn balls of the type (r 1, r2, ... , rm). where r 1 +r2 + · · · + rm = n, into k distinguishable ordered urns without any restriction is given by (9.9)
and, with the restriction that r urns are occupied, is given by (9.10)
PROOF Note that any distributionofn balls of the type (r 1, r 2, ... , rm) into k distinguishable ordered urns, with or without restrictions, may be carried out in two consecutive stages. Initially, the n balls are distributed into the k distinguishable urns, without taking into account which elements and in what order are placed into each urn. The number Q(n, k) of these distributions equals the number of the distributions of n indistinguishable balls into k distinguishable urns. This number, in the absence of any restriction, is given by
and, with the restriction that r urns are occupied, according to Corollary 9.3, is given by
Further, for each such distribution, by taking into account that the n balls are of the type (r 1 , r 2 , ... , r m) and permuting them, all possible orderings in the urns are accomplished. Since the number of permutations of the n balls of the type
9.3. ORDERED DISTRIBUTIONS AND OCCUPANCY
349
(r 1 , r 2 , ... , rm), according to Theorem 2.4, is given by
n!
applying the multiplication principle, (9.9) and (9 .I 0) are deduced.
I
The following corollary of Theorem 9.4 is concerned with the particular case of n distinguishable balls.
COROLLARY 9.4 The number of distributions of n distinguishable balls, into k distinguishable ordered urns, without any restriction, is given by
n.1
(k + nn 1) (k+n1)n, _
(9.11)
and, with the restriction that r urns are occupied, is given by (9.12)
Example 9.4 Priority lists Suppose that n persons have applied for certain vacant positions ink organizations of the public sector. Each candidate is placed on only one of k different priority lists and each organization is filling its vacant positions from the corresponding priority list. Find the number of ways of placing the n candidates on the k priority lists. Then candidates may be considered as n distinguishable balls and the kdifferent priority lists ask distinguishable ordered urns. Then, the number of ways of placing the n candidates on the k priority lists without any restriction, according to (9 .11 ), is given by
n.1
(k + nn 1) _(k + n
1)n,
and, with the restriction that each of r organizations should fill at least one position, according to (9.12), is given by
DISTRIBUTIONS AND OCCUPANCY
350
9.4
BALLS OF GENERAL SPECIFICATION AND DISTINGUISHABLE URNS
In the introduction to this chapter, the balls of a general specification were classified according to the kind to which they belong and the notation (r 1, r 2 , ... , rm), with r1 +r2+ · · ·+rm = n, where ri is the number of balls of the ith kind, i = 1, 2, ... , m, was used. These balls may also be classified according to their multiplicity and the notation [m 1 , m 2 , ... , mnJ, with m 1 + 2m 2 + · · · + nmn = n, where mi ~ 0 is the number of distinguishable kinds of balls, each including i like balls, i = 1, 2, ... , n, may be used. In this case, the partition notation (1m, 2m 2 • • • n m,.) may also be used.
THEOREM9.5 The number of distributions ofn balls of the specification [m 1, m2, ... , mn], with mi ~ 0, i = 1, 2, ... , nand m 1 +2m 2 + · · · + nmn = n, into k distinguishable urns is given by
(k)1 (k +2 1) m,
m
2
...
(k + n _ 1) n
m,.
. (9.13)
PROOF
Note first that the number of distributions of i like balls into k distinguishable urns equals (k+! 1), i = 1, 2, ... , n. Further, by the multiplication principle, the number of distributions of mi ~ 0 distinguishable kinds of balls, each including i like balls, into k distinguishable urns is given by (k+! 1 ) m,, i = 1, 2, ... , n. Applying, once more, the multiplication principle, we conclude that the number of distributions of n balls of the specification [m 1 , m 2 , ... , mn], with mi ~ 0, i = 1, 2, ... , nand m 1 +2m 2 +···+ nmn = n, into k distinguishable urns is given by the expression (9.13). I
THEOREM9.6 The number of distributions ofn balls of the specification [m 1, m2, ... , mn], with mi ~ 0, i = 1, 2, ... , nand m 1 +2m 2 +···+ nmn = n, into k distinguishable urns so that r urns are occupied is given by
PROOF
Consider the set Jl of distributions of n balls of the specification [m 1, m 2, ... , mn] into k distinguishable urns and let Ai be the subset of these distributions in which the ith cell remains empty, i = 1, 2, ... , k. Then, for any
9.4.
BALLS OF GENERAL SPECIFICATION
351
selection of s indices {i 1, i2, ... , is} out of the k indices {1, 2, ... , k }, according to Theorem 9.5,
This expression implies the exchangeability of the sets A 1 , A 2 , ... , Ak. Further, the number R(m 1 , m 2 , .•. , mn; k, r) of distributions of n balls of the specification [m 1, m2, ... , mn] into k distinguishable urns, which leave k  r urns empty (and consequently r urns occupied), equals the number Nk,kr of elements of Jl that are contained in k  r among the k sets A 1 , A2 , ... , Ak. Thus, according to Corollary 4.4, (9.14) is established. I Some particular cases of (9.14), of special interest, have been already examined in Section 9.2. The case of n distinguishable balls corresponds to m 1 = n, mi = 0, i = 2, 3, ... , n, while the case of n indistinguishable (like) balls corresponds to mi = 0, i = 1, 2, ... , n1, mn = 1. Another particular case that corresponds to m 8 = m, mi = 0, i = 1, 2, ... , s 1, s + 1, ... , n is explicitly presented in the next corollary.
COROLLARY 9.5 The number of distributions of sm balls, which belong in m distinguishable kinds with s like balls of each kind, into k distinguishable urns so that r urns are occupied is given by
(9.15)
Let us now consider the case of the existence of restrictions in the capacity of the urns. Specifically, the placement of a ball into an urn excludes the possibility of placing any other like ball into it. Thus, the balls that are allowed to be placed in any urn are all distinguishable.
THEOREM9.7 The number of distributions ofn balls of the specification [m 1, m2, ... , mn]. with mi 2: 0, i = 1, 2, ... , nand m 1 +2m 2 + · · · + nmn = n, into k distinguishable urns, each of which may accommodate at most one ball from each kind, is given by
V(m1,m2, ... ,mn;k) =
(k)1 (k)2 m,
m2
(k)
.. · n
mn
(9.16)
PROOF Note first that, since each urn may accommodate at most one ball from each kind, the number of distributions of i like balls into k distinguishable urns
352
DISTRIBUTIONS AND OCCUPANCY
equals(~), i = 1, 2, ... , n. Further, by the multiplication principle, the number of distributions of m; ;::: 0 distinguishable kinds of balls, each including i like balls, into k distinguishable urns is given by (~) mi, i = 1, 2, ... , n. Applying, once more, the multiplication principle, we conclude that the number of distributions of n balls of the specification [m 1, m 2, ... , mn], with m; ;::: 0, i = 1, 2, ... , n and m 1 + 2m 2 + · · · + nmn = n, into k distinguishable urns, each of which may I accommodate at most one ball from each kind, is given by (9.16).
Using the inclusion and exclusion principle and Theorem 9. 7, the next theorem (analogous to Theorem 9.6) is deduced.
THEOREM9.8 The number of distributions ofn balls of the specification [m1, m2, ... , mn], with m; ;::: 0, i = 1, 2, ... , nand m 1 +2m 2 +···+ nmn = n, into k distinguishable urns, each of which may accommodate at most one ball from each kind, so that r urns are occupied, is given by
COROLLARY 9.6 The number of distributions of sm balls, which belong in m distinguishable kinds with s like balls of each kind, into k distinguishable urns, each of which may accommodate at most one ball from each kind, so that r urns are occupied, is given by Q(m,8,k,r)
=G) ~(1riG) (!)m
(9.18)
Example 9.5 Formation of committees Assume that, for the 8m = n positions of m committees, each with 8 positions, there are k candidates. Find the number of ways then members of the committees can be selected so that r of the k candidates will participate in at least one committee. The 8m = n positions of the m committees, each with 8 positions, may be considered as m distinguishable kinds of balls, each with 8 balls, and the k candidates as k distinguishable urns. Thus, the placement of a ball of the ith kind into the jth urn corresponds to the selection of the jth candidate for a position of the ith committee. We assume that no candidate can be assigned to more than one position of the same committee, but can participate in more than one committee. This assumption implies that each urn may accommodate at most one ball from each kind. Consequently, according to Corollary 9.6, the number of ways then members of
9.5. GENERATING FUNCTIONS
353
the committees can be selected so that r of the k candidates will participate in at least one committee is given by
Q(m,8,k,r)
=G) ~(rrj(:) (!)m
In the particular case when each of the k candidates must participate in at least one committee, the number of ways the n members of the committees can be selected is given by
Example 9.6 Formation of committees from grouped candidates Assume that each of k different political groups submits a full priority list of n candidates to the president's office for the 8m = n positions of m committees, each with 8 positions. Find the number of ways then members of the committees can be selected so that r of the k political groups will be represented in at least one committee. As in the preceding example, the 8m = n positions of the m committees, each with 8 positions, may be considered as m distinguishable kinds of balls, each with 8 balls, and the k political groups as k distinguishable urns. In this case, there is no restriction on the number of balls of the same kind that each urn may accommodate. Consequently, according to Corollary 9.5, the number of ways then members of the committees can be selected so that r of the k political groups will be represented in at least one committee is given by P(m,8,k,r)
=G) ~(rrjG)
e+: l)
m
In the particular case when each of the k political groups must be represented in at least one committee, the number of ways the n members of the committees can be selected is given by
9.5
GENERATING FUNCTIONS
The use of generating functions in the study of distributions and occupancy leads to a more extensive development of this subject. In this section,
DISTRIBUTIONS AND OCCUPANCY
354
we present the generating functions of the cases of distinguishable or indistinguishable balls and distinguishable urns, as well as the case of balls of a general specification and distinguishable urns. Consider first the case of distinguishable balls and urns. Let us attach the occupancy indicator Xj to the jth urn, j = 1, 2, ... , k, so that the homogeneous product of order n, x~' x~ 2 • • • x~k, with r 1 + r 2 + · · · + rk = n, uniquely determines the distribution of n distinguishable balls into k distinguishable urns, with Tj balls in the jth urn, j = 1, 2, ... , k. Then the indicator (generating function) of the distributions of a single ball into k distinguishable urns is
and hence the generating function of the distributions of n distinguishable balls into k distinguishable urns, with respect to the numbers of balls in the urns, is (9.19) Note that (9.19) expanded into powers of x 1 , x2, ... , Xk by using the multinomial theorem,
2:
n! I I · · "Tk·l TI.T2·
x2 rk x r, 1 x 2 ···xk
= ( x 1 +x 2 + ···+xk )n ,
gives the number of distributions of n distinguishable balls into k distinguishable urns with ri balls in the jth urn, j = 1, 2, ... , k, r 1 + r 2 + · · · + Tk = n. The exponential generating function of the sequence 9n(x 1 , x 2, ... , xk), n = 0, 1, ... , of generating functions (9.19) is oo
G(t;xl,X2,··. ,xk)
= L9n(XI,X2,···
tn ,xk)!
n. = exp{(x1 + x2 + · · · + xk)t} n=O
Therefore, the enumerator for occupancy of k distinguishable urns is given by (9.20) where the enumerator for occupancy of the jth urn is given by x2t2
E(t;xi)
= 1 +xit +  12. 1 + ··· = e"''t,
j = 1,2, ... ,k,
(9.21)
exactly as in the case of permutations with repetition (see Section 6.3). The occupancy for each urn is specified independently of the other urns. Thus,
9.5. GENERATING FUNCTIONS
355
if the jth urn must contain at least r and at most s balls, the enumerator for its occupancy is
xrtr Er,s(t; Sj) =
xr+ltr+l
~! + Cr + 1)! + · · · +
xsts
7
(9.22)
and (9.20) is accordingly modified. In the case of n indistinguishable balls, there exists only one distribution of them into k distinguishable urns so that the jth urn contains r i balls, for j = 1, 2, ... , k and r 1 + r2 + · · · + rk = n. Hence, the generating function of the distributions of n indistinguishable balls into k distinguishable urns, with respect to the numbers of balls in the urns, is given by (9.23)
where the summation is extended over all rj 2: 0, j = 1, 2, ... , k with r 1 +r 2 +· · ·+rk = n. The function hn(x1,x2, ... ,xk) is called homogeneous product sum of weight k. Note that the corresponding generating function of the distributions of n distinguishable balls is given by
gn(x1,x2, ... ,xk) = [hl(x1,x2, ... ,xk)]n, where hi(XI,X2, ... ,xk) =XI+ X2 + · · · + Xk· The ordinary generating function of the sequence hn n = 0, 1, ... , of generating functions (9.23), since 00
00
LLx~'x~ 2 n=O
..
00
00
·x~ktn = L:Cx1tf' L:Cx2tr 2 rt=O
=hn (x1, x2, ... , Xk), ...
r2=0
L(xktfk, rk=O
is deduced as 00
H(t; XI, X2, ... , Xk) = L hn(XI, X2, ... , Xk)tn n=O 1
(9.24)
Therefore, the enumerator for occupancy of k distinguishable urns is given by
H(t; x1, x2, ... , xk) = F(t; xi)F(t; x2) · · · F(t; Xk),
(9.25)
where the enumerator for occupancy of the jth urn is given by
F(t;xi) = 1 +xit +x]t 2 + ·· · = (1 x 3t) 1 , j = 1,2, ... ,k, (9.26) exactly as in the case of combinations with repetition (see Section 6.3). The occupancy for each urn is specified independently of the other urns. Thus,
DISTRIBUTIONS AND OCCUPANCY
356
if the jth urn must contain at least r and at most s balls, the enumerator for its occupancy is · (9.27)
and (9.25) is accordingly modified. Let us finally consider the general case of n balls of the specification [m 1, m2, ... , mn], with m 1 +2m2 + · · · + nmn = n, where mi 2: 0 is the number of balls of multiplicity i (that is, the number of distinguishable kinds of balls, each including i like balls), i = 1, 2, ... , n. The generating function of the distributions of i indistinguishable (like) balls into k distinguishable urns, with respect to the numbers of balls in the urns, according to (9.23), is given by hi(x 1, x2, ... , Xk), the homogeneous product sum of weight i. Since the mi 2: 0 distinguishable kinds of balls, each of which includes i like balls, are independently distributed, the generating function of the distributions of these balls is given by [hi(x 1, x2, ... , xk)]m•, i = 1, 2, ... , n. Therefore, the generating function of the distributions of n balls of the specification [m 1, m2, ... , mnJ, mi 2: 0, i = 1, 2, ... , n, m 1 + 2m2 + · · · + nmn = n, into k distinguishable urns, with respect to the numbers of balls in the urns, is given by n
hm 1 ,m2 , ... ,mn(XI,X2,··· ,Xk)
= II[hi(XI,X2,···
,Xk)]m'.
(9.28)
i=l
Note that (9.28), for xi = 1, j = 1, 2, ... , k, yields the total number of distributions of then balls of the specification [m 1, m2, ... , mn] into k distinguishable urns. The fact that this number agrees with that of Theorem 9.5 follows from the expression hi(1, 1, ... , 1) = (k+!I), i = 1, 2, ... , n. The enumerator for occupancy of the jth cell by n balls of the specification [m1, m2, ... , mn], with mi 2: 0, i = 1, 2, ... , n, m1 +2m2+·· ·+nmn = n and m 1 + m 2 + · · · + mn = m, where m is the total number of distinguishable kinds of balls, is given by
F(t1,t2,···,tm;Xj)=
1
)
(1 Xjt1)(1 Xjt2 · · · (1 Xjtm
)'
where the term (xitd"' (xit 2)"2 • • • (xjtm)"m indicates the occupancy of the jth urn by v = v 1 + 112 + · · · + Vm balls, among which 111 are of the lth kind, l = 1, 2, ... , m. The occupancy of each urn is specified independently of the other urns and so the enumerator for occupancy of the k urns is given by
H(t1,t2,··· ,tm;XI,X2,··. ,xk) = F(t1,t2,··· ,tm;xi)F(ti,t2,··· ,tm;X2)···F(t1,t2,··· ,tm;Xk)· Two illustrative examples follow.
9.5. GENERATING FUNCTIONS
357
Example9.7 The generating function G (t), of the number of distributions of n distinguishable balls into k distinguishable urns, without any restriction, is deduced from (9.20) and (9.21) by setting Xj = 1, j = 1, 2, ... , k. Then, •
E(t; 1)
=G(t; 1, 1, ... , 1) = ekt = L kn 1tnn. , co
= et, G(t)
n=O
and hence the number of distributions of n distinguishable balls into k distinguishable urns equals In the case ofthe restriction that each urn should contain at least one ball, it follows from (9.20) and (9.22) that t2
E1(t; 1) = t +I+···= et 1, G 1(t) =: GI(t; 1, 1, ... , 1) = (et 1)k. 2.
Expanding the generating function GI(t) into powers oft by using Newton's binomial theorem, we find
Therefore, the number of distributions of n distinguishable balls into k distinguishable urns so that all k urns are occupied is given by N(n, k)
=
t( j=O
1)j
(~) (k j)n = k!S(n, k), J
in agreement with (9.1) in the particular case of r = k.
0
Example9.8 The generating function, H ( t), of the number of distributions of n indistinguishable balls into k distinguishable urns, without any restriction, is deduced from (9.25) and (9.26) by setting Xj = 1, j = 1, 2, ... , k. Then
F(t; 1) = (1 t) 1 , H(t) =:H(t;1,1, ... ,1) = (1t)k =
co 2)l)n
n=O
(k) tn, n
358
DISTRIBUTIONS AND OCCUPANCY
and hence the number of distributions of n indistinguishable balls into k distinguishable urns equals
in agreement with the results stated in Section 9.2. In the case of the restriction that each urn should contain at least one ball, it follows from (9.25) and (9.27) that
Expanding the generating function H 1 (t) into powers oft by using Newton's negative binomial theorem, we find
Therefore, the number of distributions of n indistinguishable balls into k distinguishable urns, so that all k urns are occupied, is given by
1)
n ( k 1
.
Further applications may be found in the exercises.
0
9.6 BIBLIOGRAPHIC NOTES P. A. MacMahon (1915, 1916) in his treatise on combinatorial analysis introduced and extensively studied, mostly through generating functions, the theory of distribution of balls (objects) into urns (cells, boxes). Theorem 9.5, which is concerned with the distributions of balls of a general specification into distinguishable urns, was derived by the aid of generating functions. Also, Section 9.5 on the study of distributions through enumerating generating function is based on the books of P. A. MacMahon (1915, 1916) and J. Riordan (1958). W. Feller (1968) treated the classical occupancy problem and the coupon collector's problem by using the inclusion and exclusion principle. N. L. Johnson and S. Kotz (1977) devoted a chapter to the occupancy theory containing a wealth of information and an extensive Jist of references.
9. 7. EXERCISES
9.7
359
EXERCISES
1. Let N 1 (n, k, r) be the number of distributions of n distinguishable balls into k distinguishable urns, so that r urns are occupied, each by j balls. Show that .
_
N1 (n,k,r)
(k)r ~( k
_
1)
ir
r) (kir
'
n. _ · nij (j!)i(nij)!(k z) .
2. (Continuation). (a) Show that
and hence that
=Nj(n, /l(i)(n, k,j), i 0, 1, ... , k, are the factorial moments of the = k, r), r 0, 1, ... , k, show that n' .. /l(i)(n,k,j) =(k)i(j!)i(n.ij)!(ki)n•J, i =0,1, ... ,k.
(b) If /l(i) sequence ar
= =
3. Let B(n, k) be the number of distributions of n distinguishable balls into k indistinguishable urns. Show that
B(n, k)
k = k!1 ~
where
(k) rnkj, j
k
Dk
.
= k! L (~)· z.
i=O
is the ksubfactorial. 4. Let Q1 (n, k, r) be the number of distributions of n indistinguishable balls into k distinguishable urns, so that r urns are occupied, each by j balls. Show that
Qj(n, k,
r) =(k) ~) _1 r
i=r
r) (k + n i(j ~ 1) 1).
)ir (~z r
n  ZJ
DISTRIBUTIONS AND OCCUPANCY
360
5. (Continuation). (a) Show that
and hence that 00
L
k
L Qj(n, k, r)xr tn = [(1 t) 1
+(x 1)tj]k.
n=Or=O
= k,j), 0, 1, ... , k, are the binomial moments of the =bci)(n, Qj(n, k, r), r 0, 1, ... , k, show that . (k)(k+ni(j+1)1) . ... ,k. b(i)(n,k,J)= . .. , z=0,1, z n ZJ
(b) If b(i) sequence ar
i =
=
6. Show that the number of distributions of n indistinguishable balls into k distinguishable urns, each of capacity limited to s balls, is
and, with the restriction that r urns remain empty, equals k
L(n,k,s,r)
=G) L(1)ir(:~;)L(n,ki,s). t=r
7. (Continuation) Let Lj(n, k, r) be the number of distributions of n indistinguishable balls into k distinguishable urns, each of capacity limited to s balls so that each of r urns is occupied by j balls. Show that
and, for j = s, conclude that
L 8 (n, k, s, r) =
G)
8. (Continuation). (a) Show that
L(n sr, k r, s).
9. 7. EXERCISES
361
and hence that sk
k
L L Qj(n,k,r)xrtn = [(1 t + )(1 t)8
1
1
+ (x 1)ti]k.
n=Or=O
=
(b) If b(i) b(i)(n, k, s,j), i = 0, 1, ... , k, are the binomial moments of the sequence ar Lj(n, k, s, r), r = 0, 1, ... , k, show that
=
b(i)(n,k,s,j) =
(~)L(nij,ki,s),
i=0,1, ... ,k.
9. (Continuation). Let Mj(n, k, s) be the number of distributions of n indistinguishable balls into k distinguishable urns, each of capacity limited to s balls, so that each urn is occupied by at least j balls. Show that
Mj(n, k, s) = L(n kj, k, s j) =
t,( _G) (n1)
kj
+ k ~ ~s  j + 1) 1).
1
10. Let n like balls be distributed into k distinguishable urns. (a) If hn(x) is the enumerator for the occupancy of a single urn, show that 00
H(t; x) =
L
hn(x)tn = (1 xt) 1 (1 x)k+l
n=O
and
(b) If B(n, k, r) is the number of distributions of n like balls into k distinguishable urns so that a specified urn contains r balls, r = 0, 1, ... , n and b(s) b(s) (n, k), s = 0, 1, ... , n, are the binomial moments of the sequence ar B(n, k, r), r = 0, 1, ... , n, conclude that
= =
B(n, k, r) =
k  r  2) (n+ nr
, b(s)(n, k) =
(n +ns k  1)
.
11. (Continuation). If hn(x, y) is the enumerator for the occupancy of a pair of urns, show that 00
H(t; x, y)
=L
n=O
hn(x, y)tn
= (1 xt) 1 (1 yt) 1 (1 x)k+ 2
362
DISTRIBUTIONS AND OCCUPANCY
and hence conclude that
hn(x,y) =
~~ L....tL....t
•=0 r=O n nj
= 2: 2: j=O i=O
(n
+ k r s 3)xrys nrs
(: ~ ~ =J~) (x 1)i(y 
1)j.
12. Let n like balls be distributed into k distinguishable urns, each with
s distinguishable cells of capacity limited to one ball. Show that (a) the enumerator for occupancy of the jth urn is
and (b) the generating function of the distributions with r1 balls in the jth urn, j = 1, 2, ... , k, is
where the summation is extended over all 0 r1 + r2 + · · · + rk = n.
:S r1 :S s, j
= 1, 2, ... , k, with
13. (Continuation). In the case of cells of unlimited capacity, show that (a) the enumerator for occupancy of the jth urn is
G.(t;
Sj)
=
~ (s + ~ 
= ~(1r (~
1 )xjtr 8
)x;tr = (1 x1t)•,
j
= 1,2, ...
,k
and (b) the generating function of the distributions with r1 balls in the jth urn, j = 1, 2, ... , k, is
where the summation is extended over all r1 + r2 + · · · + rk = n.
rj
2: 0, j
= 1, 2, ... , k,
with
14. Let us consider k distinguishable urns, each with s distinguishable cells of capacity limited to one ball. Show that the number of distributions
9. 7. EXERCISES
363
of n like balls into the k urns so that each of r urns contains j balls equals
and conclude that the number of distributions of n like balls into the k urns so that r urns are completely occupied is given by
R(n, k,
s, r)
=
(k) ~( _1)ir (~ r) (s(k i)). r
L....t i=r
t 
r
sk  n
15. (Continuation). (a) Show that
~
s(kr)+jr
Rj(n,k,s,r)tn=
(kr) [(;)ti] (1+t)B; r [
(
)
ti
] kr
and hence that
16. Let us consider k distinguishable urns, each with s distinguishable cells of unlimited capacity. Let Tj(n, k, s, r) be the number of distributions of n like balls into the k urns so that each of r urns contains j balls. Show that
TJ(n, k, s, r)
=
G) t( _ 1 =;) (s +; 1) (s(k i)n+_nj~ 1). )ir (:
17. (Continuation). (a) Show that 00
LTJ(n, k, s, r)tn n=O
and hence that
i
ij
DISTRIBUTIONS AND OCCUPANCY
364
18. Let us consider kr distinguishable cells, each of capacity limited to one ball, arranged in k columns and r rows. Show that the number of distributions of n like balls into the kr cells so that j from 8 predetermined columns of cells are occupied (each by at least one ball) equals
19. (Continuation). Show that the number of distributions of n like balls into the kr cells so that no row and no column of cells remain empty equals
A(n, k, r) =
t t(
1)k+rji
(~) (:) (~) J
t=OJ=O
and conclude that the number of distributions of n distinguishable balls into the kr cells so that no row and no column of cells remain empty equals n!A(n,k,r)
= tt(1)k+rji(~) J
t=O J=O
(:)(ij)n·
20. Let us consider kr distinguishable cells of unlimited capacity, arranged in k columns and r rows. Show that the number of distributions of n like balls into the kr cells so that j from 8 predetermined columns of cells are occupied (each by at least one ball) equals
T(n, k,
8,
r,j) =
G) ~(1)i G) (r(k + j: 8
i)
+ n
1).
21. (Continuation). Show that the number of distributions of n like balls into the kr cells so that no row and no column of cells remain empty equals B(n, k, r) =
t t( •=0]=0
1)k+rji
(~) (:) (ij J
+: i).
22. Let M(n, k, 8) be the number of distributions of n distinguishable balls into k distinguishable urns, each of capacity limited to 8 balls. Show that sk tn ( t2 t• ) k Mk(t) = M(n, k, 8) 1 = 1 + t + 2l + ... + 1 n=O n. . 8.
L
9. 7. EXERCISES
365
and M(n+1,k,s)=k{M(n,k,s) (:)M(ns,k1,s)}, n?.s, M(n,k,s) = kn, n
~
s.
Further, derive the vertical recurrence relation M(n, k, s) =
f: (~)M(n
j, k 1, s), m = min{n, s }.
J
j=O
23. Let Sr(n, k) be the number of distributions of n distinguishable balls into k distinguishable urns so that each urn contains at least r balls. Show that
tn
00
sk ,r (t) = '""' sr (n ' k)1 = ~ n.
n=rk
(
r1 ) k
et  1  t  ...  ,::(r _ 1)1.
and Sr(n
+ 1,k) =
k { Sr(n,k)
Sr(n, k)
= 0,
+
(r:
n
1
)sr(n r
< rk,
+ 1,k 1)},
Sr(rk, k)
=
n?. rk,
(rk)! (r!)k ·
Further, derive the recurrence relation
24. Let us consider k distinguishable urns, each with s distinguishable cells of capacity limited to one ball. Let G(n, k, s, r) be the number of distributions of n like balls into the k urns so that each urn contains at most r balls. Show that
and (n
+ 1)G(n + 1, k, s, r)
= (sk n)G(n, k, s, r)
1)
sk ( s r
G(nr,k1,s,r), n?. r,
DISTRmUTIONS AND OCCUPANCY
366
G(n,k,s,r) =
(s:),
n:; r.
Further, derive the vertical recurrence relation G(n, k, s, r) =
f: (~)
G(n j, k 1, s, r), m = min{n, s, r }.
J
j=O
25. (Continuation). Let Cr(n,k,s) be the number of distributions ofn like balls into the k urns so that each urn contains at least r balls. Show that 8
Gk,r(t)= i::kCr(n,k,s)tn= [(l+t) 1 G)t .. ·
c~1)trlr
and (n
+ 1)Cr(n + 1, k, s) = (sk n)Cr(n, k, s) 1 r1
+sk (s ) Cr(n r + 1, k 1, s), rk
Cr(n,k,s)
= 0,
Cr(sk, k, s)
n
= 1,
< rk, Cr(rk,k,s) Cr(n, k, s)
= 1,
< n < sk,
= (;) k' n
> sk.
Further, derive the recurrence relation
26. Let us consider k distinguishable urns, each with s distinguishable cells of unlimited capacity. Let H(n, k, s, r) be the number of distributions of n like balls into the k urns so that each urn contains at most r balls. Show that rk
Hk(t)
= L H(n, k, s, r)tn n=O
= [1 +
C) s
t+ (
s+1
2 ) t2 + ... +
k
s+r1 (
r
)
tr ]
and (n + 1)H(n + 1, k, s, r)
= (sk + n)H(n, k, s, r) sk ( s +r) r H(nr,k1,s,r), n 2: r,
9. 7. EXERCISES
367
1 (sk+: ), n:::;r.
H(n,k,s,r)=
Further, derive the vertical recurrence relation
~ H(n,k,s,r)=~
(s 1) +j.
i=O
.
H(nJ,k1,s,r), m=min{n,r}.
J
27. (Continuation). Let Dr(n, k, s) be the number of distributions of n like balls into the k urns so that each urn contains at least r balls. Show that 00
Dk,r(t) =
L
Dr(n, k, s)tn
n=k
=[(1t)
s
)t .. ·(
s  1  (1
s
+ r 2 r 1
)t
k r1
]
and (n
+ 1)Dr(n + 1, k, s) = (sk + n)Dr(n, k, s) +sk ( s
Dr(n, k, s)
= 0,
n
+ rr1
1)
Dr(n r
+ 1,k 1,s),
< rk, Dr(rk, k, s) = (
s
+ rr
1)
n
> rk,
k
Further, derive the recurrence relation
28. Let us consider n + r balls among which n are distinguishable and the other r are indistinguishable. Let U(n, r, k) be the number of distributions of these balls into k distinguishable urns without any restriction and W(n, r, k) the corresponding number with the restriction that no urn remains empty. Show that
and
368
DISTRIBUTIONS AND OCCUPANCY
29. (Continuation). Show combinatorially that
W(n,r,k)=~ m
(
r+J·
k
1
1)
where
. . (k)jS(n,J), m=mm{n,k},
j
S(n,j)
= ~ L( 1)jl (~)in 0
~
J. i=O
is the Stirling number of the second kind. Also, show that W(n, r, k)
= k{W(n 1, r, k) + W(n 1, r, k 1)},
W(n,O,k) = kn, W(O,r,k) = (r
30. (Continuation). Show that
and
Further, conclude that
1
+ :· ),
W(n,r,k) = 0, k
>n +r
Chapter 10 PARTITIONS OF INTEGERS
10.1
INTRODUCTION
The theory of partitions of integers, established by Euler in the 18th century and enhanced by Hardy, Rarnanujan and Rademacher, belongs to both combinatorics and number theory. As regards the combinatorial aspect of partitions, note first .that the positive integer solutions of the linear equation x 1 + x 2 + · · · + ck = n with n a positive integer, enumerated in Chapter 2, are ordered solutions. Further, the enumeration of the corresponding nonordered solutions, the partitions of the integer n, is of great interest. Note also that, the enumeration of distributions of indistinguishable balls into indistinguishable urns is merely a partition enumeration problem. In this chapter, after the introduction of the notion of a partition of a positive integer n, recurrence relations and generating functions for the total number of partitions of n and the numbers of partitions of n into k and into at most k parts are derived. Then, a universal generating function for the number of partitions of n into parts of specified or unspecified number, whose number of parts of any specific size belongs to a subset of nonnegative integers, is obtained. As applications of this generating function several interesting sequences of partition numbers are presented. Furthermore, relations connecting various partition numbers are deduced by using their generating functions. Also, after introducing the Ferrers diagram of a partition and the notion of a conjugate (and a selfconjugate) partition, additional interrelations among certain partition numbers are derived. Euler's pentagonal theorem on the difference of the number of partitions of n into an even number of unequal parts and the number of partitions of n into an odd number of unequal parts is obtained. The last section is devoted to the derivation of combinatorial identities; the Euler and GaussJacobi identities are deduced. As a complement to this subject, a collection of exercises on qnumbers, qfactorials and qbinomial coefficients, as well as on qStirling numbers of the first and second kind, is provided.
PARTITIONS OF INTEGERS
370
10.2
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
DEFINITION 10.1 A partition ofa positive integer n is a nonordered collection ofpositive integers whose sum equals n. These integers (terms ofthe sum) are called summands or parts of the partition.
In the case of a partition of n into a specified number of parts, say k, then the term partitions of n into k parts is used. Since the order of parts in a partition does not count, they are registered in decreasing order of magnitude. Thus, a partition of n into k parts is a solution in positive integers of the linear equation: (10.1)
In a partition of n, let Yi fori= 1, 2, ... , n. Then Y1
2: 0 be the number of parts that are equal to i,
+ 2y2 + · · · + nyn = n,
Yi
2: 0,
i
= 1, 2, ... , n
(10.2)
In the case of a partition of n into k parts, in addition, Y1
+ Y2 + · · · + Yn
(10.3)
= k.
If {Yi 1 , Yi 2 , ••• , Yirl, Yir }, with i1 < i2 < · · · < ir1 < ir, is the set of the positive Yi in (10.2), the corresponding partition of n is denoted by
omitting the exponents that are equal to 1. REMARK 10.1 It is important to point out the clear distinction between a partition of a positive integer and a partition of a nonempty set. Thus, a partition of the positive integer n into k parts is a solution of the equation X1
+ X2 + · · · + Xk = n,
X1
2:
X2
2: · · · 2:
Xk
2:
1
in positive integers, say {r 1, r 2, ... , r k} ,r1 2: r2 2: · · · 2: rk 2: 1, which corresponds to a distribution of n indistinguishable balls (the units) into k indistinguishable urns containing r 1, r 2, ... , rk balls. In contrast, a partition of a nonempty set B, with N(B) = n elements, into k subsets is a solution of the settheoretic equation
10.2.
RECURRENCES AND GENERATING FUNCTIONS
371
in nonempty disjoint sets, say {A 1 ,A 2 , ... ,Ak}, N(AI) 2: N(A 2 ) 2: ··· 2: N(Ak) 2: 1, Ai n Ai = 0, i =j:. j, with N(Ai) = ri, i = 1, 2, ... , k, which corresponds to a distribution of n distinguishable balls (the elements of B) into k indistinguishable urns containing r 1 , r 2 , ••• , rk balls. I Example 10.1 Let us determine the partitions of the integers I, 2, 3, 4, 5 and 6. Clearly, number 1 has only one partition with one part equal to I, number 2 has two partitions: 2, 12 and number 3 has three partitions: 3, 21, 13 . Further, since 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1, the partitions of number 4 are: 4, 31, 22 , 21 2 , 14 . Also, since 5 = 4 + 1 = 3 + 2 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1, the partitions of number 5 are: 5, 41, 32, 31 2 , 22 1, 21 3 , 15 . Finally, since 6 = 5 + 1 = 4 + 2 = 3 + 3 = 4 + 1 + 1 = 3 + 2 + 1 = 2 + 2 + 2 = 3+1+1+1=2+2+1+1=2+1+1+1+1=l+l+l+l+l+l,fue partitions of number 6 are: 6, 51, 42, 32 , 41 2 , 321, 23 , 31 3 , 22 12 , 21 4 , 16 • D
Let p(n) be the number of partitions of nand p(n, k) be the number of partitions of n into k parts. Then n
p(n)
= '2:p(n,k),
n
= 1,2, ....
(10.4)
k=l
If P(n, k) is the number of partitions of n into at most k parts, then k
P(n, k)
= '2:p(n, r),
(10.5)
r=l
p(n, k) = P(n, k) P(n, k 1)
(10.6)
P(n, k) = p(n), k 2: n.
(10.7)
and
A recurrence relation for the calculation of the number of partitions of n into k parts is given in the following theorem. THEOREM 10.1 The number p(n, k) of partitions ofn into k parts satisfies the recurrence relation m
p(n,k)='2:p(nk,r), k=2,3, ... ,n1, n=2,3, ... , r=l
with m = min {k, n  k} and initial conditions p(n, 1) = p(n, n) = 1, n = 1, 2, ... , p(n, k) = 0, k > n.
(10.8)
372
PARTITIONS OF INTEGERS
PROOF Setting in the linear equation (10.1) z; = x; 1, i = 1, 2, ... , k, we deduce the linear equation Zt
+ Zz + · · · + Zk = n  k,
Zt ~
~
Zz
· · · 2: Zk
~
0.
(10.9)
Since the transformation is one to one, the number p(n, k) of solutions of (10.1) equals the number of solutions of (10.9). If P is the set of solutions of (10.9) and Pr ~ P is the subset of solutions of (10.9) for which Zr ~ 1 and z; = 0, i = r + 1,r +2, ... ,k, r = 1, 2, ... ,m, m = min{k, n k}, then P; npi = 0, i =f. j and P = P 1 + Pz + · · · + P m. Therefore, according to the addition principle, m
p(n,k)
= N(P) = LN(Pr) r=l
and, since N(Pr) equals the number p(n k, r) of solutions of the linear equation Zt
+ Zz + · · · + Zr = n k,
Zt ~
Zz
~
· · · ~ Zr 2: 1,
I
(10.8) is established. Its initial conditions are obvious.
Using recurrence relation (10.8) and its initial conditions, Table 10.1 of the numbers p(n, k) is constructed. Summing the elements of each row we get, according to (10.4), the numbers p(n).
Table 10.1 The Numbers p(n, k) and p(n)
k
l
2
I l I l l l l l l l
l l 2 2 3 3 4 4
3
4
5
6
7
8
9
10
p(n)
n l 2 3 4 5 6 7 8 9 10
5
l l 2 3 4
5 7 8
l l 2 3
5
l l 2 3
6 9
5 7
l l 2 3
5
l l 2 3
l l 2
l l
l
l 2 3 5 7 II 15 22 30 42
REMARK 10.2 From recurrence relation (10.8) the following relations are readily deduced: fork~ n/2, whence n k:::; n/2, and by the expression (10.4), nk
p(n, k)
=L
r=l
p(n k, r)
= p(n k),
(10.10)
10.2.
RECURRENCES AND GENERATING FUNCTIONS
while, fork< n/2, whence n k
> n/2, k1
k
p(n, k)
373
= LP(n k, r) = LP(n k, r) + p(n k, k) r=l
r=l
= p(n 1, k 1)
+ p(n k, k).
(10.11)
Note that this recurrence relation may also be derived independently of the recurI rence relation (1 0.8) (see Exercise I). From Theorem 10.1, using (10.5), (10.6) and (10.7), the following corollary is deduced.
COROLLARY 10.1 The number P(n, k) ofpartitions ofn into at most k parts satisfies the recurrence relation P(n, k) = P(n, k 1) fork
= 2, 3, ...
, n  1, n
+ P(n k, k),
(10.12)
= 2, 3, ... , with initial conditions P(n, 1) = 1, P(O, k) = 1.
Using recurrence relation (10.12) and its initial conditions, Table 10.2 of the numbers P(n, k) is constructed.
Table 10.2 The Numbers P(n, k)
k
1
2
3
4
5
6
7
8
9
10
1 1 1 1 1 1 1 1 1 1
1 2 2 3 3 4 4
1 2 3 4
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
5
5 7
6 9 11 15 18 23
5 7 10 13 18 23 30
5 7 11 14 20 26 35
5 7 11 15 21 28 38
5 7 11 15 22 29 40
5 7 11 15 22 30 41
5 7 11 15 22 30 42
n 1 2 3 4
5 6 7 8 9 10
5 5 6
8 10 12 14
Bivariate and vertical generating functions for the numbers p(n, k), k = 1, 2, ... , n, n = 1, 2, ... , are derived in the next theorem.
PARTITIONS OF INTEGERS
374
THEOREM 10.2 (a) The bivariate generating function of the numbers p(n, k) of partitions ofn intokparts, k = 1, 2, ... , n = 1, 2, ... , is given by n
CXl
(X)
(10.13)
n=Dk=O
i=l
where p(O, 0) = 1. (b) The vertical generating function of the numbers p(n, k) of partitions of n into k parts, n = k, k + 1, ... , for fixed k, is given by k
(X)
Gk(t)
=L
p(n, k)tn
= tk II (1 ti) 1 .
n=k
(10.14)
i=l
PROOF (a) Using the fact that the number p(n, k) of partitions of n into k parts equals the number of nonnegative integer solutions of (10.2) and (10.3), it follows that CXl
G(t, u) =
n
L L (I: uYl+Y2+" ·+YntYl+2Y2+"·+nYn)' n=Dk=O
where in the inner sum the summation is extended over all nonnegative integer solutions of (10.2) and (10.3). Since this inner sum is summed over all k 0, 1, ... , nand n 0, 1, ... , it follows
=
=
establishing ( 10.13). (b) Interchanging the order of summation in (10.13), we get (X)
G(t, u) =
L Gk(t)uk k=O
and since
G(t,ut) = (1 ut)G(t,u), we deduce the recurrence relation
Consequently,
10.2.
RECURRENCES AND GENERATING FUNCTIONS
375
and k
Gk(t) = tk
II(l
ti)1'
i=1
I
which completes the proof of the theorem.
The bivariate generating function (10.13) of p(n, k) for u = 1 yields, by virtue of (10.4), the generating function of p(n). This is given in the following corollary of Theorem 10.2. COROLLARY 10.2 The generating function of the number p(n) of partitions of n, n = 1, 2, ... , is given by 00
G(t)
n=O
where p(O)
00
=bP(n)tn =II (1
1 ti) ,
(10.15)
i=1
= 1.
Bivariate and vertical generating functions for the numbers P(n, k), k = 0, 1, ... , n = 0, 1, ... , are derived in the following corollary of Theorem 10.2. COROLLARY 10.3 (a) The generating function of the numbers P( n, k) of partitions of n into at most kparts, k = 0, 1, ... , n = 0, 1, ... , is given by 00
00
00
(10.16) n=Dk=O
i=O
(b) The vertical generating function of the numbers P(n, k), n = 0, 1, ... , for fixed k, is given by k
00
Fk (t)
=L P(n, k)tn =II (1 nn=O
1
.
(10.17)
i=1
PROOF (a) The bivariate generating function F(t, u) of the numbers P(n, k), k = 0, 1, ... , n = 0, 1, ... , upon introducing (10.5), may be expressed as
00
00
= (1 u) 1 LLP(n,r)urtn, n=Or=O
PARTITIONS OF INTEGERS
376
which, by virtue of(l0.13), implies (10.16). (b) Expanding into powers of u both members of the relation
F(t,ut) = (1 u)F(t,u), according to 00
F(t,u) = LFk(t)uk, k=O
it follows that
Thus
Fk(t)
= (1 tk) 1 Fk! (t),
k
= 1, 2, ...
, Fo(t)
=1
and k
Fk (t)
= II (1 ti)1' i=l
which completes the proof of the corollary.
