The $A_2$ Rogers-Ramanujan identities revisited

TL;DR

This paper revisits the $A_2$ Rogers-Ramanujan identities, providing a new derivation using cylindric partitions, thereby offering fresh insights into these classical combinatorial identities.

Contribution

It introduces a novel derivation of the $A_2$ Rogers-Ramanujan identities through cylindric partitions, expanding the combinatorial understanding of these identities.

Findings

New derivation of $A_2$ Rogers-Ramanujan identities

Connection between cylindric partitions and classical identities

Enhanced combinatorial interpretation

Abstract

In this note we show how to rederive the $A_2$ Rogers-Ramanujan identities proven by Andrews, Schilling and Warnaar using cylindric partitions. This paper is dedicated to George Andrews for his $80^{th}$ birthday.

Equations

$$\sum _{n\ge 0}\frac{q^{n(n+i)}}{(q;q{)}_n}=\frac{1}{(q^{1+i};q^5{)}_{\infty }(q^{4-i};q^5{)}_{\infty }}.$$

$$\frac{1}{(q;q{)}_{\infty }}\sum _{n\ge 0}\frac{q^{n(n+1)}}{(q;q{)}_n}=\frac{1}{(q,q^2,q^2,q^3,q^3,q^4,q^5;q^5{)}_{\infty }}$$

$$\frac{1}{(q;q{)}_{\infty }}\sum _{n_1,\dots ,n_k}\frac{q^{{\sum }_{j=1}^k{n}_j^2+{\sum }_{j=i}^k{n}_j}}{(q{)}_{n_1-n_2}\dots (q{)}_{n_{k-1}-n_k}(q{)}_{n_k}}=\frac{(q^i,q^{2k+3-i},q^{2k+3};q^{2k+3}{)}_{\infty }}{(q;q{)}_{\infty }^2}.$$

$$\begin{array}{rl}\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_1+n_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right]& =\frac{1}{(q^2,q^3,q^3,q^4,q^4,q^5;q^7{)}_{\infty }},\\ \sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right]& =\frac{1}{(q,q^2,q^3,q^4,q^5,q^6;q^7{)}_{\infty }},\\ \sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1+1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_1}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1+1}{n_2}\right]& =\frac{1}{(q,q^2,q^3,q^4,q^5,q^6;q^7{)}_{\infty }},\\ \sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1+1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1+1}{n_2}\right]& =\frac{1}{(q,q^2,q^2,q^5,q^5,q^6;q^7{)}_{\infty }},\\ \sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right]& =\frac{1}{(q,q,q^3,q^4,q^6,q^6;q^7{)}_{\infty }},\end{array}$$

$$\left[\genfrac{}{}{0ex}{}{n}{k}\right]=\frac{(q;q{)}_n}{(q;q{)}_k(q;q{)}_{n-k}}.$$

$$F_{(4,0,0),n}(q)$$

$$F_{(3,1,0),n}(q)$$

$$F_{(3,0,1),n}(q)$$

$$F_{(2,2,0),n}(q)$$

$$F_{(2,1,1),n}(q)$$

$$F_c(q)$$

$$F_c(y,q)$$

$$F_{c,n}(q)$$

$$\frac{1}{(q^t;q^t)}\cdot \prod _{i=1}^k\prod _{j=i+1}^k\prod _{m=1}^{c_i}\frac{1}{(q^{m+d_{i+1,j}+j-i};q^t{)}_{\infty }}\cdot \prod _{i=2}^k\prod _{j=2}^{i-1}\prod _{m=1}^{c_i}\frac{1}{(q^{t-(m+d_{j,i-1}+i-j)};q^t{)}_{\infty }}.$$

$$\begin{array}{rl}{F}_{(4,0,0)}(q)& =\frac{1}{(q;q{)}_{\infty }(q^2,q^3,q^3,q^4,q^4,q^5;q^7{)}_{\infty }},\\ F_{(3,1,0)}(q)& =\frac{1}{(q;q{)}_{\infty }(q,q^2,q^3,q^4,q^5,q^6;q^7{)}_{\infty }},\\ F_{(3,0,1)}(q)& =\frac{1}{(q;q{)}_{\infty }(q,q^2,q^3,q^4,q^5,q^6;q^7{)}_{\infty }},\\ F_{(2,2,0)}(q)& =\frac{1}{(q;q{)}_{\infty }(q,q^2,q^2,q^5,q^5,q^6;q^7{)}_{\infty }},\\ F_{(2,1,1)}(q)& =\frac{1}{(q;q{)}_{\infty }(q,q,q^3,q^4,q^6,q^6;q^7{)}_{\infty }}.\end{array}$$

