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\title{The $q$-variations of Sylvester's  bijection \\
between odd 
and strict partitions }
\author{Jiang Zeng\\
\small Institut Girard Desargues,  Universit\'e Claude Bernard (Lyon 1)\\
\small 43, boulevard du 11 novembre 1918, 69622 Villeurbanne cedex, France\\
 \small E-mail:   \texttt{zeng@desargues.univ-lyon1.fr}}
\date{}
\begin{document}
\hsize=13cm
\maketitle
\begin{abstract}
The $q$-identities corresponding to Sylvester's bijection between odd
and strict partitions are investigated. In particular,
we show that 
Sylvester's bijection  implies 
the Rogers-Fine identity and  give a simple proof of
a partition theorem  of Fine, which does not follow
directly 
from Sylvester's bijection. Finally, the so-called $(m,c)$-analogues
of Sylvester's bijection are also discussed.
\end{abstract}

\section{Introduction} 
A \emph{partition} is any sequence of nonnegative integers 
$\la=(\la_1,\la_2,\ldots)$ in weakly decreasing order and containing only
finitely many non-zero terms, called \emph{parts}. A partition is called \emph{odd} if all the
parts  are odd, and \emph{strict} if the parts are 
distinct each other. A partition of an integer $n$ is a partition such that 
the
sum of its parts is equal to $n$.
Let ${\cal P}_{\cal O}(n)$ be the set of 
partitions into \emph{odd} parts of an integer $n$ and ${\cal P}_{\cal D}(n)$ be the
set of \emph{strict} partitions  of $n$. Then
Euler's partition theorem  states 
that $\# {\cal P}_{\cal O}(n)=\#{\cal P}_{\cal D}(n)$.
It is easy to see that Euler's result follows from the infinite 
product identity~: 
\begin{equation}
\prod_{i\geq 0}\frac{1}{1-q^{2i+1}}
=\prod_{i\geq 0}(1+q^{i+1}).
\end{equation}
In 1882 Sylvester~\cite{Syl82} published an interesting bijective proof of
Euler's result. Another classical bijection was given by
Glaisher~\cite{Gl83}, but his proof is usually
considered to  be  close to the above generating function (or g.f.) 
proof (see \cite[p.52]{Bre99}). A 
 natural question is then to  seek the generating function proof
that corresponds or is closer  to Sylvester's bijection.
In this paper we shall investigate the dual problem~: 
what kind of  $q$-identities can be
derived from Sylvester's bijection ?   

There are several ways to
 describe Sylvester's bijection~\cite{And84, Bre99, Mac84, Las00}, his 
original 
version was actually in terms of \emph{2-modular} diagrams, 
later used by Bessenrodt~\cite{Bes94}, other descriptions are 
based on \emph{fish hooks} diagrams~\cite[p. 13, vol. II]{Mac84} or Frobenius representation 
of partitions~\cite{And84}. Let us recall 
 Sylvester's bijection
$\varphi : {\cal P}_O(n)\longrightarrow {\cal P}_D(n)$ through
an  example.
If $\lambda=(9^2\,5\,3\,1)\in {\cal P}_O(27)$, then draw the 2-modular
diagram and read it through its hooks, where each node ``2'' should be read twice~:
$${\cal P}_O(27)\ni  (9^2\,5\,3\,1)\mapsto 
\small{
\begin{array}{l}
2=2=2=2-1\\
\parallel\\
2\hspace{14pt}2=2=2-1\\
\parallel\hspace{14pt}\parallel\\
2\hspace{14pt}2\hspace{14pt}1\\
\parallel\hspace{14pt}\hspace{1pt}\mid\\
2\hspace{14pt}1\\
\hspace{1pt}\mid\\
1\\
\end{array}
}\mapsto (9\,7\,6\, 4\, 1)\in {\cal P}_{\cal D}(27).
$$

Given a partition $\la=(\la_1, \, \la_2, \, \ldots)$ we shall note
$\la_1$ the largest part of $\la$ and $l(\la)$ the 
\emph{length} of $\la$, i.e. the number of its parts.  
The \emph{rank} of $\la$  is its largest part less its 
length, i.e. $\la_1-l(\la)$. The \emph{alternative length} of  $\lambda$ is 
 $l_a(\lambda)=\lambda_1-\lambda_2+\lambda_3-\cdots$. 
A \emph{chain} of  $\lambda$ is a subsequence 
$\lambda_p,\, \lambda_{p+1},\, \ldots,
\lambda_q$ such that $\lambda_i-\lambda_{i+1}\leq 1$ for $p\leq i\leq
q$ so in a strict partition, a  \emph{chain} is a maximal 
sequence of parts that are consecutive integers. 
Let $n_d(\lambda)$ be the number of \emph{distinct} parts of
 $\lambda$ and $n_c(\lambda)$ the number of chains of  $\lambda$.
Finally let $d_2(\la)$ be the size of Durfee square in the 2-modular diagram of $\la$.

For example, if $\la=(12,\, 10,\, 9,\, 8,\, 5,\, 5)$, then 
$\la$  has three chains : 
$\{12\}$, $\{10, 9, 8\}$, $\{5, 5\}$, so $n_c(\la)=3$, 
$l(\la)=6$, $\la_1=12$,
the rank is $12-6=6$, $n_d(\la)=5$, $l_a(\la)=12-10+9-8+6-5=4$ and 
$d_2(\la)=4$.

