Wilf posed the following problem: determine asymptotically as $n\to\infty$
the probability that a randomly chosen part size in a randomly chosen
composition of n has multiplicity m. One solution of this problem was given by
Hitczenko and Savage. In this paper, we study this question using the
techniques of generating functions and singularity analysis.Comment: 12 page

Carla D. Savage

Cecil Rousseau

Corteel

Knuth

Paweł Hitczenko

Rousseau

English

arXiv

Crossref

AbstractWilf posed the following problem: determine asymptotically as n→∞ the probability that a randomly chosen part size in a randomly chosen composition of n has multiplicity m. One solution of this problem has been given by two of the authors 〈http://www.csc.ncsu.edu/faculty/savage/〉. In this paper, we study this question using the techniques of generating functions and singularity analysis

Hitczenko, Paweł

Rousseau, Cecil

Savage, Carla D.

Elsevier - Publisher Connector 

Journal of Computational and Applied Mathematics 142 (2002) 107–114
www.elsevier.com/locate/cam
A generating functionology approach to a problem of Wilf
Pawe l Hitczenkoa ; ∗, Cecil Rousseaub, Carla D. Savagec
aDepartment of Mathematics and Computer Science, Drexel University, Philadelphia, PA 19104-2875, USA
bDepartment of Mathematical Sciences, University of Memphis, Memphis, TN 38152-3240, USA
cDepartment of Computer Science, North Carolina State University, Raleigh, NC 27695-8206, USA
Received 10 October 2000
Abstract
Wilf posed the following problem: determine asymptotically as n→∞ the probability that a randomly chosen part size
in a randomly chosen composition of n has multiplicity m. One solution of this problem has been given by two of the
authors 〈http:==www.csc.ncsu.edu=faculty=savage=〉. In this paper, we study this question using the techniques of generating
functions and singularity analysis. c© 2002 Elsevier Science B.V. All rights reserved.
1. Introduction
Let n be a positive integer. A composition of n with p parts is a solution of the equation
n= 1+2+· · ·+p in positive integers 1; 2; : : : ; p. We shall write = (1; 2; : : : ; p) to symbolize
the composition. For example, there are 16 compositions of 5, namely
(5) (4; 1) (1; 4) (3; 2)
(2; 3) (3; 1; 1) (1; 3; 1) (1; 1; 3)
(2; 2; 1) (2; 1; 2) (1; 2; 2) (2; 1; 1; 1)
(1; 2; 1; 1) (1; 1; 2; 1) (1; 1; 1; 2) (1; 1; 1; 1; 1):
The terms 1; : : : ; p are called the parts of the composition. The multiplicity of a part size is the
number of parts with that size. For example, in the composition (1; 1; 1; 2) the multiplicity of 1 is 3
∗ Corresponding author.
E-mail addresses: phitczenko@mcs.drexel.edu (P. Hitczenko), ccrousse@memphis.edu (C. Rousseau),
savage@csc.ncsu.edu (C.D. Savage).
0377-0427/02/$ - see front matter c© 2002 Elsevier Science B.V. All rights reserved.
PII: S 0377-0427(01)00462-9
108 P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114
and the multiplicity of 2 is 1. A partition of n with p parts is a solution of n= 1 + 2 + · · ·+ p
with 1¿ 2¿ · · ·¿ p. In [1] Corteel et al. proved that for every Exed m¿ 1, the probability that
a randomly chosen part size in a random partition of n approaches 1=(m(m + 1)) as n → ∞. Wilf
then posed the corresponding problem for compositions: determine asymptotically (as n → ∞) the
probability that a randomly chosen part size in a randomly chosen composition of n has multiplicity
m. One solution of this problem has been given by two of the authors [3]. In this note, we address
the same question using generating functions and singularity analysis.
It is well known that there are 2n−1 compositions of n. One way to arrive at this result uses
generating functions. The generating function for compositions with p parts is
(z + z2 + z3 + · · ·)p =
(
z
1− z
)p
;
and summing over p we have the generating function for all compositions:
G(z) =
∞∑
p=1
(
z
1− z
)p
=
z
1− 2z :
The coeFcient of zn in the expansion of G(z), denoted by [zn]G(z), is the number of compositions
of n, and clearly [zn]G(z) = 2n−1. From an analytic point of view, the fact that the number of
compositions of n is asymptotically (as well as exactly) 2n−1 is associated with the fact that the
generating function is a rational function for which the pole nearest to the origin (the only pole in
this case) is simple and located at z= 12 . Our solution of Wilf’s problem uses the same principle.
We shall use the following notation. The probability of the event A is denoted by P(A), and