$$c_{i}{(J)}=\left\{\begin{array}[]{ll}c_{i}-1&{\rm if}\ {i\in J}\ {\rm and}\ {(i-1)\not\in J,}\\ c_{i}+1&{\rm if}\ {i\notin J}\ {\rm and}\ {(i-1)\in J,}\\ c_{i}&{\rm otherwise.}\end{array}\right.$$

$$F_c(y,q)=\sum _{\varnothing \subset J\subseteq I_c}(-1{)}^{|J|-1}\frac{F_{c(J)}(y{q}^{|J|},q)}{1-y{q}^{|J|}}.$$

$${\lambda }^{(j)}=\{\begin{array}{ll}(m+n,{\mu }_1^{(j)},{\mu }_2^{(j)},\dots )& \text{if j∈J,}\\ {\mu }^{(j)}& \text{if j∈/J.}\end{array}$$

$$\sum _{m=0}^{\infty }{y}^{m+n}{q}^{|J|(m+n)}{q}^{|M|}=y^n{q}^{|J|n+|M|}\sum _{m=0}^{\infty }(y{q}^{|J|}{)}^m=\frac{y^n{q}^{|J|n+|M|}}{1-y{q}^{|J|}}.$$

$$\sum _{M\in {\mathcal{C}}_{c(J)}}\frac{y^{\max (M)}{q}^{|J|\max (M)+|M|}}{1-y{q}^{|J|}}=\frac{F_{c(J)}(y{q}^{|J|},q)}{1-y{q}^{|J|}},$$

$$G_c(y,q)=(yq;q{)}_{\infty }{F}_c(y,q).$$

$$G_c(y,q)=\sum _{\varnothing \subset J\subseteq I}(-1{)}^{|J|-1}(yq;q{)}_{|J|-1}{G}_{c(J)}(y{q}^{|J|},q)$$

$$\begin{array}{rl}{G}_{(4,0,0)}(y,q)& =\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}{y}^{n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_1+n_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right],\\ G_{(3,1,0)}(y,q)& =\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}{y}^{n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2+n_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right],\\ G_{(3,0,1)}(y,q)& =\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}{y}^{n_1}{q}^{n_1^2+n_2^2-n_1{n}_2}\frac{q^{n_1}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right]\\ & +\sum _{n_1=1}^{\infty }\sum _{n_2=0}^{2n_1-2}{y}^{n_1}{q}^{n_1^2+n_2^2-n_1{n}_2}\frac{q^{2n_2}}{(q;q{)}_{n_1-1}}\left[\genfrac{}{}{0ex}{}{2n_1-2}{n_2}\right],\\ G_{(2,2,0)}(y,q)& =\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}{y}^{n_1}{q}^{n_1^2+n_2^2-n_1{n}_2}\frac{q^{n_1}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right]\\ & +\sum _{n_1=1}^{\infty }\sum _{n_2=0}^{2n_1-2}{y}^{n_1}{q}^{n_1^2+n_2^2-n_1{n}_2}\frac{q^{n_2}(1+q^{n1+n2})}{(q;q{)}_{n_1-1}}\left[\genfrac{}{}{0ex}{}{2n_1-2}{n_2}\right],\\ G_{(2,1,1)}(y,q)& =\sum _{n_1=0}^{\infty }\sum _{n_2=0}^{2n_1}{y}^{n_1}\frac{q^{n_1^2+n_2^2-n_1{n}_2}}{(q;q{)}_{n_1}}\left[\genfrac{}{}{0ex}{}{2n_1}{n_2}\right].\end{array}$$