10.3
I
A UNIVERSAL GENERATING FUNCTION
The generating functions (10.13), (10.15) and (10.16) constitute particular cases of the generating function of the number of partitions of n into parts of specified or unspecified number, whose number of parts of any specific size belongs to a subset of nonnegative integers. Specifically, let A = (a;,j), i = 1, 2, ... , j = 0, 1, ... , be an infinite matrix with elements a;.,j = 0 or 1. The matrix A determines a sequence Y;, i = 1, 2, ... of subsets of nonnegative integers and vice versa as follows: for a specific i = 1, 2, ... , Y; = {j : ai,j = 1} is the set of indices j of the elements of the ith row of the matrix A that are equal to 1. Let us denote by p(n; A) the number of partitions of nand by p(n, k; A) the number of partitions of n into k parts whose number y; of parts that are equal to i belongs to Y;, i = 1, 2, .... Also, let us denote by P(n, k; A) the number of partitions of n into at most k parts whose number y; of parts that are equal to i belongs to Yi, i = 1, 2, .... Note that, in particular for a;,j = 1, i = 1, 2, ... , j = 0, 1, ... , whence Y; = {0, 1, ... }, i = 1, 2, ... , p(n; A) := p(n), p(n, k; A) := p(n, k) and P(n, k; A) := P(n, k). A bivariate generating function for p(n, k; A), k = 0, 1, ... , n, n = 0, 1, ... is derived in the next theorem.
10.3. A UNIVERSAL GENERATING FUNCTION
377
THEOREM 10.3 Let A = (ai,J), i = 1, 2, ... , j = 0, 1, ... be an infinite matrix with elements ai,j = 0 or 1 and p(n, k; A) be the number of partitions of n into k parts whose number Yi of parts that are equal to i belongs to the set Yi = {j : ai,j = 1}, i = 1,2, .... Then
(10.18)
PROOF The number p(n, k; A) equals the number of nonnegative integer solutions of (1 0.2) and (1 0.3) with Yi E
Yi = {j : ai,J = 1}, i = 1, 2, ....
(1 0.19)
Consequently, oo
G(t, u; A)=
n
L L (2: uY1+Y2+·+YntY1+2Y2+·+nYn)' n=Ok=O
where, the summation in the inner sum is extended over all nonnegative integer solutions of (1 0.2) and (1 0.3) that satisfy (1 0.19). Introducing the elements a;,j of the matrix A into this sum, we get the expression oo
n
G(t ' u·A) =""'""'("'a a2 ,y2 ···a n,yn uY1+Y2+··+YntY1+2Y2+··+nyn) ' ~~ ~ l,yl
'
n=Ok=O
where, the summation in the inner sum is extended over all nonnegative integer solutions of (1 0.2) and (1 0.3). Note that this transformation of the inner sum incorporates condition ( 10.19), adding zero terms and thus saves from the concern of selecting the solutions of ( 10.2) and ( 10.3) that satisfy ( 10.19). Since this inner sum is summed over all k = 0, 1, ... , nand n = 0, 1, ... , it follows
which is the required expression of the generating function.
I
COROLLARY 10.4 Let A = (ai,J ), i = 1, 2, ... , j = 0, 1, ... , be an infinite matrix with elements ai,j = 0 or 1. Also let p(n; A) be the number of partitions of nand P(n, k; A) be the number of partitions of n into at most k parts whose number Yi of parts that
PARTITIONS OF INTEGERS
378
are equal to i belongs to the set Yi
= {j : ai,j = 1}, i = 1, 2, ....
Then
(10.20)
and
PROOF Expressing the number p(n; A) as a sum of the numbers p(n, k; A), k =0,1, ... ,n,weget oo
00
n=O
n
n=Ok=O
and so, from (10.18) with u = 1, (10.20) is deduced. Similarly we get oooo
ooook
F(t,u;A) = LLP(n,k;A)uktn= LLLP(n,r;A)uktn n=Ok=O n=Ok=Or=O
n
oo
= (1 u) 1 L
LP(n,r; A)urtn n=Or=O
and so, by virtue of (I 0.18), (I 0.21) is deduced.
I
Several particular cases of the generating functions (10.18) and (10.20) are presented in the following examples. Example 10.2 Partitions into unequal parts Let q(n) be the number of partitions of n into unequal (different) parts and q(n, k) be the number of partitions of n into k unequal (different) parts. Determine the generating functions ex>
H(t) = L q(n)tn, H(t, u) n=O
ex>
=L
n
L q(n, k)uktn,
n=Ok=O
ex>
Hk(t)
=L
q(n, k)tn.
n=O
Note that, in this case, the number Yi of parts that are equal to i may take the values 0 or I, i = 1, 2, .... Consequently, q(n) = p(n; A), q(n, k) = p(n, k; A),
10.3. A UNIVERSAL GENERATING FUNCTION
= =0,1 and
with ai,j 1, j and (10.20),
ai,j
oo
=0, j =2, 3, ... , i =1, 2, .... Thus, by (10.18) n
00
n=Ok=O and
i=l
00
H(t)
379
00
=L
q(n)tn
n=O
=II (1 + ti). i=l
For the calculation of the third generating function, note that
H(t, u)
= (1 + ut)H(t, ut)
and since
00
=L
H(t, u)
Hk(t)uk,
k=O it follows that
Therefore
and
Hk(t)
=
00
L q(n,k)tn n=k(k+l)/2
k
k
j=l
j=l
=II ti(l tj)l =tl+2+ .. ·+k II(l ti)1.
Hence k
Hk(t) = tk(k+l)/2
II (1 tj)l.
0
j=l Example 10.3 Partitions into even and odd parts (a) LetPo(n) be the number of partitions ofn into even parts and Po(n, k) be the number of partitions of n into k even parts. Determine the generating functions oo
Go(t)
oo
= LPo(n)tn, n=O
Go(t, u)
n
=L
LPo(n, k)uktn. n=Ok=O
In this case, the number y 2 i of parts that are equal to 2i belongs to the set Y2i = {0, 1, ... }, while the number Y2il of parts that are equal to 2i 1 belongs tothesetY2 il {O},i 1,2, .... Consequently,po(n) =p(n;A),po(n,k)
=
=
=
PARTITIONS OF INTEGERS
380
p(n, k; A), with ai,O = 1, a2i,j Thus, by (10.18) and (10.20),
= 1 and a2il,j = 0, i = 1, 2, ... , j = 1, 2, .... n
oo
00
n=Ok=O
and
i=l
00
00
n=O
i=l
(b) Let Pt (n) be the number of partitions of n into odd parts and p 1 ( n, k) be the number of partitions of n into k odd parts. Then, proceeding as in the previous case, we conclude that n
oo
Gt (t, u)
=L
oo
:~::> 1 (n, k)uktn
=II (1 ut 2 il) 1
n=Ok=O
and
i=l
00
00
n=O
i=l
(c) Let q0 ( n) be the number of partitions of n into unequal even parts and q1 ( n) the number of partitions of n into unequal odd parts. Determine the generating functions 00
Ho(t)
00
=L
=L
qo(n)tn, Ht(t)
n=O
qt(n)tn.
n=O
In the first case Y2i = 0 or I, Y2i1 = 0, i = 1, 2, ... and so qo(n) = p(n; A), with a2i,j = 1, j = 0, 1, a2i,j = 0, j = 2, 3, ... , i = 1, 2, ... and a2it,o = 1, a2il,J = 0, j = 1,2, ... , i = 1,2, .... In the second case Y2il = 0 or I, Y2i = 0, i = 1, 2, ... and so q1 (n) = p(n; A), with a2il,J = 1, j = 0, 1, a2il,J = 0, j = 2, 3, ... , i = 1, 2, ... and a2i,o = 1, a2i,j = 0, j = 1, 2, ... , i = 1, 2, .... Thus, by virtue of ( 10.20), it follows that 00
Ha(t)
00
=L
qo(n)tn
n=O
=II (1 + t 2 i) i=l
and 00
Ht(t)
=L
00
qt(n)tn
n=O
= II(1 + t 2 il).
0
i=l
Example 10.4 Partitions into parts of restricted size Let R(n, k) be the number of partitions ofn with no part greater thank. Determine the generating 00
Rk(t) =
L R(n, k)tn. n=O
10.3. A UNIVERSAL GENERATING FUNCTION Note that R(n, k) = p(n; A), with ai,O 1,2, ... ,k,ai,j = O,i = k+1,k+2, ... ,j
381
= 1, i = 1, 2, ... , ai,j = 1, i = = 1,2, .... So,accordingto(l0.20),
00 k Rk(t) = L R(n, k)tn = (1  ti) 1 . i=l n=O
IT
Note that this generating function equals the generating function (1 0.17), Rk(t) = Fk(t). Consequently, the number R(n, k), of partitions of n with no part greater than k, equals the number P(n, k), of partitions of n into at most k parts, R(n, k) = P(n, k). 0
Example 10.5 Partitions into an even and an odd number of parts (a) Let P0 (n) be the number of partitions of n into an even number of parts and P 1 (n) be the number of partitions of n into an odd number of parts. Determine the generating functions 00 00 Ao(t) = LPo(n)tn, A1(t) = LP1(n)tn. n=O n=O The numbers P0 ( n) and P 1 ( n) may be expressed in terms of the number p( n, k) of partitions of n into k parts, k = 1, 2, ... , n, as [n/2]
P0 (n)
=
L
p(n,2r)
n
1
= 2 L{p(n,k) + (1)kp(n,k)} k=O
r=O
and [(n1)/2]
P 1 (n)
=
L r=O
p(n, 2r + 1)
=
1
n
L {p(n, k) ( 1)kp(n, k)}, 2 k=O
respectively. Therefore, using (10.13) with u = ±1, it follows that
and
g(l
00 1{00 At(t) = ~P1(n)tn = 2
ti) 1

g(1 + 00
ti) 1 }
.
(b) Let Q 0 ( n) be the number of partitions of n into an even number of unequal parts and Q1 (n) be the number of partitions of n into an odd number of unequal parts. Determine the generating functions 00
Ba(t)
=L n=O
00
Qo(n)tn, Bt (t)
=L
QI (n)tn.
382
PARTITIONS OF INTEGERS
The numbers Q 0 ( n) and Q1 ( n) may be expressed in terms of the number q( n, k) of partitions of n into k unequal parts, k = 1, 2, ... , n, as [n/2]
Qo(n)
=
L
q(n,2r) =
r=O
1
n
L{q(n,k)
2 k=O
+ (1)kq(n,k)}
and [(n1)/2]
Ql (n) =
L
q(n, 2r
+ 1)
=
r=O
1
n
L {q(n, k) ( 1)kq(n, k)},
2 k=O
respectively. Therefore, using the generating function H(t, u) of Example 10.2 with u = ±1, it follows that
and
Example 10.6 Partitions into specified parts In the partitions of a positive integer n, without any restriction, XJ
+ X2 + · · · + Xk = n,
X! ~ X2 ~ · · • ~ Xk ~
1, k
= 1, 2, ...
, n,
the parts Xi, i = 1, 2, ... , n belong to the set { 1, 2, ... , n}. Let us now consider the partitions of n whose parts Xi, i = 1, 2, ... , n belong to a subset {it, i 2, ... , ir} of { 1, 2, ... , n}. In this case, if Yi. is the number of parts that are equal to i 8 , s = 1,2, ... ,r, then
The number D(n; it, i 2, ... , ir) of partitions ofn whose parts belong to the subset {it, i2, ... , ir} of { 1, 2, ... , n} equals the number of nonnegative integer solutions of this equation. Determine the generating function 00
D; 1 ,i2 ... ,dt)
= LD(n;it,i2,··.
,ir)tn.
n=O
=
Note that D(n;i 1 ,i 2 , ... ,ir) p(n;A), where the elements of the matrix A= (a;,j),i = 1,2, ... , j = 0,1, ... ,aregivenbyai,O = 1,i = 1,2, ... ,
383
10.4. INTERRELATIONS AMONG PARTITION NUMBERS
=
l,i E {it,i2,···,ir},ai,j Hence, by (I 0.20),
ai,j
= O,i
¢ {it,i2,···,ir},j r
00
Di 1 .i2, ... ,ir(t)
= 1,2, ....
=L
D(n; it, i2, ... , ir)tn
n=O
=II (I ti·)
1
.
s=l
The moneychanging problem is a characteristic example requiring the calculation of partitions into specified parts. Specifically, let us consider a cashier that has an unrestricted number of coins of 5, 10, 25 and 50 cents. Calculate the number of ways of forming a given amount of, say, $ I. This number equals D(IOO; 5, 10, 25, 50) = D(20; 1, 2, 5, 10), the number of nonnegative integer solutions of the linear equation Sy5 + 10y10 + 25Y25 + SOy50 = 100 or, equivalently, the equation Yt + 2y2 + 5y 5 + 10y 10 = 20, which equals the coefficient of t 20 in the expansion of the generating function D1,2,5,10(t)
10.4
= [(1 t)(I t 2)(1 t 5)(1 t 10 )t 1.
o
INTERRELATIONS AMONG PARTITION NUMBERS
The generating functions of the numbers of partitions that satisfy certain conditions may be used to deduce relations between these numbers. Such relations are derived in the following theorems.
THEOREM 10.4 The number q(n) of partitions ofn into unequal parts equals the number p 1 (n) of partitions ofn into odd parts: (10.22) q(n) = Pt(n). PROOF The generating function of the number q(n) of partitions of n into unequal parts is given by (see Example 10.2) 00
00
H(t) = L q(n)tn n=O
= II (1 + ti) i=l
and the generating function of the number p 1 ( n) of partitions of n into odd parts is given by (see Example 10.3) 00
Gt(t) = LPt(n)tn
n=O
00
=II (I t2il)1. i=l
Using the identity (1
+ ti)(I ti) = (1 t 2i),
i
= 1,2, ...
,
PARTITIONS OF INTEGERS
384
the generating function H(t) may be written in the form 00
H(t)
00
00
=II (1 t i)(1 ti) =II (1 t i) II (1 ti)1
2
2
i=1
i=1
1
.
j=1
If, in the last product the terms with j even are separated from those with j odd, then it is written as 00
00
00
j=1
i=l
i=1
and so 00
H(t)
=II (1 t2i1 )1 = Gl (t). i=1
The last relation implies (10.22).
I
THEOREM 10.5 The number P(n, k) of partitions of n into at most k parts, which equals the number R(n, k) of partitions of n with no part greater thank, is equal to the number p( n + k, k) of partitions of n + k into k parts:
P(n, k)
= R(n, k) = p(n + k, k).
(10.23)
PROOF
The generating function of the numbers P(n, k) 0, 1, ... is (see Corollary 10.3 and Example 10.4) 00
Fk(t) = LP(n,k)tn = LR(n,k)tn = n=O
II(1 ti)
00
n=O
00
II (1 ti) 1. i=l
00
= LP(r,k)trk = LP(n+k,k)tn. r=k
The last relation implies (10.23).
n=O
I
1
,
= k, k + 1, ... , is (see
k
00
Gk(t) = LP(r, kW = tk r=k
=
i=1
while the generating function of the numbers p(r, k), r Theorem 10.2)
LP(n,k)tn
R(n, k), n
k
00
n=O
=
10.4. INTERRELATIONS AMONG PARTITION NUMBERS
385
THEOREM 10.6 The number q( n, k) of partitions of n into k unequal parts equals the number P( n  k( k + 1) /2, k) of partitions of n  k( k + 1) /2 into at most k parts, which is equal to the number p(n k(k 1)/2,k) of partitions ofn k(k 1)/2 into k parts: q(n, k) = P(n k(k PROOF
+ 1)/2, k)
= p(n k(k 1)/2, k).
(10.24)
Since (see Example 10.2 and Corollary 10.3) ~
Hk(t)=
k
L
q(n,k)tn=tk(k+l)/2IT(Iti)l
n=k(k+l)/2
and
i=l
~
Fk(t)
=
k
LP(r,kW r=O
~
= II(lti)1, i=l
~
L q(n, k)tn n=k(k+l)/2
=L
P(r, k)tr+k(k+l)/2
r=O ~
=
L P(n k(k n=k(k+l)/2
+ 1)/2, k)tn.
The last relation implies the first part of (1 0.24). Its second part is deduced from (10.23) by noting that n k(k + 1)/2 + k = n k(k 1)/2. I REMARK 10.3 The following combinatorial derivation of (10.24) possesses its own merits. The number q(n, k) of partitions of n into k unequal parts equals the number of integer solutions of the linear equation X1
(a) Putting z; to the equation Zr
(10.25) + X2 + · · · + Xk = n, Xr > X2 > · · · > Xk ~ 1. = x; (ki+ 1), i = 1, 2, ... , k, equation (1 0.25) is transformed
+ Z2 + · · · + Zk = n k(k + 1)/2,
z 1 ;:::
Z2 ;::: · · · ;::: Zk1 ;::: Zk ;::: 0,
whose number of integer solutions equals the number P(n k(k + 1)/2, k) of partitions of n  k(k + 1) /2 into at most k parts. (b) Setting Wi = xi  (k i), i = 1, 2, ... , k, equation (10.25) is transformed to the equation W1
+ w 2 + · · · + Wk
= n k(k 1)/2, Wr ;::: W2 ;::: · · · ;::: Wk1 ;::: Wk ;::: 1,
PARTITIONS OF INTEGERS
386
whose number of integer solutions equals the number p(n k(k 1)/2, k) of I partitions of n  k (k  1) /2 into k parts. A simple and instructive representation of the partition of a positive integer n is its Ferrers diagram, which is defined as follows. Ferrers diagram of a partition of a positive integer n into parts Xi, i = 1,2, ... 'k, ... ' X1
+ X2 + · · · + Xk + · · · =
n,
X1
:2:
X2
:2: · · · :2:
Xk
:2: · · · :2: 1, (10.26)
is called the diagram of (equidistant) points whose ith horizontal row of points (numbering from the bottom up) has Xi points, i = 1, 2, ... , k, .... To each partition there corresponds its conjugate partition, defined as follows. Conjugate partition of the partition (10.26) is called the partition of n into parts YJ> j = 1, 2, ... , r, ... , Y1
+ Y2 + · · · + Yr + · · · =
n,
Y1
:2:
Y2
:2: · · · :2: Yr :2: · · · :2: 1, (10.27)
whose Ferrers diagram the number YJ of points of the jth horizontal row equals the number of points of the jth column (numbering from left to right) of the Ferrers diagram of the partition (10.26). Selfconjugate partition is called a partition that coincides with its conjugate partition. FIGURE 10.1 Ferrers diagram y
6
•
5
•
•
4
•
•
•
3
•
•
•
2
•
•
•
•
•
1
•
•
•
•
•
•
0
1
2
3
4
5
6
X
Example I 0. 7 Figure 10.1 gives the Ferrers diagram of the partition 653 2 21 of 20. Clearly, its D conjugate partition is 6542 2 1.
10.4. INTERRELATIONS AMONG PARTITION NUMBERS
387
Some interesting interrelations among partition numbers are derived by considering the Ferrers diagrams of partitions and their conjugate partitions. Thus we have:
THEOREM 10.7 The number p( n, k) of partitions of n into k parts equals the number of partitions ofn into parts of which the maximum is k. PROOF Consider a partition of n into k parts and its Ferrers diagram. Associate to it its conjugate partition, which is also a partition of n into parts of which the maximum is k. Since this correspondence is one to one, the required conclusion is deduced. I
THEOREM 10.8 The number of selfconjugate partitions ofn equals the number q 1 ( n) of partitions of n into unequal odd parts. Consider a selfconjugate partition of n into parts xi. i = 1, 2, ... ,
PROOF X1
+ Xz + · · · + Xk + · · · =
n,
X1
~ Xz ~ · · · ~ Xk ~ · · · ~
1, (10.28)
and its Ferrers diagram. Note that this diagram is symmetric with respect to the line x = y. This symmetry implies that the pairs of straightline sections defined by the points {(i,i), (x;,i)} and {(i,i), (i,yi)}, i = 1,2, ... ,k .. . , since xi= y;, i = 1, 2, ... , k, ... , form isosceles angles. The number Zi ofpointsofthediagram that lie on the ith isosceles angle is odd, Zi
and since x 1
~
x2
= 2(xi 1) + 1, i = 1,2, ...
~ · • · ~ Xk ~ • • · ~ Z1
,k, ... ,
(10.29)
1, we have in addition
> Zz > · · · > Zk > · · · ~ 1.
Let us now correspond to the selfconjugate partition of n (10.28) the partition of n into unequal odd parts, Z1
+ Zz + · · · + Zk + · · · = n,
Z1
> Zz > · · · > Zk > · · · ~ 1. (10.30)
This correspondence of (1 0.28) to (I 0.30), according to (I 0.29), is onetoone and thus the number of selfconjugate partitions of n equals the number of partitions of n into unequal odd parts. I REMARK 10.4 The onetoone correspondence between the selfconjugate partitions of n and the partitions of n into unequal odd parts is further clarified by considering the following specific cases.
PARTITIONS OF INTEGERS
388 FlGURE 10.2
0
1 2 3 4 5 6
X
0
1 2 3 4 5 6 7 8 9 10 11
Z
(a) Consider the selfconjugate partition 62 53 2 2 of 25 into the parts x 1 = 6, = 6, x3 = 5, x 4 = 3, x 5 = 3, x 6 = 2. Figure 10.2 (left) represents its Ferrers diagram in which the three pairs of straightline sections: x2
({ (1, 1), (6, 1)}, {(1, 1), (1, 6)} ), ({ (2, 2), (6, 2)}, { (2, 2), (2, 6)} ), ( {(3,3), (5,3)}, {(3, 3), (3, 5)}) form isosceles angles. The number z;, of points of this diagram that lie on the ith isosceles angle, is z; = 2(x;  1) + 1, i = 1, 2, 3: z1 = 11, z2 = 9, z 3 = 5. Hence, to the selfconjugate partition 6 + 6 + 5 + 3 + 3 + 2 = 25 there corresponds the partition 11 + 9 + 5 = 25 into unequal odd parts the Ferrers diagram of which is represented in Figure 10.2 (right). FlGURE 10.3
w
y
•
5
I
x=y
4
3
•
3 2 1
0
z3
123456789z
0
1
2
3
4
5
X
(b) Consider the partition 971 of 17 into unequal odd parts: z1 = 9, z2 = 7, = 1. Figure 10.3 (left) represents its Ferrers diagram. We have x; = y; =
10.4. INTERRELATIONS AMONG PARTITION NUMBERS
389
+ (zi 1)/2, i = 1, 2, 3: x1 = Y1 = 5, x2 = Y2 = 5, x3 = y 3 = 3 and so the. Ferrers diagram of the corresponding selfconjugate partition forms three isosceles angles:
i
( {(1, 1), (5, 1)}, {(1, 1), (1, 5)} ), ( {(2, 2), (5, 2)}, {(2, 2), (2, 5)} ),
({(3,3),(3,3)}, {(3,3),(3,3)}) Note that the third angle degenerates to the point (3, 3) of the straight line x = y. Figure 10.3 (right) represents this diagram. From this diagram it follows that the I selfconjugate partition that corresponds to the partition 971 is 52 32 2 •
THEOREM 10.9 Euler's pentagonal theorem Let Qo (n) be the number of partitions of n into an even number of unequal parts and Q1 ( n) be the number of partitions of n into an odd number of unequal parts. Then, the difference E(n) = Qo(n) Q 1 (n) is given by
E(n)
(1)k, n = (3k 2 ± k)/2
={ 0,
(10.31)
otherwise.
PROOF Let us, initially, consider the Ferrers diagram F of a partition of n in any (even or odd) number of unequal parts and let: (i) A be the straightline section that starts from the easternmost point (of the first horizontal line) of the diagram F, forms an angle of 3Jr /4 with the horizontal axis and thus, passing through only the outer points ofF, ends in the northernmost of them. (ii) B be the straightline section that starts from the northernmost point (of the first column) of the diagram F, is parallel to the horizontal axis and thus, passing through only the outer points ofF that have the same ordinate, ends in the easternmost of them. It should be noted that B, as well as A, may contain only one point ofF (degenerate to a point straightline sections). Further, let N (A) = r and N (B) = s be the number of points of the diagram F that lie on A and B respectively. Now, we associate with the partition of n into unequal parts with Ferrers diagram F, the partition of n into unequal parts with Ferrers diagram G that is defined as follows: (a) If r < s, the Ferrers diagram G is deduced from F by removing the points that lie on A and adjoining them to the first columns, one to each one of them (Figure 10.4), except when A and B have a common point and r = s  1. In the latter case the corresponding partition of n into unequal parts has k = r parts and xk = s = k + 1 so that n
= 2k + (2k
1)
+ · · · + (k + 2) + (k + 1) = (3k 2 + k)/2
(10.32)
and to its Ferrers diagram F we do not associate any diagram G (Figure 10.5).
PARTITIONS OF INTEGERS
390
FIGURE 10.4 y 5 4
y
. ..... . B
F

1
0
1 2 3
5 6
7 8
X
G
1
• • • • • • • • • • • • • • • • •
.
0
1 2
7
3 2
..
4
·· • •
4
• • • • • • • • • • ' A • • • • • • • •
3 2
B'
5
3 4 5 6
A'
X
FIGURE 10.5 y 5
F
B
4
•  •  ·  • 
3
•
•
•
•
• '•, A
2
•
•
•
•
•
1
• • • • • • • '•
0
1 2 3 4 5 6 7 8
a,
'
• ..
X
(b) If r :2: s, the Ferrers diagram G is deduced from F by removing the points that lie on B and adjoining them to the first rows, one to each one of them (Figure 10.6), except when A and B have a common point and r = s. In the latter case the corresponding partition of n into unequal parts has k = r parts and Xk = s = k so that
n = (2k 1)
+ (2k 2) + · · · + (k + 1) + k =
(3k 2

k)/2
(10.33)
and to its Ferrers diagram F we do not associate any diagram G (Figure 10.7). The transition from F toG changes (increases or decreases) by one the number of parts of the corresponding partition, without altering the unequality of the parts. Thus, ifF belongs to the set of Ferrers diagrams :F0, of the partitions of n into an even number of unequal parts, then G belongs to the set of Ferrers diagrams :F1, of the partitions of n into an odd number of unequal parts, while, if F belongs to :F1, then G belongs to :F0 and vice versa. Consequently, if n f. (3k 2  k) /2 and n f. (3k 2 + k) /2, then the correspondence is one to one and so E( n) = Qo (n) Q 1 (n) = 0. Ifn = (3k 2 ± k)/2 and k is even, the Ferrers diagram of the partition (10.32) or (10.33) belongs to :Fo and has no image in :F1. while ifn = (3k 2 ± k)/2 and k is odd the Ferrers diagram of the partition (l 0.32) or (l 0.33) belongs to :F1
10.5. COMBINATORIAL IDENTITIES
391
FIGURE 10.6
y 5 3 2 1
• F • • • • • • • •', A • • • • • '41. ' • • • • • • '•
0
1 2 3 4 5 6 7
4
y 5
B
B'
X
G
··
4
3 2 1
• • • • • • • • • • • • • • • • • • •
0
1 2 3 4 5 6 7 8
A' X
FIGURE 10.7
y 5
F
B
4
····· • • • • •
3 2 1
• • • • • • ' •' • • • • • • •' ' • • • • • • • • '•
0
1 2 3 4 5 6 7 8 9
"
A
"
X
andhasnoarchetype(preimage)inF0 . HenceE(n) = Q0 (n)Q 1 (n) for n = (3k 2 ± k)/2, and the proof of the theorem is completed. I
10.5
= (1)k,
COMBINATORIAL IDENTITIES
Some interesting combinatorial identities may be deduced from the relation ()()
G(t, u; A)= L Gk(t; A)uk, k=O which connects the generating functions oo
G(t,u;A)
n
= LLP(n,k;A)uktn, n=Ok=O
oo
Gk(t;A)
= LP(n,k;A)tn, n=k
of the numbers p(n, k; A) of partitions of n into k parts, whose number of parts equal to i belongs to the set {j : ai,j = 1}, i = 1, 2, ... , where
PARTITIONS OF INTEGERS
392
=
=
=
A= (ai,j), i 1, 2, ... , j 0, 1, ... , with ai,J 0 or 1. In the literature, in expressing these identities instead of the variables t and u the variables q and x, respectively, are used. Further, such an identity is referred to as qidentity. Two typical qidentities are given in the following theorem. THEOREM 10.10 00
00
k
i=l
k=O
j=l
IT (1 xqi)1 =I: xkqk IT (1 qi)1, 00
00
k
i=1
k=O
j=1
(10.34)
IT (1 + xqi) =I: xkqk(k+1)/2 IT (1 qi)1. PROOF
(10.35)
Introducing into the relation 00
G(t,u) = 'L:Gk(t)uk, k=O expressions ( 10.13) and (10.14) of the generating functions, n
oo
G(t,u)
=L
oo
LP(n, k)uktn n=Ok=O
=IT (1 uti)
1
,
i=l k
00
IT(1
Gk(t) = LP(n,k)tn = tj) 1, j=l n=k of the numbersp(n, k) of partitions of n into k parts, and putting t
= q and u = x,
(10.34) is readily deduced. Similarly, introducing into the relation 00
H(t,u) = LHk(t)uk, k=O the expressions of the generating functions (see Example 10.2), oo
H(t, u)
=L
n
00
Hk(t)
=
oo
L q(n, k)uktn n=Ok=O
L q(n, k)tn n=k(k+l)/2
=IT (1 +uti), i=l k
= tk(k+I)/ 2 IT (1 ti) 1, j=1
of the numbers q( n, k) of partitions of n into k unequal (different) parts, and putting
t = q and u = x, (10.35) is deduced.
I
10.5. COMBINATORIAL IDENTITIES
393
Combinatorial identities may also be deduced from the relation 00
G(t; A) = L
p(n; A)tn
n=O
when the number p(n; A), of partitions of n whose number of parts equal to i belongs to the set {j : ai,j = 1}, i = 1, 2, ... , where A = (ai,j), i = 1, 2, ... , j = 0, 1, ... , with ai,j = 0 or 1, can be evaluated independently of the generating function G(t; A). A characteristic case of such qidentity is Euler's identity that is deduced in the following corollary of Euler's pentagonal theorem.
COROLLARY 10.5 00
00
i=l
k=l
(10.36)
PROOF The generating functions of the numbers Q 0 (n) and Q 1 (n) of partitions of n into an even and odd number of unequal parts, respectively, are given by (see Example 10.5)
and
Hence
00
00
n=O
i=l
Further, the difference E (n) = Q 0 ( n)  Q1 ( n) has been evaluated in Theorem 10.9 as 2 (1)k, n = (3k ± k)/2 E(n) = { 0, otherwise and so
00
LE(n)tn
00
=1+
n=O
L(1)k{tk(3kI)/2 +tk(3k+I)/2}. k=l
Consequently,
rr(l ti> 00
i=l
z:(00
= 1+
k=l
1)k{tk(3kI)/2 + tk(3k+I)/2}
PARTITIONS OF INTEGERS
394
and, putting t
= q, (10.36) is deduced.
I
COROLLARY 10.6 The number p( n) of partitions of n satisfies the recurrence relation n1
p(n)
n2
= ~)1)k
1
p(n k(3k 1)/2) + ~)1)k 1 p(n k(3k + 1)/2),
k=I
k=I
(10.37) where ni is the largest positive integer satisfying fori = 1 the inequality n 1 (3n 1 1) ::=; 2n and for i = 2 the inequality n 2(3n2 + 1) ::=; 2n. Expression ( l 0.15) may be rewritten as
PROOF
Replacing the first factor of the lefthand side by expression ( 10.36), it follows that
{1_
~( 1 )kl{tk(3kl)/2 + tk(3k+I)f2}}. {~p(n)tn} = 1
and, since p(O) = 1,
~ p(n)tn = {~ p(n)tn}. {~( 1)kl {tk(3kl)/2 + tk(3k+l)f2}} =
~ {~( 1)k p(n k(3k 1)/2) 1
+ ~(1)kIp(n k(3k + 1)/2)} tn. Equating the coefficients of tn in both members of the last identity, (10.37) is I deduced. In the next theorem, using (10.34) and (10.35), the GaussJacobi identity is derived.
THEOREM 10.11 GaussJacobi identity ~
+~
I1( 1 q2i)( 1 +xq2iI)( 1 +xIq2iI)= L i=l
k=~
xkqk
2 •
(10.38)
10.5. COMBINATORIAL IDENTITIES
395
PROOF Replacing q by q2 in ( 10.35) and, in the resulting expression, replacing xq by x, it follows that
oo
oo
n
i=l
n=O
j=l
II(l +xq2il) = "L:>nqn 2 II( 1 
q2j)l.
0: (1  q i) and using the relation 2
Multiplying both members by
1
00
n
00
i=l
j=l
j=l
we get 00
00
00
i=l
n=O
j=l
II (1 _ q2i)(1 + xq2il) = "L:>nqn 2 II (1 _ q2n+2j) 00
=
2:::: n=oo
xnqn
2
00
II (1 _ q2n+2i), j=l
where the last equation holds since the additional terms are zero because, for any negative integer n, the factor j = n in the product is zero. Also, replacing q by q2 and x by q 2 n in (10.35), we get
j=l
r=O
i=l
and, introducing it into the previous expression, we sequentially deduce
oo
II( 1 i=l
+oo
oo r 2 2 xnqn L:xrqr +r+2rn 1  q2i)l n=oo r=O i=l oo r +oo =L:(1Y(xlqfii(1q2i)l xn+rq(n+r)2 n=oo r=O i=l oo r +oo ="L::(1f(xlqfii(1q2i)l xkqk2. r=O i=l k=oo
q2i)( 1 + xq2il) =
II(
2::::
L
L
Further, replacing q by q2 in (10.34) and, in the resulting expression, replacing xq by x 1 , it follows that
Introducing it into the previous expression, we deduce (10.38).
I
PARTITIONS OF INTEGERS
396
10.6 BIBLIOGRAPHIC NOTES The roots of partitions of integers, according to L. E. Dickson (1920), may be traced in a letter from Gottfried Wilhelm Leibnitz to John Bernoulli in 1669. Their systematic study and development started in 1746 with Leonhard Euler and his twovolume work Introductio in Analysin Infinitorum. The proof of Euler's pentagonal theorem given here is due to Fabian Franklin (1881). The contribution of G. H. Hardy and Ramanujan (1917, 1918) and Rademacher (1937a,b, 1940, 1943) enhanced this subject. In his twovolume treatise, P. A. MacMahon (1915, 1916) discussed at length combinatorial aspects of partitions. For further reading on partitions of integers, refer to the books of J. Riordan (1958), L. Comtet (1974) and G. E. Andrews (1976).
10.7
EXERCISES
1. Show combinatorially that the number p(n, k) of partitions of n into k parts satisfies the recurrence relation
= p(n 1, k 1) + p(n k, k),
p(n, k)
k < n/2,
with initial conditions
p(n,1)=p(n,n)=1, n=1,2, .... 2. Show combinatorially that the number P(n, k) of partitions of n into at most k parts satisfies the recurrence relation
P(n,k)=P(n,k1)+P(nk,k), k=2,3, ... ,n, n=2,3, ... , with initial conditions P(n,1)=1, n=1,2, ... , P(O,k)=1, k=0,1, .... 3. Partitions into parts of restricted number and restricted size. Let P(n, k, r) be the number of partitions of n into at most k parts, none of which is greater than r. Show that oo
Fr(t,u)
oo
r
= LLP(n,k,r)uktn = II(l uti) 1 n=Ok=O
i=l
397
10. 7. EXERCISES
and
r
oo
Fr,k(t)
=L
P(n, k, r)tn
=II (1 tr+i)(1 ti)
n=O
1
.
i=1
4. Let r(n, k) be the number of partitions of n whose smallest part equals k. Show that 00
00
n=O
i=k
and r(n, k) = r(n 1, k 1) r(n k, k 1),
fork= 2,3, ... , [(n
+ 1)/2], n = 3,4, ... , with
r(n, 1) =p(n1),n = 2,3, ... ,r(n,n) = 1,n = 1,2, ... ,r(n,k) = O,n
< k.
5. Show combinatorially that the number q(n, k) of partitions of n into k unequal parts satisfies the recurrence relation q(n, k) = q(n k, k)
+ q(n k, k 1),
k = 2, 3, ... , [n/2], n = 4, 5, ... ,
with initial conditions q(n, 1) = 1, n
= 1, 2, ... , q(n, k)
= 0, n
< k(k + 1)/2,
q(k(k + 1)/2, k) = 1.
6. Let Q(n, k) be the number of partitions of n into at most k unequal parts. Show that 00
F(t, u) =
00
00
LL
Q(n, k)uktn
= (1  u) 1
n=Ok=O
II (1 +uti) i=1
and, using the identity
(1 u)F(t, u) = (1 u 2t 2)F(t, ut), conclude that the generating function 00
Fk(t)
=L
Q(n, k)tn
n=O
satisfies the recurrence relation
(1 tk)H(t) = Fk_I(t) tkFk2(t), k = 2,3, ... , with F0 (t) = 2, F 1 (t) = (1 t) 1 .
PARTITIONS OF INTEGERS
398
7. Partitions into even and odd unequal parts. Let q0 (n, k) be the number of partitions of n into k even unequal parts and q1 (n, k) be the number of partitions of n into k odd unequal parts. Show that n
oo
00
n=Ok=O oo
i=l
n
oo
H1 (t, u) = L L q1 (n, k)uktn n=Ok=O
=II (1 + ut iI) 2
i=l
and k
00
Ho,k(t) = Lqo(n,k)tn = tk(k+I) n=O
II(l t i)I, 2
i=l
k
00
HI,k(t) = Lql(n,k)tn = tk2 n=O
II(l
t2i)1.
i=l
8. Partitions into parts of restricted size. Let p(n, k, r) be the number of partitions of n into k parts, none of which is greater than r. Show that oo
n
r
Gr(t, u) = L LP(n, k, r)uktn n=Ok=O
=II (1 uti) 1 i=l
and conclude that (1 ut)Gr(t, u) = (1 utr+I )Gr(t, ut). Using this recurrence relation show that k
00
Gr,k(t)
=L
p(n, k, r)tn
n=k
= tk II (1 tr+il )(1 ti) 1 . i=l
9. Let 00
L Gr(t, u)gr(t)ur = G(t, u), r=O where Gr(t, u) is the generating function of the number p(n, k, r) of partitions of n into k parts, none of which is greater than rand G(t,u) is the generating function of the number p(n, k) of partitions of n into k parts. Using the relations (1 ut)G(t,u) = G(t,ut), (1 ut){Gr(t, u) + utr+ 1Gr+1 (t, u)} = Gr(t, ut),
10. 7. EXERCISES
399
show that
r
gr(t)
= tr2 II (1 ti)1 i=1
and conclude the identity r
oo
2:: II (1tr2
r=O
oo
ti)2
=II (1
i=1
ti)1.
i=1
10. Show that k1
II (1 +
2k1 t2.)
=
2:: tn
n=O
i=O
and thus conclude that 00
II(l +
2 t .)
= (1 t) 1 ,
itl
< 1.
i=O
Using the last identity, show that for every nonnegative integer n there exists a unique finite sequence ni E {0, 1}, i = 0, 1, ... , r, such that r
n
= Z::ni2i, i=O
where the positive integer r is determined by the inequalities 2r :S n
< 2r+1.
11. (Continuation). Show that the number R 0 (n) of partitions of n into an even number of parts, with values (sizes) in the set {1, 2, 22 , ••• }, equals the number R 1 (n) of partitions of n into an odd number of such parts, for n = 2,3, ....
12. Show that
and thus conclude that
Using the last identity, show that for every nonnegative integer n there exists a unique finite sequence ni E {0, 1, ... , i}, i = 1, 2, ... , r, such that r
n = Z::nii!, i=1
PARTITIONS OF INTEGERS
400
where the positive integer r is determined by the inequalities r! (r + 1)!.
~ n
<
13. Partitions into unequal parts of restricted size. Let q(n, k, r) be the number of partitions of n into k unequal parts, none of which is greater than r. Show that oo
Hr(t, u)
n
r
=L L q(n, k, r)uktn =II (1 +uti) n=Ok=O
i=l
and conclude that
(1
+ ut)Hr(t, ut) = (1 + utr+ 1 )Hr(t, u).
Using this recurrence relation, show that k
00
L
Hr,k(t) =
p(n, k, r)tn =
n=k(k+l)/2
tk(k+l)/ 2
II (1
ri+l )(1
ti)1,
i=l
fork< r, and
14. Sum of multinomial coefficients. The multinomial coefficient
is a symmetric function of the variables k1 , k2, ... , kr1, kr:
Thus, for the representative of this class of multinomial coefficients, with n and k1, k2, ... , kr fixed, we may assume that k1 2: k2 2: · · · 2: kr. (a) Show that the number of different classes of multinomial coefficients, for given positive integers nand r, equals the number P(n, r) of partitions of n into at most r parts, which, when r 2: n, equals the number p(n) of partitions of n. (b) Let n! M(n)
=L
k 1·'k 2·I ... k r·I'
where the summation is extended over all nonnegative integer solutions of the linear equation k1 + k2 + · · · + kr = n, k1 2: k2 2: · · · 2: kr 2: 0. This is the sum of the multinomial coefficients, representatives of the p(n) classes. Show that
10. 7. EXERCISES
401
15*. Perfect partitions. A partition of a positive integer n that contains just one partition for every positive integer r less than n is called perfect partition of n. Thus, an integer solution of the equation y 1 + 2y 2 + · · · + nyn = n, Yi 2 0, i = 1, 2, ... , n is a perfect partition of n, if for every r = 1, 2, ... , n  1, the equation z 1 + 2z2 + · · · + rzr = r, 0 S Zi S Yi, i = 1, 2, ... , r, has a unique solution. For example, the partitions 31 2 , 2 2 1, 15 are perfect partitions of n = 5. Show that the number of perfect partitions of n equals the number of ordered factorizations of n + 1 without unit factors. 16. Compositions of integers. A composition of a positive integer n is an ordered collection of positive integers whose sum equals n. Thus a composition of n into k parts (summands) is a solution (r 1 , r2, ... , rk) in positive integers of the linear equation x 1 + X2 + · · · + Xk = n. Show that the number c( n, k) of compositions of n into k parts equals c(n,k)=
(~=~),
ksn
and, consequently,
17. (Continuation). Let c(n, k; S) be the number of compositions of n into k parts of sizes Xi, i = 1, 2, ... , k belonging to the set S = { s 1 , s2, ... , Sr, ... }, with 1 S SJ < s2 < .. · < Sr < .. ·. Show that
18. (Continuation). Let c2 (n) be the number of compositions of n with no part greater than 2. Show that 00
1+
L
c2(n)tn
=
(1 t t 2 ) 1
n=I
and thus conclude that c2(n) = fn, n number.
= 1, 2, ... , where fn
is the Fibonacci
19. (Continuation). Let cr(n) be the number of compositions of n with no part greater than r. Setting cr(O) = 1, show that 00
Cr(t) =
L
n=O
er(n)tn = (1 t t 2  ...  nl'
PARTITIONS OF INTEGERS
402 and Cr(n)
= Cr(n 1) + Cr(n 2) + · · · + Cr(n r), Cr(n) = 2nl, n
n 2: r,
< r.
Thus, conclude that Qr(n) = 2er(n) is the number of npermutations of 2 with repetition that include no r consecutive objects alike. 20. Let p(n, k) be the number of partitions of n into k parts and q(n, k) the number of partitions of n into k unequal parts. Show that k!q(n,k) :S ( n1) k_ :S k!p(n,k) 1
and, using the relation p(n, k) = q(n + r, k), r = k(k 1)/2, conclude that k!p(n, k) :S (
n r k_
1) 1
, r = k(k 1)/2.