the expected value of a random variable X is denoted by E(X ). The natural logarithm and base 2
logarithm are denoted by log n and log2 n, respectively.
To state the problem more precisely, suppose that a composition  is selected uniformly at random
from the set of all 2n−1 compositions of n. Then out of the set of part sizes in , a part size k is
chosen uniformly at random. Let A(m)n denote the event in which k has multiplicity m. For example,
inspection of the 16 partitions of 5 shown above yields
P(A(1)5 ) = 58 ; P(A
(2)
5 ) =
3
16 ; P(A
(3)
5 ) =
1
8 ; P(A
(5)
5 ) =
1
16 ;
and otherwise P(A(m)5 ) = 0. The object is to determine P(A
(m)
n ) asymptotically as n→∞. We shall
End that P(A(m)n ) tends to 0 at the rate 1=log n. It then turns out that log nP(A(m)n ) does not have a
limit, but oscillates about the value 1=m as n→∞.
2. Results
The answer to Wilf’s question is given in the following theorem, Erst proved in [3].
Theorem 1. Let A(m)n be the event in which a randomly selected part size in a randomly
selected composition of n has multiplicity m. Then
log nP(A(m)n ) = (1 + o(1))
(
1
m
+ F({log2 n})
)
; n→∞;
P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114 109
where {a}= a− a is the fractional part of a and
F(x) =
2
m!
Re
∞∑
p=1
e−2ipx 
(
1 + i
2p
log 2
)
;
with  denoting the gamma function.
Using well-known facts about the gamma function ((1+z) = z(z) and (z)(1−z) = =sin(z)),
we obtain
F(x) =
2
m!
∞∑
p=1
(
p
sinh(p)
)1=2
cos(2px − p);
where = 22=log 2 and p is the argument of (1 + i2p=log 2). This series converges quite
rapidly, and its sum may be approximated by the Erst term. But even the Erst term is quite small
since 2(=sinh )1=2 ≈ 10−5. Thus for large n one Ends that log nP(A(m)n ) is quite close to 1=m, but
there is a residual dependence on {log2 n}. In the treatment given here using generating functions
and singularity analysis, the proof of Theorem 1 will reduce to a well-known calculation after we
have established the appropriate sequence of lemmas.
Let  be a composition of n. Then D() will denote the set of distinct part sizes in , and Mm()
will denote the set of part sizes of  that have multiplicity m.
Lemma 1. For a random composition of n;
P(k ∈Mm()) = 12n−1 [z
n]
zkm(1− z)m+1
(1− 2z + zk(1− z))m+1 :
Proof. Let Gk(z; w) be the two-variable generating function in which [znwm]Gk(z; w) is the number
of compositions of n in which k has multiplicity m. To construct such a generating function, we
Erst note that the contribution made by compositions with p (not necessarily distinct) parts is
(z + z2 + · · ·+ wzk + zk+1 + · · ·)p =
(
z
1− z + (w − 1)z
k
)p
:
Thus,
Gk(z; w) =
∞∑
p=1
(
z
1− z + (w − 1)z
k
)p
=
z + (w − 1)zk(1− z)
1− 2z − (w − 1)zk(1− z)
=
1− z
1− 2z − (w − 1)zk(1− z) − 1 =
1− z
1− 2z + zk(1− z)− wzk(1− z) − 1:
Since there are 2n−1 compositions of n, we then have
P(k ∈Mm()) = 12n−1 [z
nwm]
1− z
1− 2z + zk(1− z)− wzk(1− z)
=
1
2n−1
[zn]
zkm(1− z)m+1
(1− 2z + zk(1− z))m+1 ;
as claimed.
110 P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114
Lemma 2. The polynomial 1− 2z + zk(1− z) has precisely one zero z=  satisfying |z|6 1. This
zero is given by
=
1
2
+
1
2k+2
+ O
(
k
22k
)
; k →∞:
For all k¿ 1;
exp
(
− n
2k
)
¡
1
(2)n
¡ exp
(
− n
2k+2
)
:
Proof. For the Erst part, simply observe that if z= ei then |1 − 2z|2 = 5 − 4 cos  and |1 − z|2 =
2−2 cos , so |1−2z|¿ |1−z| for all z with |z|= 1. Apply RouchKe’s theorem. To get the approximate
location of , write = 12 +  and substitute into 1− 2 + k(1− ) = 0.
This yields
=
1
2k+2
+
k − 1
2k+1
 +
k(k − 3)
2k
2 + · · · ;
and thus the stated result by iteration. Next, we prove
1
2− 2−(k+1) ¡¡
1
2
+
1
2k+1
:
A simple calculation shows that Q(x) = 1−2x+xk(1−x) decreases on (0; 1). Set a= 1=(2−2−(k+1))
and b= 12 + 2
−(k+1). Then, we End that Q(a)¿ 0 and Q(b)¡ 0, so a¡¡b.
Then, since (1 + x)n ¡ exp(nx) for x¿− 1, we have
1
(2)n
¡
(
1− 1
2k+2
)n
¡ exp
(
− n
2k+2
)
and
1
(2)n
¿
1
(1 + 2−k)n
¿ exp
(
− n
2k
)
:
We shall show that the number of distinct part sizes |D()| of a random composition of n satisEes
|D| ∼ log2 n with probability 1 − o(1). The underlying probabilistic considerations are given in the
following lemma.
Lemma 3. Let X =
∑
Ij where (Ij) are indicator random variables. Suppose that P(Ik) =pk;n =pk .
If a and b are chosen so that both
∑
j6a (1− pj) and
∑
j¿b pj are o(1); then
P(a6X 6 b)¿ 1− o(1):
Proof. For all a6 b, we have
P(a6X 6 b) = 1− P({X ¡a} ∪ {X ¿b})¿ 1− P(X ¡a)− P(X ¿b):
Now, denoting for simplicity a set and its indicator by the same symbol,
P(X ¿b)6P