$$\begin{array}{rl}{G}_{(4,0,0)}(y)& =G_{(3,1,0)}(yq),\\ G_{(3,1,0)}(y)& =G_{(3,0,1)}(yq)+G_{(2,2,0)}(yq)-(1-yq)G_{(2,1,1)}(y{q}^2),\\ G_{(3,0,1)}(y)& =G_{(4,0,0)}(yq)+G_{(2,1,1)}(yq)-(1-yq)G_{(3,1,0)}(y{q}^2),\\ G_{(2,2,0)}(y)& =G_{(3,0,1)}(yq)+G_{(2,1,1)}(yq)-(1-yq)G_{(2,1,1)}(y{q}^2),\\ G_{(2,1,1)}(y)& =G_{(2,1,1)}(yq)+G_{(2,2,0)}(yq)+G_{(3,1,0)}(yq)\\ & -(1-yq)(G_{(2,2,0)}(y{q}^2)+G_{(2,1,1)}(y{q}^2)+G_{(3,0,1)}(y{q}^2))\\ & +(1-yq)(1-y{q}^2)G_{(2,1,1)}(y{q}^3).\end{array}$$

$$\begin{array}{rl}{G}_{(4,0,0)}(y)& =G_{(3,1,0)}(yq),\\ G_{(3,1,0)}(y)& =G_{(2,2,0)}(yq)+y{q}^2{G}_{(3,1,0)}(y{q}^3)+yq{G}_{(2,1,1)}(y{q}^2),\\ G_{(3,0,1)}(y)& =G_{(2,1,1)}(yq)+yq{G}_{(3,1,0)}(y{q}^2),\\ G_{(2,2,0)}(y)& =G_{(2,1,1)}(yq)+yq{G}_{(2,1,1)}(y{q}^2)+y{q}^2{G}_{(3,1,0)}(y{q}^3),\\ G_{(2,1,1)}(y)& =G_{(2,1,1)}(yq)+yq{G}_{(2,2,0)}(yq)+yq{G}_{(2,2,0)}(y{q}^2)\\ & +y{q}^3{G}_{(3,1,0)}(y{q}^4)+y{q}^2{G}_{(2,1,1)}(y{q}^3).\end{array}$$

$$\begin{array}{rl}(1-q^n)g_{(2,1,1)}(n)& =(q^n+q^{2n-1})g_{(2,2,0)}(n-1)\\ & +q^{4n-1}{g}_{(3,1,0)}(n-1)+q^{3n-1}{g}_{(2,1,1)}(n-1).\end{array}$$

$$g_{(2,1,1)}(n)=\sum_{n_{2}=0}^{2n}\frac{q^{n^{2}+n_{2}^{2}-nn_{2}}}{(q;q)_{n}}\left[\begin{array}[]{c}2n\\ n_{2}\end{array}\right].$$

$$\begin{array}{rl}{F}_{c,n}(q)& =[y^0]F_c(y,q)+[y^1]F_c(y,q)+\cdots +[y^n]F_c(y,q)\\ & =[y^n]\frac{F_c(y,q)}{1-y}=[y^n]\frac{G_c(y,q)}{(y;q{)}_n}\end{array}$$

$$\frac{1}{(y,q{)}_n}=\sum _{i=0}^n\frac{y^i}{(q;q{)}_i},$$

$$F_{c,n}(q)=\sum _{n_1=0}^n\frac{1}{(q;q{)}_{n-n_1}}[y^{n_1}]G_c(y,q).$$

$$F_{(3,0,1),n}(q)=\sum_{n_{1},n_{2}}\frac{q^{n_{1}^{2}+n_{2}^{2}-n_{1}n_{2}}}{(q;q)_{n-n_{1}}(q;q)_{n_{1}}}\left(q^{n_{1}}\left[\begin{array}[]{c}2n_{1}\\ n_{2}\end{array}\right]+q^{2n_{2}}(1-q^{n_{1}})\left[\begin{array}[]{c}2n_{1}-2\\ n_{2}\end{array}\right]\right)\$$

$$\sum_{n_{1},n_{2}}\frac{q^{n_{1}^{2}+n_{2}^{2}-n_{1}n_{2}+n_{1}}}{(q;q)_{n_{1}}}\left[\begin{array}[]{c}2n_{1}\\ n_{2}\end{array}\right]+\sum_{n_{1},n_{2}}\frac{q^{n_{1}^{2}+n_{2}^{2}-n_{1}n_{2}+2n_{2}}}{(q;q)_{n_{1}-1}}\left[\begin{array}[]{c}2n_{1}-2\\ n_{2}\end{array}\right]\\$$

$$=$$

Full text

To George Andrews for his $80^{th}$ birthday

1. Introduction

The Rogers-Ramanujan identities were first proved in 1894 by Rogers and rediscovered in the 1910s by Ramanujan [14]. They are

[TABLE]

with $i=0,1$ and where $(a,q)_{\infty}=\prod_{i\geq 0}(1-aq^{i})$ and $(a;q)_{n}=(a;q)_{\infty}/(aq^{n};q)_{\infty}$.