21. Using the GaussJacobi identity, show that oo
+oo
II <1 _ q2riH 1 + q2rir+sH 1 + q2rirs) = :L: i=l
2
qrk +sk, r, s ;::: 0
k=oo
and oo
II(l q2ri)( 1 
q2rir+s)( 1  q2rirs)
i=l
=
+oo
L
2
(1)kqrk +sk, r,s 2:0.
k=oo
Putting r = 3/2, s = 1/2, conclude Euler's identity:
II<1 qi) = 1 + L(1)k{qk(3k1}/2 + qk(3k+l)f2}. 00
00
i=l
k=l
Further, setting r = 1/2, s the Jacobi identity:
= E+ 1/2 and taking the limit as E t 0, conclude
00
00
i=l
k=l
II (1 qi) = 1 + :L:< 1)k(2k + 1)qk(k+l)/2. 22*. Let
00
9i(t) =
L Ui(k)tk'
i = 1, 2,
0
0
0
k=O
and a(n) =
L Pn
u1(kt)u2(k2) · · · Un(kn), n = 0, 1, ... ,
10. 7. EXERCISES
403
where the summation is extended over the set Pn of partitions of n, that is over all nonnegative integer solutions (k 1, k 2 , •.• , kn) of the equation k1 + 2kz + · · · + nkn = n. Show that 00
00
n=O
i=l
= L a(n)tn =II g;(ti)
G(t)
and conclude that 00
00
23*. (Continuation). Let
Show that
00
00
i=l
j=l
=II (1 ti)1 L Ujtj (1 t1)1 and conclude that n
L A (k1 , kz, . . . , kn; u1 , Uz, ... , Un) = L p( n  k) L Ud. Pn
k=I
In particular, for u 1
djk
= j, j = 1, 2, ... , conclude that n
np(n) = LP(n k)a(k), a(k) = Ld. dlk
k=l
24. Partitions of ordered pairs of integers. A partition of an ordered pair (m, n) of nonnegative integers, different (0, 0), is a nonordered collection of nonnegative integers (x;, y;), i = 1, 2, ... , different (0, 0), whose sum equals (m, n). In a partition of a bipartite number (m, n), let Zi,J be the number of ordered pairs that are equal to (i,j), i = 0,1, ... ,m, j = 0,1, ... ,n, (i,j) ::j:. (0,0). Then m
n
n
m
LiLZi,J = m,
LJLZi,J = n.
i=l
j=l
j=O
i=O
PARTITIONS OF INTEGERS
404
In the case of a partition of (m, n) into k parts, in addition, n
m
LL Zi,j = k. j=Oi=O (i,j)#(O,O)
Let Pm,n be the number of partitions of the bipartite (m, n) and Pm,n,k the number of partitions of the bipartite number (m, n) into k parts. Prove that oo
F(t, u, w)
=L
oo
m+n
L L
oo
Pm,n,ktmunwk
=
oo
II II (1 
wtiuj) 1,
j=O i=O (i,j)#(O,O)
n=O m=O k=O
where Po,o,o = 1, and conclude that <X>
F(t, u) =
00
00
LL
Pm,ntmun =
n=Om=O
00
II II (1
1 tiuj) ,
j=Oi=O (i,j);>!(O,O)
where Po,o = 1. 25. qnumber, qfactorial and qbinomial coefficient. Let 0 < q < 1, x be a real number and k a positive integer. The number [x]q = (1 qx)/(1 q) is called qreal number and, in particular, [k]q = (1 qk)/(1 q) is called qpositive integer. The factorial of the qnumber [x]q of order k, which is defined by [x]k,q = [x]q[x 1]q · · · [x k
+ 1]q =
(1 _ qx)(1 _ qx1) ... (1 _ qxk+l) ( _ q)k ' 1
is called qfactorial of x of order k. In particular, [k]q! = [1]q[2]q · · · [k]q = (1 q)(1 q2 ) · · · (1 qk)/(1 q)k. The qbinomial coefficient is defined by
X] [x]k (1 [k q = [k];r =
qx)(1  qx1) ... (1  qxk+1) (1 _ q)(1 _ q2) ... (1 _ qk)
For x a real number and k a positive integer, derive the recurrence relations [x1] k[x1] k k1 q +q k [x] q q and
= [x)q [X 1] [X] k q [k]q k 1 q.
10. 7. EXERCISES
405
Also, show that
[~X] q =
( _ 1 )kqkxk(k1)/2
[X+~
[~ L/q =
~
qk(xk) [
1]
q'
L
and
26. (Continuation). For n,k and m positive integers, show that (a)
and (b)
[:L [;L = [~L [:~~L = [m~kL [n;+kL, [:L [n~mL [~L [n:kL [m:kL [m~kL 27. (Continuation). For n, k and m positive integers, show that
~ rk[r1] L..,;q r=m
k1
_
q
nk[n] _q (ml)k[m1] k k
q
q
q
and conclude that
28. (Continuation). For n, k and m positive integers, show that
f)
1rkqr(r+I)/2 [
r=k
~~
n
= qk(k+l)/2 [ ~]
q
q
+( _ 1)mkq(m+I)(m+2)/2 [ and conclude that
~( _ 1rkqr(r+I)/2 [ ~ ~ ~ Jq =
qk(k+I)/2 [
~ Jq.
m+1 n
] q
406
PARTITIONS OF INTEGERS
29. qbinomial and negative binomial formulae. For n a positive integer and x a real number, prove that
and
fr(1xqi1)I=f:[n+z1] k=O
i=l
xk,
O
q
Also, prove that (qx)n
='f) _
1)kqk(k1)f2(1 _ q)k
k=O
[~]
[x]k,q· q
30. (Continuation). For n and k positive integers, show that
and
More generally, for x and y real numbers and n a positive integer, show that [ X: Y]
= q
t
q
k=O
[~] [n ~ k] q
q
or, equivalently, that
[x + Y]n,q =
L n
k=O
(nk)(xk) q
[
n] k [x]k,q[Y]nk,q· q
31. (Continuation). For n and k positive integers, show that
and
10. 7. EXERCISES
407
32*. qStirling numbers of the first kind. Consider the nth order factorial ofthe qnumber [t]q = (1 qt)/(1 q), 0 < q < 1 (nth order qfactorial oft): [t]n,q = [t]q[t 1]q · · · [t n
+ 1]q
= qn(nl)f2[t]q([t]q _ [1]q) ... ([t]q _ [n _ 1]q)· This is a polynomial of the qnumber [t]q of degree n: n
[t]n,q = qn(nl)/ 2
L s(n, klq)[t]~,
n = 0, 1, ....
k=O
The coefficient s(n, klq) is called qStirling number of the first kind. (a) Show that s(n, klq) = ( 1)nk L[it]q[i2]q · · · [ink]q,
where the summation is extended over all (n  k )com bin at ions {i 1 , i 2, ... ,
ind of then 1 positive integers {1, 2, ... , n 1}. (b) Derive the triangular recurrence relation
+ 1, klq)
s(n
+ 1, n
fork = 1, 2, ... , n
= s(n, k 1lq) {n}qs(n, klq),
= 0, 1, ... , with
s(O,Oiq}=1, s(n,klq}=O, k>n, and the vertical recurrence relation n
s(n
+ 1, k + 1lq)
=
L( 1)nr[n]nr,qS(r, klq). r=k
33*. qStirling numbers of the second kind. Consider the expansion of the nth power of the qnumber [t]q into factorials of the same number: n
L qk(kl)/ S(n, klq)[t]k,q,
[t]~ =
2
n = 0, 1, ....
k=O
The coefficient S(n, klq) is called qStirling number of the second kind. (a) Derive the triangular recurrence relation S(n
for k = 1, 2,, .. , n
+ 1, klq)
+ 1, n
= S(n, k 1lq)
+ [k]qS(n, klq),
= 0, 1, ... , with
S(O, Olq) = 1, S(n, klq) = 0, k
> n,
PARTITIONS OF INTEGERS
408 and the generating function k
00
¢>k(uiq) =
L S(n, klq)un = uk IT (1 [j)qu)n=k
1
,
k = 1, 2, ...
j=l
and conclude the expansion __ 1_ = qk(k+l)/2
[t)k+l,q
f
n=k
S(n, kiq)
~+1
[t)q
and its inverse 1 ~ =
[t)q
00
1 L qn(n+l)/2 s(n, kiq) . n=k [t)n+l,q
(b) Further, deduce the vertical recurrence relation n
S(n, kiq) = L[k]~r S(r 1, k 1lq), r=k and the expressions
where the summation in the first sum is extended over all integers rj 2: 0, j = 1, 2, ... , k, with r 1 + r2 + · · · + rk = n  k, while, in the second sum, is extended over all (n  k )combinations {i 1 , i 2 , ... , ink}, with repetition, of the k positive integers {1, 2, ... , k}. (c) Derive the orthogonal relations n
n
r=k
r=k
L s(n, riq)S(r, kiq) = 6n,k, L S(n, riq)s(r, klq) = 6n,k· 34*. (Continuation). Show that the qStirling number of the second kind S(n, kiq), k = 0, 1, ... , n, n = 0, 1, ... , is given by the sum k(k1)/2
S(n, kiq) = q [k] ! q
Alternatively,
L( k
r=O
1)kr q(kr)(kr1)/2
[k] r
[r]~. q
10. 7. EXERCISES
409
Inverting the last expression, deduce for the qStirling number of the first kind s(n, klq), k = 0, 1, ... , n, n = 0, 1, ... the explicit expression
[n]
s(n kiq) = ~( 1)rkq(nr)(nr1)/2 r ' (1 _ q)nk ~ r=k 1
q
(r). k
35*. qLah numbers. Consider the expansion n
[t]n,q1
= qn(n1)/2 L qk(k1)f2 L(n, klq)[t]k,q· k=O
Since [t]n,q1 = [1]~[t+n1]n,q and setting IL(n, klq)l = [1]~L(n, kiq), it can be written as n
[t + n 1]n,q = qn(n1)/2
L qk(k1)f21L(n, kiq)i[t]k,q· k=O
The coefficients L(n, klq) and IL(n, klq)l are called qLah and signless qLah numbers, respectively. Show that
IL(n, klq)l = qn(n1)f2+k(k1)f2 ~~l::
[~:::: ~]
q.
Chapter 11 PARTITION POLYNOMIALS
11.1
INTRODUCTION
The partition polynomials, introduced by E. T. Bell (1927), are multivariable polynomials that are defined by a sum extended over all partitions of their index. Such a sum had been previously used by F. Faa di Bruno (1855) to express the derivatives of a composite function in terms of the derivatives of the component functions. The later partition polynomials have found many applications in combinatorics, probability theory and statistics, as well as in number theory. Since they are quite general polynomials, they include as particular cases the exponential polynomials and their inverses, the logarithmic polynomials, as well as the potential polynomials, owing their particular names to the form of their generating functions. In this chapter, the partition polynomials are presented along with some of their combinatorial and probabilistic applications. Specifically, the exponential and partial Bell partition polynomials are examined first. Generating functions and recurrence relations are derived. As an example, the general arithmetical function, that led E. T. Bell to introduce the exponential partition polynomials is presented. Then, the general partition polynomials are examined; the Faa di Bruno formula is expressed in terms of these polynomials. Their use in expressing the probability function of a compound discrete distribution in terms of the component probability functions is discussed. The logarithmic and potential polynomials, constituting particular cases of the partition polynomials, are further explored. The Lagrange formula for the inversion of a power series is derived by the aid of a generalization of Faa di Bruno formula. J. Touchard (1939), in studying the problem of enumeration of permutations with a given number of partially ordered cycles introduced the partition polynomials that were later called Touchard polynomials. These polynomials are briefly presented at the end of this chapter.
PARTITION POLYNOMIALS
412
11.2
EXPONENTIAL BELL PARTITION POLYNOMIALS
DEFINITION 11.1 The polynomial Bn XI , x2, . . . , Xn defined by the sum
= Bn(xl, x2, ... , Xn) in the variables (11.1)
where the summation is extended over all partitions ofn, that is, over all nonnegative integer solutions (k1, k2, ... , kn) of the equation ki + 2k2 + · · · + nkn = n, is called exponential Bell partition polynomial.
=
DEFINITION 11.2 The polynomial Bn,k Bn,k (XI, x2 ... , Xn) in the variables X1, x2, ... , Xn of degree k defined by the sum ( 11.2)
where the summation is extended over all partitions of n into k pans, that is, over all nonnegative integer solutions (ki, k2, ... , kn) of the equations ki + 2k2 + · · · + nkn = n, ki + k2 + · · · + kn = k, is called (exponential) panial Bell panition polynomial. Definitions 11.1 and 11.2 directly imply the following relations: (11.3) ( 11.4) n
Bn(XI' X2, ... 'Xn)
=L
Bn,k(Xl' X2, ... 'Xn), n
= 0, 1,... .
(11.5)
k=O
REMARK 11.1 The number of monomials contained in the exponential Bell partition polynomial ( 11.1) equals the number p( n) of partitions of n and in the partial Bell partition polynomial (11.2) equals the numberp(n, k) of partitions ofn into k parts. These remarks justify the term partition that is included in the name of polynomials (11.1) and (11.2). The term partial that is used for polynomial (11.2) in contrast to polynomial (11.1) is justified by the fact, that in (11.2), the summation is extended over a subset of the partitions of n and, specifically, over
11.2.
EXPONENTIAL BELL PARTITION POLYNOMIALS
413
the partitions of n into k parts. The exponential form of their generating functions (see Theorem 11.1 that follows) justifies the term exponential that is included in the name of polynomial ( 11.1) and, occasionally, in the name of polynomial ( 11.2). The coefficients of polynomials ( 11.1) and ( 11.2) are positive integers. Specifically, the coefficient of the general term of the exponential Bell partition polynomial ( 11.1) equals the number of partitions of a finite set of n elements into subsets, among which ki ~ 0 subsets include i elements each, fori = 1, 2, ... , n. Also, the coefficient of the general term of the partial Bell partition polynomial ( 11.2) equals the number of partitions of a finite set of n elements into k subsets, among which ki ~ 0 subsets include i elements each, fori = 1, 2, ... , n. Thus, polynomials ( 11.1) and ( 11.2) are the multivariable generating functions of the numbers of partitions of a finite set of n elements into any number of subsets and into k subsets, respectively, with respect to the numbers ki ~ 0 of subsets including i elements each, i = 1, 2, ... , n. I Bivariate and vertical generating functions for the partial Bell partition polynomials are derived in the following theorem.
THEOREM 11.1 (a) The bivariate generating function of the partial Bell partition polynomials Bn,k(x1, x2, ... , Xn). k = 0, 1, ... , n, n = 0, 1, ... , with Bo,o 1. is given by
=
oo
B(t, u) =
n
tn
L L Bn,k(Xt, X2, ... , Xn)uk n! = exp{u[g(t) xo]}. (11.6) n=Ok=O
(b) The vertical generating function of the partial Bell partition polynomials Bn,k(Xt, x2, ... , Xn), n = k, k + 1, ... ,for fixed k, is given by
where (11.8)
PROOF . (a) Multiplying (11.2) by uktn fn! and summing the resulting expression fork = 0, 1, ... , nand n = 0, 1, ... , it follows that
414
PARTITION POLYNOMIALS
where the summation in the inner sum is extended over all nonnegative integer solutions (k1, k2, ... , kn) of k1 +2k2 +· · ·+nkn = n andk1 +k2 + · · ·+kn = k. Since this inner sum is summed over all k = 0, 1, ... , n and n = 0, 1, ... , it follows that the summation is extended over all kr = 0, 1, ... , for r = 1, 2, ... , and so 00
E(t, u) =
00
II L r=l
1 (
kf
Xr~ tr) r.
k,=O r·
00
k,
=
II exp
(
Xr~
tr)
r.
r=l
L tr) r! . 00
= exp
(
u
Xr
r=l
Using (11.8), the last expression implies (11.6). (b) Interchanging the order of summation, ( 11.6) may be written as 00
00
~
00
E(t,u) = L LEn,k(xl,X2,··· ,xn)uk n! = LEk(t)uk k=On=k k=O and, since E(t, u) = exp{u[g(t) xo]}, it follows that
Equating the coefficients of uk in both sides of the last expression, ( 11. 7) is deduced. I The generating function of the exponential Bell partition polynomials, by virtue of (11.5), may be deduced from (11.6) by setting u = 1.
COROLLARYII.l The generating function of the exponential Bell partition polynomials En _ En(XI,X2, ... , Xn). n = 0, 1, ... , with Eo= 1, is given by (11.9)
Recurrence relations for the exponential and the partial Bell partition polynomials are derived in the following theorem.
THEOREM 11.2 (a) TheexponentialBellpartitionpolynomialsEn
=
En(x 1 ,x2,··· ,xn). n
=
11.2.
EXPONENTIAL BELL PARTITION POLYNOMIALS
0, 1, ... , with B 0
415
= satisfy the recurrence relation 1,
(11.10)
(b) The partial Bell partition polynomials Bn,k = Bn,k(xl, X2, ... , Xn). k = 0, 1, ... , n, n = 0, 1, ... , with Bo,o 1, satisfy the recurrence relations
=
nk ( Bn+l,k+l = L r=O
)
~ Xr+l Bnr,k
(11.11)
nk(n+1) r + 1 Xr+1Bnr,k, r=O
(11.12)
and
Bn+l,k+l = k
1
+1 L
fork = 0, 1, ... , n, n = 0, 1, .... PROOF (a) Differentiating the generating function ( 11.9) and expanding the resulting expression, d d
dt B (t)
= B (t) dt g(t),
into powers oft it follows that 00
00
l[ l n) l 00
tn = [ LXr+l! tr · LEn! tn LBn+l! n=O n. r=O r. n=O n. and so 00
00
L
[ n L Bn+l 1tn = L ( n=O n. n=O r=O r
tn Xr+lBnr 1· n.
Equating the coefficients of tn fn! in both sides of the last expression, (11.10) is deduced. (b) Differentiating the generating function ( 11. 7) and expanding the resulting expression, d d dtBk+l(t) = Bk(t) dtg(t), into powers oft, it follows that
416
PARTITION POLYNOMIALS
Equating the coefficients oftn /n! in both sides of the last expression, (ll.ll) is deduced. Further, the generating function (11.7) may be written in the form
(k + 1)Bk+I(t) = [g(t) x 0 ]Bk(t), which, expanded into powers oft, yields 00
l[
00
00
tn+1 [ tr+1 E(k+1)Bn+1,k+1(n+ 1)! = ?;xr+l(r+ 1)! 00
[nk =E?;
(n + 1) r+ 1
tnl · EBn,kn!
Xr+1Bnr,k
l
tn+1 (n+ 1)!"
Equating the coefficients of tn+l /(n + 1)! in both sides of the last expression, ( 11.12) is deduced. I Using recurrence relation (11.11), a table of partial Bell partition polynomials Bn,k Bn,k(x 1,x2 ... ,xn) can be constructed. Table 11.1 gives the polynomial Bn,k fork= 1, 2, ... , n, n = 1, 2, ... , 10. From this table, using the relation Bn = 2::::~= 1 Bn,k, the exponential Bell partition polynomials may be deduced.
=
Example 11.1 An arithmetical function The Bell partition polynomials were introduced by E. T. Bell (1927) as new arithmetical functions, unifying a diversity of such functions. Let S be any set of positive integers and
Xn=aLdunfd, n=1,2, ... , d
where the summation is extended over all divisors d of n that belong to S. Consider the polynomials Bn(x 1, 1!x 2, ... , (n 1)!xn). n = 1, 2, .... Show that oo
n
n=O
n.
L Bn(x1, 1!x2, ... , (n 1)!xn)!, = IJ (1 ul)a. kES
The generating function of the sequence Xn, n = 1, 2, ... , 00
00
h(t) = L Xntn =aLL dunfdtn, n=1 n=1 d on replacing the variables (d, n) by the variables (k, r), with k = d, r = n/d, and summing first for r = 1, 2, ... and then fork E S, may be obtained as 00
h(t) =a L k L(utkr =a L kutk(l utk) 1. kES r=1 kES
11.2.
EXPONENTIAL BELL PARTITION POLYNOMIALS
Table 11.1
417
Partial Bell Partition Polynomials Bn,k
Bt ' 1 = Xt,· B2 ' 1 = x2; B2 ' 2 = x 21,· B3 ' 1 = X3; B3 ' 2 = 3xtx2; B3 ' 3 = x 31·, B4 1 = X4; B4 2 = 4xtX3 + 3x 22; B4 3 = 6x 12x2; B4 4 = x 14; ' ' ' ' B5,1 = x5; B5,2 = Sx1x4 + l0x2x3; B5,3 = 10xix3 + 15xtx~; B5,4 = 10xrx2; B5,5 = x~; B6,l = x5; B6,2 = 6x1x5 + lSx2x4 +lOx~; B6,3 = 15xix4 + 60x1x2x3 + 15x~; B6,4 = 20xrx3 + 45xix~; B6,5 = ISxix2; B6,6 = x~; B1,1 = X7; B1,2 = 7x1x6 + 21x2x5 + 35x3x4; B7,3 = 21xix5 + 105xtX2X4 + 70x1x~ + 105x~x3; B7,4 = 35xtX4 + 105xtx~ + 210xix2x3; B7,5 = 35xfx3 + 105xrx~; B7,6 = 21x~x2;B7,7 =xi; Bs,1 = xs; Bs,2 = 8xtX7 + 28x2x6 + S6x3x5 + 35x~; Bs, 3 = 28xix6 + l68x1x2x5 + 280x1x3x4 + 210x~x4 + 280x2x~; Bs,4 = 56xrx5 + 420xix2x4 + 280xix~ + 840xtx~x3 + 105xt Bs, 5 = 70x{x4 + S60xrx2x3 + 420xix~; Bs,6 = 56x~x3 + 210xfx~; Bs,7 = 28x~x2; Bs,s = x~;. B 9,1 = x 9 ; B9,2 = 9x1X8 + 36x2x7 + 84x3x6 + 126x4x5; B 9,3 = 36xix7 + 252xtX2X6 + 504XtX3X5 + 378x~x5 + 315XtX~ +1260x 2x3x4 + 280x~; B9,4 = 84xrx6 + 756xix2x5 + 1260xix3x4 + 1890x1x~x4 +2520X1X 2 X~ + 126Qx~x3; B 9,5 = 126xfx 5 + 126oxrx2x4 + 840xrx~ + 3780xix~x3 + 945xtx~; B9,6 = 126x~x4 + 126oxix2x3 + 1260xrx~; B9,7 = 84x~x3 + 378x~x~; B9,8 = 36xi x2; B9,9 =xi; B10,1 = x~ 0 ; B10,2 = l0x1x 9 + 45x2X8 + 120x3X7 + 210x4x6 + 126x;; Bl0,3 = 45xixs + 360xtX2X7 + 840xtX3X6 + 630x~x6 + 1260x1X4X5 +2520x2X3X5 + 1575x2x~ + 2100x~x4; BI0,4 = 120xrx7 + 1260xix2x6 + 2520xix3x5 + 3780xtx~x5 + 1575xix~ +126QQx 1x 2X3X4 + 315Qx~X4 + 28QQx 1 x~ + 63QQx~x~; BI0,5 = 210xfx6 + 2520xrx2x5 + 4200xrx3x4 + 9450xix~x4 + 126Q0xix 2 x~ + 126QQx 1 X~X3 + 945x~; B10, 6 = 252x~x5 + 3150x{x2x4 + 2100x{x~ + 12600xrx~x3 + 4725xix~; B 10 ,7 = 210x~x 4 + 2520x~x2x3 + 3150x{x~; B10,8 = 120xix3 + 630x~x~; B10,9 = 45x~x2; B10,10 = x~ 0 .
PARTITION POLYNOMIALS
418
Differentiating the logarithm of the function
G(t)
= IT (1 utk)a, kES
we get
and so
G'(t) G(t)
h(t)
t
Integrating the last equation in the interval (0, t) and using the initial condition = 1 we find the relation
G(O)
00
tr) ,
G(t) = exp ( ~Xr;
from which, by virtue of (11.9), the required expression is deduced.
0
Example 11.2 Dinner choices in a Chinese restaurant A Chinese restaurant offers x 1 different dinner choices for one person eating alone, x 2 different choices for two persons eating together, x 3 still different choices for three persons eating together, etc. Suppose that a group of n persons arrive at this restaurant to have dinner. In how many different ways may they choose their dinner? The set (group) of then persons {p 1 , p 2 , ... , Pn}, according to Theorem 2.10, may be partitioned into subsets (subgroups), among which ki 2: 0 include i persons each, i = 1, 2, ... , n, in
different ways. Further, there are xi different choices for each subset (subgroup) of i persons eating together, i = 1, 2, ... , n. Thus, for fixed ki 2: 0, i = 1, 2, ... , n, with k1 + 2k2 + · · · + nkn = n, there are
different ways for the n persons to choose their dinner. Summing these ways for all ki 2: 0, i = 1, 2, ... , n, with k 1 + 2k2 + · · · + nkn = n, using (11.1), it follows that there are Bn (x 1 , x 2 , ... , xn) different ways n persons may choose their dinner. 0
11.3. GENERAL PARTITION POLYNOMIALS
11.3
419
GENERAL PARTITION POLYNOMIALS
DEFINITION11.3
ThepolynomialPn:: Pn(x1,X2,··· ,xn;c1,c2,··· ,en) in the variables x1, x2, ... , Xn and parameters c1, c2, ... , Cn defined by the sum
( 11.13) where k = k 1 + k2 + · · · + kn and the summation is extended over all partitions of n, that is, over all nonnegative integer solutions ( k1, k2, . . . , kn) of the equation k1 + 2k2 + · · · + nkn = n, is called partition polynomial.
The partition polynomial Pn(XI.x2, ... , Xn; c1, c2, ... , en), using the symbolic calculus, may be expressed by the exponential Bell partition polynomial Bn(XI, x2, ... , Xn) as Pn(XI' X2, ... 'Xn; C}' c2, ...
'Cn)
= Bn(CX1' CX2, ... 'CXn), ck
=
Ck' (11.14)
since
=
ck. This symbolic expression of the partition polynomial will be with ck used in the sequel. Further, the partition polynomial Bn(cxi, cx2, ... , cxn), ck ck, k = 1, 2, ... , n, can be expressed as a linear combination of the partial Bell partition polynomials Bn,k(x 1 , x 2 , ... , xn), k = 1, 2, ... , n with coefficients ck, k = 1, 2, ... , n. Specifically, from Definitions 11.3 and 11.2 and putting B 0 = eo, it follows directly that
=
n
Bn(CXI,CX2,··· ,cxn) = LCkBn,k(X1,X2,··· ,xn),
(11.15)
k=O
for n = 0, 1, .... A generating function and a recurrence relation for the partition polynomials may be deduced (at least symbolically) from the generating function and the recurrence relation for the exponential Bell partition polynomials by using the symbolic expression (11.14). Specifically, the following theorem is deduced from (11.9) and (11.10).
PARTITION POLYNOMIALS
420
THEOREM 11.3 (a) The generating function of the sequence of the partition polynomials ck, n = 0, 1, ... , with Eo =eo. is given by
En(cx 1 , cx 2, ... , cxn). ck
where g(t) =
=
I:::o Xrtr /r!, or symbolically by
exp(tE)
= exp[c(etx xo)],
En= Bn, ck =: Ck, xr
(b) The partition polynomials En(cx 1 , cx2, ... , cxn). ck with E 0 eo. satisfy the recurrence relation
=
=
=Xr·
Ck.
n
= 0, 1, ...
En+ I (ex I' cx2, ... 'CXn+d
=
t (~)
Xr+JCBnr(CXI, CX2, ... , CXnr), Ck :: Ck.
(11.17)
r=O
In the next theorem, the derivative of order n of a composite function is expressed as a partition polynomial of the derivatives up to order n of the component functions.
THEOREM 11.4 Faa di Bruno formula Letftu) and g(t) be two functions of real variables for which all the derivatives,
9r =
[d~~t)]
, r = 0, 1, ... , /k =
[dd~u)]
t=a
, k = 0, 1, ... , u=g(a)
U
exist. Then the derivatives of the composite function h(t) = f(g(t)),
hn=
[d:h~t)] t
'n=0,1, ... ' t=a
are given by n
hn
= LfkEn,k(9I,92,···
,gn)
k=O
= En(f9I, Jg2, · · · ,/gn),
fk:: fk·
PROOF By successive differentiation of the composite function h( t) it follows that
(11.18)
= f (g( t) ),
421
11.3. GENERAL PARTITION POLYNOMIALS
and, in general,
z= n
hn =
!khn,k, n =
o, 1, ... ,
(ll.l9)
k=O
where ho,o = 1 and the polynomials hn,k = hn,k (91, 9 2 , ... , 9n) of the derivatives 9 1 ,92 , ... , 9n, k = 1, 2, ... , n, n = 1, 2, ... , do not depend on the derivatives h, h, ... , f n. Hence they may be determined by an arbitrary choice of the function f. The exponential function
with c an arbitrary parameter, which is the most convenient, entails
fk
=
[dk f(u)] k du u=g(a)
=
[dkecu] dk U
= u=g(a)
k cg(a) ce '
and
h = [dnh(t)] n dtn
t=a Thus, according to ( 11.19), it follows that ecg(a) [.!!:_ecg(t)] dtn t=a
=
t
hn,kCk, n
= 0, 1, ....
k=O
Further,
ecg(a) [ d: ecg(t)] = [ dnn ec[g(s+a)g(a)]] , dt t=a ds s=O with
9(s+a) 9(a)
sr sr ='"' [dr9(t)] ·dtr r! ='"'9rr! t=a oo
oo
L....
L....
r=1
r=1
and so 00
8
dn { r }] [ dsn exp c ~9r;:T r1
, n
= 0, 1, ....
s=O
Multiplying this expression by sn /n! and summing for n = 0, 1, ... we get the generating function
PARTITION POLYNOMIALS
422
which, compared to (11.6), implies
hn,k = hn,k(9I, 92, ... , Yn) = Bn,k(9I, 92, ... , Yn), n = 0, 1, ... and, according to ( 11.19), the first part of ( 11.18) is deduced. The second part of (11.18) is a direct consequence of (11.15). I
Example 11.3 Compound discrete distributions Consider a discrete distribution assuming nonnegative integer values, with probability function Pk. k = 0, 1, ... and probability generating function 00
f(u)
= L:>kuk. k=O
Further, consider another discrete distribution assuming nonnegative integer values, with probability function qr, r 0, 1, ... and probability generating function
=
00
g(t) =
L qrtr. r=O
Then, the distribution with probability g((nerating function the composite function h(t) = f(g(t)) is called compound distribution. Its probability function Pn, n = 0, 1, ... , may be expressed in terms of Pk· k = 0, 1, ... and qr, r = 0, 1, .... Indeed, since
Pn
=~
n.1
[dnh(t)] dtn
t=o
, n = 0, 1, ... ,
using Theorem 11.4, with
9r
= [d~~t)]
= rlqr, r
= 1, 2, ... ,
Yo
= g(O) = qo,
t=O
and
fk
=
dkf(u)] [ ;t;;k
. ..Jk =~ L..J (J )kPj'lo , u=g(O)=qo
k
= 0, 1, ...
,
j=k
we deduce the expression
Pn =
~!
t
fkBn,k(qi, 2!q2, ... , n!qn)
k=O
=
~Bn(fqi,f2!q2,···
n.
,Jn!qn), fk
=fk,
for n = 0, 1, .... The factorial moments of the compound distribution, 00
M(n) = LU)nPj, n j=n
= 1, 2, ...
,
11030 GENERAL PARTITION POLYNOMIALS
423
may be expressed in terms of the factorial moments 00
00
= L(J)rqj,
JL(r)
= 1, 2, o o o ,
r
= L(j}kpj,
ffi(k)
j=r
= 1, 2, o o o ,
k
j=k
on the distributions Pk. k = 0, 1,
0
0
and qr, r = 0, 1,
0
o
0
M _ [dnh(t)] dtn ' n = 1, 2, (n) 
0
0
Indeed, since
0
0
0
'
t=l
using Theorem 11.4, with
[d~~t)]
gr =
= JL(r)•
r
= 1, 2, o o o ,
= 1,
go= g(1)
t=l
and
fk
= [dd~u)] U
= m(k)•
k
= 1, 2,
°
0
0
,
u=g(l)=l
we deduce the expression n
M(n)
= Lm(k)Bn,k(JL(l),Jl(2),ooo
,JL(n))
k=O = Bn(ffiJL(l),ffiJL(2),ooo ,mJL(n)), mk =::=m(k)• for n = 1, 2,
0
0
0
0
0
REMARK 11.2
The discrete distribution with probability function
a·Oi Qi = /(O), j = 0, 1, where the coefficients ai, j = 0, 1,
0
0
0
0
0
0
,
0 < 0 < p,
are independent of() and
00
f(O) = Lai()i, 0 < () < p, j=O
is called power series distribution with parameter() and series function f(O)o In the preceding example, if q0 = 0, then
fk while if q0
= [ddu~u)] u=qo=O = k!pk,
= 0, 1,
0
o
o
,
> 0, then 00
!k
k
.
j
= q0 k J(qo) L(j)k JP(1 q0) = q0 k J(qo)m(kl (qo), i=k
qo
k
= o, 1,
0
0
0
,
where me k l ( q0 ) is the factorial moment of order k of the power series distribution with parameter q0 and series function f(qo) = 'L.'f'=o P1q6°
I
PARTITION POLYNOMIALS
424
11.4
LOGARITHMIC PARTITION POLYNOMIALS
The logarithmic partition polynomials, which are the inverse of the exponential Bell partition polynomials, are defined as follows.
DEFINITION 11.4 The polynomial Ln X1, X2, ... , Xn defined by the sum
= Ln(XI, x
2 , ... ,
Xn) in the variables
(11.20)
where k = k1 + k2 + · · ·+ kn and the summation is extended over all partitions of n, that is, over all nonnegative integer solutions ( k1, k2, ... , kn) of the equation k1 + 2k2 + · · · + nkn = n, is called logarithmic partition polynomial. The logarithmic partition polynomial Ln (x 1, x 2, ... , xn) constitutes a particular case of the partition polynomial Bn(cx 1, cx 2, ... , cxn), ck ck. Specifically, for n = 0, 1, ... ,
=
(11.21) with Co = 0, Ck = ( 1)kl (k 1)!, k = 1, 2, ... , n. Note that B 0 =Co and so L 0 = 0. Further, Definitions 11.4 and 11.2 imply, for n = 1, 2, ... , the relation n
Ln(XI' X2, ... 'Xn) = ~) 1)kl (k 1)!Bn,k(XI' x2, ... 'Xn) (11.22) k=l The generating function of the logarithmic partition polynomials may be deduced from the generating function (11.16) of the partition polynomials by setting ck = (1)k 1(k1)!, k = 1,2, ... , and using (11.21) and the expansion of the logarithm, log(1 + t) = L:;~ 1 (1)kItk/k. Thus the following theorem is deduced.
THEOREM 11.5 The generating function of the logarithmic partition polynomials Ln(XI, x2, ... , Xn), n = 1, 2, ... , with L 0 0, is given by
=
(11.23)
11.4. LOGARITHMIC PARTITION POLYNOMIALS
425
REMARK 11.3 The recurrence relation for the partition polynomials, (11.17), in the particular case of eo = 0, ck = (1)kl (k 1)!, k = 1, 2, ... , does not yield a corresponding recurrence relation for logarithmic partition polynomials. This is because, according to the symbolic calculus and (11.15), n
cBn(cx!,CX2,··· ,cxn)
= LCk+lBn,k(xl,X2,···
,xn),
k=O a relation, which for expression
eo=
0, Ck
=
(1)k 1 (k 1)!, k
=
1,2, ... , yields the
n
cBn(CX!,CX2,· .. ,cxn)
= L(1)kk!Bn,k(X!,X2,···
,xn)
k=O n
= L( 1)kBn,k(Xl, x2, ... 'Xn)· k=O Consequently,
for
eo= 0, ck = (1)kl (k
1)!, k
= 1, 2, ...
, n, where
n
Cn,s(Xl' X2, ... 'Xn)
= L(s)kBn,k(Xl' X2, ...
'Xn), n
= 0, 1, ...
'
k=O with s a real number, is the potential partition polynomial examined in the next section. Thus, recurrence relation (11.17) for c0 = 0, Ck = (1)k 1 (k 1)!, k = 1, 2, ... , upon using ( 11.21 ), yields Ln+l(xl,X2,··· ,Xn+i) =
t (~)
Xr+1Cnr,l(xl,x2,··· ,Xnr),
r=O
for n
= 0, 1, ... , with Co,1 = 1.
I
A recurrence relation for the logarithmic partition polynomials may be obtained by differentiating generating function (11.23). Specifically, the following theorem is derived.
THEOREM 11.6
=
The logarithmic partition polynomials Ln Ln(Xl, x2, ... , Xn), n = 1, 2, ... , with L 1 = x 1 , satisfy the recurrence relation
Ln+l
= Xn+l
t
r=l
(~) XrLnr+l,
n
= 1,2, ....
(11.24)
PARTITION POLYNOMIALS
426
PROOF Differentiating generating function ( 11.23) and multiplying the resulting expression by 1 + [g(t)  x 0 ], we get the relation
d d d dt L(t) = dtg(t) [g(t) x 0 ] dt L(t). Expanding it into powers oft, it follows that 00
00
tn LLn+l! n=O n.
l[
00
00
and so 00
l
tn [ LXr! tr · LLn+l! tn = LXn+l!n=O n. r=l r. n=O n.
00
n) XrLnr+ll1· tn
00
tn tn [ n ( L Ln+I 1 = L Xn+I 1  L L n=O n. n=O n. n=l r=I r
n.
Equating the coefficients of t 11 /n! in both sides of the last expression, (11.24) is I deduced.
Example 11.4 Connection of cumulants and moments The cumulant generating function 00
tTl
K(t) = Lll:n! n. n=l is usually defined through the moment generating function
by K(t) =log M(t). Thus
Comparing this relation with (11.23), it follows that ~~:n
= Ln(JL~,JL~, . ..
,JL~), n
= 1,2, ....
The logarithmic partition polynomials, according to (11.9) and (11.23), are the inverse of the exponential Bell partition polynomials and so
11.4. LOGARITHMIC PARTITION POLYNOMIALS
427
Example 11.5 Elementary and power sum symmetric functions A polynomial P(x 1, x 2 , ••• , Xn) inn variables is called symmetric function if, for any permutation (h , )2, . . . , j n) of the n indices { 1, 2, ... , n}, it holds that P(x]t,Xh, . .. ,Xj,) = P(xt,X2,· .. ,x 11 ). The elementary symmetric function ak = ak (x 1 , x 2 , ... , Xn) is defined by the sum where the summation is extended over all kcombinations {i 1 , i 2 , ... , ik} of the
n indices {1, 2, ... , n }. Therefore, with ao = 1, n
L
n
ak(Xt, X2,
... ,
X11 )tk
=II (1 + X;t).
k=O
(11.25)
i=l
The power sum symmetric function sk sum
= sk(Xt, x2, ... , Xn) is defined by the n
sk(Xt,X2,··· ,xn) =LX~, k
= 1,2, ....
i=l
Its generating function is readily deduced as
n
=
log
II (1  x;t).
(11.26)
i=l
Express the elementary symmetric functions ak = ak (x 1 , x2, ... , x 11 ), k = 1, 2, ... , n in terms of the power sum symmetric functions sk = sk (x1, x2, ... , X 11 ), k = 1, 2, ... , and vice versa. Note first that, in (11.25) and (11.26), n may be regarded as being as large as may be necessary. Then,
and 00
00
tk =log ( 1 + ~artr ) . t;(1)klSkk Therefore, according to (11.9) and (11.23), it follows that
ak = ( 1)k Bk(s1, s2, ... , (k 1)!sk)/k! and
PARTITION POLYNOMIALS
428
11.5
POTENTIAL PARTITION POLYNOMIALS
DEFINITION 11.5 The polynomial Cn,s =: Cn,s (Xt, X2, ... , X11 ) in the variables Xt, x 2 , ••• , Xn defined for a real (or complex) numbers by the sum
where k = kt + k2 + · · · + kn and the summation is extended over all partitions of n, that is, over all nonnegative integer solutions ( k1 , k2 , •.. , kn) of the equation kt + 2k 2 + · · · + nkn = n, is called potential partition polynomial.
The potential partition polynomial Cn,s(x 1 , x 2 , ••• , Xn) constitutes a particular case of the partition polynomial Bn (cx 1 , cx 2 , .•• , cxn), ck ck. Specifically, for n = 0, 1, ... ,
=
(11.28) with Ck = (s)k, k = 0,1, ... ,n. Further, Definitions 11.5 and 11.2 imply the relation n
Cn,s(Xt' X2, ... , Xn)
= L(s)kBn,k(Xt, x2, ...
, Xn),
n
= 0, 1, .... (11.29)
k=O
The generating function of the potential partition polynomials may be deduced from the generating function of the partition polynomials, (11.16), by setting Ck = (s)k, k = 0, 1, ... , and using (11.28) and the expansion
Thus the following theorem is deduced. THEOREM 11.7 The generating function of the potential partition polynomials Cn,s(Xt, X2, ... , X11 ), n = 0, 1, ... , is given by oo
c.(t)
=
L
tn Cn,s(Xt, X2, ... 'Xn) n!
n=O
where g(t) = L~o Xrtr /r!.
= [1 + (g(t) Xo)]",
(11.30)
11.5. POTENTIAL PARTITION POLYNOMIALS
429
REMARK 11.4 The recurrence relation for the partition polynomials, (I I. I 7), in the particular case of Ck = (s )k, k = 1, 2, ... , and since n
CBn(CXt,CX2, ... ,CXn) = L(s)k+IBn,k(Xt,X2, ... ,Xn) k=O n
s L(s 1)kBn,k(Xt, X2, ... , Xn) k=O = sCn,s1 (xi, X2, ... , Xn), =
yields for the potential partition polynomials Cn,s 0, 1, ... , the recurrence relation
Cn+l,s =
S
t (~)
= Cn,s(x 1, x 2 , .•. , Xn), n =
Xk+l Cnk,s1•
(11.31)
k=O
Note that this recurrence relation expresses the potential partition polynomial
Cn+I,s· with parameters, in terms of the potential partition polynomials ck,s1. k = 0, 1, ... , n, with parameters 1. I A recurrence relation for the potential polynomials, in which s remains constant, is obtained in the next theorem. THEOREM 11.8 The potentia/partition polynomials Cn,s Cn,s(Xt, x2, ... ,xn). n = 0, 1, ... , with Co,s = 1, satisfy the recurrence relation
=
PROOF
Differentiating generating function (I I .30) and multiplying the resulting expression by 1 + [g(t)  x 0 ], we get the relation
dC.(t) dt
= sC
s
(t)dg(t) _ [g(t) _ xo]dC.(t). dt dt
Expanding it into powers oft, it follows that
l[ l
= tn = s [ LXr+I! = F · LCn,s! = tn LCn+I,s! n=O n. r=O r. n=O n. 
= tr+l ] [ = tn1 ] [~Xr+l (r + l)! · ~ Cn,s (n _ 1)!
PARTITION POLYNOMIALS
430
and so
Equating the coefficients of tn fn! in both sides of the last expression, ( 11.32) is deduced. I The potential partition polynomial Cn,s (X1, X2, ... , Xn), n = 0, 1, ... , in the case of a positive integers, may be expressed as a partial Bell partition polynomial. Specifically, we have the following theorem. THEOREM 11.9 If sis a positive integer, then for n = 0, 1, ... ,
PROOF
For Y1 = 1, Yr =
TXr1.
r = 2, 3, ... ,
If s is a positive integer, expanding the lefthand side according to ( 11.30) and the righthand side according to (II. 7), we get the relation
which implies (11.33).