⋃
j¿b
Ij

6∑
j¿b
P(Ij) =
∑
j¿b
pj;
P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114 111
and
P(X ¡a)6P
(⋃
j6a
I cj
)
6
∑
j6a
(1− P(Ij)) =
∑
j6a
(1− pj):
Hence, if a and b are chosen so that both
∑
j6a (1− pj) and
∑
j¿b pj are o(1) we get
P(a6X 6 b)¿ 1− o(1):
Lemma 4. Let  be a random composition of n. As n → ∞ the number of distinct part sizes
|D()| satis;es |D()| ∼ log2 n with probability 1− o(1).
Proof. As a special case of Lemma 1, the probability that k has multiplicity 0 in the random
composition  is
P(k ∈M0()) = 12n−1 [z
n]
1− z
1− 2z + zk(1− z) ; n¿ 1:
Hence,
P(k ∈D()) = 1− 1
2n−1
[zn]
1− z
1− 2z + zk(1− z) :
From Lemma 2, the rational function (1− z)=(1− 2z + zk(1− z)) is analytic for |z|6 1 except for
a simple pole at z=  ≈ 12 . The residue is −(1 − )=(2 + (k + 1)k − kk). By standard arguments
[6, Section 5:2],
[zn]
1− z
1− 2z + zk(1− z) =
(
1− 
2 + (k + 1)k − kk−1
)
1
n+1
+ O(1):
By Lemma 2,
1− 
2 + (k + 1)k − kk−1 =
1
4
(
1 + O
(
k
2k
))
:
Hence we have
1
2n−1
[zn]
1− z
1− 2z + zk(1− z) =
1
(2)n+1
(
1 + O
(
k
2k
))
:
Using the general bound from Lemma 2
exp
(
− n
2k
)
¡
1
(2)n
¡ exp
(
− n
2k+2
)
;
we see that
P(k ∈D()) = 1− 1
(2)n+1
(
1 + O
(
k
2k
))
6 1− exp
{
− n + 1
2k
}(
1 + O
(
k
2k
))
6
n + 1
2k
+ exp
{
−n + 1
2k
}
O
(
k
2k
)
;
112 P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114
so that letting b= log2 n+ log log n we get∑
k¿b
P(k ∈D()) = O
(
1
log n
)
:
Similarly,
P(k ∈D())¿ 1− exp
{
− n + 1
2k+1
}(
1 + O
(
k
2k
))
;
that is
1− P(k ∈D())6 exp
{
− n + 1
2k+1
}(
1 + O
(
k
2k
))
:
Consequently, for any positive a,∑
16k6a
(1− P(k ∈D()))6C
∑
16k6a
exp
{
− n + 1
2k+1
}
= C
∑
06r¡a
exp
{
−2r n + 1
2a+1
}
6C
∑
r¿0
exp
{
−(r + 1)n + 1
2a+1
}
= C
exp{−(n + 1)=2a+2}
1− exp{−(n + 1)=2a+2} ;
and thus∑
16k6a
(1− P(k ∈D())) = O
(
1
log n
)
;
provided a6 log2 n − log log n. Hence, by Lemma 3 applied to Ik = {k ∈D()}, |D()| ∼ log2 n
with probability 1− o(1).
Given a random composition , the probability that a randomly selected part thereof has multi-
plicity m is |Mm()|=|D()|. Lemma 4 greatly simpliEes the basic problem. Since doing so,
amounts to the neglect of a set of compositions with total probability measure o(1), we may assume
that as n → ∞ the randomly selected composition  satisEes |D()| ∼ log2 n. Thus, P(A(m)n ) ∼
E(|Mm|)=log2 n.
Now we wish to study the asymptotic behavior of P(k ∈Mm()), with the aim of estimating
E(|Mm|) =
∑
k
P(k ∈Mm()):
Lemma 5. The expected value of |Mm| is given by
E(|Mm|) = (1 + o(1))n
m
m!
∑
k
2−km exp(−n=2k):
P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114 113
Proof. As we found in Lemma 1, the relevant generating function is
G(z) =
1
2n−1
P(z)
Qm+1(z)
where P(z) = zkm(1− z)m+1; Q(z) = 1− 2z + zk(1− z):
Recall that Q has a simple zero at z=  ≈ 12 and no other zeros in {z: |z|6 1}. In a deleted
neighborhood of , we have the Laurent expansion
P(z)
Qm+1(z)
=
m+1∑
r=1
c−r
(z − )r +
∞∑
s=0
cs(z − )s:
The asymptotic behavior of [zn]P(z)=Qm+1(z) is governed by the principal part, more speciEcally by
the r =m + 1 term. In view of
[zn](1− z)− =
(
n + − 1
n
)
;
a simple calculation gives
P(k ∈Mm()) =
(
n + m
m
)
2P()
(−Q′())m+1
1
(2)n
(
1 + O
(
1
n
))
:
Set
q(n) =
log n− log log n− log(4(m + 1))
log 2
;
and note that if k ¡q(n) then 2k ¡n=(4(m + 1) log n), so
nm
(2)n
¡nm exp
(
− n
4 · 2k
)
¡nm exp(−(m + 1) log n) = 1
n
:
Hence, we have∑
k6q(n)
P(k ∈Mm()) = O
(
log n
n
)
; n→∞:
In view of the fact just noted, in estimating
∑
k P(k ∈Mm), we can limit ourselves to cases where
k ¿q(n). In that case
1
(2)n
= exp
(
− n
2k+1
)(
1 + O
(
(log n)3
n2
))
; n→∞:
Now
P() = 2−km 2−(m+1)
(
1 + O
(
log n
n
))
and Q′() =− 1 + O
(
log n
n
)
;
so
P(k ∈Mm()) =
(
1 + O
(
log n
n
))
nm
m!
2−(k+1)m exp
(
− n
2k+1
)
:
It is now evident that the contribution to the sum
∑
k P(k ∈Mm()) from those terms with
k ¿ log2 n + log log n is o(1), so there are O(log log n) terms in the sum that make a nontrivial
114 P. Hitczenko et al. / Journal of Computational and Applied Mathematics 142 (2002) 107–114
contribution. Thus, the bound on the error for an individual term suFces to give the correct asymp-
totic result for the sum. Replacing k + 1 by k in the sum, we have the stated result.
Proof of Theorem 1. The computational problem that remains is the asymptotic evaluation of
nm
m!
∞∑
k=1
2−km exp(−n=2k):
Problems of this kind occur frequently in probability theory and the analysis of algorithms, and
now there are diNerent methods available for their study, and these methods are described in several
excellent references [2, chapter 7]. We sketch an approach due to N.G. de Bruijn, which is described
in [4, pp. 131–134] and elsewhere. A special case (m= 1) of the above sum is treated in [5]. The
starting point is Mellin’s formula
exp(−w) = 1
2i
∫ &+i∞
&−i∞
w−z(z) dz; w; &¿ 0:
Substituting this representation (with &=m − 12) for exp(−n=2k) and using uniform convergence,
one obtains
∞∑
k=1
2−km exp(−n=2k) = 1
2i
∫ &+i∞
&−i∞
n−z(z)
2m−z − 1 dz:
Then by the residue theorem,
nm
m!
∞∑
k=1
2−km exp(−n=2k)
=
1
m! log 2