There have been many attempts to give combinatorial proofs of these identities and the first one is due to Garsia and Milne [11]. Unfortunately, it is not simple, and no simple combinatorial proof is known. Recently in [6], the first author presented a new bijective approach to the proofs of the Rogers-Ramanujan identities via the Robinson-Schensted-Knuth correspondence as presented in [13]. The bijection does not give the Rogers-Ramanujan identities but the Rogers-Ramanujan identities divided by $(q;q)_{\infty}$; namely

[TABLE]

where $(a_{1},\ldots,a_{k};q)_{\infty}=\prod_{i=1}^{k}(a_{i};q)_{\infty}$. This proof uses the combinatorics of cylindric partitions [10]. We interpret both sides as the generating function of cylindric partitions of profile $(3,0)$ and the bijection is a polynomial algorithm in the size of the cylindric partition. The idea to use cylindric partitions is due to Foda and Welsh [12] in a more general setting: the Andrews-Gordon identities [2]. For $k>0$ and $0\leq i\leq k$, these identities divided by $(q;q)_{\infty}$ are

[TABLE]

Foda and Welsh interpret the sum side as a generating function for (what they call) decorated Bressoud paths and the product side is interpreted as the generating function of cylindric partitions of profile $(2k+1-i,i)$,and they provide a bijection between the two objects. See [12] for more details.

In this note, we take the idea of applying cylindric partitions to Rogers-Ramanujan type identities a step further, by using them to give an alternative proof of the $A_{2}$ Rogers-Ramanujan identities due to Andrews, Schilling and Warnaar [3].

Theorem 1.1**.**

We have

[TABLE]

where the Gaussian polynomial $\genfrac{[}{]}{0.0pt}{}{n}{k}$ is defined by

[TABLE]

All but the fourth of these identities were obtained in Theorem 5.2 of [3], while the fourth was conjectured in Section 2.4 of [8]. Note that the second and third expressions are equal.

In this note, we prove the following theorem, giving the generating functions $F_{c,n}(q)$ of cylindric partitions indexed by compositions $c$ of 4 into 3 parts, with largest entry at most $n$:

Theorem 1.2**.**

[TABLE]

This result gives a finite version of the sum side of the $A_{2}$ Rogers-Ramanujan identities.

In the $n\to\infty$ limit, we recover the sum side of the identities of Theorem 1.1 divided by $(q;q)_{\infty}$. We also explain how to get the product side. It is a corollary of a result of Borodin [4]. In Section 2 below, we start by defining cylindric partitions and then obtain the product sides of particular cylindric partitions. These yield the right hand sides of the expressions in Theorem 1.1. The sum expressions on the left hand sides are computed in Section 3.

Acknowledgments. SC was in residence at MSRI (NSF grant DMS-1440140) and was visiting the Mathematics department at UC Berkeley during the completion of this work. TW acknowledges partial support from the Australian Research Council. The authors wish to thank Omar Foda for his interest in this work and useful discussions. The authors also wish to thank the anonymous referee for her excellent suggestions and careful reading.

2. Cylindric partitions and the product side

Cylindric partitions were introduced by Gessel and Krattenthaler [10] and appeared naturally in different contexts [4, 5, 7, 9, 12, 15]. Let $\ell$ and $k$ be two positive integers. In this note, we choose to index cylindric partitions by compositions of $\ell$ into $k$ non negative parts.

Definition 2.1**.**

Given a composition $c=(c_{1},\ldots,c_{k})$, a cylindric partition of profile $c$ is a sequence of $k$ partitions $\Lambda=(\lambda^{(1)},\ldots\lambda^{(k)})$ such that :

• •

$\lambda^{(i)}_{j}\geq\lambda^{(i+1)}_{j+c_{i+1}},$**

• •

$\lambda^{(k)}_{j}\geq\lambda^{(1)}_{j+c_{1}}$.

for all $i$ and $j$.