I
An interesting property of the potential partition polynomial is given in the following theorem. THEOREM 11.10 LetCn,s = Cn, 8 (XI,x2, ... ,xn), n
=
0,1, ... , bethepotentialpartitionpoly
11.5. POTENTIAL PARTITION POLYNOMIALS
431
nomial. Then, for any real (or complex) numbers,
C
PROOF
_ n,s
n ~
(
s r)
(
ns ) C n r n,r·
(11.34)
Generating function ( 11.30) may be expressed as
C8 (t) = [1
+ (g(t) xoW
=
f: (Z)
[g(t) xo]k.
(11.35)
k=O
Note that the function
has the factor tk and, thus, for the determination of Cn,s• which is the coefficients of tn In!, the first n terms of the sum in the righthand side of (11.35) are sufficient. Consequently, the potential partition polynomial Cn,s is the coefficient of tn In! in the expansion
E(Z)
[g(t) xo]k =
E~(1)kr (Z) (;) Cr(t)
=
E~(1)kr (Z) (;) ~Cn,r:~
=
~ {E~(1)kr (Z) (;) Cn,r} :~.
Hence we get
Cn,s =
=
E~(1)kr (Z) (;) Cn,r
~{t(1)kr (Z) (;) }cn,r
and, since, by the horizontal recurrence relation of binomial coefficients (2.9),
PARTITION POLYNOMIALS
432
I
( 11.34) is deduced.
REMARK 11.5 The following formulation of Theorem 11.10 is of interest. For any real (or complex) numbers and for every function h(t), with h(to) = 1, for which the derivatives at t 0 exist, ( 11.36) Indeed (11.30), with h(t) 11
d (h(tW] n [ dt
_ 
= 1 + g(t t 0 )  g(O) and g(t) = L~o Xrtr jr!, yields
[d (1+{g(t)xo})'] n 11
d
t=to
C ( ) n,s Xr, X2, ... , Xn
t
t=O
I
and so (11.36) is readily deduced from (11.34).
Example 11.6 Homogeneous product sum symmetric function The homogeneous product sum symmetric function hk = hk(x 1 , x2, ... , X11 ) is defined by the sum hk(xr,x2,··· ,xn)
= Lxr
1
x;
2
where the summation is extended over all n that r 1 + r 2 + · · · + r 11 = k. Therefore, with h 0 oo
Lhk(XI,X2, ... ,xn)tk
x~n, k
= 1,2, ... , = 0, 1, ... , k, i = 1, 2, ... •••
, n, such
= 1,
oo
n
noo
k=O
i=l
i=l r;=O
=LLII(xiW• =II L(xity'
k=O
and so n
oo
L
hk(Xr, X2,
k=O
... ,
Xn)tk
=II (1 Xit)
1
•
( 11.37)
i=l
Express the homogeneous product sum symmetric functions hk = hk(x 1 , x2, ... , x 11 ), k = 1, 2, ... in tenns of (a) the elementary symmetric functions ak = ak(x 1 ,x2 , ••• ,x 11 ), k = 1,2, ... and (b) the power sum symmetric functions Sk = sk(x 1 , x 2 , .•• , X11 ), and vice versa (see Example 11.5). Generating functions ( 11.25) and ( 11.26) are connected with ( 11.37) by the relations
I+ t,(l)'h,f' ~ (1 +~act")_,, I+ t,(l)'a,t' ~(I+ ~hct") _,,
11.6. INVERSION OF POWER SERIES
433
and
Therefore, according to (11.9), (11.23) and (11.30), it follows that
and hk = Bk(sl, s2, ... , (k 1)!sk)/k!,
Sk = Lk(hl, 2!h2, ... 'k!hk)/(k 1)!.
11.6
D
INVERSION OF POWER SERIES
Consider the power series (11.38)
which has no constant term, and let (11.39)
be its inverse series, ¢ 1 ((t)) = ( 1 (t)) = t. Since 1
[
d (u)] du
u=¢(t)
. [d(t)] = 1 ' dt
it follows that 7/J 1 = ¢1 1 . The coefficients 7/Jn, n = 2, 3, ... , may be expressed as potential polynomials of the coefficients n, n = 2, 3, .... Interesting on its own a preliminary result, is derived in the following lemma, which constitutes a generalization of Faa di Bruno formula.
434
PARTITION POLYNOMIALS
LEMMA 11.1 Let I (u) and g( t) be two functions of real variables for which all the derivatives,
[~ l(u)]
I = gn = [dng(t)] dtn t=O ' n exist. Further, let h( t) =
dU n
u=g(O)
'
n = 1, 2, ... '
I (g( t)) and
hn = [dnh(t)] dtn
' n t=O
= 1' 2, . . . .
Then, for r = 1, 2, ... , n and n = 1, 2, ... , n
Bn,r(ht,h2,··· ,hn) = LBk,r(/t,/2, ... ,/k)Bn,k(gl,g2,· .. ,gn)· k=r (11.40) PROOF
According to (11.7), oo tn LBn,r(ht,h2,··· ,hn), n. n=r
=
[h(t)  h(OW I ' r r.
= 1,2, ....
Further,
f.Bk,r(ft,h, ... k=r where
,lk)(uk~o)k
= [l(u)
In= [~1~)] u=ua
U
,
~t(uoW,
r = 1,2, ... ,
n=1,2, ....
Consequently, with u = g(t), u 0 = g(O) and, since ~ tn ~Bn,k(Yt,g2,··· ,gn)n!
=
[g(t) g(O)]k k! , k
= 1,2, ... ,
n=k it follows that
[l(g(t))
~/(g(O)W
=f.
Bk,r(ft, h. ... 'lk) [g(t)
~~g(O)jk
k=r oo
=L
k=r
oo
Bk,r(ft, h .... ' /k)
L
n=k
tn Bn,k(gl,g2, ... ,gn) n!
11.6. INVERSION OF POWER SERIES
435
Thus
and, equating the coefficients of tn j n! in both members of this relation, ( 11.40) is I deduced. Note that expression (11.40) for r = 1, since Bn,1 (h1, h2, ... , hn) = hn, Bk,1 (!I, h, ... , /k) = fk, reduces to Faa di Bruno formula (11.18).
THEOREM 11.11 Inversion formula of Lagrange Let ¢(t) be power series ( 11.38), which has no constant term, and ¢ 1(u) its inverse power series (11.39). Then fork= 1, 2, ... , nand n = 1, 2, ... ,
dn _1 k] [ dun(¢ (u)) u=O = k(n
1)k1
PROOF Applying Lemma 11.1 with f(u) h(t) = f(g(t)) = t, and since
[ dnk (¢(t)) dtnk t
nl
t=O.
(11.41)
= ¢(u), g(t) = ¢ 1 (t), whence
1 [ dn ] _ [dr¢1(t)] Bn,k('I/J1' 'I/J2, ... ''1/Jn) = k! dtn (¢1 (t))k t=O' '1/Jr dtT t=O' it follows that
Thus, for ( 11.41) to hold, it suffices to prove that
Cn,r
_ n k = Ln k=r
(n) [dnk (¢(t))n] dtnk t
1 [dk r] _ . ;:! dtk (¢(t))   On,r·
k
t=O
t0
PARTITION POLYNOMIALS
436
Using the relation
Cn,r may be written as
1
n (n) [dnk (¢(t))n] k dtnk tt=O
= n(r 1)! ~
Cn,r
d: {t(<jJ(t)r1 d<jJ(t) }]
X [
ill
ill
t=O
and so, according to the formula of the nth derivative of the product of two functions, C = 1 [!!!:___ { (¢(t)) n t( ¢(t)r1 d¢(t) }] n,r n(r 1)! dtn t dt t=O
[!!!:___ {tn+1(¢(t))n+r1 d¢(t) }]
1
=
n(r 1)! dtn
dt
. t=O
In particular, for r = n, Cn
,n
= _!_
[!!!:___ {tn+l(¢(t))1 d¢(t) }]
n!
dtn
= _!_
t=O
[!!!:___ {tn+l dlog¢(t) }]
n! and so, setting x 0 it follows that
dt
dtn
dt
t=O
= ¢ 1 , Xi = ¢1 1¢i+t/ (i + 1), i = 1, 2, ... , and using (11.23),
1 [ dn { t n+l dlog(xot) cn,n _ n! dtn dt
+ t n+l dtd log
( 1 + Loo x·ti) }] 'i! i=l t=O
~
1 [ dn { n n+l d ( ) tJ }] _ n! dtn t +t dt~Lj Xt,X2,··· ,Xj Jl t=O
Jl 00
= n!1
[ dn { dtn
tn
+ ~ Lj(Xt, X2, ...
,
tn+j }] Xj) (j _ 1)!
= 1. t=O
Jl
Also, for r < n,
_
1
Cn,r   n(n r)(r 1)!
[ dn { n+l d (,!.,( ))n+r }] dtn
t
dt
'1'
t
t=O
11.6. INVERSION OF POWER SERIES
437
and so, setting Xo = ¢1, Xi = ¢1 1
I
Therefore Cn,r = On,r and the proof of (Il.4I) is completed.
COROLLARY 11.2 Let ¢(t) be power series ( 11.38), which has no constant tenn, and ¢ 1(u) its inverse power series ( 11.39). Then the nordercoefficient'l/Jn,k• n = k, k of the kth power of ¢ 1 (u), oo
+ 1, ...
n
"""' u = [
= 1,2, ... ,
is given by
1/Jn,k
= k(n 1)k_1 C (
In particular (fork = 1),
1/Jn
PROOF
.
(Il.43)
According to ( II.38),
¢(t))n ( t
.!h._ ~) 1/Jr cn 1,n (.!h._ 2¢1' 3¢1 ' ... 'n¢1
= __!:__
( =
00
tn1)n
~
1 ( =
00
1
+~
(r
¢r+ 1 tr)n + 1)¢1 · r!
and so, using (11.30),
¢(t)) n 1 ~ (
438
PARTITION POLYNOMIALS
Hence dnk [ dtnk
(¢(t)) nl t
= _!_C
¢~ nk,n
u=O
(.!h._
.!h._ f/Jnk+l ) 2¢1' 3¢1 '· .. ' (n k + 1)¢1
and by (11.41) we get '1/Jn,k
d" 1 = [ ~(¢ (u)) k] U
=
u=O
k(n 1)k_1
¢~
Cnk,n
( ¢2 ¢3 f/Jnk+1 ) 2¢1' 3¢1 ' ... ' (n k + 1)¢1 '
I
which is the required expression (11.42).
COROLLARY 11.3
If
with s a positive integer, then 1
¢ (u)
=u
1 + ~ Yn oo
(
:! , sn)
where Yn = Cn,sn1(X1,X2,··. ,Xn)/(sn + 1) n
= L( 1)k(sn + k)k1Bn,k(X1, X2, ... , Xn)·
(11.44)
k=1
PROOF The inversion formula of Lagrange ( 11.41 ), in the particular case of k = 1 and for sn + 1 instead of n, reduces to
n! Yn = (sn
+ 1)!
n! = (sn
+ 1)!
[ d•n+1
1
]
du•n+I ¢
[ d"n dt•n
(u)
(¢(t))
u=O 1
•n ]
t
t=O
~n!:[:.: (1 + f: t;~) •n(sn + 1)! Xr
r=
1
1 ]
t=O
11.6. INVERSION OF POWER SERIES
439
Thus, using (11.30), it follows that
Yn= (sn~ 1 )! [::s:~Cj,sn1(x1,x2, ... ,xn)~:]
,
1
t=O
yielding the first part of (11.44). Further, from (11.29) and since
(sn 1)k sn + 1
=
(sn 1)(sn 2) · · · (sn k) sn + 1 (1)k(sn + k)k1,
I
the second part of ( 11.44) is readily deduced.
In the following theorem two useful forms of Lagrange formula are derived.
THEOREM 11.12 Let f(t) = 2::~ 0 fktk /k! and u
J(t) = /(0)
= ¢(t) = L~o aktk jk!, with ao =/: 0.
+~
[~~11
(gn(t)
Then
d~~)) L=O. :~
( 11.45)
and
d(t))J(t) ( 1 U_!!__d t with u
1
00
=L n=O
[dn d n (gn(t)J(t)) ]
t
t=O
n · :!:.._,
= tjg(t).
PROOF Introducing into the expansion of f (t) the inverse series t and expanding the resulting expression into powers of u, we have
Using (11.41) and putting u
f(t) = /(0)
~ ~
00
 /(O)
+L
~1
{
= ¢ 1 ( u)
= ¢(t) = tj g(t), we get
(n1) [dkdtkf(t)] . [dnkgn(t)] . un kdtnk t=O t=O 1) [dnkgn(t)] . [~ df(t)] } . Un Ln (nk 1 dtnk dtk1 dt
+~~
k=1 n=k _
(11.46)
n!
~1
n!
1
~
~
n!
PARTITION POLYNOMIALS
440
and, according to the formula of the (n 1)th order derivative of the product of two functions, ( 11.45) is deduced. Differentiating formula ( 11.45) with respect to u, we get
Further,
df(t) du
= df(t) dt
I
du dt
and, from ug(t) = t, by differentiation with respect tot, it follows that
+ u dg(t) = 1.
g(t) du dt
dt
Therefore
df(t) du
= g(t) df(t) dt
(1 u dg(t)) dt
1
and
g(t)df(t) ( 1 udg(t))dt dt
1
=
f
[!£... (gn+ 1(t)df(t))] un dt t=O . n! .
n=O dtn
I
Replacing g(t)df(t) / dt by f(t), (11.46) is deduced.
Example 11.7 Determine the inverse power series of (a) ¢(t) (a) By (11.41), with k = 1, it follows that
[.!E__¢1(u)J dun
=[~(¢(t))n] t dtn 1
= tet and (b) 1,/!(t) = t(1 + t) r.
1
u=O
t=O
1 =[dn ent] dtn 1
=nn1 t=O
and so the inverse of ¢(t) = tet is the power series
(b) Again by (11.41), with k = 1, it follows that
~
[
dun 1,/1
_ 1
]
[
(u) u=O =
dn1 dtn1 dn1
= [d
t
n 1
(1,/l(t)) nl t
(1
+ tYn
t=O
] t=O
= (rn)n1
441
11.6. INVERSION OF POWER SERIES
and so the inverse of 1/J(t)
= t(1 + t)r is the power series
Example 11.8 Determine the factorial moments of the sequence of probabilities Pk = eAk
(.\k)k1 k!
, k = 1, 2, ... , 0
< ,\ < oo.
The generating function of this sequence of probabilities may be obtained as
where
is the inverse power series of <jJ(t) = tct (see Example 11.7). The generating function of the factorial moments JL(n)• n = 1, 2, ... , oo
B(z)
n
= 1 + LJL(n)~, n. n=1
is connected with the generating function P(z) by B(z) = P(1
+ z) and so
¢ 1 (u) is the inverse of u = <jJ(t) = tet, it follows that (1 + z)Xe>.. = tet and z = (t/X)e>..[(t/>..) 1]  1. Putting w = (t/X) 1 and f(w) = (1 + w)e>..w  1, we have z = f(w) and
Since t =
where w = / 1 (z) is the inverse of z = f(w) = (1
r
oo 1 (z)
+ w)e>..w n
" ' Jl(n)' z = 1 + 'L...., n=1
n.
 1. Therefore
PARTITION POLYNOMIALS
442
and n
oo
f(w) = (1
+ w) L( 1)n_xn; 1 n.
n=O n
oo
oo
m+l
= "'( 1)n_xn~ + "'( 1)n _xn~ ~ n! ~ n! n=I
n=O
n
oo
= "'(1tl(n A)AnI~. ~ n!
n=I
The coefficients are obtained as /l(n) =
Jl(n)•
n = 1, 2, ... , of w = f 1 (z), using (11.43) and (11.29),
nI (1)nI(n)k_xnI _ ,X)n+k 1 k=O
L
(
In particular, for n Jl
=
BnI,k(1
A/2, .X/3, ... , 1 Ajn).
= 1, 2, /l(l)
= (1 .X) 1 ,
fl(2)
= .X(2 .X)(1 .X) 3
and
11.7
TOUCHARD POLYNOMIALS
=
DEFINITION 11.6 The polynomial Tn,k Tn,k(XI, X2, ... , Xn; YI,Y2, ... , Yn) in the variables XI, x2, . .. , Xn and y 1 , y 2 , . .. , Yn defined by the sum
where the summation is extended over all nonnegative integer solutions (k1, k2, ... ,kn,ri,r2,··· ,rn) oftheequations n
n
i=l
i=l
L i(ki + ri) = n, L ki = k, is called Touchard polynomial.
(11.48)
443
11.7. TOUCHARD POLYNOMIALS
Note that
and
Tn,k(XI,X2,··· ,xn;O,O, ... ,0) = Bn,k(XI,X2,··· ,xn), where Bn(YI,Y2, ... ,Yn) and Bn,k(XI,x2, ... ,xn) are the exponential and the partial Bell partition polynomials, respectively. Also
Tn,k(abxi,a 2bx2,··· ,anbxn;ayi,a2y2, ... ,anyn)
= anbkTn,k(XI,X2,···
,xn;YI,Y2,··· ,yn)·
Further, summing the Touchard polynomials fork= 0, 1, ... , nand setting Si = ki + ri, i = 1, 2, ... , n, it follows that n
LTn,k(xi,X2,··· ,xn;YI,Y2···· ,Yn) k=O
L =
fi k7'~:o
n! SI !s2! · · · sn! i=I
(Si) (Xi)k; (Yi)s;k; ki i! i! 81
="'"' n! (XI+ YI ) L..,.s1!s2!···sn! 1!
82
(X2 + Y2) 2!
•••
(Xn + Yn) n!
Sn'
where the summation is extended over all partitions of n, that is, over all nonnegative integer solutions (SI, s 2, ... , sn) of the equation si + 2s2 + · · · + nsn = n. Hence n
LTn,k(Xt,X2,··· ,xn;YI,Y2,··· ,Yn) k=O
=Bn(Xt +Yt,X2+Y2,··· ,xn+Yn), n=0,1, .... This relation may also be deduced from the bivariate generating function T(t, u) of the Touchard polynomials (see Theorem 11.13 that follows) by putting u = 1 and comparing the resulting generating function with the generating function of the exponential Bell partition polynomials (11.9). REMARK 11.6 The coefficients of the Touchard polynomials (11.47) are positive integers. Specifically,
ktl(1!)k'k2!(2!)k 2
• • •
n! kn!(n!)knrtl(1!)T 1 r2!(2!Y 2
• •
·rn!(n!)Tn'
with E~=l i(ki + ri) = n and E~=l ki = k, equals the number of partitions {AI, A2, ... , Ak, BI, B2, ... , Br, ... } of a finite set Wn of n elements, where
444
PARTITION POLYNOMIALS
~ W n into k subsets, among which ki ~ 0 subsets includei elements each, i 1, 2, ... , n, and {Bt, B2, ... , Br, ... } is a partition of the set Wn U into subsets, among which ri ~ 0 subsets include i elements each, i = 1, 2, ... , n. I
{ A 1 , A 2 , •.• , Ak} is a partition of a set U
=
Bivariate and vertical generating functions for the Touchard polynomials are derived in the following theorem. THEOREM 11.13 (a) The bivariate generating function of the Touchard polynomials, Tn,k _ Tn,k(XI,X2, ... ,xn;YbY2, ... , Yn), k = 0, 1, ... , n, n = 0, 1, .. . , with To,o 1, is given by
=
oo
~
n
T(t,u) = LLTn,k(Xt,X2, ... ,xn;Yt,Y2, ... ,yn)uk n! n=Ok=O = exp{u(g(t) xo] + (h(t) yo]}.
(11.49)
(b) The vertical generating function of the Touchard polynomials Tn,k = k, k + 1, ... ,for fixed k, is given by
=
Tn,k(Xt, x2, ... , Xn; Yt, Y2, ... , Yn), n
oo tn Tk(t) = L:Tn,k(Xt' X2, ... 'Xn; Yt' Y2, ... 'Yn) n! n=k [g(t) xo]k = k! exp(h(t) yo], k = 0, 1, ... ,
( 11.50)
where (11.51)
PROOF (a) Multiplying (11.47) by uktn jn! and summing the resulting expression fork = 0, 1, ... , nand n = 0, 1, ... , it follows that oo
T(t,u) =
~
n
L LTn,k(Xt,X2, ... ,xn;Yt,Y2, ... ,yn)uk n! n=Ok=O oo
n
L
= L L n=Ok=O X (
1
kt !k2! .. ·
kn~~l !r2! .. · rn!
X~~t) k, ... ( X:~tn) kn
( y;n r, ... (
y~~n
rn '
where the summation in the inner sum is extended over all nonnegative integer solutions (k~, k2, ... , kn, r1, r2, ... , rn) of 2::~ 1 i(ki + ri) = nand 2::~ 1 ki = k.
445
11.7. TOUCHARD POLYNOMIALS
Since this inner sum is summed over all k = 0, 1, ... , n and n = 0, 1, ... , it follows that the summation is extended over all ki = 0, 1, ... , ri = 0, 1, ... , for i = 1, 2, . . . . Therefore,
00
=IT exp i=l
(
Xi~ z.
ti
ti) +~ .
z.
( 'L:Xi:;ti + 00
= exp
u
i=l
z.
00
ti)
LYi7j
z.
i=l
.
Using (11.51 ), the last expression implies (11.49). (b) Interchanging the order of summation ( 11.49) may be expressed as 00
T(t,u)
~
00
= LLTn,k(xi,X2,···
,XniYI,Y2···· 1 Yn)uk n!
k=O n=k
and, since T(t,u) = exp{u[g(t) x 0 ] + [h(t) y 0 ]}, it follows that
t k=O
Tk (t)uk = exp[h(t) Yo]
t
[g(t)
~ xo]k uk.
k=O
Equating the coefficients of uk in both sides of the last expression, (11.50) is deduced. I The Touchard polynomial is expressed as a finite sum of Bell partition polynomials in the following theorem. THEOREM 11.14 The Touchard polynomial Tn,k(XI,x2, ... 1 XniYI,Y2 ... ,yn) is expressed in terms of the partial Bell polynomials Br,k(x 1 , x 2 , .•• , Xr) and the exponential Bell polynomials Bnr(YI, Y2, ... , Ynr).for r = k, k + 1, ... , n, as
(11.52)
PROOF
Expanding the righthand side of (11.50) into powers oft, using (11. 7)
PARTITION POLYNOMIALS
446 and (11.9), we get the relation
which, by virtue of oo
Tk(t)
= LTn,k(X!,X2,···
tn ,XniYl,y2,··· ,yn)n!'
n=k
implies ( 11.52).
I
In the next theorem a recurrence relation for the Touchard polynomials is deduced. THEOREM 11.15 The TouchardpolynomialsTn,k Tn,k(Xl,x2,··· ,xn; Yl,Y2,··· ,yn), k 0, 1, ... , n, n = 0, 1, ... , with To,o 1, satisfy the recurrence relation
=
=
PROOF Differentiating the generating function ( 11.50) and expanding the resulting expression,
into powers oft, it follows that
Equating the coefficients oftn jn! in both sides of the last expression, (11.53) is deduced. I
11.8. BIBLIOGRAPHIC NOTES
11.8
447
BIBLIOGRAPHIC NOTES
Francesco Faa di Bruno (1855) expressed the derivatives of a composite function as a multi variable polynomial of the derivatives of the component functions. These polynomials are essentially partition polynomials. A history of the problem of deriving a general formula for the derivatives of a composite function was outlined by Eugene Lukacs (1955). The expressions of the cumulants in terms of the (power) moments, and vice versa, were also given in that paper (see Example 11.4). In the classical book of Charles Jordan (1939a) a brief sketch of proof of Faa di Bruno formula, using Taylor series, is quoted. Eric Temple Bell (1927), aiming to the unification ·of a variety of arithmetical functions, introduced and studied the partition polynomials. Further, Bell (1934a,b) examined additional arithmetical and other properties along with certain particular cases of these polynomials. It is worth noting that R. Frucht (1969), in a combinatorial approach to the Bell polynomials, furnished a nice combinatorial interpretation of these polynomials; Example 11.2 was discussed in that paper. John Riordan (1958, 1968) named the partition polynomials after E. T. Bell and extensively studied them. The derivation of the expression of the derivatives of a composite function as partition polynomials of the derivatives of the component functions provided by J. Riordan (1958) is based on umbra! calculus. A combination of the derivation of Faa di Bruno formula provided by Ch. Jordan (1939a) and J. Riordan (1958) that avoids the use of umbra! calculus is adopted in this book (see Theorem 11.4). The reader interested in a rigorous umbra! calculus derivation of this formula is referred to the paper of Steven Roman (1980). Louis Comtet (1974) devoted much of the chapter on identities and expansions to a thorough presentation of the Bell polynomials and provided an updated and rich bibliography of the subject. The exponential, logarithmic and potential polynomials, which are the most striking partition polynomials, have a direct use in combinatorics in expressing the elementary symmetric functions and the homogeneous product sum symmetric functions in terms of the power sum symmetric functions and vice versa. Examples 11.5 and 11.6, in which such expressions are derived, were taken from P. A. MacMahon (1915, 1916). Their use in expressing the probability function and factorial moments of a compound discrete distribution in terms of the probability functions and factorial moments of the component distributions was demonstrated by Ch. A. Charalambides (1977b ). Apart from the classical convolution formula, G. P. M. Heselden (1973) derived an interesting convolution formula for the exponential Bell partition polynomials; Exercise 2 is based on this paper. The bipartitional polynomials and their applications in combinatorics and statistics, which
448
PARTITION POLYNOMIALS
appear in the exercises, were examined by Ch. A. Charalambides (1981). The derivatives of a bivariate composite function may be expressed as bipartitional polynomials of the derivatives of the component functions. L. Comtet (1968) expressed the derivatives of an implicit function in terms of certain bipartitional polynomials.
11.9 EXERCISES 1. Convolution of Bell polynomials. Let Bn(x 1 , x2, ... , xn) be the exponential Bell partition polynomial. Show that Bn ( X1
=
+ Yr, X2 + Y2, · . ·
t (~)
, Xn
+ Yn)
Bk(xr,x2,··· ,xk)Bnk(Yr,Y2,··· ,Ynk)·
k=O
2*. Heselden's convolution of Bell polynomials (Heselden, 1973). For any real or complex number a such that a k ::/: 0, k = 0, 1, ... , n, show that
~a~k (~)Bk((ka)zr,(ka)z2,··· ,(ka)zk) xBnk((a k)z1, (a k)z2, ... , (a k)znk) = cl"n,O·
Using this relation, derive the following convolution formula
~a~ k ( ~) Bk((k a)z1, (k a)z2, ... , (k a)zk) xBnk(Yr kzr,Y2 kz2,··· ,Ynk kznk)
= Bn(Yr azr,Y2 az2, ...
,yn azn).
Further, show that
t a~ ~) = t a~ (~)Bk(xr k (
Bk(xr
+ kz1,x2 + kz2, ...
,xk
+ kzk)
k=O
xBndYr  kzr, Y2  kz2, ... , Yyk  kznk) k
+ azr,X2 + az2, ...
,xk
+ azk)
k=O
XBnk (Yt  az1, Y2  az2, ... , Ynk  aznk)·
11.9. EXERCISES
449
3. Hermite and Laguerre polynomials. Show that the exponential Bell partition polynomial Bn(x 1 ,x2,··· ,xn) in the particular case of x 1 = x, x2 = 1, Xr = 0, r = 3, 4, ... , reduces to
Bn(x, 1) =in Hn(ix), i = where
H (x) n
H,
[n/ 2]
= (1)nex2f2Dnex2f2 = n''"' · ~ k=O
x
( 1)k n2k
X (n 2k)!k!2k
is the Hermite polynomial. Also, show that Bn (x 1 , x2, .. . , xn) in the particular case of Xr = r!xr+ 1 , r = 1, 2, ... , reduces to
1 n+l) _ (n _ 1)1.X n+1£(1) B n (1 1 ., 21.X 1 , ... , n.X n 1 (  X) , where
is a Laguerre polynomial. 4. Generalized Hermite polynomials (Bell, 1934b). Show that the exponential Bell partition polynomial Bn(x 1 , x2, ... , xn) in the particular case of Xr = (s)raxsr, r = 1,2, ... ,s, Xr = 0, r = s + 1,s + 2, ... , with sa positive integer, reduces to
where
H n (x; a, s )
= e ax•Dnxe ax•
is a generalized Hermite polynomial. Further, show that n
H (x· a s)  '"'C(n k· s)akxskn n
''
~
''
'
k=D
where
C(n, k; s)
k = k!1 ~( 1r (k)r (sr)n
is the generalized factorial coefficient. 5. Chebyshev polynomials of the first kind. Show that the logarithmic partition polynomial Ln(x 1 , x2, ... , xn) in the particular case of x 1 = x, x2 = 1, Xr = 0, r = 3, 4, ... , reduces to
Ln(x, 1)
= Tn(x)jn,
450
PARTITION POLYNOMIALS
where [n/2]
(
Tn(x)=n!L:>1)kn:k
n~
k)
k=O
T
n2k
is the Chebyshev polynomial of the first kind. 6. Gegenbauer polynomials. Show that the potential partition polynomial Cn,s(XI,X2, ... ,xn) in the particular case of X1 = x, X2 = 1, Xr = 0, r = 3, 4, ... , reduces to
Cn,s(x, 1)
= G~s)(x),
1/2 < s < 0 or 0 < s < oo,
where
is the Gegenbauer polynomial. Note that Gh / ) (x) = Pn(x) is the Legendre 1 polynomial and Gh )(x) = Un(x) is the Chebyshev polynomial of the second kind. Also lims>o s 1nGh8 )(x) = Tn(x) is the Chebyshev polynomial of the first kind. 1 2
7. Let Bn,k(x 1, x 2, ... , Xn) be the partial Bell partition polynomial. Show that, for r = 1, 2, ... ,
with Yi
= Xr+if(r + i)r, i = 1, 2, ....
Further, show that
Bn,k (0, · .. , 0, Xr+l, ... , Xn) k
~"'( 1)J j!(r!)i (n)rJ B nrj,kj (O , ... , o,Xr, ... ,XnrjXr )J )=0
and
Bn,k (0, ... , 0, Xr. ... , Xn) (n)rj ( )i ( =~ ~ ;. Bnrj kj 0, .. · , 0, Xr+I,. · · , Xnrj :Lr· 1 . J.1 r.)J , J=O
8. (Continuation). Show that
2,
') _
1)
B n,k (1.,1 ., ... 'n.  n!  1 k! ( nk
11.9. EXERCISES
451
and
9. (Continuation). Show that
1)
( )I) _ (2r)! ( r B 2r,k (0 ' 21 ., ..• '0, 2r .  k! k  1 B2rl,k(O,
'
2!, ... , (2r 2)!, 0) = 0
and
B2r,2s(l.,1 0, ... , (2r
_
_(2r)!(r+s1) 1).,1 0)  ( s)! , 28 _ 1 2
B2r,2s1 (1!, 0, ... , (2r 1)!, 0) = 0, (2r1)! (r+s2) ' B2rl 2st(1!, 0, ... , (2r 1)!, 0) = ( )I 2s 2 2s 1 . ' B2r1,2s(1!, 0, ... , 0, (2r 1)!) = 0. 10. Let Cn,s mial. Show that
(
=
Cn,s(Xt,
n+s)c n n,s
x2, ... , Xn) be the potential partition polyno
=~(n+s) L....t n + r
(ns) n  r (n+r)c n n,r
r=O
with s any real (or complex) number. 11 *. (Continuation). Setting Xt
1! 1
Yn
X2 2!
X3
Xt
X2 2!
1!
3!
Xt
Yn
= Cn, I( x1 , x2, ... Xn2
Xn3
Xn2
Xn
n! Xn1
(n 3)! (n 2)! (n 1)! Xn4
1
0
0
0
1
0
0
0
0
= (1)nn!
Xn1
(n 2)! (n 1)!
0
1!
, Xn), show that
Xn3
Xn2
(n 4)! (n 3)! (n 2)! X!
1! 1
X2 2! Xt
1!
PARTITION POLYNOMIALS
452
12. Let u(n)
=L
d be the sum of the divisors of n, n
= 1, 2, ....
Show
din
that
00
00
n=l k=l Further, show that the generating function of the number p( n) of partitions of n, which is given by
G(t)
00
00
n=O
k=l
= LP(n)tn = IT (1 tk) 1 ,
may be expressed as 00
G(t) = exp
(
tr) ~u(r);:
and thus conclude that p(n)
= Bn(o(1), 1!u(2), ...
, (n 1)!u(n))/n!.
13. Show that the factorial moments J.L(n), n of probabilities
= 1, 2, ... , of the sequence
_ k! 1 ( k rk Pk_ ) p k( 1 p )rkk+1 , k _ 1, 2, ... , 1
where r is a positive integer and 0 < p < 1/r orr is a negative real number and 1/r < p < 0, are given by pn1
J.L(n)
= (1 rp )n Cn1n(XI,Xz, ... '
where
x1
,Xn1), n= 1,2, ... '
= (r) 1 1(n+j)p/(j+1). , J = 1, 2, .... 1 rp
=
14. Laguerre polynomials. Show that the Touchard polynomial Tn,k Tn,k(x1,xz, ... ,xn;Yl,Yz, ... ,Yn) in the particular case ofxr = r!xr+l, Yr = r!xr+l, r = 1, 2, ... , reduces to 1 n+1.11 1 n+1) _ T n,k (1 1 ., 21.X 1 , ... ,n.X ' ., 21.X 1 , ... ,n.x for n
= k, k +
1, ... , k
= 1, 2, ... , where
nl L(k 1) (x)
· k!xnk nk
'
11.9. EXERCISES
453
is the Laguerre polynomial.
15. Let Tn,k (xi, X2, ... , Xn; Yl, Y2, ... , Yn) be the Touchard polynomial. Show that, for r = 1, 2, ... ,
Tn,k(O,O, ... ,O,xr+l,Xr+2,··· ,xn;YI,Y2,··· ,Yn) n! ( )I Tnrk k(ZI, Z2, · · · , Znrk; Yl, Y2, · · · , Ynrk), n rk. ' with Zi = Xr+i/(r
+ i)r, i
= 1, 2, .... Further, show that
Tn,k(O, ... ,O,xr+l,··· ,xn,;YbY2,··· ,Yn)
~
· (n)rj
·
= L..,..(1)':( ).Tnrjk;(0, ... ,O,Xr,··· ,Xnrj;YI,Y2,··· ,Ynrj)X~ J.1 r! J , j=O
and
Tn,k(O, ... ,O,xr,··· ,xn,;YI,Y2,··· ,yn) k
 ~ (n)rj . ·(0 0 .. ·) j  L..,.. j!(r!)i TnrJ,kJ ' ... ' 'Xr+l' ... 'XnrJ' Yl' Y2, ... 'YnrJ xr. J=O
16. (Continuation). Show that
1
'2'., ... 'n.,'·Of.r, 1'.r, ... ' (n 1)')n!(n+rT n,k (1 ., .r  k! k + r  1 ) and
Tn,k(1, 2, ... , n; r, 0, ... , 0) = (
~)
(k
+ r)nk.
17. Let f(u), 9(t) and h(t) be functions of real variables for which the derivatives
9r =
[d'it~t)]
[drdht~t)]
' hr = t=a
' fr = t=a
[dr1~u)] U
' u=g(a)
for r = 0, 1, ... , exist. Show that the derivatives of the composite function 1/J(t) = f(9(t)) exp[h(t) h(a)],
1/Jn
= [~d1jJ~t)] t
, n
= 0, 1, ...
,
t=a
are given by n
1/Jn =
L
fkTn,k (91, 92, · · · , 9n; h1, h2, · .. , hn)
k=O
= Bn(/91
+ h1, /92 + h2, · · ·, f9n + hn), fk:::::: fk·
454
PARTITION POLYNOMIALS
=
18*. Exponential bipartitional polynomials. The polynomial Am,n Am,n(xo,t,Xt,a, ... ,xm,n) in the variables Xo,t,Xt,o, ... ,xm,n defined by the sum
where the summation is extended over all partitions of the bipartite number (m,n), that is, over all nonnegative integer solutions (k0 , 1 ,k1 ,0 , ... ,km,n) of the equations m
n
n
L:i:L:k;,j i=l j=O
m
=m, L:i:L:ki,j =n, j=l
i=O
is called exponential bipartitional polynomial. The polynomial Am,n;k _ Am,n;k(xo,l, Xt,o, ... , Xm,n) in the variables xo,l, Xt,o, ... , Xm,n of degree k, defined by the sum Amn·k ' '
=""' ~
m!n! (Xo,l)ko,l (Xl,O)kl,O ... (Xm,n)k~,n ko,t!kt,o! · · · km,n! 0!1! 1!0! m!n! '
where the summation is extended over all partitions of the bipartite number (m, n) into k parts, that is, over all nonnegative integer solutions (k0 , 1 , k1 ,0 , ... , km,n) of the equations m
n
n
m
i=l
j=O
j=l
i=O
n
m
2::: i 2::: k;,j = m, 2::: j 2::: k;,j = n, 2::: L:ki,j = k, j=Oi=O i+#O
is called (exponential) partial bipartitional polynomial. Show that
Am,n;k(bcxo,l, aCXt,o, ... , ambnCXm,n) =ambnckAm,n;k(xo,l,Xl,O,··· ,Xm,n),
and m+n Am,n(Xo,t,Xl,O, ... ,Xm,n) =
L Am,n;k(Xo,l,xl,O,··· ,Xm,n)·
k=l
Further, show that 00
A(t, u) =
00
LL n=Om=O
tm un Am,n m! ~ = exp[g(t, u)  xo,o],
11.9. EXERCISES
455
oo
A(t, u, w)
oo
=L
m+n
LL
tm n Am,n;kWk m! ~!
= exp{ w[g(t, u) xo,o]}
n=Om=O k=O
and
A (t u) =~~A . tm un k , L...t L...t m,n,k 1 1 n=Dm=O m. n.
= [g(t,u)xo,o]k kl , '
where 00
g(t,u)
=L
tm un
00
L Xm,n11· n=Om=O m. n.
19*. (Continuation). Show that
and
tf (7) (;)
Am+l,n;k+l =
Xi+l,jAmi,nj;k·
J=O •=0
20*. General bipartitional polynomials. The polynomial Pm,n :::::: Pm,n(xo,I, Xt,o, ... , Xm,n) in the variables xo,I, Xt,o, ... , Xm,n and parameters Ct, C2, . . . , Cm+n, defined by the sum
Pmn '
=L
1)ko,t
m!n! (Xo Ck 'ko,t!kt,o! · · · km,n! 0!1!
(Xl,O 1!0!
)kt,O ... (Xm,n)km,n , m!n!
where k = k0,1 + k 1 ,o + · · · + km,n and the summation is extended over all partitions of the bipartite number (m, n), that is, over all nonnegative integer solutions (k0 , 1 , k 1 ,o, ... , km,n) of the equations m
n
n
m
L:iL:ki,j =m, L:iL:ki,j =n, i=l j=O j=l i=O is called bipartitional polynonial. Note that it can be expressed by the exponential bipartitional polynomial Am,n(xo, 1 ,xt,o, ... ,xm,n) as
Pm,n(Xo,I' XJ,O, ... 'Xm,ni Ct' c2, ... 'Cm+n) = Am,n(Cxo,l' CXt,o, ... 'CXm,n), ck
=Ck·
Show that
m+n Am,n(Cxo,l, CXt,o, ... , CXm,n) =
L
k=O
CkAm,n;k(Xo,l, XJ,o, ... , Xm,n)
PARTITION POLYNOMIALS
456
and 00
A(t,u;c1,c2,···) =
tm un
00
'2:: '2:: Am,n(cxo,l,cxl,O,··· ,CXm,n)m! n! n=Om=O
~ [g(t, u) xo o]k k = ~ Ck k! ' = exp{c[g(t, u) xo,o]}, c
k=O where 00
g(t,u) =
=
Ck,
tm un
00
'2:: '2:: Xm,n;1· m.n.
n=Om=O Further, show that
m+n hm,n =
2::
fkAm,n;k (Yo,l, Y1,0, · · · , Ym,n) k=O = Am,nUYo,l, fgl,O, · · · , fgm,n), Jk
=fk,
where
hm,n
=[
8m+nh(t, u)] 8 n8tm U
t=a,u=b
' Ym,n
= [8m+ng(t, u)] 8 n8tm
t=a,u=b
U
'
and
f _ [dk f(w)] k
dwk
w=g(a,b) .
21 *. (Continuation). Compound bivariate discrete distributions. Consider a discrete distribution with probability function Pk, k = 0, 1, ... and probability generating function f( w) = ~':=o PkWk. Further, consider a bivariate discrete distribution with probability mass function Qi,j, i = 0, 1, ... , j = 0, 1, ... and probability generating function g(t, u) = ~;: 0 ~:o Qi,JtiuJ. Then the distribution with probability generating function the composite function h(t, u) = f(g(t, u)) = ~~=O ~:'=o Pm,ntmun is called compound bivariate distribution. Show that
Pm,n
=
1 m.n. 1 1
m+n
2:: fkAm,n;k(Qo,l,Ql,O,··· ,Qm,n)
k=O
1  Am,nUQo,l, fql,O, · · · , fqm,n), = m.n. 1 1
J
k
= fk
with fk = ~:k(i)kPiQtok, and
m+n 1 M(m,n) =  1 1 ll(k)Am,n;k(J.t(O,l),J.t(l,O)' · · · ,J.t(m,n)) m.n. k=O
L
1
k 
= m!n!Am,n(IIJ.t(O,l),IIJ.t(l,O),··· ,IIJ.t(m,n)),v = ll(k)
11.9. EXERCISES
00
M(m,n)
457
00
00
=L
L(i)m(j)nPi,j, J.l(m,n) j=ni=m
=L
00
L(i)m(j)nQi,j·
j=n i=m
22*. (Continuation). Logarithmic biparlitional polynomials. These polynomials constitute a particular case of the bipartitional polynomials: Lm,n(Xo,I,XI,o,XI,I,··· ,xm,n) = Am,n(cxo,I,CXI,O,··· ,cxm,n), ck with
eo = 0,
=
Ck,
Ck = ( 1)kI (k 1)!, k = 1, 2, .... Show that 00 00 tm un L(t, u) = """' """'Lm n(Xo I, XI o, ... , Xm n) ~~ m,n, 1 1 n=Om=O m+ni"'O = log{1 + [g(t, u) xo,o]}, I
where
I
I
I
00 00 tm un g(t,u) = L L Xm,n;1· n=Om=O m. n.
Further, show that Lm+I,n = Xm+I,n
~ ~ (;)
(
7)
Xi,jLmi+I,nj, LI,O = XI,O
i+#O
and Ym,n = Am,n(Xo,I, XI,o, · · · , Xm,n),
if and only if Xm,n = Lm,n(Yo,I, YI,O, · · · , Ym,n)·
23*. (Continuation) Cumulants and moments of bivariate distributions. Let oo tm n L J.L'm,n,;, n=O m=O m. n. oo
M(t,u)
=L
oo tm n L 1\;m,n;; n=O m=O m. n. oo
K(t,u)
=L
be the generating functions of the moments J.L'm n' m, n = 0, 1, ... and the cumulants ~~;m,n, m, n = 0, 1, ... , respectively, ~f a bivariate distribution. Using the relation K(t, u) =log M(t, u), show that
and conclude that
PARTITION POLYNOMIALS
458
24*. (Continuation). Potential bipartitional polynomials. These polynomials constitute a particular case of the bipartitional polynomials:
Cm,n,s(Xo,l,Xl,O,··· ,Xn,r) =Am,n(cxo,l,CXl,O,··· ,cxm,n), ck =:ck, with Ck
= (s)k, k = 0, 1, ....