(m− 1)! + 2 Re
∞∑
p=1
e−2ip log2 n
(
m + i
2p
log 2
)
 (1 + o(1)):
The stated result for log nP(A(m)n ) follows.
References
[1] S. Corteel, B. Pittel, C.D. Savage, H.S. Wilf, On the multiplicity of parts in a random partition, Random Structures
Algorithms 14 (1999) 185–197.
[2] P. Flajolet, R. Sedgewick, Analytic Combinatorics, to be published. In part available at 〈http:==pauillac.
inria.fr=algo=flajolet=Publications=publist.html〉.
[3] P. Hitczenko, C.D. Savage, On the multiplicity of parts in a random composition of a large integer, preprint. Available
at 〈http:==www.csc.ncsu.edu=faculty=savage=〉.
[4] D.E. Knuth, The Art of Computer Programming. Vol. III: Sorting and Searching, Addison-Wesley, Reading, MA,
1973.
[5] C.C. Rousseau, Solution of Problem 322, Canad. Math. Bull 26 (1983) 375–377.
[6] H.S. Wilf, Generatingfunctionology, 2nd Edition, Academic Press, San Diego, 1994. Available at
〈http:==www.cis.upenn.edu=∼wilf=〉.


A generating functionology approach to a problem of Wilf 

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A generatingfunctionology approach to a problem of Wilf

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