For example, the sequence $\Lambda=((3,2,1,1),(4,3,3,1),(4,1,1))$ is a cylindric partition of profile $(2,2,0)$. One can check that for all $j$, $\lambda^{(1)}_{j}\geq\lambda^{(2)}_{j+2}$, $\lambda^{(2)}_{j}\geq\lambda^{(3)}_{j}$ and $\lambda^{(3)}_{j}\geq\lambda^{(1)}_{j+2}$ for all $j$. Note that this definition implies that cylindric partitions of profile $(c_{1},\ldots,c_{k})$ are in bijection with cylindric partitions of profile $(c_{k},c_{1},\ldots,c_{k-1})$.

Our goal is to compute generating functions of cylindric partitions of a given profile $c$ according to two statistics. Given a $\Lambda=(\lambda^{(1)},\ldots,\lambda^{(k)})$, let

• •

$|\Lambda|=\sum_{i=1}^{k}\sum_{j\geq 1}\lambda^{(i)}_{j}$, the sum of the entries of the cylindric plane partition, and

• •

$\max(\Lambda)=\max(\lambda^{(1)}_{1},\ldots\lambda^{(k)}_{1})$, the largest entry of the cylindric plane partition.

Going back to our example, we have $|\Lambda|=24$, and $\max(\Lambda)=4$.

Let ${\mathcal{C}_{c}}$ be the set of cylindric partitions of profile $c$ and let ${\mathcal{C}_{c,n}}$ be the set of cylindric partitions of profile $c$ and such that the largest entry is at most $n$. We are interested in the following generating functions.

[TABLE]

A surprising and beautiful result is that for any $c$, the generating function $F_{c}(q)$ can be written as a product. Namely, with $t=k+\ell$,

Theorem 2.2**.**

[4]** The generating function $F_{c}(q)$ is equal to

[TABLE]

where $d_{i,j}=c_{i}+c_{i+1}+\ldots+c_{j}$.

The original result is written is a different but equivalent form.

For what follows, we restrict attention to the case $\ell=4$ and $k=3$. As cylindric partitions of profile $(c_{1},\ldots,c_{k})$ are in bijection with partitions of profile $(c_{k},c_{1},\ldots,c_{k-1})$, we need only compute the generating functions for the compositions $(4,0,0)$, $(3,1,0)$, $(3,0,1)$, $(2,2,0)$, and $(2,1,1)$. We now apply the previous theorem:

Corollary 2.3**.**

[TABLE]

Note that these five products are precisely those in Theorem 1.1 divided by $(q;q)_{\infty}$.

3. The sum side

We first prove a general functional equation for $F_{c}(y,q)$ for any profile $c$. Suppose that $k>1$ and $c=(c_{1},\ldots,c_{k})$. Let $I_{c}$ be the subset of $\{1,\ldots,k\}$ such that $i\in I_{c}$ if and only if $c_{i}>0$. For example if $c=(2,2,0)$ then $I_{c}=\{1,2\}$. Given a subset $J$ of $I_{c}$, we define the composition $c{(J)}=(c_{1}(J),\ldots,c_{k}(J))$ by

[TABLE]

Here we set $c_{0}=c_{k}$.

Proposition 3.1**.**

For any composition $c=(c_{1},\ldots,c_{k})$,

[TABLE]

with the conditions $F_{c}(0,q)=1$ and $F_{c}(y,0)=1$.

Proof.

The proof make use of an inclusion-exclusion argument.

First, for fixed $J$ such that $\emptyset\subset J\subseteq I_{c}$, we require the generating function of cylindric partitions $\Lambda$ of profile $c$ such that $\lambda_{1}^{(j)}=\max(\Lambda)$ for all $j\in J$.

Let $M=(\mu^{(1)},\ldots,\mu^{(k)})$ be a cylindric partition of profile $c{(J)}$, and set $n=\max(M)$. Then, for a fixed integer $m\geq 0$, create a cylindric partition $\Lambda=(\lambda^{(1)},\ldots,\lambda^{(k)})$ using the following recipe:

[TABLE]

It is easily checked that $\Lambda$ is a cylindric partition of profile $c$ and that $\max(\Lambda)=m+n$. Moreover, $\lambda_{1}^{(j)}=\max(\Lambda)$ for all $j\in J$. The generating function for all cylindric partitions $\Lambda$ obtained from $M$ in this way is

[TABLE]

Then the generating function for all cylindric partitions $\Lambda$ obtained in this way from any cylindric partition $M$ of profile $c(J)$ is

[TABLE]

making use of the definition (11).