Show that
00 00 tm un c.(t, u) = ""' ""'Cm n .(xo 1' X} o, ... 'Xm n), , .L...t ~ ' ' ' ' ' m. n. n=Om=O = [1 + {g(t,u) xo,o}]",
where
00 tm un L Xm,nll' m. n. n=Om=O 00
g(t,u) Also, with Cm,n,s
Cm+l,n,s
=L
= Cm,n,s(Xo,l, X1,o, ... , Xm,n),
t f (]) (7) tf (~) (7)
=S
show that
Xi+l,jCmi,nj,s
J=O •=0
Xi,jCmi+l,nj,s, Co,o,s = 1.
J
j=Oi=O i+#O
=
25*. Symmetric functions. The symmetric functions ak,r ak,r(x 1, x2, ... ,Xn,Yl,Y2, .. · ,yn), Sk,r =: Sk,r(XI,X2, ... ,Xn,Yl,Y2, ... ,Yn), hk,r =: hk,r(x1,x2, ... ,xn,Yl,y2, ... ,yn), with respect to the variables x 1,x2, ... , Xn and with respect to the variables y 1 , y 2 , ••. , Yn, have generating functions: n nr
=L
n
=II
L ak,rtkui (1 + xit + YiU), i=l k=O oo oo tk r n S(t, u) = L L(k + r  1)!sk,r k! ~! = log (1 Xit Yiu), r=Ok=O i=l k+r;iO A(t, u)
r=O
II
oo
H(t, u)
=L
oo
L hk,rtkur r=Ok=O
n
=II (1 Xit y;u)
1
.
i=l
Show that (1)k+r k!r! Ak,r(so,l,sl,o, ... ,(k+r1)!sk,r) (l)k+r k!r! ck,r,1 (ho,l, ht,o, ... 'k!r!hk,r ),
11.9. EXERCISES
459 (1)k+r1 )! Lk,r(ao,1, a1,o, ... , k!r!ak,r) 1
Sk,r = (k + r _
(k
+: _
1)! Lk,r(ho,1, h1,o, ... , k!r!hk,r)
and hk,r =
( 1)k+r k!r! Ck,r,1 (ao,1, a1,0, · .. , k!r!ak,r) 1
= k!r!Ak,r(so,1,SI,o, ... ,(k+r1)!sk,r).
Chapter 12 CYCLES OF PERMUTATIONS
12.1
INTRODUCTION
The permutations of a finite set of n elements may be distinguished and enumerated according to certain characteristics they may possess. Thus, in Chapter 5, the enumeration of the permutations of n with a given number of fixed points and the permutations of n of a given rank was carried out. Also, the (linear) permutations of n and the circular permutations of n with a given number of successions of points were enumerated. Further, the permutations of a finite set of n elements may be distinguished according to their decomposition into disjoint cycles. A host of interesting enumeration problems emerges from this consideration. The present chapter is devoted to the enumeration of the permutations of a finite set of n elements that are decomposed into cycles of specified or unspecified number, whose number of cycles of any specific length belongs to a subset of nonnegative integers. Specifically, after the introduction of the necessary basic notions, the permutations that are decomposed into a given number of cycles are enumerated. In particular, a combinatorial interpretation of the signless Stirling numbers of the first kind is provided. Then, the permutations are classified into even and odd, according to the number of transpositions into which they can be decomposed. The enumeration of the even and odd permutations by cycles is reduced to the preceding problem of enumeration of the permutations by cycles. Also, considering the set of possible orderings of the elements of a cycle, the cycles are distinguished by the subset of orderings allowed for their elements. The enumeration of the permutations with a given number of partially ordered cycles is carried out. It is worth noting that the problem of enumeration of the permutations of a finite set by the number of the totally ordered cycles, into which they can be decomposed, is merely a problem of enumeration of the partitions of a finite set by the number of their subsets.
462
12.2
CYCLES OF PERMUTATIONS
PERMUTATIONS WITH A GIVEN NUMBER OF CYCLES
A permutation of a finite set W n = {WI, w2, ... , Wn} of n elements, as defined in Chapter 2, is an ordered ntuple (Wi 1 , Wi 2, ... , Win) with Wi. E W n, r = 1, 2, ... , nand wi. ::/: wi. for r ::/: s. Such a permutation can be equivalently considered as a rearrangement of a fixed ordering. The replacement of the fixed ordering (WI, w2, ... , Wn) by (wi,, Wi 2, ... , Win) is denoted by (12.1)
which defines the rule (mapping) O'(wr) = Wir of the replacement of the point Wr by the point Wir, r = 1, 2, ... , n. Recall that a point Wr for which Wir = Wr is called fixed point of this permutation. The columns in (12.1) corresponding to the fixed points may be omitted. Consider a subset wk = { Wi, 'Wi2' ... 'Wik} of the set Wn· The permutation (12.2)
in which the replacements 0'( wiJ = Wi 2 , 0'( Wi 2) = Wi 3 , ••• , 0' (wik_,) = wi~, O'(Wik) = wi, close a cycle is called cycle. The notation (12.2) of a cycle may be abbreviated by writing in parentheses the rule of succession of the permuted (nonfixed) points, (12.3)
The number k of elements of cycle (12.3) is called length of it. The representation (12.3) of a cycle is not unique. The k permutations
which are formed by putting as first any of the k elements and keeping the same rule of succession, correspond to (represent) the same cycle. It is conventional to write a cycle with its smallest element in the first position, making its representation (12.3) unique. Evidently, there are (n)k/k cycles of length k. A cycle of length n, which includes all the elements of the set Wn, is particularly called circular {cyclic) permutation. Thus, there are (n)n/n = (n  1)! circular permutations of the set Wn. A transposition is a cycle of length k = 2. The number of trans.positions of the set Wn equals n(n 1)/2. Clearly, any permutation of the set Wn is either a cycle or can be decomposed into disjoint cycles. This decomposition is unique up to a rearrangement of the cycles.
12.2. PERMUTATIONS WITH A GIVEN NUMBER OF CYCLES
463
Example 12.1 Consider the following permutation ofthe set W10 = {0, 1, ... , 9}:
0,1,2,3,4,5,6,7,8,9)· ( 2, 1,0,5,9, 7,8,6,3,4 Note that the element 1 is a fixed point of this permutation. Further, (0, 2) and (4, 9) are two cycles of length 2 and (3, 5, 7, 6, 8) is a cycle of length 5. Thus, this permutation may be expressed as a product of cycles as
(1 )(0, 2)( 4, 9) (3, 5, 7, 6, 8). The permutation
0,1,2,3,4,5,6,7,8,9) ( 3,9,5, 7,6, 1,8,2,0,4 is a cycle (cyclic permutation) (0, 3, 7, 2, 5, 1, 9, 4, 6, 8).
0
The permutations of a finite set of n elements may be distinguished according to their type of decomposition into disjoint cycles. Specifically, we introduce the following definition DEFINITION 12.1 A pennutation of a finite set Wn. ofn elements, which is decomposed into ki 2:: 0 cycles of length i, i = 1, 2, ... , n, so that ki + 2kz + · · · + nkn = n, is said to be of type [ki, kz, ... , kn]· Note that the number of different types in which then! permutations of n may be classified equals the number p( n) of partitions of n.
Example 12.2 The 24 permutationS of the Set W4 = {WI, Wz, W3, W4}, With WI < WI < W3 w4, written as products of cycles can be classified in the following types: (a) The only permutation of the type [4, 0, 0, OJ is
(b) The permutations of the type [2, 1, 0, 0] are the following six:
(wi )( wz) (w3, w4), (wt) (W3 )( Wz, W4), (wt) (w4 )( w2, w3),
(Wz )( W3 )(WI, W4), (Wz )( W4 )(WI, W3), (w3)( W4 )(WI, Wz). (c) The permutations of the type [1, 0, 1, 0] are the following eight:
<
CYCLES OF PERMUTATIONS
464
(w3)(w1, w2, w4), (w3)(w1, w4, w2), (w4)(w1, w2, w3), (w4) (w1, w3, w2). (d) The permutations of the type [0, 2, 0, OJ are the following three:
(e) The permutations of the type [0, 0, 0, 1] are the following six:
The next theorem is concerned with the enumeration of the permutations of a finite set by their decomposition into cycles. THEOREM 12.1 (a) The number ofpermutations ofa finite set ofn elements that are decomposed into k cycles, among which ki 2:: 0 are of length i, i = 1, 2, ... , n, equals (12.4)
with k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. (b) The number ofpermutations ofa finite set ofn elements that are decomposed into cycles, among which ki 2:: 0 are of length i, i = 1, 2, ... , n, equals (12.5)
with k1
+ 2k2 + · · · + nkn
= n.
PROOF (a) Consider a partition { A 1 , A 2 , •.. , Ak} of a finite set Wn, with N (Wn) = n, into k subsets, among which ki 2:: 0 subsets include i elements each, i = 1, 2, ... , n, where k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. To it there correspond
permutations of the set W n that are decomposed into k cycles, among which ki 2:: 0 are of length i, i = 1, 2, ... , n; these permutations are formed by permuting the i  1 elements of each of the ki 2:: 0 subsets with i elements (excluding the smallest element, which, by convention, occupies the first position) in [( i  1) !Jk• ways, i = 1, 2, ... , n. Therefore Cn,k(ki, k2, ... , kn) = [(1 1}!]k' [(2 1}!]k2
• • •
[(n 1}!]k" B(n, k; k1, k2, ... , kn},
12.2. PERMUTATIONS WITH A GIVEN NUMBER OF CYCLES
465
where B(n, k; ki, k2 , ... , kn) is the number of partitions of the set Wn into k subsets, among which ki 2:: 0 subsets include i elements each, i = 1, 2, ... , n, where ki + 2k 2 + · · · + nkn = nand ki + kz + · · · + kn = k. This number, according to Theorem 2.1 0, is given by
with ki + 2kz + · · · + nkn = n, ki + kz + · · · + kn = k and so (12.4) is deduced. (b) Clearly, the summation of the numbers Cn,k(ki, k2 , ... , kn) in (12.4)for all k = 0, 1, ... , n is equivalent to keep the total number of cycles ki + k 2 + · · · + kn unspecified and so the number of permutations of a finite set of n elements that are decomposed into cycles, among which ki 2:: 0 are of length i, i = 1, 2, ... , n, is given by (12.5). I The number of permutations of a finite set of n elements that are decomposed into k cycles may be obtained by summing the numbers (12.4) for all ki 2:: 0, i = 1, 2, ... , n, with ki + 2kz + · · · + nkn = n, ki + kz + · · · + kn = k. Then, by virtue of the expression (8.24) of the signless Stirling numbers of the first kind, we deduce the following corollary of Theorem 12.1.
COROLLARY 12.1 The number ofpennutations of a finite set of n elements that are decomposed into k cycles equals
ls(n, k)l, the signless Stirling number of the first kind. REMARK 12.1 The sum of the numbers Cn,k(ki, kz, ... , kn) in (12.5) for all ki 2:: 0, i = 1, 2, ... , n, with ki + 2k 2 + · · · + nkn = n, gives the total number n!, of the permutations of a finite set of n elements. Thus, we deduce Cauchy identity:
~
1
 1
~ ki !1 kJ kz !2k2 ... kn!nkn 
'
where the summation is extended over all integers ki 2:: 0, i = 1, 2, ... , n with ki + 2kz + · · · + nkn = n. I
Example 12.3 Restricted permutations with a given number of cycles Consider a finite set W n+r = {WI, Wz, ... , Wn+r}, with WI < Wz < · · · < Wn+r, and its subset Wr = {WI, Wz, ... , Wr}. Determine the number of permutations of the set Wn+r that are decomposed into k + r cycles such that the r elements of the set Wr belong in r distinct cycles.
CYCLES OF PERMUTATIONS
466
The k cycles of such a permutation rna y contain a total of j elements from the set Wn+r Wr = { Wr+l, Wr+2> ... , Wn+r }, ofn elements, for j = k, k + 1, ... , n. The j elements can be chosen in (n) different ways. Further, the number of permutations of a set of j elements that are decomposed into k cycles, according to Corollary 12.1, equals the signless Stirling number of the first kind ls(j,k)l. Also, the remaining n j elements of the set Wn+r Wr can be placed into the r distinct cycles in (r + n j  1)nj = r(r + 1) · · · (r + n j  1) different ways. Indeed, the first element can be placed in any of the r places between the r elements w 1, w 2, ... , Wr or after the last element. After the first element is placed, the second element can be placed in any of the r + 1 places between the r + 1 elements or after the last element. Finally, after then j  1 elements are placed, the last element can be placed in any of the r + n  j  1 places between the r + n  j  1 elements or after the last element.. Consequently, the number of permutations of the set Wn+r that are decomposed into k + r cycles, such that the r elements of the set Wr belong in r distinct cycles, is given by ls(n, k; r)l =
t (~) j=k
(r
+ n j 
1)njls(j, k)l,
J
which is the noncentral signless Stirling number of the first kind (see Section
s.s).
D
The multivariate generating functions of the numbers of permutations of a finite set that are decomposed into cycles of unspecified or specified number with respect to cyclelength numbers may be expressed in terms of the exponential and partial Bell partition polynomials. Thus, from Theorem 12.1, using (11.1) and (11.2), we deduce the following corollary.
COROLLARY 12.2 (a) The generating function of the sequence cn(k 1 , k 2 , ••• , kn), ki 1, 2, ... , n with k 1 + 2k 2 + · · · + nkn = n, for fixed n, is given by Cn(X], X2,
0
0
0
'Xn) =
L Cn(kl' k2,
0
0
'kn)X~ 1 X~ 2
0
0
0
0
2:
0, i
=
x~n (12.6)
= Bn(XI, 1!x2, ... , (n 1)!xn),
where Bn is the exponential Bell partition polynomial. (b) The generating function of the sequence Cn,k(kl, k2, ... , kn). ki 2: 0, i = 1, 2, ... , n with k 1 + 2k2 + · · · + nkn = nand k1 + k2 + · · · + kn = k, for fixed n and k, is given by Cn,k(Xl' X2, .. 'Xn) 0
=
L Cn,k(ki. k2,
0
..
'kn)x~· X~ 2
0
= Bn,k(XI, l!x2, ... , (n 1)!xn),
where Bn,k is the partial Bell partition polynomial.
..
x~n (12.7)
12.2. PERMUTATIONS WITH A GIVEN NUMBER OF CYCLES 467
Further, using the generating functions of the Bell and the partial Bell partition polynomials (11.6) and (11.9), we deduce the following corollary.
COROLLARY 12.3 (a) The generating function of the sequence Cn(k1, k2, ... , kn), ki 1, 2, ... , n, with k1 + 2k2 + · · · + nkn = n, n = 0, 1, ... , is given by
~ 0,
i =
(12.8)
(b) The generating function of the sequence Cn,k(ki, k2, ... , kn), ki ~ 0, i = 1, 2, ... , n, with k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k, k = 0, 1, ... , n, n = 0, 1, ... , is given by
(12.9)
Example 12.4 Let Cn (k; r) be the number of permutations of the set Wn for which their decomposition into cycles includes, among other cycles, k cycles of length r. Determine a generating function for the sequencecn(k; r), k = 0, 1, ... , [n/r], n = 0, 1, ... , and, expanding it, deduce an explicit expression for these numbers. The number cn(k; r) is given by the sum
=
=
where kr k and the summation is extended over all ki ~ 0, i 1, 2, ... , n, i :f. r with L:~=I,i# iki = n rk. Consequently, from (12.8), with xi = 1, i = 1, 2, ... , i :f. r, and Xr = x, we get the bivariate generating function
Loo [n/r] L cn(k; r)xk 1n.tn = exp { (x 1)trr + log(1 t)
n=O k=O
= ( 1  t) 1 exp { ( x  1)
~} .
1
}
468
CYCLES OF PERMUTATIONS
Expanding it into powers oft,
oo
[n/r]
tn
L L Cn(k; r)xk 1n. =
n=O k=O
1)i}
 oo {[n/r] (x
I: I: ., n=O
j=O
j
n
t,
J.T
we deduce the generating function [njr]
[njr] (x _
L Cn(k;r)xk = n! L k=O
}=0
)j
1
"!ri
.
J
Further, expanding it into powers of x, [njr]
L
k=O
[njr] cn(k; r)xk =
n!
j
L L(
j=O
[njr]
~
k
.
1)}k
(~) ~ri J
k=O
n! { [njr] ( 1)jk } k!
~
(jk)!ri
k
x'
we conclude for cn(k; r) the expression n! [njr] ( 1)1k (j k)!ri.
Cn(k; r) = k!
L
J=k
Note that cn(k; 1) = Dn,k is the number of permutations of the set Wn that have k fixed points (see Section 5.2). 0
12.3
EVEN AND ODD PERMUTATIONS
As previously noted, any permutation of a finite set W n is either a cycle or can be decomposed into disjoint cycles; this decomposition is unique up to a rearrangement of the cycles. Further, each cycle can be generated by successive transpositions. Indeed, a cycle (wi) of length 1 (fixed point) can be generated as (wi) = (wi,w1 )(w1 ,wi), i =f. j, while a cycle (w;,, Wi 2 , ••• , w;kt, wik) of length k 2': 2 can be generated by k  1 successive transpositions as
12.3. EVEN AND ODD PERMUTATIONS
469
Consequently, any permutation of a finite set Wn can be decomposed into transpositions. But while the decomposition of a permutation into disjoint cycles is unique up to a rearrangement of the cycles, its decomposition into transpositions, among which two or more have elements in common, may be done in different ways. Nevertheless, in these different decompositions of a permutation, the number of transpositions is either even or odd. This fact is used for the following discrimination of the permutations. DEFINITION 12.2 A permutation of a finite set W n• of n elements, is called even if it can be decomposed into an even number of transpositions, while is called odd if it can be decomposed into an odd number of transpositions.
The enumeration of the even and odd permutations is facilitated by the following characteristic property. LEMMA 12.1 A permutation of a finite set W n• of n elements, is even or odd if and only if, in its decomposition into cycles, the number of cycles of even length is even or odd, respectively. PROOF Consider a permutation of a finite set Wn, of n elements, that is decomposed into k cycles. Assume that its decomposition includes r cycles of even lengths, ii, h, ... , Jr. and k r cycles of odd lengths, ir+l, ir+2• ... , ik· Since each cycle of length j 8 can be decomposed into j 8  1 transpositions, this permutation can be decomposed into r
k
n k
k
= L(j• 1) = :~::::>• + •=l
L
(j. 1) r
•=l
transpositions. Note that the summands j 8 , s = 1, 2, ... , r and j 8  1, s = r + 1, r + 2, ... , kin the sums of the righthand member are even numbers and so the sum r
k
s=l
s=r+l
L:J. + L:
(j. 1)
is always an even number. Consequently, the number n  k of transpositions into which the considered permutation can be decomposed is even or odd if and only if the number r of cycles of even length is even or odd, respectively. I Let us denote by an,k (k1, k2, ... , kn) and bn,k ( k1, k2, ... , kn) the number of even and odd permutations of the type [k1 , k2, ... , knJ, respectively, so that k1 +2k2+· · ·+nkn =nand k1 +k2 +· · ·+kn = k, with k2+k4+· · ·+k2m
CYCLES OF PERMUTATIONS
470
an even and odd number, respectively, where m
= [n/2].
Clearly,
(12.10) and bn,k(kl,k2,•··,kn)=
1 ( 1Y
2
Cn,k(kl,k2,···,kn),
(12.11)
where r = k 2 +k4+· · ·+k2m, m = [n/2]. Utilizing these expressions, generating functions for the numbers an,k(kl, k2, ... , kn) and bn,k(k 1, k2, ... , kn) may be deduced from the corresponding generating functions for the numberscn,k(k1,k2,··· ,kn)·
COROLLARY 12.4 (a) The generating function of the sequence an,k(k1, k2, ... , kn), k; ~ 0, i = 1, 2, ... , n, with k1 + 2k2 + · · · + nkn =nand k1 + k2 + · · · + kn = k, for fixed n and k, is given by '"' k2 ~ an,k (k 1, k 2, ... , k n ) x kt 1 x2
· ··
Xnkn
1 = 2Bn,k(x1, 1!x2, ... , (n 1)!xn)
+~Bn,k(X1,1!x2,···
,(1)n 1(n1)!xn),
(12.12)
where Bn,k is the partial Bell partition polynomial. (b) The generating function of the sequence an,k ( k1, k2, ... , kn), k; ~ 0, i = 1, 2, ... , n, with k1 + 2k2 + · · · + nkn =nand k1 + k2 + · · · + kn = k, k = 0, 1, ... ,n, n = 0, 1, ... , is given by
(12.13)
PROOF (a) The required generating function, using (12.12) and since r k2 + k4 + · · · + k 2m, m = [n/2], may be expressed as
L an,k (k1, k2, ... , kn)X~ x~ 1
2
• • •
x~n
1 '"' k2 ... Xn kn = 2 ~ Cn,k (k 1' k 2' ... ' k n ) X2ki X2
+~ LCn,k(k1,k2,··· ,kn)(1)k,+k•+···+k 2 mx~ 1 X~ 2 ···X~n.
=
12.4. PERMUTATIONS WITH PARTIALLY ORDERED CYCLES 471
Then, by virtue of (12. 7), (12.12) is deduced. (b) Multiplying (12.12) by uktn jn! and summing for k = 0, 1, ... , n, n = 0, 1, ... , upon using the generating function (11.6) of the partial Bell partition I polynomials, (12.13) is readily obtained.
Example 12.5 Find the number an,k of even permutations of a finite set Wn of n elements that are decomposed into k cycles. Clearly, setting Xi = 1, i = 1, 2, ... in (12.13), a bivariate generating function of the sequence an,k. k = 0, 1, ... , n, n = 0, 1, ... , is obtained as
L Ln an,kuk tnn! = 21 exp[u log(1 t)oo
1
n=Dk=O
1 ]
+
2 exp[u log(1 + t)]
1( )u +1+t. 1( )u =1t
2
2
Thus, according to (8.14) and (8.5), it follows that 1 an,k = 2ls(n, k)l
1
+ 2s(n, k)
=
1+(1)nk ls(n, k)l, 2
where is(n, k)l is the signless Stirling number of the first kind.
12.4
D
PERMUTATIONS WITH PARTIALLY ORDERED CYCLES
In the preceding enumeration of permutations, cycles are distinguished only by their length. In Theorem 12.1, the number Cn,k(k1,k2, ... ,kn) of permutations of a finite set Wn, of n elements, that are decomposed into k cycles, among which ki 2: 0 are of length i, i = 1, 2, ... , n, has been deduced from the number B(n, k; k 1 , k2 , ..• , kn) of partitions of the set Wn into k subsets, among which ki 2: 0 include i elements, i = 1, 2, ... , n. In that case (i 1)! permutations were formed from any subset of i elements by permuting the i  1 elements (excluding the smallest element which, by convention, occupies the first position). It is possible to go further by considering the set of possible orderings of the elements within the cycles. Then the cycles may also be distinguished by the subset of orderings allowed for their elements. Only orderings independent of the particular elements of any cycle are considered. Note that, in the extreme case of (totally) ordered cycles with all their elements in a specific order, for example from the smallest to the largest, only one cycle of length i is formed from a
CYCLES OF PERMUTATIONS
472
subset of i elements. In the enumeration of permutations by the number of partially ordered cycles, the number ai of cycles of length i that are formed from a subset of i elements, where 1 :::; ai :::; (i  1)!, i = 1, 2, ... , n, is needed. Then, Theorem 12.1 is readily extended as follows. THEOREM 12.2 (a) The number ofpermutations ofa finite set ofn elements that are decomposed into k partially ordered cycles, among which ki 2: 0 are of length i and ai cycles are formed from any subset ofi elements, i = 1, 2, ... , n, equals
with k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. (b) The number ofpermutations ofa finite set ofn elements that are decomposed into partially ordered cycles, among which ki 2: 0 are of length i and ai cycles are formed from any subset ofi elements, i = 1, 2, ... , n, equals
with k1
+ 2k2 + · · · + nkn = n.
PROOF (a) Consider a partition {A 1 ,A 2 , ... ,Ak} of finite set Wn, with N (Wn) = n, into k subsets, among which ki 2: 0 subsets inc! ude i elements each, i = 1,2, ... ,n,wherek1 +2kz+···+nkn = nandk 1 +k2 +···+kn = k. Toit there correspond a~ 1 a~ 2 • • • a~n, permutations of the set Wn that are decomposed into k partially ordered cycles, among which ki 2: 0 are of length i and ai cycles are formed from each subset of i elements, i = 1, 2, ... , n; these permutations are formed by permuting the i  1 elements of each of the ki 2: 0 subsets with i elements (excluding the smallest element, which, by convention, occupies the first position) in all a~; permissible ways, fori = 1, 2, ... , n. Therefore
where B(n, k; k 1, k2 , .•• , kn) is the number of partitions of the set Wn into k subsets, among which ki 2: 0 subsets include i elements each, i = 1, 2, ... , n, where k 1 + 2k2 + · · · + nkn =nand k 1 + k2 + · · · + kn = k. Thus, introducing into it the expression of B(n, k; k 1, k2 , .•. , kn). (12.14) is obtained. (b) Clearly, the summation of the numbers (12.14) for all k = 0, 1, ... , n is equivalent to keep the total number of cycles k 1 + k2 + · · · + kn unspecified and
12.4. PERMUTATIONS WITH PARTIALLY ORDERED CYCLES 473
so the number of permutations of a finite set of n elements that are decomposed into cycles, among which ki ~ 0 are of length i, i = 1, 2, ... , n, is given by o2.1s).
I
The number of permutations of a finite set of n elements that are decomposed into k partially ordered cycles may be obtained by summing the numbers (12.14) for all ki ~ 0, i = 1, 2, ... , n with k 1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. Also, the number of permutations of a finite set of n elements that are decomposed into partially ordered cycles may be obtained by summing the numbers (12.15) for all ki ~ 0, i = 1, 2, ... , n with k1 + 2k2 + · · · + nkn = n. These sums, according to Definitions 11.1 and 11.2, are the partial and the exponential Bell partition polynomials. Thus the following corollary is deduced. COROLLARY 12.5 (a) The number ofpennutations of a finite set of n elements that are decomposed into k partially ordered cycles, such that ai cycles are fanned from each subset of i elements, equals the partial Bell partition polynomial. (b) The total number of pennutations of a finite set of n elements that are decomposed into partially ordered cycles, such that ai cycles are fanned from each subset ofi elements, equals
the exponential Bell partition polynomial.
Further, using the generating functions of the Bell and the partial Bell partition polynomials (11.6) and (11.9), we deduce the following corollary. COROLLARY 12.6 (a) The generating function of the sequence Cn(kt, k2, ... , kn; a1, a2, ... , an). ki ~ 0, i = 1, 2, ... , n, with k 1 + 2k2 + · · · + nkn = n, n = 0, 1, ... , is given by
(12.16)
(b) The generating function of the sequence Cn,k(kt, k2, ... , kn; a1, a2, ... , an). ki ~ 0, i = 1, 2, ... , n, with k1 + 2k2 + · · · nkn = n, k1 + k2 + · · · + kn = k,
474
CYCLES OF PERMUTATIONS
k = 0, 1, ... , n, n = 0, 1, ... is given by
00
= exp
(
F)
u ~ arXr r!
. (12.17)
Example 12.6
Find the number Cn,k of permutations of a finite set W n of n elements that are decomposed into k totally ordered cycles. Clearly, according to the first part of Corollary 12.5, setting ai = 1, i = 1, 2, ... and using expression ( 11. 2), the number Cn,k is obtained as
where the summation is extended over all nonnegative integer solutions (k 1 , k2 , ... , kn) of the equations kt + 2k2 + · · · + nkn =nand kt + k2 + · · · + kn = k. Thus, by virtue of (8.25), Cn,k = S(n, k), where S(n, k) is the Stirling number of the second kind. Further, the total number of permutations of a finite set W n of n elements that are decomposed into totally ordered cycles is given by n
En= LS(n,k) k=l
the Bell number. Note that the numbers of permutations of a finite set Wn that are decomposed into a specified or unspecified number of totally ordered cycles coincide with the corresponding numbers of partitions of Wn into a specified or unspecified number of subsets (see Exercises 2.43 and 2.46). This is a consequence of the onetoone and onto correspondence between the sets of the above permutations and partitions of Wn. 0 Example 12.7 Restricted permutations with a given number of or
dered cycles Consider a finite set Wn+r = {Wt, w2, ... , Wn+r }, with Wt < w2 < · · · < Wn+r• and its subset Wr = {w 1 , w2, ... , Wr}. Determine the number of permutations of the set W n+r that are decomposed into k + r ordered cycles, such that the r elements of the set Wr belong in r distinct cycles. Note first that the required number equals the number of partitions of the set Wn+r into k + r subsets, such that the r elements of the set Wr belong in r
12.4. PERMUTATIONS WITH PARTIALLY ORDERED CYCLES 475
distinct subsets, since each permutation of a finite set that is decomposed into ordered cycles uniquely corresponds to a partition of this set into subsets. Further, the k subsets of such a partition may contain a total of j elements from the set Wn+r Wr = {wr+l, Wr+2, ... , Wn+r }, ofn elements, for j = k, k + 1, ... , n. The j elements can be chosen in (n) different ways. Also, the number of partitions of a set of j elements into k su6sets equals the Stirling number of the second kind S(j, k) (see Example 12.6), while the remaining n j elements of the set Wn+r  Wr can be distributed into the r distinct subsets in rnj different ways. Consequently, the number of partitions of the set Wn+r into k + r subsets, such that the r elements of the set Wr belong in r distinct subsets, is given by
S(n, k; r)
=
t
j=k
(~)rnj S(j, k), J
which is the noncentral Stirling number of the second kind (see Section 8.5). D The next theorem is concerned with the more general problem of enumeration of permutations of a finite set by partially ordered cycles, where in a specified number of cycles the orderings belong in a given subset and in the other cycles the orderings belong to another given subset.
THEOREM 12.3 (a) The number of permutations of a .finite set Wn ofn elements that are decomposed (i) into k partially ordered cycles, with elements from a set U ~ W n• such that k; :2: 0 are of length i and a; cycles are formed from any subset of i elements, i = 1, 2, ... , n, and (ii) into other partially ordered cycles, with elements from the set Wn U, such that r; :2: 0 are of length i and b; cycles are formed from any subset ofi elements, i = 1, 2, ... , n, equals Cn,k(ki, r1, k2, r2, ... , kn, rn; a1, b1, a2, b2, ... , anbn) nlak1bT1ak2br2,, ·aknbTn '1122 nn kl!rl!(l!)kl+rlk2!r2!(2!)k2+r2. ··kn!rn!(n!)kn+rn'
(12.18)
with L~=I i(k; + ri) = nand L~=l ki = k. (b) The number ofpermutations ofa .finite set W n ofn elements that are decomposed (i) into k partially ordered cycles, with elements from a set U ~ W n. such that a; cycles are formed from any subset ofi elements, i = 1, 2, ... , n, and (ii) into other partially ordered cycles, with elements from the set Wn U, such that bicycles are formed from any subset ofi elements, i = 1, 2, ... , n, equals
the Touchard polynomial.
476
CYCLES OF PERMUTATIONS
PROOF (a) Consider a partition { A 1, A2, ... , Ak, B1, B 2, ... } of a finite set W n, of n elements, where { A1, A2, ... , Ak} is a partition of a set U ~ W n into k subsets, among which ki ~ 0 include i elements each, i = 1, 2, ... , n, and { B 1, B 2, ... } is a partition of the set W n  U into subsets among which ri ~ 0 include i elements each, i = 1, 2, ... , n, so that I:~= I i(ki + ri) = nand L:~ 1 ki = k. To it there correspond a~ 1 b~ 1 a~ 2 b~ 2 · · ·a~"b~", permutations of the set Wn of n elements that are decomposed (i) into k partially ordered cycles, with elements from the set U ~ Wn. such that ki ~ 0 are of length i and ai cycles are formed from any subset of i elements, i = 1, 2, ... , n, and (ii) into other partially ordered cycles, with elements from the set Wn  U, such that ri ~ 0 are of length i and bi cycles are formed from any subset of i elements, i = 1, 2, ... , n. Specifically, these permutations are formed by permuting the elements of each of the ki ~ 0 and each of the Ti ~ 0 subsets with i elements in all af; and W permissible ways fori = 1, 2, ... , n, respectively. Therefore
Cn,k(kl, r1, k2, r2, ... , kn, rn; a1, b1, a2, ~ •... , an, bn) _ alk1br1 1 a2k2br2 2 ... ank,.br"B( n n, k·' k I,TI, k 2,r2,···' k n,Tn,. a1, b1, a2, b2, ... , an, bn), with B(n,k;kl,r1,k2,r2,··· ,kn,rn;al,bl,a2,b2,··· ,an,bn) the number of partitions {A1, A 2, ... , Ak, B 1, B 2, ... } of a finite set W n of n elements; { A1, A 2, ... , Ak} is a partition of a set U ~ Wn into k subsets, among which ki ~ 0 include i elements each, i = 1, 2, ... , n, and { B 1, B 2, ... } is a partition of the set Wn  U into subsets among which Ti ~ 0 include i elements each, i = 1, 2, ... , n, so that L:~ 1 i(ki + ri) = n and L:~ 1 ki = k. Clearly, this number is given by
B(n, k; k1, r1, k2, r2, ... , kn, rn; a1, b1, a2, ~ •... , an, bn) n!
with L:~ 1 i(ki + ri) = n, L:~=l ki = k and so (12.18) is deduced. (b) The number of permutations of a finite set Wn of n elements that are decomposed (i) into k partially ordered cycles, with elements from the set U ~ Wn, such that ai cycles are formed from any subset of i elements, i = 1, 2, ... , n, and (ii) into other partially ordered cycles, with elements from a set Wn U, such that bi cycles are formed from any subset of i elements, i = 1, 2, ... , n, is obtained bysummingthenumbers(12.18)fora11ki ~ Oandallri ~ O,i = 1,2, ... ,n, with L:~ 1 i(ki + ri) = nand L:~ 1 ki = k. According to the Definition 11.6, I this sum is the Touchard polynomial and so the proof is completed. Further, using the generating function of the Touchard polynomial (11.50), we deduce the following corollary.
12.4. PERMUTATIONS WITH PARTIALLY ORDERED CYCLES 477
COROLLARY 12.7 The generating function of the sequence Cn,k Cn,k(kl, r1, k2, r2, ... , kn, rn; a!,b!,a2,b2,··. ,an,bn), for ki ~ 0, ri ~ 0, i = 1,2, ... ,n, with I:~= I i(ki + ri) = n, I:~=I ki = k, n = k, k + 1, ... , and for fixed k, is given by
=
(12.19)
Example 12.8 Find the number R( n, k) of permutations of a finite set W n of n elements that are decomposed into k (nonordered) cycles, with elements from a set U ~ W n and into other (totally) ordered cycles, with elements from the set Wn U. A generating function for the sequence R( n, k ), n = k, k + 1, ... , for fixed k, may be obtained from (12.19) by setting am = (m 1)!, bm = 1 and Xm = 1, Ym = 1, m = 1, 2, .... Then
~ R( ~
n,
k)tn = [log(1 t)]k ( t  1) k' exp e . n.1 .
n=k Expanding the righthand member into powers oft, using the generating function of the signless Stirling numbers of the first kind ls(n, k) 1.
~I( k)ltn=[log(1t)jk ~
n=k
s n,
k'
n.1
.
and the generating function of the Bell numbers Bn =
,
I:Z=o S (n, k ),
we get
L R(n, k) 1tnn. (L ls(n, k)l 1n.tn) (L Br:"jti) 00
00
00
n=k
j=O
=
n=k
=
f {t (~)
n=k Therefore
R(n, k) =
J=O
:t (~) j=O
J
ls(n
J.
j,k)IBi}
J
ls(n j, k)IBi.
~
D
CYCLES OF PERMUTATIONS
478
12.5
BIBLIOGRAPHIC NOTES
A systematic and complete study of the theory of enumeration of permutations by cycles is provided by the elegant paper of Jacques Touchard (1939). In this study several partition polynomials are examined. The Touchard partition polynomials, which were briefly presented in Chapter 11, were so named in recognition of his fundamental contribution to this subject. Credit for its illuminating introduction to the theory of enumeration of permutations by cycles belongs to J. Riordan (1958). Examples 12.3 and 12.7 on the enumeration of restricted permutations by cycles are based on the paper of A. Z. Broder (1984).
12.6
EXERCISES
1. Show that the number of permutations of n that are decomposed into k cycles of length no smaller than 2 is given by
is2(n,k)l
= "f)1) 1 (~)is(nj,kj)l, J
j=O
where ls(n, k)l is the signless Stirling number of the first kind. The number is2(n, k)l is the signless associated Stirling number of the first kind (see Exercise 8.10). 2. (Continuation) Show that the number of permutations of n that are decomposed into k cycles of length no smaller than r is given by signless rassociated Stirling number of the first kind lsr(n, k)l for which
isr+l (n, k)l
="f) 1)1(~):j isr(n rj, k j)l, i=O
r
= 1, 2, ... ,
J.r
with ls 1 (n, k)l = is(n, k)l the signless Stirling number of the first kind (see Exercise 8.11). 3. Let cn(k; r) be the number of permutations of a finite set Wn for which their decomposition into cycles includes, among other cycles, k cycles of length r. Show that the generating function [n/r]
Cn(x;r)
=L
k=O
en(k;r)xk
12.6. EXERCISES
479
satisfies the recurrence relation
Cn(x; r) nCn1 (x; r) = c5n,rj
n!(x 1)i j!ri '
and conclude that n!( 1)jk Cn(k; r) nCn1 (k; r) = c5n,rj k!(j _ k)!ri, where c5n,rj = 1 for n = rj and c5n,rj = 0 for n :j; rj is the Kronecker delta. 4. (Continuation) Assume that a permutation is randomly chosen from the set of the n! permutations of a finite set of n elements. The probability that its decomposition into cycles includes, among other cycles, k cycles of length r is given by
. ) _ Cn(k;r) Pk (n,r , k = 0, 1, ... , [n/r]. n.1 Show that its factorial moments [n/r]
J.L(j)(n;r) = L(k)jPk(n;r), j = 1,2, ... ,[n/r] k=j are given by
/l(j) (n; r)
= rj,
j
= 1, 2, ... , [n/r].
5. Let dn(k;r) be the number of permutations of a finite set Wn that are decomposed into k cycles none of which is of length r. Show that the generating function
[njr] Dn(x;r) = L dn(k;r)xk k=O
satisfies the recurrence relation
Dn+I(x;r) = (n + x)Dn(x;r) x(n)r1Dnr+1(x;r) + x(n)rDnr(x;r), for n = r, r + 1, ... and Dn(x; r) = (x + n 1)n, for n = 0, 1, ... , r 1. Further, conclude the recurrence relation
dn+I (k; r) = ndn(k; r)
+ dn(k
1; r) (n)r1dnr+1 (k 1; r)
+(n)rdnr(k  1; r), fork= 1,2, ... ,n+ 1, n = r,r+ 1, ... , and dn(k;r) = ls(n,k)l, fork= 0, 1, ... , n, n = 0, 1, ... , r 1.
CYCLES OF PERMUTATIONS
480
6. (Continuation) Let dn(r) be the number of permutations of the set Wn that are decomposed into cycles none of which is of length r. Show that [nfr) ( 1)j dn(r) = n! ""' .. , 1 L J.rJ J=O
and
dn(r)  ndnl (r) where 6n,rj
= 1 for n = rj
and 6n,rj
= 6n ,rj
n!(1)1 .1 . , J.rJ
= 0 for n "I rj is the Kronecker delta.
7. Let Q(n; k, r) be the number of permutations of a finite set Wn for which their decomposition into cycles includes, among other cycles, k cycles of length 1 and r cycles of length 2 and consider the generating function [n/2]
Qn(x, y)
=
n2r
L L
Q(n; k, r)xkyr, n
= 1, 2, ....
r=O k=O Show that
tn
oo
L Qn(x, y),n. = (1 
t) 1 exp((x 1)t + (y 1)t2 /2]
n=O
and conclude that
8. (Continuation) Show that oo
(1 t)
tn
L Qn+l(x,y),n. n=O oo
tn
n=O
n.
= {1 + (x 1)(1 t) + (y 1)(1 t)t} L Qn(x, y), and conclude the recurrence relation
Qn+l (x, y) = (n
+ x)Qn(x, y) + n(y x)Qnl (x, y) +n(n 1)(1 x)Qnz(x, y),
for n = 2, 3, ... , with initial conditions Q0 (x, y) Qz(x, y) = x 2 + y.
= 1,
Q 1 (x, y)
=x
and
481
12.6. EXERCISES
9. Cayley identity. Show that
~ ~
( 1)k1+k2+···+kn
k 1·1 ' k1 k 2!2 k2
••
·kn!n kn = 0, n = 2, 3, ... '
where the summation is extended over all integers ki ~ 0, i = 1, 2, ... , n, with kt +2k2 +· · ·+nkn = n, and conclude that the number of permutations that are decomposed into an odd number of cycles equals the number of permutations that are decomposed into an even number of cycles. 10. (Continuation) Show that
where the summation is extended over all integers ki ~ 0, i = 1, 2, ... , n, with k1 + 2k 2 + · · · + nkn = n. Note that, in particular for x = 1, it reduces to the Cauchy identity (see Remark 12.1), while for x = 1, it reduces to the Cayley identity. 11. Let an,k and bn,k be the numbers of even and odd permutations, respectively, of a finite set Wn that are decomposed into k cycles and consider the generating functions n
n
An(u) = ~ an,kUk, Bn(u) = ~ bn,kUk, n = 1, 2, .... k=O k=O
Show that
and conclude that nAn+2(u) = (2n + 1)uAnH(u) + (n + 1)(n 2

u 2 )An(u), n = 1, 2, ... ,

u 2 )Bn(u), n = 1, 2, ... ,
with A 1 (u) = u, A 2 (u) = u 2 and nBn+2(u) = (2n + 1)uBn+l(u) + (n + 1)(n 2
with Bt(u) = 0, B 2 (u) = u. 12. Let 9n,k and hn,k be the numbers of even and odd permutations, respectively, of a finite set Wn that are decomposed into k cycles of length no smaller than 2 and consider the generating functions n
Gn(u)=L9n,kUk, k=O
n
Hn(u)=Lhn,kUk, n=0,1, .... k=O
CYCLES OF PERMUTATIONS
482
Show that
and
= nHn(u) + nuHn1 (u), Hn+l (u) = nGn(u) + nuGn_I(u),
Gn+l (u)
= 1, 2, ... n = 1, 2, ...
n
, .
Deduce the recurrence relations
Gn+4(u)
= (n + 2)(n + 3)Gn+2(u) + (n + 3)(2n + 3)uGn+I (u) +(n
for n
= 0, 1, ... , with G0 (u) = 1, GI(u) = 0, G2(u) = 0, G3(u) = 2u and Hn+4(u) = (n
for n
+ 1)(n + 3)u 2Gn(u),
+ 2)(n + 3)Hn+2(u) + (n + 3)(2n + 3)uHn+l(u) +(n + 1)(n + 3)u 2 Hn(u),
= 0, 1, ... , with
H0 (u)
= 0,
HI(u)
= 0,
H 2(u)
= u,
H3(u)
= 0.
13. (Continuation) (a) Show that
where
an,k
=
1 + (1)nk 2
ls(n, k)l, bn,k
=
1
( 1)nk 2
ls(n, k)l
are the numbers of even and odd permutations, respectively, of a finite set of n elements that are decomposed into k cycles. (b) Setting n
Gn
=L
n
gn,k, Hn
k=O
=L
hn,k, n
= 0, 1, ...
'
k=O
show that
Gn
= nGn1 + (l)n Hn
[1
(~)],
= nHn1 + (1)n(~),
n
n
= 2,3, ... ,
= 2,3, ... ,
H1
G1
= 0,
=0
483
12.6. EXERCISES
and Gn
+ Hn = Dn,
Gn  Hn
= (l)nl (n
1), n
= 1, 2, ...
