A combinatorial proof of a partition identity of Andrews and Stanley

In his paper,"On a Partition Function of Richard Stanley,"George Andrews proves a certain partition identity analytically and asks for a combinatorial proof. This paper provides the requested combinatorial proof.


Introduction
In the October 2002 issue of American Mathematical Monthly, Richard Stanley posed a problem on partitions [6]. In [2], George Andrews studies the function S(r, s; n) = the number of partitions π of n such that π has r odd parts and the conjugate π ′ of π has s odd parts, and proves a result from which he solves Stanley's problem as a corollary.
Andrews also states some additional interesting corollaries, including Theorem 1.
n,r,s≧0 S(r, s; n)z r y s q n = ∞ j=1 (1 + yzq 2j−1 ) which he proves analytically as the limiting case of a certain polynomial identity [2,Theorem 1]. At the end of the paper, Andrews states that (1) "cries out for a combinatorial proof." The purpose of this paper is to present such a proof.

Definitions and Notation
The following defintions and notations follow Macdonald [4].
A partition π of an integer n is a nonincreasing sequence of nonnegative integers containing only finitely many nonzero terms such that the sum of the terms is n. Thus, π = (π 1 , π 2 , π 3 , . . . ), with and π i ∈ Z. Since the tail of such a sequence must, by definition, be an infinite string of zeros, it will be convenient to suppress this in the notation. Thus the partition (2, 2, 1, 1, 1, 0, 0, 0, 0, 0, . . . ) is normally written (2, 2, 1, 1, 1). In any event, no distinction shall be drawn among sequences written with or without any number of trailing zeros. The nonzero terms π i in (2) are called the parts of the partition π. The number of parts of π is called the length of π and is denoted l(π). Additionally, I will follow Stanley [6] and Andrews [2] and let O(π) denote the number of odd parts in the partition π.
The set of all partitions is denoted P.

The Plan
I will start with the set of all partitions P counted in such a way so that it is easily seen to be generated by the infinite product on the right hand side of (1). I will then map P bijectively onto itself while keeping track of enough of the internal counting to see that (1) holds. In order to keep the presentation as straightforward as possible, the mapping will be presented in three simple steps (the maps α, β and γ), and the bijection will be the composition of these three maps.

The Generating Function
By standard elementary reasoning (see [ is the generating function for partitions into parts ≡ 0 (mod 4), is the generating function for partitions into parts ≡ 2 (mod 4), where each part is counted with a weight of z 2 , and is the generating function for partitions into odd parts where each part is counted with weight y and each distinct integer of odd multiplicity is counted with weight x. Thus, where R(n) = the number of partitions of n with s odd parts, ρ parts congruent to 2 (mod 4) and t different odd integers of odd multiplicity.
Upon setting x equal to z and renaming 2ρ + t as r, we see that the infinite product of (1) generates partitions with exactly s odd parts and such that twice the number of parts congruent to 2 (mod 4) plus the number of different odd integers of odd multiplicity is exactly r. Thus, by mapping each of these partitions to a partition with exactly r odd parts and s odd parts in its conjugate, we will have a bijective proof that (1) holds.

The Mapping
Start with an arbitrary partition κ = 1 m 1 (κ) 2 m 2 (κ) 3 m 3 (κ) · · · . Define a map where A is the set of all partitions with no repeated odd parts and B = P \ A = the set of partitions with only odd parts of even multiplicity, by and Informally, α separates κ into two partitions λ and µ, where λ gets all of the even parts of κ and one copy of each odd part of odd multiplicity in κ, and µ gets what is left over, i.e. even quantities of odd parts. Notice that and Define the map β 1 : or equivalently, m i (ζ) = P (m 2i−1 (κ)) + 2m 2i (κ) + P (m 2i+1 (κ)).
Informally, this means that every even part 2j of λ is mapped by β 1 to a pair of j's and each and every odd part 2j + 1 is mapped to the pair j + 1, j. So, C is the set of partitions ζ = (ζ 1 , ζ 2 , . . . ) such that for all i ∈ Z + , Define the map β 2 : B → D so that ξ := β 2 (µ) = µ ′ . Note that D is the set of partitions ξ such that for all i ∈ Z + , ξ 2i = ξ 2i+1 , and ξ i is even.
Proof of Claim 2. To see this, notice that by (8), the only way that ζ ′ 1 can be odd is if 1 is a part of λ (which of course means that m 1 (κ) is odd). This is so because every part of λ other than a 1 is mapped to a pair of parts, but 1 is mapped to (1, 0), which, by definition is only one part since 0 does not count as a part.
Next, the only way that ζ ′ 2 can be odd is if λ contains a 3; and in general, the only way that ζ ′ i can be odd is if λ contains 2i − 1 as a part (which means that m 2i−1 (κ) is odd).
Finally, define a map γ : C × D → P, by π = γ(ζ, ξ) := ζ + π. Thus, we have O(π) = O(ζ) + O(ξ) = 2ρ + t =: r where the third to last equality follows from the fact that ⌊ λ i +1 2 ⌋ − ⌊ λ i 2 ⌋ equals 0 if i is even, and equals 1 if i is odd. Now let us consider the invertibility of each of the maps. It is easy to see that α is invertible; The invertibility of γ, however, is more subtle. Given any partition π, we need to "split it" into the sum of two partitions ζ ∈ C and ξ ∈ D. It is sufficient to recover ξ 1 and ξ 2i for i ≧ 1 since ξ 2i = ξ 2i+1 for all i ≧ 1, and once we have ξ, we know ζ since ζ + ξ = π. Of course, we need to do this given only the partition π.
The easiest case is finding ξ 1 since we know that ξ 1 = s − t. Both s and t are readily available to us in π: Besides counting the number of odd parts in κ, s counts the number of odd parts in π ′ . Now t counts the number of odd parts of odd multiplicity in κ which translates into the number of odd parts in λ, which translates into the number of pairs (ζ 2i−1 , ζ 2i ) in ζ whose sum is odd, which in turn translates to the number of pairs (π 2i−1 , π 2i ) in π whose sum is odd (since all parts of ξ are even). Thus we can recover ξ 1 from π. Now that we have both π and ξ 1 in hand, we can recover the rest of ξ: Proof of Claim 4.
Note: I have written a Maple package which performs all of the mappings described in this paper, and also provides additional examples. Please visit http://www.math.rutgers.edu/~asills/papers.html to download it.

Original Presentation
This paper provides the full account of the talk I presented in the special session on q-series at the AMS sectional meeting in Baton Rouge on March 15, 2003 [5].

Subsequent Developments
It has come to my attention that Cilanne Boulet [3] and Ae Ja Yee [7] have independently discovered additional combinatorial proofs of Theorem 1.