Let $\Lambda=(\lambda^{(1)},\ldots,\lambda^{(k)})$ be an arbitrary cylindric partition of profile $c$, and let $p=\max(\Lambda)$. Because $\lambda_{1}^{(i-1)}\geq\lambda_{1}^{(i)}$ whenever $i\notin I_{c}$, it must be the case that $p=\lambda^{(j)}_{1}$ for some $j\in I_{c}$. Then, if $J\neq\emptyset$ is such that $p=\lambda^{(j)}_{1}$ for each $j\in J$ (this $J$ might not be unique), we see that $\Lambda$ is one of the cylindric partitions enumerated by (16). However, because $\Lambda$ can arise from various different $J$, the generating function for cylindric partitions of profile $c$ is obtained via the inclusion-exclusion process. This immediately gives (14). ∎

Now, for each composition $c$, define

[TABLE]

In terms of this, the previous result translates to

[TABLE]

with $G_{c}(0,q)=G_{c}(y,0)=1$.

Theorem 3.2**.**

[TABLE]

Proof.

In this proof, we abbreviate $G_{c}(y,q)$ to $G_{c}(y)$ for convenience. Applying the form (18) of Propostion 3.1 to the case $\ell=4$ and $k=3$ yields

[TABLE]

By manipulating these equations, we obtain

[TABLE]

We claim that this system of equations (19) together with the boundary conditions $G_{c}(0,q)=G_{c}(y,0)=1$ for each composition $c$, is uniquely solved by the expressions stated in the theorem. This is proved using an induction argument involving all five expressions.

We use induction on the exponents of $y$. For each composition $c$, let $g_{c}(n)$ denote the coefficient of $y^{n}$ in the solution $G_{c}(y)$ of (19). The boundary conditions $G_{c}(0,q)=1$ imply that each $g_{c}(0)=1$. This holds for the expressions of the theorem. So now, for $n>0$, assume that $g_{c}(n_{1})$ agrees with the coefficient of $y^{n_{1}}$ in the statement of the theorem for each $n_{1}<n$ and each composition $c$. We must check that $g_{c}(n)$, as determined by the expressions (19), is equal to the coefficient of $y^{n}$ in the statement of the theorem for each $c$.

The fifth expression in (19) implies that

[TABLE]

Using the expressions for $g_{c}(n)$ implied by the induction hypothsis then yields

[TABLE]

This expression, along with the other expressions for $g_{c}(n)$ implied by (19), enables us to compute, in turn, $g_{(2,2,0)}(n)$, $g_{(3,0,1)}(n)$, $g_{(3,1,0)}(n)$ and finally $g_{(4,0,0)}(n)$. Because the expressions that result agree with the corresponding coefficients of $y^{n}$ in the statement of the theorem, the induction argument is complete. ∎

Proof of Theorem 1.2. Comparing (11) and (12) leads to

[TABLE]

having used (17). By the $q$-binomial theorem [1], we have

[TABLE]

and therefore it follows that

[TABLE]

Applying this to the expressions of Theorem 3.2 then yields those of Theorem 1.2. ∎

Proof of Theorem 1.1. Let us now comment on how to get the end of the proof of Theorem 1.1. The right hand side of the five identities correspond to Corollary 2.3. To get the sum side; we first let $n$ tend to infinity in Theorem 1.2 and multiply by $(q;q)_{\infty}$. We get directly the left hand side of the first, second and fifth identities. Let us show how to get the left hand side third identity. Theorem 1.2 states that the generating function of the cylindric partitions of profile $(3,0,1)$ and largest entry at most $n$ is:

[TABLE]

We let $n\rightarrow\infty$ and multiply by $(q;q)_{\infty}$ in (20), we get

[TABLE]

After the change $(n_{1},n_{2})\rightarrow(n_{1}+1,n_{2}-1)$ in the second sum, we get

[TABLE]

using the well known recurrence relation

[TABLE]

This is the left hand side of the third identity of Theorem 1.1.

We leave the computation for the left hand side of the fourth identity to the reader. One needs to show that \displaystyle{\sum_{n_{1},n_{2}}\frac{q^{n_{1}^{2}+n_{2}^{2}-n_{1}n_{2}+n_{2}}}{(q;q)_{n_{1}}}\left[\begin{array}[]{c}2n_{1}+1\\ n_{2}\end{array}\right]} equals

[TABLE]

∎