,
where Dn is the number of derangements of a finite set of n elements (see Section 5.2). 14. Let b(n, k) be the number of permutations of a finite set of n elements that are decomposed into k odd (of even length) cycles and n
bn(u)
=L
b(n, k)uk, n
= 0, 1, ....
k=O
Show that
and conclude that bn(2u)
=
(2r)!(u {
and b(n,k) =
+ r
r!
0,
1)r
,n n
= 2r,
= 0, 1, ...
.
= 2r + 1, r = 0, 1, ...
,
r
'n:
(2r)!ls(r, k)l r!2k 2r, r : 0, 1, ... ' { 0, n 2r + 1, r 0, 1, ... ,
where ls(r, k)l is the signless Stirling number of the first kind. 15. (Continuation) Let a(n, k) be the number of permutations of a finite set of n elements that are decomposed into k even (of odd length) cycles and n
an(u)
=L
a(n, k)uk, n
= 0, 1, ....
k=O
Show that
and conclude that
and [n/2] k
a(n, k) =
L ?=< l)n+ki ~~:r ls(n 2r, k j)l·ls(r,j)l. r=O J=O
484
CYCLES OF PERMUTATIONS
16. (Continuation) Derive the recurrence relation
an+2(u) = uan+l(u)
+ n(n + 1)an(u),
n = 0, 1, ... ,
with a0 (u), a 1 (u) = u, and ~r+2(u)
= (2r + 1)(2r + u)b2r(u), r = 0, 1, ... ,
with b0 ( u) = 1. Deduce the particular expressions
and
17. (Continuation) Let a( n) and b( n) be the numbers of permutations of a finite set of n elements that are decomposed into even (of odd length) and into odd (of even length) cycles, respectively. (a) Derive the recurrence relations a(n + 2) = a(n + 1) + n(n + 1)a(n), n = 0, 1, ... , with a(O) = 1, a(1) = 1, and
b(n+2)=(n+1) 2b(n), n=0,1, ... , with b(O) = 1, b(1) = 0. (b) Show that
b(2r)=
[~~;~!r,
b(2r+1)=0, r=0,1, ...
and
a(2r)=
[~~;~!r, a(2r+1)=(2r+1)[~~;~!r,
r=0,1, ....
18. A universal generating function. Let A= (ai,j), i = 1,2, ... , j = 0, 1, ... be an infinite matrix with elements ai,J = 0 or 1 and c(n, k; A) be the number of permutations of a finite set Wn, of n elements, that are decomposed into k cycles such that the number of cycles of length i belongs to the set {j : ai,J = 1}. Show that
G(t,u;A)
oo n tn = LLc(n,k;A)uk nl
n=O k=O
00
=IT i=l
00
[
ui Lai,j1 j=O
J
(ti)j] i ·
485
12.6. EXERCISES
19. Let cn(r) be the number of permutations of n that are decomposed into cycles none of which is of length longer than r. Show that
2: cn(r) In. = exp 2::ti) 00
tn
(
n=O
r
i=l
t
and conclude the recurrence relation r1
Cn+l(r)
= 2:(n)JCnJ(r). j=O
20. Let T(n, k) be the number of permutations of n that are decomposed into cycles among which k are of length 2 and the other n 2k are of length 1 (fixed· points). Show that the generating function [n/2)
Tn(x)
= 2: T(n, k)xk k=O
satisfies the recurrence relation Tn(x)
= Tn!(x) + (n 1)xTn2(x),
with T 0 (x) = 1, T 1 (x) rence relation T(n, k)
= 1 and
n
= 2,3, ... ,
deduce for the numbers T(n, k) the recur
= T(n 1, k) + (n 1)T(n 2, k 1),
for n = 2k, 2k+ 1, ... , k n < 2k.
= 1, 2, ... , with T(O, 0) = T(1, 1) = 1, T(n, k) = 0,
Chapter 13 EQUIVALENCE CLASSES
13.1
INTRODUCTION
The enumeration of the number of kcombinations of a finite set Wn = {w 1 , w 2 , ... , wn}, of n elements, is usually carried out in two stages. Specifically, ifCk(Wn) and Pk(Wn) are the sets of kcombinations and kpermutations of Wn, respectively, then, at the first stage, to each kcombination Ak = { a 1, a2, ... , ak} that belongs to Ck(Wn) we uniquely correspond the set of its permutations Pk(Ak), which is a subset of Pk(Wn)· Note that Pk(Ak) and Pk(Bk) are disjoint sets for Ak and Bk different kcombinations from Ck(Wn) and also that each kpermutation from Pk(Wn) belongs in one of the sets Pk(Ak), Ak E Ck(Wn)· Thus, the set {Pk(Ak) ~ Pk(Wn) : Ak E Ck(Wn)} constitutes a partition of the set Pk(Wn)· Further, according to the definition of a permutation, N(Pk(Ak)) = N(Pk(Wk)) for every Ak E Ck(Wn) and so, by the addition principle,
AkECk(Wn)
At the second stage we enumerate the number P(n, k) = N(Pk(Wn)) of kpermutations of the set Wn and conclude the number P(k) P(k, k) = N(Pk(Wk)). Then the number C(n, k) = N(Ck(Wn)), of kcombinations of the set Wn, is obtained as C(n, k) = P(n, k)/ P(k). This technique has been employed for the derivation of this number in Theorem 2.5. Note that the sets Pk(Ak), Ak E Ck(Wn) constitute equivalence classes of the kpermutations that belong to the set Pk(Wn)· The equivalence relation is defined as follows: A permutation (a 1, a 2 , .•• , ak) that belongs to Pk(Wn) is equivalent to a permutation (b 1 , b2 , •.• , bk) that also belongs to Pk (Wn) if there exists a permutation (it, h, ... , jk) of the set of the k indices { 1, 2, ... , k} such that a 8 = b3,, s = 1, 2, ... , k. In other words, two kpermutations of Wn are equivalent if they contain the same k elements. Thus, the enumeration of the number of kcombinations of a finite set of n
=
488
EQUIVALENCE CLASSES
elements is merely a problem of enumeration of the equivalence classes into which the kpermutations of the finite set of n elements are distributed. This chapter is devoted to the study of the general problem of enumerating equivalent classes of a finite set with respect to a group of its permutations. After a brief presentation of groups of permutations and their cycle indicator, we proceed to the enumeration of the orbits (equivalence classes) of the elements of a finite set, under a group of permutations of it. In this respect Burnside's lemma is shown. The study is then focused on the enumeration of models of colorings of the elements of a finite set, under a group of permutations of it; P6lya's fundamental counting theorem is derived.
13.2
CYCLE INDICATOR OF A PERMUTATION GROUP
Consider a finite set Wn = { w1 , w2 , ... , wn}, of n elements, and let Pn = Pn(Wn) be the set of its n! permutations. To each pair of permutations in Pn
there corresponds the permutation aT in P n defined by
which is called their product. If
is another permutation in Pn, then
and
Thus, the product of permutations is associative, (ap)p = a(Tp). Further, the identity permutation
13.2. CYCLE INDICATOR OF A PERMUTATION GROUP
489
is the neutral element for the product, ca = ac = a. Also, for every permutation a in Pn, there exists its inverse permutation in Pn, a1
= (wi,,wi2, ...
,win)
WI, W2, ... ,Wn
so that aa 1 = a 1 a = €. Consequently, the set Pn of permutations of Wn equipped with the operation of multiplication constitutes a group called symmetric group of permutations. This group is of order N(Pn) = n! and degree N(Wn) = n. Let us consider a subset 9n of permutations of Wn that is closed under the operation of multiplication. This set, as a subset of the finite group Pn, is a subgroup and is called group of permutations of Wn. Further, consider a permutation a in 9n, the decomposition of which into disjoint cycles includes ki = ki(a) ~ 0 cycles of length i, i = 1, 2, ... , n, so that k1 (a) + 2k2(a) + · · · + nkn (a) = n. Let us attach the indicator Xi to any cycle of length i, i = 1, 2, ... , n. Then the homogeneous product x~,(u) x~ 2 (u) · · · x~n (u), of degree k1 (a) + k2 (a) + · · · + kn (a) = k( a), is the cycle indicator of the permutation a and the sum C(x1,x, ... ,xn; 9n)N(9n) =
L
X~ 1 ( 17 )X~ 2 (u} · · ·x~n(u),
(13.1)
uE9n
where the summation is extended over all permutations a in 9n, is the cycle indicator of the permutations contained in the group 9n· The polynomial C(x 1 , x 2 , ... , x; 9n) is called cycle indicator of the group of permutation 9n· Introducing the number c( k 1, k 2 , ••• , kn; 9n) of permutations contained in 9n, for which the decomposition into cycles includes ki ~ 0 cycles of length i, i = 1, 2, ... , n, so that k 1 + 2k2 + · · · + nkn = n, the cycle indicator may be written as
where the summation is extended over all partitions of n, that is, over all nonnegative integer solutions (k~, k2, ... , kn) of the equation k 1 + 2k 2 + · · · + nkn = n. In the following examples several specific groups of permutations of a finite set are considered and their cycle indicators are calculated. Example 13.1 Identity group of permutations The identity permutation c = (wi)(w 2 ) · · · (wn) of the set Wn = {w 1,w2 , ... , Wn} is clearly a group. Since this permutation includes n cycles of length 1, the cycle indicator of the identity group {€}, according to ( 13.1 ), is given by
C(xi,X2,··· ,xn;{€}) = xf.
D
EQUIVALENCE CLASSES
490
Example 13.2 Symmetric group of permutations The number of permutations that are contained in the symmetric group of permutations Yn = P n and are decomposed into ki 2: 0 cycles of length i, i = 1, 2, ... , n, with k 1 + 2k2 + · · · + nkn = n, is given by (see Theorem 12.1)
Hence, the cycle indicator of the symmetric group of permutations, according to (13.2) and since N(Pn) = n!, is given by
where the summation is extended over all partitions of n, that is over all nonnegative integer solutions ( kt, k2, ... , kn) of the equation k 1 + 2k2 + · · · + nkn = n. This polynomial, using (ll.l ), may be expressed as
where Bn(x1 , x 2 , ... , xn) is the exponential Bell partition polynomial.
0
Example 13.3 Alternating group of permutations Consider the set An C Pn of even permutations (which can be decomposed into an even number of transpositions) of the set W n. Note that the product of two even permutations a = ti, ti 2 • • • ti 2 m and T = t j 1 t h · · · t h·, where ti, , ti 2 , ••• , th,. and tit, tj 2 , • • • , tj 2 , are transpositions of elements of Wn, is also at:l even permutation aT = ti, ti 2 • • • ti 2 m tit ti2 · · · t32 •. Therefore, the set An is closed under multiplication and, as a subset of the finite symmetric group of permutations Pn, is a subgroup called alternating group of permutations. Further, the number of permutations that are contained in the alternating group of permutations Yn = An and are decomposed into ki 2: 0 cycles of length i, i = 1, 2, ... , n, with k, + 2k2 + · · · + nkn = n, is given by
where r = k2 + k 4 + · · · + k2m, m = [n/2]. Then, the cycle indicator of the alternating group of permutations, by (13.2) and since N(An) = n!/2, is given by
1 C(x,, x2, ... , xn; An) = 1Bn(xd!x2, ... , (n 1)!xn) n.
+~Bn(Xt,1!x2,··· n. where Bn(x 1 , x 2 ,
•.• ,
,(1)n 1 (nl)!xn),
xn) is the exponential Bell partition polynomial.
0
13.2. CYCLE INDICATOR OF A PERMUTATION GROUP
491
Example 13.4 Cyclic group of permutations
Consider a permutation a of the symmetric group P n of permutations of the set W n· The powers of a arerecursivelydefined by the relations ai =a ail, j = 2, 3, ... , a 0 = c and ai = (a 1 )i, j = 2, 3, ... , with a 1 the inverse permutation of a. Since the symmetric group of permutations Pn has a finite number of elements, N(Pn) = n!, it follows that there exists a positive integer r such that ar = c because, otherwise, Pn would have an infinite number of elements. The smallest positive integer r with this property is called the order of the permutation a. In this case, the r powers aj = ai, j = 1, 2, ... , r are all different because, if we assume that ai = ai for 1 ::; i < j ::; r, then aii = c for 1 ::; j  i < r, which contradicts the hypothesis that r is the smallest positive integer with this property. Also, any positive, zero or negative power ak of a coincides with one of the r different powers a j = ai, j = 1, 2, ... , r, since the integer k can be written in the form k = sr + j, with 0 ::; j < r, whence ak = asr+j = (ar)sai = cai = ai. The set 9n,r = { c, a, a 2 , ••• , arl }, whose elements are powers of a permutation a of order r, is closed under multiplication and, as a subset of the finite symmetric group of permutations P n• is a subgroup called cyclic group of permutations generated by the permutation a. Note that, for j = 1, 2, ... , rand d the greatest common divisor of j and r, the permutations aj = ai and ad= ad, which belong to the same cyclic group of permutations 9n,r• generate the same cyclic subgroup of permutations 9n,r 1d of order rId. The orbit of a point w of the set W n under a permutationaoforderristheset{w,a(w),a 2 (w), ... ,arl(w)} ~ Wn,where the length r is the smallest positive integer that satisfies the relation ar (w) = w. The cycle indicator of the cyclic group of permutations 9n,r is determined as follows: let d be a divisor of rand Hn,rld = {aj E 9n,r : d the greatest common divisor of j and r} = {a i E 9n,r : j I d a positive integer relatively prime to r ld}. Then, if sis a divisor of r different from d, the sets Hn,rls and Hn,rld are disjoint. Also, every permutation a j = ai, which belongs to the cyclic group of permutations 9n,r, is contained in one of the sets Hn,r ld and specifically in the set for which j I dis relatively prime to rId. Therefore the sets Hn,r ld for all divisors d of r constitute a partition of the cyclic group of permutations 9n,r. 9n,r
= LHn,rld· dlr
Note that the permutation ad = ad clearly belongs to the set Hn,rld· Further, for every other permutation ai = ai that belongs to Hn,rld• dis the greatest common divisor of j and r, whence the permutations aj and ad generate the same cyclic subgroup of permutations 9n,rld of order rid. Hence, the orbit of a point w of the set Wn under a permutation ai that belongs to Hn,rld is {w, ai (w), aJ( w), ... , a;ldl (w)}, with length rl d. Consequently, the permutation aj is decomposed into d cycles, each of length rl d. The number N(Hn,rld), of permutations that belong to the set Hn,r 1d· which is the number of positive integers less than or equal and relatively prime to rId, is equal to the Euler's function
EQUIVALENCE CLASSES
492
¢(r/d) (see Example 4.4). Thus, by (13.1) and since N(Qn,r) = r, the cycle indicator of the cyclic group of permutations 9n,r is given by
D Example 13.5 Let us consider the set W6 = {1, 2, 3, 4, 5, 6} of the six faces of a cube. Its faces are numbered so that {1, 6}, {2, 5} and {3, 4} are nonordered pairs of opposite faces. Let 96 be the group of permutations of W6 that are induced by rotations of the cube around its axes of symmetry. Determine the cycle indicator of the group
96·
The group 96 contains the following permutations written in the form of product of cycles: (a) The identity permutation
(1)(2)(3)(4)(5)(6). (b) The permutations induced by 1r /2, 1r and 3n /2 rotations of the cube around the axes connecting the centers of opposite faces in the counterclockwise direction:
(1)(2, 4, 5, 3)(6), (1)(2, 5)(3, 4)(6), (1)(2, 3, 5, 4)(6), (1, 3, 6, 4)(2)(5), (1, 6)(2)(3, 4)(5), (1, 4, 6, 3)(2)(5), (1, 5, 6, 2)(3)(4), (1, 6)(2, 5)(3)(4), (1, 2, 6, 5)(3)(4). (c) The permutations induced by 1r rotations of the cube around the axes connecting opposite edges in the counterclockwise direction:
(1, 2)(3, 4)(5, 6), (1, 3)(2, 5)(4, 6), (1, 4)(2, 5)(3, 6), (1, 5)(2, 6)(3, 4), (1, 6)(2, 3)( 4, 5), (1, 6)(2, 4)(3, 5). (d) The permutations induced by 3n /2 rotations of the cube around the axes connecting opposite vertices in the counterclockwise direction:
(1, 2, 3)(4, 5, 6), (1, 2, 4)(3, 6, 5), (1, 3, 2)(4, 5, 6), (1, 3, 5)(2, 6, 4), (1, 4, 2)(3, 5, 6), (1, 4, 5)(2, 6, 3), (1, 5, 3)(2, 4, 6), (1, 5, 4)(2, 3, 6). Consequently, the cycle indicator of the group 96 , according to (13.2) and since N (Q 6 ) = 24, is given by
493
13.3. ORBITS OF ELEMENTS OF A FINITE SET
13.3
ORBITS OF ELEMENTS OF A FINITE SET
Consider a group 9n ~ Pn, of permutations of a finite set Wn = {w 1 , w2 , ... , wn} and two elements u and w from Wn. If there exists a permutation T belonging to 9n such that r(w) = u, then the element u is said to be equivalent to the element w under the group 9n; this is denoted by u w. Clearly, this relation is (a) reflexive: u u, (b) symmetric: u w implies w u and (c) transitive: u v and v w imply u w, and so it is an equivalence relation.
=
=
= =
=
=
=
DEFINITION 13.I The set Tw = {u E Wn : r(w) = u, r E 9n}, w E Wn, which contains the elements of W n equivalent to w, constitutes an equivalence class that is called orbit of the element w under 9nNote that, for any two different elements u and w of Wn, the equivalence classes Tu and Tw either coincide (when the two elements u and w are equivalent under the group 9n) or are disjoint. Thus, the set T = {Tw,,, Tw, 2 , ••• , Tw,k }, of the different orbits of the elements of Wn under the group 9n, constitutes a partition of Wn.
DEFINITION 13.2 The set Sw = {a E 9n : u(w) = w}, w E Wn, which contains the permutations of 9n that keep the element w fixed and is a subgroup ofQn is called stabilizer ofw. For each subgroup Sw of the group 9n and for each permutation T belonging to 9n, the left coset is defined by rSw = {TO" : u E Sw} and the right coset by SwT ={or: u E Sw} Clearly, N(rSw) = N(Swr) = N(Sw)Further, for any two different permutations p and T that belong to 9n, the left cosets pSw and rSw (as well as the right cosets Swp and Swr) either coincide or are disjoint. Hence, the set {rj,Sw,Tj 2 Sw, .. ,TJrSw} of distinct left cosets (as well as the set {Sw Tj 1 , Sw Tj 2 , ••• , Sw TJr } of distinct right cosets) constitutes a partition of 9nWe are now prepared to prove a theorem concerning the number of elements of the orbit of an arbitrary element of a finite set under a group of permutations of it.
THEOREM 13.1 The number N(Tw), of elements of the orbitTw of an element w of a finite set Wn,
EQUIVALENCE CLASSES
494
under a group gn of permutations ofWn, is given by (13.3)
where Sw is the stabilizer of the element w. PROOF Let u and v be two elements of the orbit Tw of w, under gn· Then, there existtwo permutationsT andp that belong to gn such that T(w) = u andp(w) = v. Further, u = v if and only if p(w) = T(w) or, equivalently, TI (p(w)) = w. The last relation is equivalent to a = TIp being in the stabilizer Sw of w, which in turn is equivalent top= Ta being in the left coset TSw. Note that the permutation T belongs to the left coset TSw, since T = Tc and the identity permutation c belongs to the stabilizer Sw. Hence, u = v if and only if the permutations T and p belong to the same left coset TSw. As a consequence, there exist a onetoone and onto correspondence of the orbit Tw to the set of the left co sets of the stabilizer Sw in which the element u, with T( w) = u, corresponds to the left coset TSw. Therefore, the number N(Tw) of elements of the orbit Tw equals the number of the left cosets of the stabilizer Sw. Since, the set of the left co sets of Sw constitutes a partition of gn and N(TSw) = N(Sw) for every left coset TSw of Sw, according to the addition principle, it follows that
N(gn) = N(Tw)N(Sw), from which (13.3) is deduced.
I
The next theorem, which is concerned with the number of orbits of the elements of a finite set, under a group of permutations of it, is due to W. S. Burnside.
THEOREM 13.2 The number of orbits of the elements of a finite set Wn, under a group gn of permutations of it, is given by 1
N(T)
= N(Qn)
L
(13.4)
k1 (a),
uE!/n
where k 1 (a) is the number of cycles Q/ length 1 of a permutation a. PROOF The number k 1 (a) of cycles of length 1 of a permutation a and the number N(Su) of the elements of the stabilizer Su of an element u of the set Wn, on using Kronecker's delta function, J(u, v) = 0 for u =/: v and J(u, u) = 1, may be expressed as
k1 (a) =
L
J(u, a(u)), N(Su) =
L
J(u, a(u)).
13.3. ORBITS OF ELEMENTS OF A FINITE SET
495
Thus
2:
k1(a) =
2: 2:
J(u,a(u)) =
2: 2: J(u,a(u)) = 2:
N(Su)
and, utilizing the fact that the set T of the different orbits of the elements of Wn under a group 9n constitutes a partition of Wn, we get the expression
2: k1(a) = 2: 2:
N(Su)·
Tw ET uETw
uEIJn
Further, the orbit Tu of every element u E Tw coincides with the orbit Tw and, according to (13.3), it follows that
N(Qn) N(Qn) N(Su) = N(Tu) = N(Tw) = N(Sw)· Hence
2:
N(Su)
= N(Tw)N(Sw) = N(Qn)
uETw
and
The last expression implies ( 13.4 ).
I
Example 13.6 The alternating subgroup A 4 of the group P 4 of permutations of the set W 4 = { 1, 2, 3, 4} contains the following permutations: (1)(2)(3)(4), (1, 2, 3)(4), (1, 2,4)(3), (1, 3, 2)(4), (1, 3, 4)(2), (1,4, 2)(3), (1, 4, 3)(2), (1)(2, 3, 4), (1)(2, 4, 3), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3). The orbits of the elements of the set W 4 , under the group A 4 , are
and the stabilizers of the elements of w4 are
s1
= {(1)(2)(3)(4),(1)(2,3,4),(1)(2,4,3)},
s2 = {(1)(2)(3)(4), (1, 3, 4)(2), (1, 4, 3)(2)},
s3 = {(1)(2)(3)(4),(1,2,4)(3),(1,4,2)(3)}, s4 = {(1)(2)(3)(4),(1,2,3)(4),(1,3,2)(4)}.
EQUIVALENCE CLASSES
496
Note that N(A 4 ) = 12, N(Si) = 3, i = 1, 2, 3, 4 and N(Ti) in agreement with Theorem 13.1. Also, N(T) = 1 and 1
 ( A) N
4
LA
uE
kl ((J) •
= 4, i = 1, 2, 3, 4,
1
= 12 (4 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) = 1,
a result that agrees with Theorem 13.2.
D
Example 13.7 Models of coloring the faces of a cube Calculate the number of models of coloring the six faces of a cube with six different colors, each face with a different color, when two colorings are considered to be of the same model if they can be transformed to each other by rotation of the cube. Any coloring of the six faces of a cube with six different colors, each face with a different color, constitutes a permutation w, = (c, 1 , c52 , ••• , C86 ) of the set C = {c 1 , c2 , ... , c6 } of the six colors, where Cs; is the color of the ith face, i = 1, 2, ... , 6. Hence, the set W720 of colorings contains 6! = 720 elements. Let 972o be the group of permutations of the set W720 that are induced by rotations of the corresponding colored cube around the axes of symmetry. Also, let 11. 6 be the group of permutations of the set :F6 = { 1, 2, ... , 6}, of the six faces of the cube, which are induced by rotations of the cube around the axes of symmetry. To the permutation O" E 972 0 , in which the coloring W 8 = (c 81 , c52 , ••• , c56 ) is replaced by the coloring 0"( W 8 ) = (csr 1 , C 8 • 2 , •• • , C 8 • 6 ), W 8 E W72o, we correspond the permutation T E 11. 6, in which the face i is replaced by the face T(i) = ri, i = 1, 2, ... '6. For example, the identity permutation c = (wl)(w2) · · · (w 720 ) E W720 , in which the coloring W 8 = (csu c 82 , . . • , c 56 ) is replaced by the same coloring c(ws) = w. = (cspCs, ... ,c.6), Ws E w720· we correspond the identity permutation c = (1)(2) · · · (6) E 11. 6 , in which the face i is replaced by the same face c(i) = i, i = 1, 2, ... , 6. The permutation O" E 9 720 , in which the coloring W 8 = (C51 , c52 , C 83 , C54 , C85 , C86 ) is replaced by the coloring O"( W 8 ) = (Cs 1 , c, 3 , Cs 4 , Cs 5 , Cs 2 , C 86 ), we correspond the permutation T = (1 )(2, 3, 4, 5)(6) E 1/. 6 , in which T(1) = 1, T(2) = 3, T(3) = 4, T(4) = 5, T(5) = 2 and T(6) = 6. This correspondence of the group of permutations 9 720 to the group of permutations 11. 6 is onetoone and onto and so N(9720 ) = N(1i6)· Further, according to the analysis of Example 13.5, N(9720 ) = 24 permutations, the identity permutation c = (wl) (w 2 ) · · · ( w72 0 ) E W720 retains fixed the k1 (c) = 720 elements of W720 . Also, none of the other 23 pennutations retains fixed any of the elements ofW720 , kl(O") = 0, since no cube whose six faces are colored with six different colors, each face with a different color, is left fixed by rotations around any of the axes of symmetry. The required number of models of colorings is equal to the number N (T) of orbits of the elements of W 720 , under the group of permutations
13.3. ORBITS OF ELEMENTS OF A FINITE SET
Q72 0 ,
497
which, according to Theorem 13.2, equals
N(T) =
720 = 30. 24
D
A simplified expression for the number of orbits of the elements of a finite set, under a cyclic group of permutations of it, is deduced in the following corollary of Theorem 13.2. COROLLARY 13.1 The number of orbits of the elements of a finite set Wn, under a cyclic group 9n,r ofpennutations of it, is given by (13.5)
where k 1 (ud) is the number of cycles of length 1 of a pennutation ud and cjJ(r /d) is Euler's function.
PROOF Consider a divisor d of the order r of the cyclic group of permutations 9n,r = {c, u, u 2, ... , url} and let 1in,rfd = {ui E 9n,r: jfdapositive integer relatively prime tor/ d}. Then, if sis a divisor of r different from d, the sets 1in,rfs and 1in,rf dare disjoint. Also, every permutation u 1 that belongs to the cyclic group of permutations 9n,r is contained in one of the sets 1in,rfd and, specifically, in the set for which j /dis relatively prime tor/ d. Therefore, the sets 1in,rfd for all divisors d of r constitute a partition of the cyclic group of permutations 9n,r• 9n,r
=L
1in,rfd•
dlr
and, since N(Qn,r)
= r, expression (13.4) may be written as
Note that the permutation ud is contained in the set 1in,rfd and, for any other permutation u 1 contained in 1in,rfd• dis the greatest common divisor of j and r. Then, the permutations ui and ud generate the same cyclic subgroup of permutations 9n,rfd of order r / d and so there exist positive integers m and s such that ui = udm and ud = ui 8 • Thus, if an element w of Wn is left fixed by the permutation ui, ui ( w) = w, then
498
EQUIVALENCE CLASSES
and w is also left fixed by ud and inversely. Consequently, k 1 (ui) every permutation ui contained in 1ln,rfd and so
= k(ud)
for
1""
N(T) =;: ~ k1(u d )N(Hn,rjd)· dlr
Since the number N(1in,rjd) of permutations that belong to the set 1ln,rfd• which is the number of positive integers less than or equal and relatively prime to r / d, is equal to Euler's function¢( rId) (see Example 4.4), the required expression ( 13.5) is established. I
Example 13.8 Models of necklaces Calculate the number K r,s of models of necklaces with r beads of s different colors when two necklaces are considered to be of the same model if they can be transformed to each other by rotation. Let Wn = { w 1 , w 2 , ... , wn} be the set of necklaces and consider a permutation u1 of the n necklaces that rotates each necklace by 2nj lr degrees in the counterclockwise direction, j = 1, 2, ... , r. Then, with u 1 = u, we have u1 = ui, j = 1, 2, ... , r, ur = c, and the set 9n,r = {c, u, u 2, ... , url} constitutes a cyclic subgroup of permutations of Wn of order r generated by u. Since a permutation ud, with d a divisor of r, rotates any necklace by 2ndlr degrees, and as a result, any d consecutive beads are moved to the positions of the next d consecutive beads, a necklace is left invariant, under ud, if and only if any of the rId blocks of d consecutive beads has the same color arrangement. In other words, a model of necklaces is completely determined once the color of d consecutive beads is specified and so k1(ud) = sd. Hence, according to (13.5), the number Kr,s = N(T) of models of necklaces is given by 1 Krs=, r
L.: c/>(rjd)s. d
dlr
When the number r of the beads is a prime, this expression reduces to
Krs= (r1)s+sr , r Note that the number Kr,s equals the number of cyclic rpermutations of the s colors with repetition. Using Table 4.1 of Euler's function, we get the following few values of the number Kr,s:
K
_s(s+l) 2,s 
2
K
,
_s(s 2 +2) 3,s 
4
3
,
K
_s(s 3 +s+2) 4,s 
4
_ s(s 5 + s 2 + 2s + 2) ' 6 5 3 6 7 _ s(s + s + 2s + 4) K _ s(s + 6) K 7,s, 8,s· 7 8
K
_ s(s
5,s
+ 4) ,
'
K
6,s
0
13.4. MODELS OF COLORINGS OF A FINITE SET
499
13.4 MODELS OF COLORINGS OF A FINITE SET Consider a group Yn ~ Pn of permutations of a finite set Wn = { w 1 , w2 , ... ,wn} and a finite set of colors Zr = {zJ,Zz, ... ,zr}· The notion of coloring the elements of the set Wn is introduced in the next definition. DEFINITION 13.3 A function f defined on the set Wn and taking values from the set Zr is called a coloring (of the elements) of the set Wn with colors from the set Zr, in which an element Wi ofWn is painted with color f(wi)from Zr, i = 1,2, ... ,n.
Note that a coloring f of the set Wn, with f(wi) = Zs;, i = 1, 2, ... , n, constitutes an npermutation (z 81 , Z 82 , ••• , Zsn) of the set Zr with repetition. Thus, the set :F of the functions f from Wn to Zr, called set of colorings of the set Wn with colors from the set Zr, contains N (:F) = rn functions (colorings). Further, iff is a coloring that belongs to :F and a is a permutation that belongs to Pn, then /(u) is also a coloring that belongs to :F, in which the element Wi is painted with color f(u(wi)). Let f and g be two colorings that belong to :F. If there exists a permutation u that belongs to the group Yn such that f(u) = g, the coloring g is called equivalent to the coloring f, under the group Yn, and is denoted by g f. Clearly, this relation is (a) reflexive: f (b) symmetric: g f implies f g and (c) transitive: g h and h f imply g f and so it is an equivalence relation. Using this equivalence relation, the notion of a model of colorings is introduced in the following definition.
=
=
=/, =
=
=
=
DEFINITION 13.4 The set Mt = {g E :F : /(u) = g, u E Yn}, f E :F, which contains the colorings in :F that are equivalent to a coloring f, under the group Yn, is called model off, with respect to the group Yn·
Note that, for any two different colorings f and g in :F, the models Mt and M 9 either coincide (when the colorings f and g are equivalent under Yn) or are disjoint. Thus the set M.r = {M!t,Mfo,··· ,M,k} of the different models of colorings, under the group Yn, constitutes a partition of :F. As regards the number of models of coloring the elements of a finite set, we first prove the following theorem. THEOREM 13.3 Let :F be the set of colorings (of the elements) of a finite set W n with colors from afinitesetZr. ThenthenumberNn(ii,iz, ... ,jr) ofmodelsinM.r, underthe
EQUIVALENCE CLASSES
500
group 9n of permutations of Wn, in which j. with color Zs from Zr, s = 1, 2, ... , r, with }1
2': 0 elements of Wn are painted
+ }2 + · · · + Jr = n, is given by
Nn(}I,}2, ... ,jr)
= N(~) n
Lc(kl,k2,··· ,kn;9n)Nn(jl,}2, ... ,jr;kl,k2,··· ,kn), (13.6)
where the summation is extended over all nonnegative integer solutions (kt, k2, ... , kn) ofthe equation k1 + 2k2 + · · · + nkn = n, c(kt, k2, ... , kn; 9n) is the number of permutations of the type [k1, k2, ... , kn] that belong to the group 9n and N n (}1, h, ... , Jr; k1, k2, ... , kn) is the number of colorings in F for which j. 2': 0 elements of Wn are painted with color Z 8 from Zr, 8 = 1, 2, ... , rand are constant on every cycle of the permutations of the type (k1, k2, . . . , kn] that belong to the group 9n· PROOF Let F)I ,12, ... ,j, ~ F be the set of colorings in which is 2': 0 elements of Wn are painted with color z 8 from Zr for 8 = 1, 2, ... , r and let P(F)I,h, ... ,j,) be the symmetric group of permutations of F)I,J,, ... ,j,· A subgroup ofP(F3d 2 , ... ,j,) can be constructed as follows. Consider a permutation a in 9n and let a be a mapping of Fj 1 ,j 2 , ... ,j, into itself defined by a(!) = f(a). If !I "I h then !I (a) "I h (a"), a(ft) "I rr(ft) and so a is a onetoone mapping of F)I ,12, ... ,j, onto itself. Also, since the set Fj 1 ,)2, ... ,J, is finite, jj is a permutation that belongs to the set P(Fh ,12, ... ,j, ). The set Q of all permutations a for a in 9n. which is a subset of P(F)I ,h, ... ,j, ), is closed, with respect to the multiplication of permutations. Indeed, if a and f belong to Q and p = rrf, then rr(f) = f(a), f(/) = f(r) and p(!) = a(f(/)) = a(f(r)) = f(r(a)) = f(p), with p = ar a permutation in 9n. whence p = rrf belongs to Q. Since Q is a subset of the symmetric group of permutations P(Fh ,h, ... ,j.). it is a subgroup of it. Further, the correspondence of the permutation a in Q to the permutation a in 9n is onetoone since, for a "IT, f(a) ::p f(r) and so a(!) "If(/), whence jj ::p f and inversely. Clearly, this is a correspondence of Qonto 9n and so N(Q) = N(Qn)· Let f be any coloring in ;:31 ,)2, ... ,J, and consider its orbit T 1 , under the group Q, and its model Mj, under the group 9n· Then, T 1 = {g E F)l,J2, ... ,j. : rr(/) = g, rr E Q} = {g E Fj,j 2 , ... ,)r : f(a) = g, a E 9n} = M 1 and so the number N n (it, h, ... , Jr) of models in M.r, under the group 9n. equals the number N(T), of orbits of the elements of F)I,]2, ... ,Jr• under the group Q. Thus, according to Theorem 13.2 and since N(Q) = N(Qn). we deduce the expression
where k 1 (a) is the number of colorings in F)I ,)2, ... ,j, which are kept fixed by the permutation a. Introducing the set Sit ,h, ... ,i. (a) = {! E Fil ,h. ... ,ir : f (a) =
13.4. MODELS OF COLORINGS OF A FINITE SET
501
!} for a in gn and using Kronecker's delta function, 6(!, g) = 0 for f =/: g and 6(!, f) = 1, the sum in the righthand side of the last expression may be written as
6(!, a(f))
=I: and so (13.7)
Further, note that, if w; is an element of Wn and {w;,a(w;),a 2(w;), ... , am 1 (w;)}, with am(w;) = w;, is its orbit, under the permutation a, then for any coloring fin s)J,h, ... ,j.(a) it holds f(ai(w;)) = f(a(ail(w;))) = f(ai 1 (w;)) f(w;),j 1,2, ... ,m1. Hence,everycoloringfin SjJ,) 2 , .•. ,Jr (a) is constant on every cycle of the permutation a (painting with the color f(w;) = Zs; the elements of the cycle of the permutation a in which w; is contained). Inversely, every coloring fin :Fh ,)2, ... ,j. (a) that is constant on every cycle of the permutation a belongs to Sh ,)2, ... ,Jr (a) since every element w; ofWn is contained in a cycle of a in which the element a(w;) of Wn is also contained and so f(w;) = f(a(w;)). Consequently, the number N(S)J,j,, ... ,1.(a)) equals the number of colorings in :Fh ,)2, ... ,Jr (a) that are constant on every cycle of the permutation a. Partitioning the group gn, according to the permutations type,
=··· =
=
where the summation is extended over all nonnegative integer solutions ( k 1 , k2, ... , kn) of the equation k1 + 2k2 + · · · + nkn = n and gn[k 1 ,k2 , .•. ,kn] includes all the permutations of the type [k1 , k 2, ... , kn] that belong to the group gn, we deduce from (13.7) the required expression (13.6). I The expression (13.6) of the number Nn(it, h, ... , ir) of models in MF, under the group gn of permutations of Wn, in which j 8 2:: 0 elements of Wn are painted with color Zs from Zr, s = 1, 2, ... , r, with j1 + h + · · · + ir = n, is unwieldy. A multivariate generating function of sequence of these numbers, which in general constitutes a powerful tool, is more manageable. In order to construct such a generating function let us assign to each element Z 8 of the set of colors Zr = { z 1 , z2, ... , Zr} its weight w( Z 8 ), s = 1, 2, ... , r. The weight of a coloring f that belongs to :F, denoted by W (f), is defined by W(f) = w(f(wt))w(f(w2)) · · · w(f(wn)).
EQUIVALENCE CLASSES
502
When in this coloring is 2: 0 elements of Wn are painted with the color z 8 , s = 1, 2, ... , r, its weight takes the form W(f)
= [w(z!)Jl' [w(z2)]i2 · · · [w(zr)Ji·.
The multivariate generating function of the numbers Nn(j1,i2, ... ,ir), is 2: 0, s = 1, 2, ... , r, with i1 + i2 + · · · + ir = n, is expressed in terms of the corresponding cycle indicator in the following P6lya counting theorem.
THEOREM 13.4 The multivariate generating function of the numbers Nn(jl, h, ... ,ir) of models in M.r, under the group gn of permutations ofWn, in which is 2: 0 elements of Wn are painted with color Zsfrom Zr, s = 1, 2, ... , r, with it+ i2 + · · · + ir = n, is given by (13.8) where r
Xj = Xj(t1,t2, . .. ,tr) = Lt~, i = 1,2, ...
,n,
(13.9)
s=l
and C(x1, x2, ... , Xn; Qn) is the cycle indicator of the group gn of permutations ofWn. PROOF Expression (13.6), of the numbers Nn(it,h, ... ,ir). is 2: 0, s = 1, 2, ... , r, with it + i2 + · · · + ir = n, suggests first deriving the generating function
9k,,k2, ... ,kn(tl,t2, ... ,tr)
= LNn(it,h, ...
,jr;k1,k2, ... ,kn)t{'t~
2
"·#.",
in terms of which the required generating function may easily be expressed. Thus, consider a permutation a of the type [k 1 , k 2 , ... , kn] that belongs to the group gn of permutations of W n. The set { A1, A2, . . . , Ak} of the subsets of W n that contain the elements of the k 1 + k 2 + · · · + kn = k cycles of the permutation a constitutes a partition of Wn and so, setting n; = N(A;), i = 1, 2, ... , k, we have n 1 + n 2 + · · · + nk = n. Let S(a) be the subset of colorings fin :F that are constant on any cycle of the permutation a, painting with the color f(ai) = Zs; then; elements of the set A; in which a; belongs, i = 1, 2, ... , k. Note that a coloring fin S(a) is completely determined by thenpermutation u E Un(Zr), of the set of colors Zr = {z 1 , z 2, ... , Zr} with repetition, in which the color Zs; appears n; times, fori = 1, 2, ... , k, with n 1 + n 2 + · · · + nk = n. Introducing the weights of the colors by w( Zs) = ts, s = 1, 2, ... , r, the generating function 9k 1 ,k 2,... ,kn (it, t2, ... , tr) is expressed as 9kt,k2, ... ,kn(lt,t2, ... ,tr) =
L /ES(rr)
W(f)
13.40 MODELS OF COLORINGS OF A FINITE SET
503
and since
L
L L
=
W(f)
/ES(CT)
w(f(wl))w(f(w 2)) 000w(f(wn))
fES(CT)
=
[w(f(a 1 ))tt [w(f(a 2 ))]n2
o
o
0
[w(f(ak))]nk
/ES(CT)
L
[w(zst)jnt [w(zs2)]n2 000[w(zsk)]nk
uEUk(Zr)
= (t[w(z.)]nt)
(t[w(zs)]n2) ooo (t[w(z.)]nk) ,
it reduces to
Note that ki ;::: 0 among the exponents n 1 , n 2 , 000 , nk are equal to i, fori 1, 2, 000 , n, with k 1 + 2k2 + 000+ nkn = n, and so
Now, multiplying both members of expression (1306) by t{ 1 ~2 000t{r and summing for j. = 0, 1, 000, s = 1, 2, 000 , r, with }I + h + 000+ ir = n, we get ~ ~
N n (}1, }2' 000 'Jr )tttth tJr 1 2 000 r o
o
•
1 = N(Qn) Lc(k1,k2,ooo ,kn;9n)Yk 1 ,k 2,... ,k,.(tl,t2,ooo ,tr),
where in the second sum the summation is extended over all partitions of n, that is, over all nonnegative integer solutions (k1, k2, 000 , kn) ofthe equation k1 + 2k2 + 000+nkn = no Introducing into it the last expression of Yk 1 ,k 2 , ... ,k,. (t1, t2, oo0 , tr) and using the cycle indicator ( 1302) of the group of permutations 9n. we deduce (13o8)o I The total number of models of coloring the elements of a finite set may be deduced from (1308) by setting t 8 = 1, s = 1, 2, 000 , r, whence Xj = r, j = 1, 2, 000 , no This number is given in the following corollary of Theorem 13.40 COROLLARY 13.2 Let F be the set of colorings (of the elements) of a .finite set Wn with colors from a .finite set Zro Then the number N(MF) of models of F. under the group 9n of
504
EQUIVALENCE CLASSES
permutations ofWn, is given by N(My) = C(r, r, ... , r; 9n),
(13.10)
where C(x1, x2, ... , xn; 9n) is the cycle indicator ofthe group 9nofpermutations ofWn. Example 13.9 Combinations with restricted repetition The number of ncombinations of r with restricted repetition may be derived, by an application of Theorem 13 .4, as follows. Consider the sets En ( Zr) and Un ( Zr) of ncombinations and npermutations of the set Zr = {z 1 , z 2 , ... , Zr} with repetition, respectively. Let us correspond to each combination Xn = {z 81 ,Z 82 , ••• ,Zsn} that belongs to En(Zr) the set of its permutations Pn(Xn), which is a subset of Un(Zr)· Note that Pn(Xn) and Pn(Yn) are disjoint sets for Xn and Yn different combinations from En(Zr) and also that each permutation from Un ( Zr) belongs in one of the sets P n (Xn), Xn E En(Zr)· Thus, the set {Pn(Xn) ~ Un(Zr) : Xn E En(Zr)} constitutes a partition of the set Un(Zr). Further, each permutation (z 81 , Z82 , ••• , Zsn) that belongs to Un ( Zr) completely determines a coloring f of the elements of the set Wn = {w1, w2, ... , wn} with colors from the set Zr = {z1, Z2, ... , Zr }, in which the element Wi oHV'n is painted with the color f( wi) = zs., i = 1, 2, ... , n. Thus, the set Un ( Zr) completely determines a set F of colorings of the set W n with colors from the set Zr. Also, two colorings f and gin F, which are respectively determined by the permutations (Zs 1 , Zs 2 , • • • , Zsn) and (z 8 ; 1 , Zs; 2 , • • • , Zs;n) in Pn(Xn), are equivalent under the permutation a in the symmetric groupPn = Pn(Wn) with a( wk) = Wik, k = 1, 2, ... , n. Indeed, from f(wi) = Zs, and g(wk) = z 8 ,k, it follows that f(a(wk)) = f(wik) = g(wk), k = 1, 2, ... , n, and so f(a) = g. Hence the subset Pn(Xn) ofUn(Zr) uniquely corresponds to the model Mj. Consequently, the number En(j 1 , )2, ... , ir) of ncombinations of the set Zr with repetition that include is ~ 0 times the element Zs for s = 1, 2, ... , r, with i 1 + )2 + · · · + ir = n, equals the number N n (i 1 , h, ... , ir) of models Pn(Wn), in which in My, under the symmetric group of permutations Pn is ~ 0 elements of Wn are painted with color Zs from Zr for s 1, 2, ... , r, with i1 + h + · · · + ir = n. The multivariate generating function of this sequence of numbers, according to Theorem 13.4 and using the cycle indicator of the symmetric group derived in Example 13.2, is given by
=
where
=
r
Xj = Xj(tt,t2,··· ,tr) = 'L:t~, j = 1,2, ... ,n, s=l
13.4. MODELS OF COLORINGS OF A FINITE SET
505
and Bn(x 1 , x 2 , ... , Xn) is the exponential Bell partition polynomial. The number E(r, n) of ncombinations of r with repetition equals the number of models in M.r under the symmetric group of permutations Pn = Pn(Wn) and, according to Corollary 13.2, is given by
1 E(r,n) = ,Bn(r, 1!r, ... , (n 1)!r). n. The generating function of the exponential Bell partition polynomials (see Corollary 11.1 ),
for
Xj
= (j  1)!r yields
tn
oo
2::: Bn(r, 1!r, ... , (n 1)!r),n. = exp[rlog(1 t)
1
]
= (1 t)r.
n=O
Therefore
2::: Bn(r, l.r, ... , (n oo
n=O
I
_
1
1).r)
_ tn _ 1
n.

00
2:::
(
n=O
r+n
1) t n
n
and 1 ... , (n E(r,n ) _ ..!._ Bn(r, l.r, n!
_
1)I.r ) _
(r + n 1) . n
D
Example 13.10 Cyclic permutations with restricted repetition The number of cyclic npermutations of r with restricted repetition may be derived, by using Theorem 13.4, as follows. Note that annpermutation (z 81 , Z82 , •• • , Zsn) of the set Zr = {Z1, Z2, ... , Zr} with repetition, in which Z8 >+ 1 follows Zs,, i = 1, 2, ... , n  1, and Z81 follows Zsn, is called a cyclic npermutation of r with repetition. Consider the sets Kn (Zr) and Un (Zr) of the cyclic and the linearnpermutations, respectively, of the set Zr. Let us correspond to each cyclic permutation Cn = (z 8 " Z 82 , •• • , Zsn) that belongs to Kn (Zr) the set of the corresponding linear permutations L:n(cn). which is a subset ofUn(Zn)· Clearly, the set {L:n(cn) ~ Un ( Zr) : Cn E Kn (Zr)} constitutes a partition of the set Un (Zr). Further, each (linear) permutation ( Z81 , Z82 , ••• , Zsn) that belongs to Un ( Zr) completely determines a coloring f of the elements of the set Wn = { w1 , w 2 , .•. , wn} with colors from the set Zr = {z 1 , z 2 , .•• , Zr }, in which the element Wi of Wn is painted with the color f(wi) = z 8 , , i = 1, 2, ... ,n. Thus, the set Un(Zr) completely detennines a set :F of colorings of the set Wn with colors from the set Zr.
EQUIVALENCE CLASSES
506
Also, two colorings f and gin :F, which are respectively determined by the permutations (Zs 1 , Zs 2 , ... , Zsn) and (zs~+l, Zs~+ 2 , ... , Zsn, Zs 1 , ••• , Zs~) in .Cn (en), are equivalent under the permutation CTm in the cyclic group 9n,n• of order nand degree n, with CTm(wi) = Wm+i• i = 1, 2, ... , n m and CTm(wi) = Wmn+i• i = n  m + 1, n  m + 2, ... , n. Indeed, from f(wi) = Z 8 ;, i = 1,2, ... ,nand g(Wi) = Zs~+i' i = 1,2, ... ,n m, g(wi) = Zs~n+i' i = nm+1,nm+2, ... ,n,itfollowsthatf(am(wi)) = f(wm+i) = g(wi), i = 1,2, ... ,n m, f(am(wi)) = f(wmn+i) = f(wi), i = n m + 1, n m + 2, ... , nand so f(a) =g. Hence the subset .Cn(cn) ofUn(Zr) uniquely corresponds to the model M 1. Consequently, the number Kn(h,i 2 , ... ,ir) of cyclic npermutations of the set Zr with repetition that include is 2 0 times the element z 8 for s = 1, 2, ... , r, with ]r + jz + · · · + ir = n, equals the number Nn(it,iz, ... ,ir) of models in M:F, under the cyclic group 9n,n• of order n and degree n, in which is 2 0 elements of Wn are painted with color z 8 from Zr for s = 1, 2, ... , r, with i 1 + iz + · · · + ir = n. The multivariate generating function of this sequence of numbers, according to Theorem 13.4 and using the cycle indicator of the cyclic group of permutations derived in Example 13.4, is obtained as
"'""'K ( . . ... ,Jr . )tj' L....t n]I,Jz, 1 th 2
 1 "'""'A..(d) (td1 ··· tJr r ;L....t'+'
+ tdz+··· + td)n/d r
din
and so
K (.
. .) n Jt,Jz, ... ,Jr
1"'""'
=; ~¢
(d)
(n/d)!
(]r/d)!(jz/d)!···(ir/d)!'
where i is the greatest common divisor of the nonzero numbers among the r numbers j 1 ,jz, ... ,jr and ¢(d) is Euler's function. The number Kn,r of cyclic npermutations of r with repetition equals the number of models in M:F. under the cyclic group of permutations 9n,n• of order n arid degree nand, according to Corollary 13.2, is given by
Example 13.11 Models of colorings of a cube with r colors Calculate the number of models of coloring the six faces of a cube with r different colors, when two colorings are considered to be of the same model if they can be transformed to each other by rotation of the cube. Let W6 = {1, 2, 3, 4, 5, 6} be the set of the six faces of a cube and Zr = { z 1 , z 2 , ... , Zr} the set of the r different colors. Note that the six faces of a usual cube are numbered so that {1, 6}, {2, 5} and {3, 4} are nonordered pairs of opposite faces. Further, let 96 be the group of permutations of W6 that are induced
13.5. BIBLIOGRAPHIC NOTES
507
by rotations of the cube around its axes of symmetry. The cycle indicator of the group 96 , derived in Example 13.5, is given by C(x1, x2, ... , x6; 96)
= 214 (x~ + 3xix~ + 6xix4 + 6x~ + 8xD.
The set :F of colorings of the set W 6 = {1, 2, 3, 4, 5, 6}, of the six faces of a cube, with colors from the set Zr is partitioned, under the group 96 , into a set of models M.r. The numberC6,r N(M.r) of models of coloring the six faces of a cube with r colors, according to Corollary 13.2, is given by the cycle indicator C (x 1 , x 2 , ... , x 6 ; 96 ) for x j = r, j = 1, 2, ... , 6 and so
=
1 6 C6,r = (r 24
+ 3r 4 + 12r3 + 8r 2 ).
The following few values of the number C 6 ,r may help to get an idea of its magnitude: C6,2 = 10, C6,3 = 57, C6,4 = 234, C6,5 = 800,
C6 , 6
13.5
= 2226,
C6,7
= 5390,
C 6 ,s
= 11,712.
0
BIBLIOGRAPHIC NOTES
W. S. Burnside, in 1897, essentially enumerated the orbits of the elements of a finite set, under a group of permutations of it. This result, known as Burnside's lemma, is included in Burnside (1911). The great Hungarian mathematician George P6lya provided fundamental techniques for counting equivalence classes. The famous P6lya's counting theorem was derived in G. P6lya (1937) and its combinatorial significance was demonstrated by J. Riordan (1958) and R. C. Read (1987). An elegant and useful generalization of it is given by N. G. DeBruijn (1964). The article of A. Tucker (1974) provides an elementary presentation of P6lya's counting method through examples. A rigorous treatment of this method can be found in C. Berge (1971) and L. Comtet (1974).
13.6
EXERCISES
1. Let W3 = {1, 2, 3} be the set of the corners of an equilateral triangle numbered circularly in the counterclockwise direction. (a) Determine the group 93, of permutations of w3, that is generated by rotation or reflection
508
EQUIVALENCE CLASSES
of the triangle around its axes of symmetry and show that its cycle indicator is given by
C(x1, x2, x3; ~h)
= 61 (x~ + 3xlx2 + 2x3).
(b) Calculate the number of models of coloring the three corners of the triangle with a set of r colors when two colorings are considered to be of the same model if they can be transformed to each other by rotation or reflection of the triangle around any of its axes of symmetry. (c) Show that, if r is a positive integer, then the positive integer r(r 2 + 3r + 2) is divisible by 6. 2. Let W 4 = {1, 2, 3, 4} be the set of the four corners of a square. Determine the dihedral group 94, of permutations of W 4 , that is generated by rotation or reflection of the square around its axes of symmetry and show that its cycle indicator is given by
3. (Continuation) Calculate (a) the number K 4 (j) of models of coloring the four corners of a square with j corners red and the other 4  j corners black, for j = 0, 1, 2, 3, 4 and (b) the number K 4 ,r of models of coloring the four corners of a square with a set of r colors when two colorings are considered to belong to the same model if they can be transformed to each other by rotation or reflection of the square around any of its axes of symmetry. 4. Let W4 = { 1, 2, 3, 4} be the set of the sides of a pyramid (tetrahedron). The base bears number 1 and the other sides are numbered circularly 2, 3 and 4 in the counterclockwise direction. (a) Determine the group 94 , of permutations of W 4 , which is generated by rotation of the pyramid around the vertical axis that connects the apex with the center of the base and show that its cycle indicator is given by 1 4 ) C(x1,x2,x3,x4;94) = 3(x 1 +2xlx3.
(b) Calculate the number of models of coloring the four sides of the pyramid with a set of r colors when two colorings are considered to belong to the same model if they can be transformed to each other by rotation of the pyramid around its axis of symmetry. 5. (Continuation) Calculate (a) the number K 4 (j) of models of coloring the four sides of a pyramid with j sides red and the other 4  j sides black, for j = 0, 1, 2, 3, 4 and (b) the number K 4 (1, 1, 1) of models of coloring
509
13.6. EXERCISES
the four sides of a pyramid with a set of four colors when in both cases two colorings are considered to belong to the same model if they can be transformed to each other by rotation of the pyramid around its axis of symmetry. 6. Let W 4 = {1, 2, 3, 4} be the set of the sides of a regular tetrahedron. (a) Determine the group ~h, of permutations of W4, which is generated by rotation of the tetrahedron around its axes of symmetry and show that its cycle indicator is given by
7. (Continuation) Calculate (a) the number K 4 (j) of models of coloring the four sides of a regular tetrahedron with j sides red and the other 4 j sides black, for j = 0, 1, 2, 3, 4 and (b) the number K4,r of models of coloring the four sides of a regular tetrahedron with a set of r colors when, in both cases, two colorings are considered to belong to the same model if they can be transformed to each other by rotation of the tetrahedron around any of its axes of symmetry. 8. (Continuation) Calculate the number of models of coloring the four sides of a regular tetrahedron with four different colors, each side with a different color, when two colorings are considered to belong to the same model if they can be transformed to each other by rotation of the tetrahedron around any of its axes of symmetry. 9. Let W 8 = {1, 2, ... , 8} be the set of the eight vertices of a cube. Determine the group 98 , of permutations of W 8 , that is generated by rotation of the cube around its axes of symmetry and show that its cycle indicator is given by
10. (Continuation) Let N 8 (i,j,k) be the number of models of coloring the eight vertices of a cube with i red, j green and k = 8  i  j black vertices, when two colorings are considered to be of the same model if they can be transformed to each other by rotation of the cube. Determine the generating function gs(t, u, w) =
LN
8 (i,j,
and show that N 8 (1, 3, 4) = 13, N 8 (2, 2, 2) Ns,3 = "£ Ns(i,j, k) = 333.
k)tiuiwk
=
22, N 8 (2, 3, 3)
24 and
EQUIVALENCE CLASSES
510
11. Let W 12 = {1, 2, ... , 12} be the set of the 12 edges of a cube. Determine the group 912 , of permutations of W 12 , that is generated by rotation of the cube around its axes of symmetry and show that its cycle indicator is given by C(x1, x2, ... , x12i 9I2)
1
= 24 (x~ 2 + 6xix~ + 3x~ + 8x~ + 6x~).
12. (Continuation) Calculate (a) the number of models of coloring the 12 edges of a cube with 12 different colors, each edge with a different color and (b) the number of models of coloring the 12 edges of a cube with a set of r colors, when in both cases two colorings are considered to belong to the same model if they can be transformed to each other by rotation of the cube. 13. Let W 6 = { 1, 2, ... , 6} be the set of the six vertices of a regular octahedron. Determine the group 96 , of permutations of W 6 , that is generated by rotation of the octahedron around its axes of symmetry and show that its cycle indicator is given by C(x1, x2, ... , x6; 96) =
1 (x~ 24
+ 6xix4 + 3xix~ + 6x~ + Bx~).
14. (Continuation) Calculate (a) the number K 6 (j) of models of coloring the six vertices of an octahedron with j vertices red and the other 6  j vertices black, for j = 0, 1, ... , 6 and (b) the number K 6 ,r of models of coloring the six vertices of an octahedron with a set of r colors when in both cases two colorings are considered to belong to the same model if they can be transformed to each other by rotation or reflection of the octahedron. 15. Models of coloring a roulette. The disc of roulette is divided into n equal sectors and can be rotated around its axis. Evaluate the number of models of coloring the n sectors of the disc of the roulette with a set of r colors. 16. (Continuation) Let Kn(i) be the number of models of coloring the n sectors of the disc of roulette with j red and n  j black sectors. Show that
Kn(j)=~
L ¢(d)(;j~,
j=0,1, ... ,n, n=1,2, ... ,
dlmj,n
where mo,n = n, m 1,n, j = 1, 2, ... , n, n = 1, 2, ... , is the greatest common divisor of j and n, and ¢(d) is Euler's function.
511
13.6. EXERCISES
1 7*. Let W n = {1, 2, ... , n} be the set of the n vertices of a regular ngon. Determine the dihedral group 9n, of permutations of Wn, that is generated by rotation or reflection of the ngon around its axes of symmetry and show that its cycle indicator is given by
for n = 2m
+ 1 and
for n = 2m, where ¢(d) is Euler's function. 18. Models of necklaces. Evaluate the number of models of necklaces with n beads of r different colors, when two necklaces are considered to be of the same model if they can be transformed to each other by rotation or reflection with respect to the diameter. 19*. Let 1ln = {h1,h2, ... ,hn} be a finite group of order N(1ln) = n and consider a fixed point h in 1ln. Show that (a) the mapping ah of1ln into itself, defined by ah(hr) = hhr, r = 1, 2, ... , n, constitutes a permutation of the elements of 1ln, and (b) the group 9n of the permutations ah for all h in 1ln is isomorphic to 1ln, that is, a gh = a 9 ah for any g and h in 1ln. Also show that (c) the cycle indicator of the group 9n is given by
where N(1ln;d) is the number of elements h in 1ln of order d.
20*. Partitions and compositions of integers. Let P n,k be the set of partitions of n into at most k parts and Cn,k the set of compositions (ordered partitions) of n into at most k parts. Let us correspond to each partition Rk = {rl,r2,··· ,rk}, r1 :::: r2 :::: ··· :::: rk :::: 0, that belongs to Pn,k the set of its permutations Pk(Rk), which is a subset of Cn,k· The set {Pk(Rk) : Rk E Pn,k} constitutes a partition of the set Cn,k· Using P6lya's counting theorem, show that the generating function of the sequence of numbers P(n, k) = N(Pn,k), n = 0, 1, ... , for fixed k, is given by 00
Fk(t}
=L n=O
P(n, k)tn
= Ck((1 t) 1, (1 t 2 ) 1 , ...
,
(1 tk} 1 ; Pk),
512
EQUIVALENCE CLASSES
where C(x 1 ,x2 , ••. ,xk;Pk) is the cycle indicator of the symmetric group Pk of order k! and conclude the identity k
ck ( (1  t) 1, (1  t2) 1, ... , (1  tk) 1; Pk)
= II (1 i=1
ti) 1.
Chapter 14 RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
14.1
INTRODUCTION
In the classical Newcomb's problem, cards are successively drawn one after the other from a deck of m numbered cards of a general specification. The cards are placed on a pile as long as the number Xj of the card drawn at the jth drawing is greater than or equal to the number XjI of the card drawn in the previous drawing; a new pile is started whenever the number XJ is less than XjI, for j = 1, 2, ... , m. The enumeration of the number of different arrangements of the cards that lead to a given number of k piles is of interest. Clearly, this is a problem of counting the permutations of n kinds of elements, with s 1 , s 2 , ... , sn elements respectively, which include k nondecreasing sequences of consecutive elements. This chapter is devoted to the enumeration of the permutations of the set {1, 2, ... , n} that have k increasing sequences of consecutive elements and the npermutations of the set {1, 2, ... , s} with repetition that have k nondecreasing sequences of consecutive elements. The Eulerian and the Garlitz numbers, which enumerate these permutations, are presented in the first two sections.
14.2
EULERIAN NUMBERS
The nth power may be expressed as a sum of binomials (binomial coefficients) of the nth order. Specifically, we successively get the expressions
t
0
=
(t)O't (t)1 ' t t (t + 1) +2 (t 1) (t +2 1) +(t) 2' 1
=
2
=
=
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
514
and, generally,
Then, the following definition is introduced.
DEFINITION 14.1 The coefficient A(n, k) of expansion ( 14.1), of the nth power into binomials of the nth order, is called Eulerian number. Clearly, this definition implies A(n, k) = 0, k > n. The numbers A(n, k), k = 0, 1, ... , n, n = 0, 1, ... , attributed to Euler are called Eulerian to distinguish them from the Euler numbers E 2 n (see Exercise 18). Expansion (14.l),uponreplacing t by t and using the relation
REMARK 14.1
is transformed to
tn =
t
A(n, k)
C+ ~
1 ), n = 0, 1, ... ,
k=O
or equivalently to
tn
=
t
A(n, n
j + 1) (t + ~ j),
n
= 0, 1, ....
J=O
Consequently
A(n,k)=A(n,nk+1), k=0,1, ... ,n, n=0,1, .... The number
A(n,k)=A(n+k+1,k+1), k=0,1, ... ,n, n=0,1, ... , (14.2) which satisfies the symmetric property A(n, k) = A(k, n) is called symmetric Eulerian number. I
515
14.2. EULERIAN NUMBERS
An explicit expression and a recurrence relation for the Eulerian numbers are derived in the following two theorems.
THEOREM 14.1 The Eulerian number A(n, k), k
= 0, 1, ...
, n, n
= 0, 1, ... , is given by the sum (14.3)
PROOF From expansion (14.1), replacing the variable k by j and putting t = k  r, for r = 0, 1, ... , k, we deduce the expression
=
(k r)n
1:
A(n,j) (n
+ k: r 1), r = 0, 1, ... , k,
J=O
which, upon using the relation
(
n1·)•
1
n+krj) = (n+kr~j) =( )krj( n krJ krJ
may be written as
(k r)n
= ~ A(n,j)( 1)krj ( L....t
j=O
1
n .) , r krJ
= 0, 1, ...
, k.
Multiplying both sides of the last expression by ( 1 Y (n~ 1 ) and summing for r = 0, 1, ... , k, we get
l) 1r ~ 1) (n
(k rt =
r=O
t
1:A(n,j)( 1)kJ (n
~ 1) (kn r1 .) J
r=OJ=O
k
=
kj
~A(n,j)(1)kj ~ (n;
1) (knj1r)
and since, by Cauchy's formula,
kLJ (n
+
r=O
r
1) ( n 1r )  ( 0 )
expression (14.3) is deduced.
k j 
I

k j
= cSk,j,
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
516
THEOREM 14.2
TheEuleriannumbersA(n,k), k gular recurrence relation
= 0,1, ...
,n, n
= 0,1, ... , satisfy the trian
+ 1, k) = kA(n, k) + (n k + 2)A(n, k 1), , n + 1, n = 0, 1, ... , with initial conditions
(14.4)
A(n fork= 1, 2, ...
A(O,O)
= 1,
= 0,
A(n,O)
n > 0, A(n,k)
= 0,
PROOF Expanding both members of the identity tn+I = coefficients, by the using ( 14.1 ),
k > n.
un
into binomial
~ A(n + 1, k) ( t + : ~ ~ + 1) = ~A( n, k) ( t + : k) t and, since
(
t + n k) t n
= (t + n k) k(t + n k + 1) + (n k + 1)(t k)
(t
n+1
n
= k + n k + 1) + (n _ k + 1) n+1
(t +n+1
n k),
we get the expression
I: k=O
A(n + 1, k) (t + n k + 1) n+1
=t k=O
kA(n, k) (t + n k + 1) n+1
+ t ( n  k + 1)A(n, k) k=O
(t + +
n k).
n
1
Equating the coefficients of binomials C+~:;_~+ 1 ) in both sides of this expression we deduce (14.4). The initial conditions follow directly from (14.1). Note that the initial conditions may be replaced by A(n, 1) = 1, n > 0, A(n, k) = 0, k > n, which can be checked in the applications. I The Eulerian numbers can be tabulated by using recurrence relation
(14.4) and its initial conditions. Table 14.1 gives the numbers A(n, k), k
= 1, 2, ... , n, n = 1, 2, ... , 9.
The derivation of a bivariate generating function of the Eulerian numbers is facilitated by an expression of the Eulerian polynomial, n
An(t) =
L
A(n, k)tk, n
k=O deduced in the following lemma.
= 0, 1, ... ,
(14.5)
14.2. EULERIAN NUMBERS
517
Table 14.1 Eulerian Numbers A(n, k)
k
n I 2 3 4 5 6 7 8 9
2
3
4
5
6
7
I I I I 4 I 11 1 26 57 I I I20 I 247 I 502
I II 66 302 II9I 4293 I4608
I 26 302 24I6 I56I9 88234
I 57 II9I I56I9 I56190
I I20 4293 88234
I 247 I4608
I
LEMMA 14.1 The Eulerian polynomial An(t), n
8 9
I 502
I
= 0, 1, ... , is alternatively expressed as 00
An(t)
= (1 t)n+l Ljntj,
n
= 0, 1, ....
(14.6)
j=O
PROOF Expression (14.5) of the Eulerian polynomial, upon introducing the explicit expression ( 14.3) of the Eulerian numbers, may be written as
Note that the inner sum, in agreement with A(n, k) = 0, k > n, vanishes for k = n + 1, n + 2, ... , and so in the outer sum the summation may be extended to infinity without altering its value. Thus
An(t) = =
~t,(1r(n~ 1 )(kr)ntk ~(1r(n;1)tr~(kr)ntkr
and, since
~ ( 1 r (n ~ 1) tr = expression (14.5) is deduced.
I
( 1  t) n+ 1 '
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
518
THEOREM 14.3 The bivariate generating function of the Eulerian numbers A(n, k), k ... , n, n = 0, 1, ... , is given by oo
n
1
n
t
= 1 teu(lt).
'"''"' ~ A(n, k)t kun! g(t, u) = ~ n=Ok=O
= 0, 1, (14.7)
PROOF The bivariate generating function of the Eulerian numbers, using expression (14.6) of the Eulerian polynomials, may be expressed as
g(t,u) =
f(l t)n+l fjntj:~ n=O
= (1 t) Etj
j=O
j=O
f
[ju(1n~ t)]n
n=O
and, since
~ [ju(1~ t)]n
~ n=O
= eiu(lt),
~[teu(It)Jj
~ j=O
=
1
1 teu(It)'
I
expression (14.7) is readily deduced.
REMARK 14.2 The bivariate generating function of the shifted Eulerian numbers A(n, k + 1), k = 0, 1, ... , n 1, n = 1, 2, ... , 00
h(t, u) = 1 +
n1
LL
n=l k=O
un A(n, k + 1)tk 1 , n.
which emerges in their applications, is closely connected with generating function (14.7). Specifically, since A(n, 0) = 0, n > 0, it follows that
h(t,u) = 1 + g(t,u;  1 and so oo
h(t, u)
=1+ L
n
L
:!
n
A(n, k)tkl
n=lk=l
which is the required generating function.
1t = eu(lt) _ t'
(14.8)
I
The Eulerian numbers are connected with the Stirling numbers of the second kind as shown in the following theorem.
14020 EULERIAN NUMBERS
519
THEOREM 14.4 The Eulerian numbers A(n, k), k = 0, 1, 000 ,n, n = 0, 1, 000' are expressed in terms of the Stirling numbers of the second kind S(n, k), k = 0, 1, 000 , n, n = 0, 1, 000' by (1409)
and inversely S(n, r)
= ;:!1 ~ L;
(n k)
r _ k A(n, k)o
(14010)
k=O PROOF
Expression (1406) of the Eulerian polynomials, upon expanding the powers of j into factorials of it by using (803), may be written as
and, since
it reduces to n
An(t) =
L r!S(n, rW(l t)nro
(l4oll)
r=O Thus, by virtue of (1405),
Equating the coefficients of tk in the first and last member of this relation we conclude (l4o9)o Again, from (14011) and setting u = t/(1 t), whence t = u/(1 + u), we get, by virtue of (1405), n
n
'L:r!S(n,r)ur = LA(n,k)uk(l +u)nko r=O k=O
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
520
Thus
n
~ r!S(n, r)ur
=
=
t; n
A(n, k)uk
n (
~ ~
=kk) urk
~{~ G=~)A(n,k)}ur
and, equating the coefficients of u r in the first and last member of this relation. we conclude (14.10). I
Example 14.1 Power series with coefficients powers of fixed order The first application of the Eulerian polynomials is in the expression of the power series with coefficients powers of fixed order. Specifically, solving (14.6) with respect to the power series, we get the expression
~ ·n j _ An(t) _ L. J t  (1  t)n+l ' n  0, 1, .... J=O
In particular, for n
= 1, 2, 3, 4 and using Table 14.1, we get the expressions 00
"
00
. j 
{=: Jt ~ ·3 L.,..J
j 
t
t(l

t " (1 t)2'
+ 4t + t 2 ) ~
(1t)3
J=l
·2 j 
{=: J ·4
'L.,..J
t
j 
t

t(1
(1 ) t +t (1  t)2'
+ llt + llt 2 + t3 ) (1t)4
.
J=l
A probabilistic problem that requires the evaluation of such a power series is the following. Consider a sequence of throws of a coin and assume that at any throw the probability of falling heads is p and so the probability of falling tails is q = 1  p. Then the probability that j tails preceded the first appearance of heads is given by Pi =pqi, j = 0,1, .... This is the probability function of a geometric distribution. Calculate the moments J.L~, r = 1, 2, ... , of the sequence of probabilities Pi• j = 0, 1, .... The rth order moment is given by 00
00
J.L~=l:FPJ=Pl:Fqi, r=1,2, ... j=l
and so
j=l
14.2. EULERIAN NUMBERS
521
In particular, the mean and the variance are obtained as
P'Q
I
J.t = Itt =
2
I
q
2
a = J.t2  J.t = p2 ·
0
Example 14.2 Sample moments Consider a sample {i 1 , i 2 , ... , iN} of N observations from a population (set) of nonnegative integers and let Ni be the frequency (number of appearances) of the number j for j = 0, 1, ... , n, so that N 0 + N 1 + · · · + Nn = N. The rth order sample moment 1 n mr= NLrNi j=O
can be expressed in terms of the successive partial sums, n
Nr,k =
L Nr1,j,
r
= 1, 2, ... , k = 0, 1, ... , n,
j=k
where No,j = Nj. j = 0, 1, ... , n. Clearly, the doubleindex sequence Nr,k. = 0, 1, ... , k = 0, 1, ... , n, satisfies the triangular recurrence relation
r
for r = 1, 2, ... , k = 1, 2, ... , n, with No,k = Nk> k = 0, 1, ... , n. Setting r = i, k = j + 1 and multiplying the resulting relation by  ~  i + 1) = (j + r  ~  i) + (j + r  ~  i + 1) , (j + rrz+1 rz rz+1
fori= 1,2, ... ,r,weget
(j + rr 
k  i + 1) _ N·1 3· i +1 ' '
(j + rr  i) N· · + (j +r r  + 1 i) N _(j + rri+1 + 1) ki
k
'• 3
i
k i
·
t,J
N .
t,J+I,
fori = 1, 2, ... , r. Summing this expression for j = k, k + 1, ... , n and since (/i1 1 ) = 0, Ni,n+l = 0, we get
(j + ri+1 k i + 1) N·_ . = ~ (j + k i) N· . ' ri '• ~ (j + r  k  i) N· . _ ~ (j + r  k  i + 1) N· . + rz. + 1 rz + 1
~
r
~ j=k
~
j=k+l
13 '
t,J
~ j=k
~
j=k
r
3
.
t,J+l,
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
522
fori = 1, 2, ... , rand so
n 2:
(j
j=k
2:n
+ r · k  i + 1) N·_ .= , 13 ri+1 '
(j
j=k
+ r  k  i) N·, . , i = 1, 2, ... , r. ri '1
Applying it successively, we get
In particular, for i = r we deduce the expression Nr+l,k
~ = {;;:.
(j
+rr
k) Nj.
Using (14.1), the rth order sample moment mr. is expressed as mr
= ~ tFNJ = ~ ttA(r,k)(j +~ k)Nj J=Ok=O
J=O
=
~
t
A(r, k)
k=O
t (j
+ ~ k) NJ
J=k
and so
1
mr
T
= N 2: A(r, k)Nr+l,k·
0
k=O
14.3
CARLITZ NUMBERS
Consider the generalized binomial of t of order n and scale parameter s, st) _ (st)n _ st(st 1) · · · (st n + 1) _ (st) _ , n 1, 2 , ... ,  1, (n n.1 n.1 0 with s a real number. It can be expressed as a sum of binomials oft+ n k of order n fork = 0, 1, ... , n with coefficients depending on the nonnegative integers n and k and the parameter s. Specifically, we get successively the expressions
523
14.3. GARLITZ NUMBERS
and, generally, (14.12)
Then, the following definition is introduced. DEFINITION 14.2 The coefficient B(n, k; s) of expansion (14.12), of the nth order generalized binomial oft with scale parameters into binomials of the nth order, is called Carlitz number.
Clearly, this definition implies B(n, k; s) REMARK 14.3 using the relation
= 0, k > n.
Expansion ( 14.12), upon replacing t by  t and s by  s and
is transformed to
or, equivalently, to
Consequently,
B(n,k;s)
= (1)nB(n,n k + 1; s)
fork= 0, 1, ... , nand n = 0, 1,....
(14.13)
I
An explicit expression and a recurrence relation for the Garlitz numbers are derived in the following two theorems. THEOREM 14.5 The Carlitz number B(n, k; s), k = 0, 1, ... , n, n = 0, 1, ... , is given by the sum
(14.14)
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
524
From expansion (14.12), replacing the variable k by j and putting
PROOF
t = k  r, for r = 0, 1, .. . , k, we deduce the expression
which, upon using the relation
(
n+krj) = n
(n+kr~j) =( 1)krj( krJ
n1.),
krJ
may be written as
Multiplying both sides of the last expression by (1)r(n~ 1 ) and summing for r = 0, 1, ... , k, we get
and since, by Cauchy's formula,
~ ~
(n + 1) (
1) (kj 0 ) =l5k,j,
n kjr
r
expression (14.14) is deduced.
=
I
THEOREM 14.6
The Carlitz numbers B(n, k; s ), k = 0, 1, ... , n, n = 0, 1, ... , satisfy the triangular recurrence relation (n
+ 1)B(n + 1, k; s) = (sk n)B(n, k; s) +[s(n k + 2) + n]B(n, k
fork= 1, 2, ... , n
+ 1, n
1; s),
= 0, 1, ... , with initial conditions
B(O,O;s) = 1, B(n,O;s) = 0, n
> 0, B(n,k;s)
= 0, k
> n.
(14.15)
525
14.3. GARLITZ NUMBERS
PROOF
Expanding the generalized binomials in both sides of the identity
by using (14.12),
~ ~(n + 1)B(n + 1, k; s) (t+nk+1) n+1
k=O
=~ f='oB(n,k;s) (t+nn and, since
e
+: k) (st n)
k) (stn)
= (sk n) (t +: ~ ~ + 1) +[s(nk+1)+n] (
t
+n+ n
1
k) ,
we get the expression
~ (t+n k+ t:'o(n + 1)B(n + 1, k; s) n+ 1
=
1)
~(skn)B(n,k;s)C+:~~+ 1 ) + ~[s(n k + 1) + n]B(n, k; s)
(t:: ~
k).
Equating the coefficients of binomials c+:~~+l) in both sides of this expression we deduce (14.15). The initial conditions follow from (14.12).
I
The Garlitz numbers can be tabulated by using the recurrence relation
(14.15) and its initial conditions. Table 14.2 gives the numbers B(n,k;s), k = 1, 2, ... , n, n = 1, 2, ... , 4. The derivation of a bivariate generating function of the Garlitz numbers is facilitated by an expression of the Garlitz polynomial, n
Bn(t;s)=l:B(n,k;s)tk, n=O,l, ... , k=O
deduced in the following lemma.
(14.16)
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
526
Table 14.2
Garlitz Numbers B(n, k; s)
k
3
2
1
4
n 1
2
3
4
G) e) G)
c;1) 4c;1) c;2) (:) C: 2) + wC: 1) 10 (s: 2) + c: 1) (s: 3)
LEMMA 14.2 The Carlitz polynomial Bn (t; s ), n
Bn(t; s)
= 0, 1, 2, ... , is alternatively expressed as
= (1 t)n+I
f: (sj)
i=O
ti, n
= 0, 1, ....
(14.17)
n
PROOF Introducing in polynomial (14.16) the explicit expression (14.14) of the Carlitz numbers, we get
Note that the inner sum, in agreement with B(n, k; s) = 0, k > n, vanishes for k = n + 1, n + 2, ... and so in the outer sum the summation may be extended to infinity without altering its value. Thus
Bn(t; s)
= ~ ~(
1r (n ~ 1) (s(k: r)) tk
= ~( 1 r(n~1)tr~ (s(k:r))tkr
14.3. GARLITZ NUMBERS
527
and, since
~(1r(n~ 1)tr = expression ( 14.17) is deduced.
(1 t)n+l,
I
THEOREM 14.7 The bivariate generating function of the Carlitz numbers B(n, k; s), k = 0, 1, ... , n, n = 0, 1, ... , is given by
=?; ,{; B(n, k; s)tkun =1 _ t[1 1t + u( 1 _ t)jB. n
oo
g(t, u; s)
(14.18)
PROOF The bivariate generating function of the Carlitz numbers, using expression (14.17) of the polynomials ( 14.16), may be expressed as
g(t,u;s) =
~(1 t)n+l ~ (~)tiun
= (1 t) ~ti ~ (~)[u(1 t)t and, since
~ ( ~) [u(1  t)t = ~o (t [1 + u(1 f='
t)
]s)j
[1
+ u(1 1
+ u(1 
1 t[1
8
t)] i,
t)]s,
I
expression (14.18) is readily deduced.
REMARK 14.4 The bivariate generating function of the shifted Carlitz numbers
B(n, k
+ 1; s), k = 0, 1, ... , n 1, n = 1, 2, ... , oo n1
h(t,u; s)
= 1+ L
L
n=lk=O
n
B(n, k
+ 1; s)tk;, n.
which emerges in their applications, is closely connected with the generating function (14.18). Specifically, since B(n, 0; s) = 0, n > 0, it follows that
g(t, u; s) 1 . )_ h( t,u, s  1 + t
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
528
and so <X>
h(t, u; s)
n
un
=1+ L
L B(n, k; s)tkl n! n=lk=l
which is the required generating function.
1 t
= [1 + u( 1 _
t)]s _ t' (14.19)
I
The Carlitz numbers are connected with the generalized factorial coefficients (see Section 8.4) as shown in the following theorem. THEOREM 14.8 The Carlitz numbers B( n, k; s ), k = 0, 1, ... , n, n = 0, 1, ... , are expressed in terms of the generalized factorial coefficients C(n, k; s), k = 0, 1, ... , n, n = 0, 1, by 0
0
.,
(14.20)
and inversely
C(n,r; s)
= n!r!
k)
r (nL r _ k B(n, k; s). k=O
(14.21)
PROOF Expression ( 14.17) of the polynomial ( 14.16), upon expanding the generalized binomial of j of order n and scale parameters into factorials of j by using (8.39), may be written as <X> n I ( .) Bn(t;s)=(1tt+ 1 LLr;c(n,r;s) 1 ti j=O r=O n. r
and, since
it reduces to n
Bn(t;s)
=L
r=O
I
r. C(n,r;s)r(1 t)nr.
n.1
(14.22)
14.3. GARLITZ NUMBERS
529
Thus, by virtue of (14.16),
r!
(n
n n n r) {;B(n,k;s)tk = ~n!C(n,r;sW~(1)kr kr tkr
=
E{t,(
1)kr :\
(~ =~) C(n, r; s)} tk
and, equating the coefficients of tk in the first and last member of this relation, we conclude (14.20). Again, from (14.22) and setting u = t/(1  t), whence t = u/(1 + u), we get, by virtue of (14.16), n
L r=O
n
1
r. C(n,r;s)ur n.1
= LB(n,k;s)uk(1 + u)nk. k=O
Thus
and, equating the coefficients of ur in the first and last member of this relation, we I conclude (14.21). A limiting expression of the Carlitz numbers as s + the following theorem.
±oo
is derived in
THEOREM 14.9 Let B(n, k; s) be the Carlitz number. Then
lim snn!B(n, k; s) 8>±oo
= A(n, k),
( 14.23)
where A(n, k) is the Eulerian number. PROOF Multiplying both members of expression (14.14) by snn!, taking the limit ass+ ±oo and since lims>±oo sn(s(k r))n = (k r)n, we get
~(
1r lim snn!B(n, k; s) = s>±oo L.... r=O
=
(n +
1 ) [ lim sn(s(k r))n] r 8>±00
~(1r(n; 1)(k r)n.
Thus, by virtue of (14.3), the limiting expression (14.23) is established.
I
530
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
14.4 PERMUTATIONS WITH A GIVEN NUMBER OF RUNS The permutations of a finite set can be classified according to the number and the length of the increasing (or decreasing) sequences of consecutive elements they include. In this respect, the following definition is introduced.
DEFINITION 14.3 Let (jt,h, ... ,jn) be apennutation of the set {1, 2, ... , n}. The sequence of consecutive elements
Um,im+l, ... ,im+rd,
with
im < im+l < · · · < im+r1,
is called ascending run of length r, for 2 :S m :S n  1, 1 :S r :S n  2, if Jm1 > im and Jm+r1 > im+r: in particular, form = 1, 1 :S r :S n 1, if Jr > Jr+b form = n r + 1, 1 :S r :S n 1, if inr > Jnr+b and for m = 1, r = n, without any additional condition. This sequence is called descending run of length r if the corresponding conditions are satisfied with the reverse inequalities.
Example 14.3 Consider the permutations (j 1 , )2, )3, j 4 ) of the set {1, 2, 3, 4}. Clearly these 24 permutations can be classified according to the number of their ascending runs as follows: (a) The permutation (1,2,3,4) is the only one that has one ascending run of length 4. (b) The permutations with two ascending runs are the following II:
(1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (2,1,3,4), (2,3,1,4), (2,3,4,1), (2,4,1,3), (3,1,2,4), (3,4,1,2), (4,1,2,3). (c) The permutations with three ascending runs are the following II:
(1, 4, 3, 2), (2, 1, 4, 3), (2, 4, 3, 1), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3,4,2,1), (4,1,3,2), (4,2,1,3), (4,2,3,1), (4,3,1,2). (d) The permutation
(4, 3, 2, 1) is the only one that has four ascending runs, each of length I.
D
THEOREM 14.10 The numberofpennutations of the set {1, 2, ... , n} with k ascending runs equals A(n, k), the Eulerian number.
14.4. PERMUTATIONS WITH A GIVEN NUMBER OF RUNS
531
PROOF Note that, from any permutation of the set { 1, 2, ... , n }, by attaching the element n + 1 in any of the n + 1 possible positions (one before the first element, n  1 between the n elements and one after the last element) n + 1 permutations of the set { 1, 2, ... , n + 1} are constructed. Further, this attachment either keeps unchanged or increases by one the number of runs. Specifically, (a) from each permutation of the set { 1, 2, ... , n} with k ascending runs, by attaching the element n + 1 after the last element of any of these runs, k permutations of the set { 1, 2, ... , n + 1} with k ascending runs are constructed. Also, (b) from each permutation of the set { 1, 2, ... , n} with k  1 ascending runs, by attaching the element n + 1 in any of the possible positions except after the last element of any of these runs (n + 1)  (k  1) = n k + 2 permutations of the set {1, 2, ... , n+ 1} with k ascending runs are constructed. Consequently, the number a( n, k) of permutations of the set { 1, 2, ... , n} with k ascending runs satisfies the recurrence relation
a(n
+ 1, k)
= ka(n, k)
+ (n k + 2)a(n, k
fork = 2, 3, ... , n
+ 1, n
= 1, 2, ... , with
1),
a(n, 1) = 1, n > 0, a(n, k) = 0, k > n. Comparing this recurrence relation and its initial conditions with the recurrence relation (14.4) of the Eulerian numbers and its initial conditions, we conclude that a(n, k) = A(n, k). I REMARK14.5 Considerapermutation(j 1 ,h, ... ,in)oftheset{1,2, ... ,n} with k ascending runs. Then, the permutation (i 1 , i 2, ... , in) of the set {1, 2, ... , n }, with im = inm+ 1 , m = 1, 2, ... , n, which is the reverse permutation, has k descending runs. Since this correspondence is onetoone, it follows from Theorem 14.10 that the number of permutations of the set {1, 2, ... , n} with k descending runs equals the Eulerian number A(n, k). I REMARK 14.6 Rises and falls of permutations. The ascending or descending runs of a permutation of the set {1, 2, ... , n} are connected with its rises or falls. Specifically, consider a permutation (j 1 , h, ... , in) of the set {1,2, ... ,n}. The pair Ur,ir+l) is called a rise if ir < ir+l and a fall if ir > ir+l• r = 1, 2, ... ,n 1. If the permutation (iJ,i2 , ... ,in) has k rises (or k falls) {(jr 1 ,ir 1 +!), Ur,ir 2 +I), ... ,(jrk,irk+J)}, then it has k + 1 descending runs (or k+ 1 ascending runs) { (j1, h, ... , ir 1 ), Ur 1 +1, ir 2 +2, ... , ir2 ), Urk + 1, irk+2, ... , in)} and inversely. Hence, by Theorem 14.10, the number of permutations of the set { 1, 2, ... , n} with k rises (or k falls) equals the shifted Eulerian number A(n, k + 1). I
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
532
Example 14.4 Consider an urn containing n numbered balls { 1, 2, ... , n} and suppose that the balls are randomly drawn one after the other, without replacement. If im is the number drawn at the mth drawing for m = 1, 2, ... , n, then (it, h, ... , in) represents a drawing of then balls. Determine (a) the probability p(k; n) that the drawing of then balls has k ascending runs, fork = 1, 2, ... , n, and (b) the rth factorial moment J.L(r) of the sequence of probabilities p(k; n), k = 1, 2, ... , n, for r = 1, 2, .... (a) The number of permutations (j 1 , )2, ... , in) of the set {1, 2, ... , n} with k ascending runs, according to Theorem 14.10, is given by the Eulerian number A(n, k). Thus, the probability p(k; n) that the drawing of then balls has k ascending runs, on using the Laplace's classical definition of probability, is given by A(n, k) p(k; n) = , k = 1, 2, ... , n. 1
n.
(b) The generating function of the sequence of probabilities p( k; n), k
= 1, 2,
... , n, by (14.5), is given by P(t)
= tp(k;n)tk = An~t) n.
k=1
and so the factorial moment generating function, B(t)
= P(t + 1), (see Section
6.4) is
tr _ An(t + 1) _ ~ B( t )  ~ J.L(r) I I . r=O r. n. Further, by (14.7), the exponential generating of the Eulerian polynomial An(t), n = 0, 1, ... , may be written as
t
un
oo
~ An (t) n! = 1  t + 1=[eu,.(t,_,,1),1],.../:(1t)" Expanding it into a geometric series of z = (eu(tl) 1)/(1t) and then expanding the powers of eu(t 1 )  1 into a series of u( t 1), by using the generating function of the Stirling numbers of the second kind, (8.17), we get oo
n
oo
L An(t) :, = 1 + t L[eu(t 1 ) n=O

1]k(t 1)k
k=1 oo
= 1+t
n
oo
LL
k!S(n, k)(t 1tk;
k=I n=k = 1+
oo ~
{
n.
t ~ k!S(n, k)(t 1)nk n
}
n
:,
14.5.
PERMUTATIONS WITH REPETITION
533
and so n
An(t)
=t L
k!S(n, k)(t 1)nk
k=l
n
=L
k!{(k + 1)S(n, k + 1) + S(n, k)}(t 1tk.
k=O
Thus, on using the triangular recurrence relation of the Stirling numbers of the second kind, (8.27), we deduce the relation n
An(t) =
L k!S(n + 1, k + 1)(t 1)nk k=O
whence 00
B(t)
r
n
= Lll(r)! =L r. r=O
r=O
(
)'
n,r "S(n+1,nr+1W. n.
Consequently, ll(r)
and ll(r) = 0 for r
14.5
= S(n+1,nr+1)/(~). > n.
r
= 1,2, ...
,n,
D
PERMUTATIONS WITH REPETITION AND A GIVEN NUMBER OF RUNS
The permutations of a finite set with repetition can be classified according to the number and the length of the nondecreasing (or nonincreasing) sequences of consecutive elements they include.
DEFINITION 14.4 Let (h, jz, ... ,in) be annpermutation of the set {1, 2, ... , s} with repetition. The sequence of consecutive elements
is called nondescending run of length r; for 2 ::; m ::; n  1, 1 ::; r ::; n  2, if im1 > im and im+r1 > im+r; in particular; form = 1, 1 ::; r ::; n 1, if ir > ir+t.for m = n r + 1, 1 ::; r ::; n 1, if inr > inr+t. and for m = 1, r = n, without any additional condition.
534
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
This sequence is called nonascending run of length r conditions are satisfied with the reverse inequalities.
if the
corresponding
Example 14.5 Consider the permutations (j 1 , jz, h) of the set {1, 2, 3} with repetition. Clearly these 27 permutations can be classified according to the number of their ascending runs as follows. (a) The permutations with one nondescending run are the following 10:
(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3), (1, 3, 3), (2, 2, 2), (2, 2, 3), (2, 3, 3), (3, 3, 3). (b) The permutations with two nondescending runs are the following 16:
(1, 2, 1), (2, 1, 1), (1, 3, 1), (3, 1, 1), (2, 1, 2), (2, 2, 1), (3, 1, 3), (3, 3, 1), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (2, 3, 2), (3, 2, 2), (3, 2, 3), (3, 3, 2). (c) The permutation
(3,2,1) is the onlx one that has three nondescending runs. Each of these runs is of length one. U The enumeration of the permutations of a finite set with repetition and a given number of nondescending runs is facilitated by distinguishing them according to the last element they include. A bivariate generating function of the number of such permutations is derived in the following theorem.
THEOREM 14.11 Let Q(n, k; s, r) be the number of npermutations of the set {1, 2, ... , s} with repetition and last element r, which have k nondescending runs. Then
~ ~ Q( k· ) k n ~~ n, ,s,r t u
t)(1 u(1 t)Jr 1t(1u(1t)]8
= ut(1
(14.24)
n=lk=l
PROOF Distinguishing the npermutations of the set { 1, 2, ... , s} with repetition and last element r that have k nondescending runs, according to the element before the last, we conclude the expressions r
8
Q(n,k;s,r)=LQ(n1,k;s,j)+ L j=l
j=r+l
Q(nl,k1;s,j),
14.5.
535
PERMUTATIONS WITH REPETITION
for r = 1, 2, ... , s  1, k = 1, 2, ... , n, n = 2, 3, ... , and 8
Q(n,k;s,s) = LQ(n1,k;s,j), j=l fork= 1,2, ... ,n, n = 2,3, ... , with Q(n,O;s,r) = 0, n = 1,2, ... , Q(1, 1; s, r) = 1 and Q(1, k; s, r) = 0, k = 2, 3, .... Multiplying these expressions by tk and summing for k = 1, 2, ... , n, we get for the generating function n
fn(t;s,r) = LQ(n,k;s,r)tk, n = 1,2, ... , r = 1,2, ... ,s, k=l the recurrence relations T
fn(t;s,r)
S
= Lfnt(t;s,j) +t j=l
for r
= 1, 2, ...
L fnt(t;s,j), j=s+l
, s  1, n = 2, 3, ... , and s
fn(t;s,s)
= Lfnt(t;s,j), j=l
for n = 2, 3, ... , with ft (t; s, r) = t. Taking the difference of these relations with respect tor, we deduce the recurrence relation.
fn(t; s, r 1) = fn(t; s, r) for r = 2, 3, ... , s, n = 2, 3, ... , with multiplying by
c~
1 ) (t _ 1)j =
G)
+ (t
1)fn1 (t; s, r),
ft (t; s, r)
(t _ 1)j
and summing the resulting expression for j
= t. Replacing n by n j,
G=~)
(t 1)j
= 0, 1, ... , i, we get
~ (i ~ 1) (t 1)j fnj(t; s, r 1) = 't (i.) (t 1)j fnj(t; s, r) ;=0
J
;=0
't (~ =~) j=l
(t 1)1 fnj(t; s, r)
J
+~
(i ~ 1) (t 1)1+1fnj1 (t; s, r) J
j=O
and so
t (i.) j=O
J
(t 1)i fnj(t; s, r) =
~ (i ~ 1) (t 1)j fnj(t; s, r 1). j=O
J
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
536
Applying successively the last relation, we deduce the expression
fn(t; s, r
i) = t (i.) j=O
(t 1)1 fnj(t; s, r)
J
for n = 1, 2, ... , i = 0, 1, ... , r 1, r = 1, 2, ... , s. Multiplying both members of this expression by un and summing for n = 1, 2, ... , we find for the bivariate generating function oo
f(t,u;s,r)
=L
oo
n
LQ(n,k;s,r)tkun
= Lfn(t;s,r)un,
n=lk=l
n=l
the relation
f(t, u; s, r i)
= [1 u(1 t)]i f(t, u; s, r),
fori= 0, ... ,r 1, r = 1,2, ... , s. In particular, for r = s,
f(t, u; s, s i) = [1 u(1 t)]i f(t, u; s, s), i = 0, 1, ... , s 1. (14.25) Summing it fori = 0, 1, ... , s 1, we get s1
"'""'
[
.
~f(t,u;s,sz)= i=O
tW f(t,u;s,s).
1  1  u(1 ( ) U 1 t
Also s1
~f(t,u; s,s i)
s1 oo
oo {s1
= ~~ fn(t;s,s i)un = ~
and, since In+ I (t; s, s) =
}
~fn(t; s,s i) un
2:j= 1 fn(t; s, j),
s1
"'""'!( . _ .) _ ~ t,u,s,s z i=O
f(t, u; s, s)  ut
.
U
Therefore
(1 t)f(t, u; s, s) ut(1 t) = {1 [1 u(1 t)"]} f(t, u; s, s) and
f( t u· s s) 
' ' '
ut(1  t)
 (1  ut(1  t)]•  t ·
Introducing the last expression into (14.25) and setting r i = s r, we conclude that
!( t, u; s, r )
=
ut(1  t)[1  u(1  t)]r 1  t [1  u (1 t) ]s
s  i, whence
14.6. BIBLIOGRAPHIC NOTES
537
and the derivation of generating function (14.24) is completed.
I
COROLLARY 14.1 Let Q(n, k; s) be the number of npermutations of the set { 1, 2, ... , s} with repetition, which have k nondescending runs. Then oo
n
L L Q(n, k; s)t
k1
u
n=l k=l
n
(
1 t)
= [1u1t ( )]
8
t .
(14.26)
PROOF Notice that the attachment of the element s after the last element of an npermutation of the set { 1, 2, ... , s} with repetition, which has k nondescending runs uniquely yields an (n +I)permutation of the set {1, 2, ... , s} with repetition and last elements, which has k nondescending runs. Consequently, Q(n, k; s) = Q(n + 1, k; s, s) and introducing this expression into (14.24), we deduce (14.26). I
Clearly, comparing the generating function (14.26) with the generating function (14.19) of the Carlitz numbers we conclude the following corollary. COROLLARY 14.2 The numberofnpermutations of the set {1, 2, ... , s} with repetition, which have k nondescending runs is given by
Q(n,k;s) = IB(n,k;s)i = (l)nB(n,k;s), where B(n, k; s) is the Carlitz number.
14.6
BIBLIOGRAPHIC NOTES
The problem of evaluating the power series with coefficients powers of fixed order led Euler to the introduction of Eulerian polynomials and numbers (see Example 14.1). Expression (14.1), of the nth power of a number as sum of binomial coefficients of the nth order with coefficients the Eulerian numbers, is due to J. Worpitzky (1883). Other properties of the Eulerian numbers and polynomials can be found in L. Carlitz (1959). Motivated by the problem of expressing the power moments of a frequency distribution in terms of cumulative totals (see Example 14.2), P. S. Dwyer (1938, 1940) introduced and studied the cumulative numbers, which are noncentral Eulerian numbers. The Carlitz numbers were so named in honor of Leonard Car!itz for his stimulating contribution on the Eulerian
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
538
numbers, their generalizations and combinatorial applications. These numbers were first studied in L. Carlitz (1979) as degenerate Eulerian numbers. Ch. A. Charalambides (1982) further examined them as composition numbers. The general Simon Newcomb's problem was studied through generating functions by P. A. MacMahon (1915, 1916). J. Riordan (1958) and Ch. A. Charalambides (2002) examined it as a problem of enumeration of permutations with restricted positions by using power and factorial rook polynomials. An interesting combinatorial treatment of the same problem was provided by J. E. Dillon and D. P. Roselle (1969). The presentation of the section on the enumeration of permutations with repetition and a given number of runs (or rises) is based on the paper of L. Carlitz, D. P. Roselle and R. A. Scoville (1966).
14.7
EXERCISES
1. Noncentral Eulerian numbers. Consider the expansion
~ (t+r) n =L..,A(n,k;r)
(t + n k) n
, n=O,l, ....
k=O
The coefficient A(n, k; r) is called noncentral Eulerian number or Dwyer number. Derive (a) the explicit expression
A(n, k; r) =
t(
~ 1) (k + r it
l)j (n
J
j=O
and (b) the triangular recurrence relation
A(n + 1, k; r) = (k for k
= 1, 2, ...
,n
+ r)A(n, k; r) + (n
k r
+ 1, n = 0, 1, ... , with initial
+ 2)A(n, k 1; r),
conditions
A(O,O;r) = 1, A(n,O;r) = rn, n > 0, A(n,k;r) = 0, k > n.
2. (Continuation). Show that n
An(t; r) = L
oo
A(n, k; r)tk = (1 t)n+l L(i
k=O
and oo
g(t, u; r)
n
L A(n, k; r)t n=Ok=O
=L
+ r)nti
i=O
k
un n!
=
(1 _ t)eru(lt) _ teu(lt) · 1
14. 7. EXERCISES
539
3. (Continuation). Show that the noncentral Eulerian numbers A(n, k; r), k = 0, 1, ... , n, n = 0, 1, ... are connected with the noncentral Stirling numbers of the second kind S(n, k; r), k = 0, 1, ... , n, n = 0, 1, ... by A(n,k;r) =
t(1)kj(~=~)j!S(n,j;r) J
j=O
and S(n,j;r)
1
= :r Lj
J k=O
(
n k k ) A(n,k;r).
._
J
4. Noncentral Garlitz numbers. Consider the expansion
(
st n+
r)
= ~ L...t B(n, k; s, r)
(t + k) nn 
, n = 0, 1, ....
k=O
The coefficient B(n, k; s, r) is called noncentral Carlitz number. Derive (a) the explicit expression
and (b) the triangular recurrence relation (n + 1)B(n + 1, k; s, r) = (sk + r n)B(n, k; s, r) +[s(n k + 2) + n r]B(n, k 1; s, r), for k
= 1, 2, ...
,n
+ 1, n
= 0, 1, ... , with initial conditions
B(O,O;s,r)=1, B(n,O;s,r)=(:). n>O, B(n,k;s,r)=O, k>n. 5. (Continuation). Show that
Bn(t; s,
r)
=
~ B(n, k; s, r)tk =
(1 t)n+l
t, (
sj:
r)
tj
and g(t,u;s,r) =
oo
Ln
k
L B(n,k;s,r)t u n=Ok=O
n=
(1  t)[1 + u{1  t}Y ( )] [ 1t 1 +u 1t s
6. (Continuation). Show that
lim snn!B(n,k;s,r)=A(n,k;p), if
s+±oo
·
limr/s=p
s±oo
540
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
and
n
LB(n,k;s,r) = sn. k=O
7. (Continuation). Show that the noncentral Carlitz numbers B(n, k; s, r), k = 0, 1, ... , n, n = 0, 1, ... are connected with the noncentral generalized factorial coefficients C(n, k; s, r), k = 0, 1, ... , n, n = 0, 1, ... by k
B(n,k;s,r)=L(1) j=O
and C(n,j;s,r) =
k· ( nJ") J.., . J k. n!C(n,J;s,r) J
n! (n k)
l" Lj
J k=O
. _ k B(n,k;s,r). J
8*. q Eulerian numbers. Consider the expansion of the nth power of the qnumber [t]q into qbinomial coefficients of order n:
[t]~ =
t
qk(kl)/2 A(n, klq) [t + ~ k] , n = 0, 1, ....
k=O
q
The coefficient A(n, klq) is called qEulerian number. (a) Show that A(n,klq)=A(n,nk+1lq), k=0,1, ... ,n, n=0,1, ...
and (b) derive the explicit expression A(n,klq) =
qk(kl)/2
~(1rqr(rl)/2
[
n;
1
L
[k
r]~,
for k = 0, 1, ... , n, n = 0, 1, .... 9*. k
(Continuation). Show that the qEulerian numbers A(n,klq), , n, n = 0, 1, ... , satisfy the triangular recurrence relation
= 0, 1, ...
A(n
for k
= 1, 2, ...
+ 1, klq) , n, n
= [k]qA(n, klq)
+ [n k + 1]qA(n, klq),
= 0, 1, ... , with initial conditions
A(O, Olq) = 1, A(n, Olq) = 0, n
> 0 A(n, klq) = 0, k > n.
10*. Records. Let (]t,h, ... ,jn) be a permutation of {1,2, ... ,n}. The element ir, for 2 :S r :S n, is called a record of this permutation if Jr > j 8 , s = 1, 2, ... , r  1. Especially, the first element h is regarded,
14. 7. EXERCISES
541
by convention, as a record. Show that the number of permutations of the set {1, 2, ... , n} that have k records equals ls(n, k)l, the signless (absolute) Stirling number of the second kind. 11. Inversions of permutations. Let (j1,h, ... ,jn) be a permutation of the set { 1, 2, ... , n }. If ir >is for 1 ::; r < s ::; n, this permutation has an inversion at the pair of points (r, s). Show that the number b(n, k) of permutations of the set { 1, 2, ... , n} with k inversions satisfies the horizontal recurrence relation k
b(n,k) =
L
b(n 1,i), m = max{O,k n
+ 1},
j=m
for k = 0, 1, ... , n(n 1) /2, n = 1, 2, ... , with initial conditions
b(n, 0) = 1, b(O, k) = 0, k
~
1, b(n, k) = 0, k
> n(n 1)/2
and conclude that
b(n, k)
= b(n, k 1) + b(n 1, k),
fork= 1, 2, ... , n 1, n = 1, 2, ... and
b(n, k)
= b(n, k 1) + b(n 1, k) b(n 1, k n),
fork= n,n + 1, ... ,n(n 1)/2, n = 3,4, .... 12. (Continuation). Show that
~n(t) =
n(n1)/2
L
n
b(n, k)tk
k=O
1
j
= IJ 1 =. tt ,
n
= 1, 2, ...
j=l
and conclude that
b(n, n(n 1)/2 k) = b(n, k), n(n1)/2
L
k=O
n(n1)/2
b(n, k) = n!,
L
(1)kb(n,k) = 0.
k=O
13. Local maxima and minima. Let (j 1 , h, ... , in) be a permutation of the set { 1, 2, ... , n}. The element ir, for 2 ::; r ::; n  1, is called a local maximum or a peak at the point r of this permutation if ir > irI and ir > ir+l· Especially, the element ii is a peak if i1 > h, while the element Jn is a peak if in > in I· The element ir is a local minimum if the conditions are satisfied with the reverse inequalities. Show that the number T1 (n, k) of
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
542
permutations of the set {1, 2, ... , n} with k local maxima (minima) satisfies the triangular recurrence relation T1 (n + 1, k) = (2k
+ 2)T1 (n, k) + (n
2k + 1)TI(n, k 1),
for k = 1, 2, ... , n = 2, 3, ... , with initial conditions T1 (n,0)=2nI, n=2,3, ... , TI(2,k)=O, k=1,2, .... 14. Show that the number T2(n, k) of permutations of the set { 1, 2, ... , n} with k ascending runs, each of length greater than one, satisfies the triangular recurrence relation
T2(n
+ 1, k)
= (2k
+ 1)T2(n, k) + (n 2k + 2)T2(n, k
1),
for k = 1, 2, ... , n = 2, 3, ... , with initial conditions T2(n, 0) = 1, n = 2, 3, ... , T2(2, 1) = 1 T2(2, k) = 0, k = 2, 3, .... 15*. Consider the double sequence of numbers T(n, k), k = 0, 1, ... , n = 2,3, ... , where T 1 (n,k) = T(n,2k + 1) and T2(n,k) = T(n,2k) are the numbers of permutations of the set { 1, 2, ... , n} with k local maxima (minima) and with k ascending runs of length greater than one, respectively. (a) Show that
T(n
+ 1, k)
= (k
+ 1)T(n, k) + (n k + 2)T(n, k 2),
for k = 2, 3, ... , n = 2, 3, ... , with T(n, 0) = 1, T(n, 1) = 2nI, n = 2, 3, ... ,
T(2, 2) = 1, T(2, k) = 1, k = 3,4, .... (b) Setting T(1, k)' = 1, k = 0,1 and T(1, k) = 0, k = 2, 3, ... , derive the generating function ~~
g(t, u) =
k
un1
1  t2
~ ~ T(n, k)t (n 1)! = t(cosh z 1)'
with z = u(1 t 2) 1 12 + arccosh(1/t), where coshz = (ez + ez)/2 is the hyperbolic cosine and arccoshw is the arch of the hyperbolic cosine of w. 16*. Show that the number R(n,k) of permutations of {1,2, ... ,n} with k (ascending or descending) runs of length greater than one satisfies the recurrence relation
R(n
+ 1, k)
= kR(n, k)
+ 2R(n, k
1) + (n k + 1)R(n, k 2),
14.7. EXERCISES
543
for k = 2, 3, ... , n = 2, 3, ... , with initial conditions
R(n, 0) = 0, R(n, 1) = 2, n = 2, 3, ... , R(2, k) = 0, k = 2, 3, .... 17*. Updown and downup permutations. A permutation (j 1 , h, ... , in) of the set { 1, 2, ... , n} is called updown permutation if hr 1 < hr for r = 1, 2, ... , [n/2] and hr > hr+ 1 for r = 1, 2, ... , [n/2]  1, while it is called downup permutation if hr 1 > hr for r = 1, 2, ... , [n/2] and hr < hr+l for r = 1, 2, ... , [n/2] 1. (a) Show that the number An of updown permutations of the set { 1, 2, ... , n}, which is equal to the number of downup permutations of the same set, satisfies the recurrence relation
with An = 1, n = 0, 1, 2 and (b) deduce the generating function
L Ann!tn =tan ( 2t + 47r) . 00
A(t) =
n=O
18*. (Continuation). Euler numbers. The sequence of Euler numbers E 2 n, n = 0, 1, ... , (E2 n+ 1 = 0, n = 0, 1, ... ) has generating function
(a) Show that
IE2nl
= (1)nE2n =
A2n,
where A 2 n is the number of updown permutations of the set {1, 2, ... , 2n }. (b) Derive the expression
IE2nl
= i)1)nd 2
where
R(n,k)=
:)!
2
k=1
R(2n,2k)
~!t,(1fG) (~j)n
is the CarlitzRiordan number of the second kind (see Exercise 8.22). 19*. (Continuation). Tangent coefficients. Consider the sequence T2 n+l, n = 0, 1, ... , (T2 n = 0, n = 0, 1, ... ) with generating function
oo
T(t) =
t2n+1
1_
e2t
~ T2n+1 (2n + 1)! = 1 + e 2 t
= tanh t.
544
RUNS OF PERMUTATIONS AND EULERIAN NUMBERS
(a) Show that IT2n+ll = (1)nT2n+l = A2n+l
1
where A 2n+l is the number of updown permutations of the set {1, 2, ... , 2n + 1}. (b) Derive the expression y.
l 2n+l
I=
~(1)nk (k + 1)(2k)! R(2n 2k) L.J k ' , 2
k=l
where R(n, k) is the CarlitzRiordan number of the second kind. 20*. (Continuation). Genocchi numbers. The sequence of Genocchi numbers Gn, n = 1, 2, ... , has generating function
G(t)
CXl
tn
2t
n=l
n.
+e
= LGn 1 = 1  t
and so G 1 = 1, G 2n+I = 0, n = 1, 2, .... (a) Show that
where A 2nl is the number of updown permutations of the set { 1, 2, ... , 2n  1}. (b) Derive the expression 2nl IG2nl = 2n
kl
L (1)nk 2 ~ S(2n 1, k),
k=l
where S(n, k) is the Stirling number of the second kind.
HINTS AND ANSWERS TO EXERCISES
CHAPTER I
= A 3 = {(t,t,t),
(t,t,h), (t,h,t), (h,t,t), (t,h,h), (h,t,h), (h,h,t), (h,h,h)}, F((t, t, t)) = 0, F((t, t, h)) = F((t, h, t)) = F((h, t, t)) = 1, F((t, h, h)) = F((h, t, h)) = F((h, h, t)) = 2, F((h, h, h)) = 3.
1.
s3
2. (a) N(S2) = 36, (b) A 2 = {(1, 1)}, A 3 = {(1, 2), (2, 1)}, A 4 = {(1, 3), (2, 2), (3, 1)}, A 5 = {(1,4), (2,3), (3,2), (4,1)}, A 6 = {(1,5), (2,4), (3,3), (4,2), (5,1)}, A 7 = {(1,6), (2,5), (3,4), (5,2), (6, 1)}, A 8 = {(2,6), (3,5), (4,4), (5,3), (6,2)}, A 9 = {(3,6), (4,5), (5,4), (6,3)}, Aw = {(4,6), (5,5), (6,4)}, Au= {(5,6), (6,5)}, A12 = {(6,6)}. 3. W = {1, 2, 3, 4, 5, 6}, S3 = W 3 , N(S3) = [N(W)j3 = 6 3 = 216, A= {(w,b,r) E S3: w+b+r > 10}, A'= {(w',b',r') E S3: w' +b' +r' :S 10}. The element (w, b, r) in A uniquely corresponds to the element (w', b', r') in A', with w' = 7 w, b' = 7 b, r' = 7 r. Indeed, w' + b' + r' = 21 (w + b + r) :S 10 if and only if w + b + r > 10.
4. N(L5 xR5)
= N(L5)N(R 5 ) = 25,
N(L5 xR4)
= N(L5)N(R4) = 20.
5. The set Bn of ndigit binary sequences (1, a1, a 2, ... , anI) is equivalent to the Cartesian product A 1 x A2 x ··· x AnI, with Ai = {0,1}, i = 1, 2, ... , n  1, and so N(Bn) = N(A1)N(A2) · · · N(An_I) = 2nl, n = 2,3, ... , N(BI) = 2. Also, N(Tn) = L~=l N(Br) = 2 +I:~,:; 2rl = 2n. Alternatively, the set Tn of binary sequences of at most n digits is equivalent to the Cartesian product A 1 x A 2 x · ·· x An, with Ai = {0, 1}, i = 1, 2, ... , nand so N(Tn) = 2n. 6. The set Bn of sequences of n symbols (a 1 , a 2 , ... , an) is equivalent to the Cartesian product A 1 x A2 x · · · x An, with Ai = {0, 1}, i = 1, 2, ... , n,
HINTS AND ANSWERS TO EXERCISES
546
where 0 corresponds to a "dot" and 1 to a "dash." Thus N(Bn) = 2n, n = 1,2, ... , and the required number is L~= 1 N(Br) = 2n+1. 7. (a)8+9=17,
(b)2·8·9+9·9=225,
(c) 1+17+225=243.
8. (a)9·5=45, (b)9·10·5=450, (c)5+45+450=500. 9.(a)n 3, (b)n(n1)(n2), (c)3·n 2. 10. (a) 4 · 3 · 2 = 24,
(b) 33 = 27,
(c) 3 · 22 = 12.
11. Note that any factor of the number N is of the form p~ 1 p;2 • • • p~n , where ir E Ar = {0, 1, ... , kr }, r = 1, 2, ... , n, and apply the multiplication principle. 12. 4n. 13. (a) 3 13 = 1,594,323, (b) 25 · 32 = 288. 14. 143 . 9 . 103 = 24,696,000.
15. kn. 16. Apply the multiplication principle (a) with Ar = {1, 2,} and (b) with Ar = {1, 2, ... , k} for r = 1, 2, ... , n. Then the number of divisions (a) in two subsets is 2n and (b) ink subsets is kn.
17. Note that a map f from X into Y corresponds to an ordered ntuple (Yit,Yh,··· ,Yjn) with Yir = f(xr), r = 1,2, ... ,nand apply the multiplication principle. 18. Suppose that each cell contains at most one object and conclude that this hypothesis leads to a contradiction. Use the fact that each integer k; E K can be written in the form ki = 2r; · s;, where r; is a nonnegative integer and s; is a positive odd integer less than 2n, and apply the pigeonhole principle.
19. Consider the partial sums s1 = k 1, s2 · · · + kn and apply the pigeonhole principle.
= k1
+ k2, ... , sn = k1 + k2 +
20. Each integer i from the set {1, 2, ... , 10} can be written in the form i = 2r; · s;, where r; is a nonnegative integer and s; belongs to the set
{1, 3, 5, 7, 9} of five positive odd integers. Apply the pigeonhole principle and conclude that between the six chosen numbers there are two having the same s.
CHAPTER2 1. (a) Let the four boys {bt, b:!, b3, b4 } seat first. Then the three girls {9 1, 92, 93} may take any three among the five permissible positions
CHAPTER2
547
{w 1,w2,w3,w4,w5 }, where Wi is the position before the ith boy fori= 1, 2, 3, 4 and w5 is the position after fourth boy. Apply the multiplication principle and conclude that the required number is 4!(5)3 = 480. (b) 7! = 5040. 2. 16. 15 = (2. 15) . 8 = 240. 3. 4!(51. 3! . 1. 1) = 17,280. 4. 4 3 = 64, 3. 3 2 = 27. 5. (7)4 = 840. 6. Consider the set Aj that contains the permutations of {j,j+1,j+2}, j = 1, 2, 3,4, and conclude that the required number is N(A 1 ) + N(A2) + N(A3) + N(A4) = 4 · 3! = 24.
7. (a) 2 · 33 =54, (b) 9 · 102 · 5 = 4500, (c) 9 · 10 2 · 5 8. (a) (10)4 = 5040, (b) 104 = 10,000.
= 4500.
9. Consider the two like elements of each of the 2 kinds {w1 , w2} as two single elements and conclude that the required number is (2 . 3 + 2 )! = 8! = 7' = 5040 23 . 12 23 . . 10. 10!/2 2 = 907,200. 11. N(A) = 3 · 3! + 2 · 3 + 1 = 25, N(fl) = 63 = 216.
N(B) = 3 · 3! + 3 · 3 = 27,
12. Note that each circular permutation (i 1 , i2, ... , in) of the set {1, 2, ... , n} corresponds ton (linear) permutations of the same set and conclude the required number. 13. (a) 2 · (5)3 · 2 · 3 = 720, 3
14. (a) (8 + 3)! · 2
(b) (5)3 · 2 = 120
= 319,334,400,
(b) (8 + 3  1)! · 23
15. (a) N(A 4) = 42 = 16, (b) N(A3) = 32 = 9, (c) N(B4) = N(A 4  A 3) = N(A 4) N(A 3) 16. N(D4)
= N(C4 C5) = N(C4) N(C5) 2  (6 5 + 1) = 5.
= (6 4 + 1) 2
17.
(7)
18. 2
2 + 7=
(7 + 2 1) 2
G) G)+(~)
19. (a)
G).=
= 28.
2
10, (b)
= 472.
G) =
10.
= 42 
= 29,030,400 32
= 7.
HINTS AND ANSWERS TO EXERCISES
548
20. (a) (;) (;),
(b)
(~)2 5 ,
21. (a) 510 = 9,765,625,
22.
23. (a) c2°) 8!
C+ ~~
(b)
G) G) + G) G)
C;).
(c)
1 ) = 1001.
= 420.
= ~! = 1,814,400, 1
(b) c3°) 7!
1
= 3~! = 604,800,
1  "22 10!  907,200. (c) ( 10) (8) 6.2 2
10! 24. 1!2!3!4! = 12,600. 6! 25. (a) ( !) 3 = 90, 2 10! 26. (a) 25"
6! (b) 31 ( 2!) 3 = 15.
= 113,400,
(5!) 2
= 14,400,
(b ) 10!  945, 5.1  120. 5125
11
27. (a) 3 11 = 177,147,
(b)
L 3 · 1i2lli = 3(2
6

1) = 189.
j=6
28. The required number is the number of paths from the point (0, 0) to the point (5, 3) and equals = 56.
m
29. Consider the inverse procedure starting with n oneelement sets and, at each stage merging together any two subsets derive, by applying the multiplication principle, the required number. 30. (a) The least number of lockers needed is the number of different "restricted" lockers needed and equals an = (;), which is the number of kcombinations of the n generals, with k = [n/2] the integer part of nj2. (b) The number of keys a general should have equals bn = (n; 1 ), which is the number of kcombinations of n  1 generals. 31. (a) The number of ncombinations of a set Wr+s, of r + s elements, which equals (r~•), may equivalently be evaluated by selecting k elements from a subset Ar ~ Wr+s, of r elements, and n  k elements from the complementary set Wr+s Ar of s elements, fork= 0, 1, ... , n, and using the principles of multiplication and addition. (b) Similarly, evaluate the number of ncombinations of Wr+s with repetitions. 32. (a) Use the expression
(~)
=
k!(n~ k)!"
(b) The pair of sets (A, B), with A~ B ~ Wand N(A) = k, N(B) = s, N(W) = n, uniquely corresponds to the pair of sets (A, C), where C =
CHAPTER2
549
B  A, with A ~ fl and C ~ A' = W  A. Also, the same pair of sets (A, B) uniquely corresponds to the pair of sets (A, D), where D = B A, with D ~ fl and A~ fl D. Evaluate the pairs of sets (A, B), (A, C) and (A, D) and conclude the first two equalities. The other two equalities are similarly deduced with s = r + k. 33.
Multiply the expression
.
(n) k
and deduce the first relat10n. Express
n'
= k!(n ~ k)! by n = (n  k)
(n) (s+(n+s)) k
=
k
+ ·k
as a sum of
binomial coefficients, by using the first expression in Exercise 31; using the second part of Exercise 32, deduce the required formula. n) n! (n+1)k+(nk) 34. Multiply the expression ( k = k!(n _ k)! by n = k+ 1 and deduce the first relation. Work as in the second part of Exercise 33. 35. Use Pascal's triangle and, splitting the sum into two sums, conclude the first formula. Using the first part of Exercise 32 and the first part of this exercise, derive the second formula. 36. The use of Pascal's triangle splits the sum into two sums, which reduce to the required expression.
37. Verify that each injective map f of X into Y, with f(xr) = Yir, r = 1, 2, ... , k, uniquely corresponds to a kpermutation (Yh, Yh, ... , Yik) of the set Y. 38. Verify that each map f of {1,2, ... ,k} into {1,2, ... ,n} uniquely corresponds to a kcombination (a) without repetition if f is strictly increasing and (b) with repetition if f is increasing.
39. Consider the set C of kcombinations of the set Wn+I in which the element w 0 appears at most s times and the subset Cj ~ C of these combinations in which the element w 0 appears exactly j times, for j 0, 1, ... , m, with m = min{k, s }, and apply the addition principle. 40. Consider the set E of kcombinations of n, with repetition, in which each element appears at most twice and the subset Cj ~ E of these combinations in which exactly j elements appear twice, for j = 0, 1, ... , m, with m = min{n, [k/2]} and apply the addition principle. 41. Consider the set A of kcombinations of n + r with repetition, in which each of n specified elements appears at most twice, while each of the other r elements appears at most once. Further, consider (a) the subset Ai ~ A of these combinations in which exactly j of the n specified elements appear twice, for j = 0, 1, ... , m, with m = min{n, [k/2]}, and apply the
550
HINTS AND ANSWERS TO EXERCISES
addition principle. Also, consider (b) the subset Bi ~A of these combinations in which exactly i of the r elements appear, fori = 0, 1, ... , s, with s = min {r, k}, and apply the addition principle. 42. Consider the set K of the kcombinations of n elements belonging in r kinds with k 1 , k2 , ... , kr elements, respectively. Further, consider the subset Ki ~ K of these combinations which contain exactly j of the k 1 elements of the first kind, for j = 0, 1, ... , k 1 and apply the addition principle. 43. (a) Use Theorem 2.10. (b) Consider an element w of the set Wn, of n elements, and enumerate the partitions of Wn in k subsets in which w belongs in a unit set and the partitions of Wn in k subsets in which w belongs in a subset that is not a unit set. Applying the addition principle, conclude the required recurrence relation. (c) Enumerate the partitions of Wn in k subsets in which w belong in a subset with a total of j elements, for j = 1, 2, ... , n and apply the addition principle. 44. (a) Show that the required number equals the product of the number of partitions of X ink subsets and the number of kpermutations ofY. (b) Use the result of Exercise 1.17. (c) Use the first part of Exercise 32. 45. Note that, to each partition of a set Wn in k (nonempty) subsets, there correspond k! divisions of Wn in k nonempty subsets, and use the result (c) of Exercise 44. 46. (a) Use Theorem 2.10. (b) Consider an element w of the set Wn, of n elements, and enumerate the partitions of Wn in which w belongs in a subset with a total of j elements, for j = 1, 2, ... , n, and apply the addition principle. 47. Consider the set P of partitions of the set Wn, of n elements, and the subset Pk ~ P of partitions of Wn in k subsets, for k = 1, 2, ... , n, and apply the addition principle and the result (a) of Exercise 43. Further, introduce the expression (c) of Exercise 44 into the preceding expression. 48. The required number equals (~:::~), the number of positive integer solutions of the linear equation x 1 + X2 + · · · + Xn = k. 49. Select first the s positions for the s runs of zeros among the n k + 1 positions between the n  k ones or before the first and after the last one. Then (a) allocate the k zeros in the selected s positions, placing at least one zero in each of these positions. (b) Choose the sr positions of the runs of length r, for r = 1, 2, ... , k, among the s selected positions. 50. To each kcombination {i 1 , i 2 , ..• , ik} of the set {1, 2, ... , n} correspond the permutation (a 1 , a 2 , ... , an) of k zeros and n k ones in which the element ai,, ai 2 , ••• , aik are zeros. Verify that this correspondence is onetoone and use the result of Exercise 49.
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51. To each (circular) kcombination of the first n integral numbers displayed on a circle, correspond a circular permutation of k zeros and n k ones in which one of the n elements is marked with a star corresponding to number 1.
52. (a) Use the onetoone transformation Yi = Xis, i = 1, 2, ... , n, and conclude that the required numbers equals (n+z=:~ 1 ). (b) Use the one to one transformation Zi = m xi, i = 1, 2, ... , n and conclude that the required number equals (n+:~=z 1 ). 53. To each partial derivative of order k of an analytic function of n variables, correspond a nonnegative integer solutions of a linear equation 1 r1 + r2 + · · · + rn = k and conclude that the required number is ).
(n+z
54. Verify that the number of monomials in the most general polynomial in n variables X1, x2, ... , Xn of degree k equals the number of nonnegative integer solutions of a linear inequality r 1 + r 2 + · · · + rn ::; k and conclude that an,k = (ntk). 55. Correspond to each kcombination of {1, 2, ... , n }, satisfying the required property, an integer solution of linear equation ji + h + · · · + ik + jk+ 1 = n, satisfying the restrictions j 1 ~ 1, h ~ s + 1, h ~ s + 1, ... , jk ~ S + 1, jk+1 ~ 0. 56. Using the result of Exercise 55 enumerate the kcombinations of {1, 2, ... , n }, displayed on a circle, which satisfy the required property and (a) do not contain the elements n s + 1, n s + 2, ... , n and (b) contain at least one of the s elements n s + 1, n s + 2, ... , n and apply the addition principle.
57. (a) The evaluation of the number C(n, k; skI) may be reduced to the enumeration ofthe kcombinations {j1 , h, ... , jk} of {1, 2, ... , nskd by using the transformation j 1 = ii, im = im Sm1> m = 2, 3, ... , k. (b) Using the same transformation, the evaluation of the number B(n, k; skI, r) may be reduced to the enumeration of the kcombinations {j 1 ,h, ... ,jk} of the set {1, 2, ... , n skd, which satisfy the restriction jk ::; n sk1 + (ii  r 1). 58. The number Qn,k equals the number of ways of selecting k positions for the zeros among the (nk1)+2 = nk+1 positions between the nk ones or before the first and after the last one. Apply the addition principle to deduce Qn· Use Pascal's triangle to deduce the recurrence relation. 59. The required number is given by Cn+I = N(AB) = N(A)N(B), where A is the set of lattice paths from the point (0, 0) to the point (n, n) and B is the set of lattice paths from the point (0, 0) to the point (n, n) that do not touch or intersect the straight line x = y  1.
HINTS AND ANSWERS TO EXERCISES
552
60. (a) Note that each lattice path from (0, 0) to (n, k) with r diagonal steps requires, in addition, k r vertical and n r horizontal steps. Select the r diagonal and k r vertical steps among the total of n + k r steps. (b) Exclude the r diagonal steps and enumerate the number of lattice paths from (0, 0) to (n  r, k  r) that do not touch or intersect the straight line x = y. Such a path passes through n + k  2r points (not including the origin). Select the r starting points of the diagonal steps among the n + k  2r points. 61. Use Laplace's classical definition of probability and the results of Exercise 60. 62. The probability that number r is drawn at the kth drawing is given by Pn,k = (n 1)kt/(n)k· 63. Consider the set fh of possible outcomes of k drawings and the subset Ak of these outcomes in which a white ball is drawn for the first time at the kth drawing. Then Pn,r,k = P(Ak)/ P(ilk), k = 1, 2, ... , r + 1. Use the fact that A 1 + A2 + · · · + Ar+l = ilk to conclude the required expression.
65. Use Laplace's classical definition of probability and the result of Exercise 58.
CHAPTER3 1. Use Vandermonde's formula to get Sn,r
= (n)r(a)r(a + b r)nr·
2. Express (x + y + n)n into factorials of x and y + n, using Vandermonde's formula, and then multiply both members by (Y)n = 1/(y + n)n· 3. Express (Y)n = ( 1)n(x + (x y 1 + n))n into factorials of x and ( x y 1 + n) and then divide both members by (x + y)n· 4. Rewrite the expression to be shown as
f.(~=
D(r; ~: j) = (r; s),
which follows from the expansion of the identity [tn(l t)n][trn(I t)r+n1] = tr(l t)r1 into powers oft.
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5. Use the result of the first part of Exercise 4 to get sn = ns/(r+ 1)n+I and Sn,m = (n + m 1)m(s)m/(r + m)n+m· 6. Use Newton's binomial formula to get sn,r = (n)rPr·
1. Use Newton's negative binomial formula to get Sn,r = (n+r1)r}r· 8. Differentiate a suitable expression of Newton's binomial formula (see Example 3.4). 9. Integrate a suitable expression of Newton's binomial formula (see Example 3.5).
10. Use the result of Exercise 2.32 and Newton's binomial formula. 11. Use Pascal's triangle to split the sum and derive the recurrence relation. 12. Use Newton's binomial formula (3.10) with x = t, y = 1 and formula l:::~= 1 uk 1 = (1 u) 1 (1 un) with u = 1 t. 13. Integrate, with respect to t, in the interval [0, u] the identity derived in the first part of Exercise 12 and then integrate, with respect to u, in the interval [0, 1] the resulting expression to deduce the required relation. 14. Use Pascal's triangle to derive the recurrence relation. 15. Use the result of Exercise 14 with z = 1/2 and z = 1/2. 16. Use the relation
1  _1_ (1 +
rk+1r+1
k
)
rk+1 to derive the recurrence relation and iterate it to get the required expression.
17. Express ('"!k) into binomials of x and k, using Cauchy's formula, and then use the result of Exercise 10. 18. Use Cauchy's formula with x = y = n = 2r and x to derive the first and the second sums, respectively.
= y = n = 2r + 1
19. Express (~!~) into binomials of n and r, using Cauchy's formula and then use the result of Exercise 10. 20. Use the relation kG) = r(~=:~) and evaluate the resulting expression by expanding the identity (x + 1r 1 (x + 1) 8 = (x + 1r+s 1 into powers of X.
21. Expand the identity (1 W(1t)s 1 = (1t)s+r 1 into powers of t and equate the coefficients of tn in both sides of the resulting expression. 22. Use the relation
(~)(n;r)
1
(r:~~k)(n:r)
1
HINTS AND ANSWERS TO EXERCISES
554
and then the result of Exercise 21. 23. Expand into powers oft the identity (1 t)r 1 (1 t)(sn) 1 = (1 t)(r+sn+ 1l 1 and equate the coefficients of tn in both sides of the resulting expression. 24. Using Newton's binomial formula, show that
tr(l t)r1
rs _ rsk ( r ~ k) ts(l t)sk1_ = 2) 1 k=D
Expand this identity into powers of t and equate the coefficients of tn in both sides of the resulting expression. 25. Expand into powers oft the identity (1 + tr(l + t)s = (1 + ws and equate the coefficients of tn in both sides of the resulting expression. 26. Use for part (a) the expression
X) X(X  1) · ·k!· (X  k + 1) , and (k =
for parts (b) and (c) the "tr