 Let  ${\cal P}_{\cal O}$ denote the set of odd partitions  and 
${\cal  P}_{\cal D}$ the set of strict partitions. 
We now summarize the
most remarkable
properties of Sylvester's bijection as following 
(cf. \cite[Proposition~2.2]{Bes94} and \cite[p. 51-52]{Bre99}). 
\begin{thm}[Sylvester-Bessenrodt]
Let $\varphi : {\cal P}_O\longrightarrow {\cal P}_D$ 
be Sylvester's bijection and $\mu=\varphi(\la)$, 
then $|\la|=|\mu|$, and 
\begin{itemize}
\item $d_2(\la)=[(l(\mu)+1)/2]$,
\item $n_d(\la)=n_c(\mu)$,
\item $l(\la)=l_a(\mu)$,
\item $l(\la)+(\la_1-1)/2=\mu_1$.
\end{itemize}  
\end{thm}
Consider the corresponding generating functions of odd and strict partitions ~:
\begin{eqnarray}
A&=&\sum_{\la\in {\cal P}_O}x^{n_d(\la)}y^{l(\la)}z^{l(\la)
+(\la_1-1)/2}t^{d_2(\la)}q^{|\la|}\label{eq:A}\\
B&=&\sum_{\mu\in {\cal P}_D}x^{n_c(\mu)}
y^{l_a(\mu)}z^{\mu_1}t^{\lfloor(l(\mu)+1)/2\rfloor}q^{|\mu|}\label{eq:B}.
\end{eqnarray}
Il follows from  Theorem~1 that  $A=B$.
Therefore one may expect to derive  some interesting $q$-identities 
by calculating $A$ and $B$ separately.
Surprisingly, the two analytical expressions
thus obtained coincide, so we will not get  a true $q$-identity but,
instead,  an 
\emph{independent} proof of the identity $A=B$.

Throughout this paper  we shall use  the standard notations of $q$-series~:
$$
(a)_n:=(a;\,q)_n=\prod_{i=1}^n(1-aq^{i-1}),\qquad 
(a)_\infty:=(a;\,q)_\infty=\prod_{i=1}^\infty(1-aq^{i-1}).
$$

\begin{thm} The generating functions (\ref{eq:A}) and (\ref{eq:B}) are both equal to
\begin{eqnarray}\label{eq:gfsyl}
&&\ x\sum_{n\geq  
0}(ty)^{n+1}z^{2n+1}q^{(n+1)(2n+1)}\,
\left[1-(1-x)yz^2\,q^{4n+3}\right]\nonumber\\
&&\qquad \times \,
\frac{((1-x)zq^2;\,q^2)_{n}((1-x)yzq;\,q^2)_{n}}{
(zq^2;\,q^2)_{n+1}(yzq;\,q^2)_{n+1}}.\label{eq:gf}
\end{eqnarray}
\end{thm}
Interesting $q$-identities will  appear when we specialize some of the 
parameters.
For example,   
we shall derive from the above theorem the so-called 
Hence Sylvester's bijection turns out to provide
 a combinatorial proof
of the Rogers-Fine identity. Note that
Andrews~\cite{And72} has given a different 
combinatorial proof of
  (\ref{eq:RF}) using Durfee rectangles. 

The rest of this paper is organized as follows.
In section~2 we shall prove  Theorem~2.
In section~3 we will derive some special cases of Theorem~2, which include
Rogers-Fine identity and an extension of
Lebesgue's identity, which corresponds to
Sylvester's original partition result. In section~4 we will review two 
partition theorems of Fine and show how
a partition theorem of 
Fine~\cite{Fine48},
which  does not follow from Sylvester's bijection, is  related to 
Sylvester's bijection through generating function point of view.
Finally, in section~5 we consider the extension of Sylvester's bijection when 
the odd partitions are replaced by partitions whose parts are in an arithmetical 
sequence~$\{c+mk\}_{k\geq 0}$.
In this context
Theorem~4  generalizes  the previous results
of Bessenrodt~\cite{Bes94}
and  Pak-Postnikov~\cite{PP98}, while Theorem~5 is an extension of Theorem~2.
However, as we will show in Proposition~1, the (m,c)-analogue of Rogers-Fine identity is actually
equivalent to the ordinary one.
 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem~2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let's first consider the generating function (\ref{eq:A}). 
We shall  apply
 the Durfee square dissection in the 2-modular diagram of an odd partition $\lambda$
 and distinguish two cases according to the element $c(\la)$
at the southeast corner of the Durfee square.
Let $d_2(\la)=r$ with $ r\geq 1$.

i) If $c(\la)=2$, then we decompose the 2-modular diagram of 
odd partitions as in Figure~1 and compute the generating functions of 
partitions corresponding to the diagrams A, B, C, D.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\unitlength=1cm
\begin{figure}
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\put(9,3.4){\line(0,-1){1.2}}\put(9.4, 2.6){A}\put(10.2,3.4){\line(0,-1){1.2}}
\put(10.4,3.4){\line(0,-1){1.2}}\put(10.8, 2.6){B}\put(11.4,3.4){\line(0,-1){1.2}}
\put(11.6,3.4){\line(0,-1){1.2}}\put(11.9, 3){C}
\put(9,2.2){\line(1,0){1.2}}
\put(10.4,2.2){\line(1,0){1}}\put(11.61, 2.21){\line(1,1){1.19}}
\put(9,2){\line(1,0){1.2}}
\put(9,2){\line(0,-1){1.2}}\put(9.4, 1.6){D}
\put(9, 0.81){\line(1,1){1.19}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\put(7, 1.6){\vector(1,0){1}}
\put(0,3.4){2$\kern 12pt \cdots\kern 18pt  
2 \kern 8pt 2\kern 8pt \cdots\kern 8pt  2 \kern 8pt 
2 \kern 2pt \cdots\kern 5pt $ 2 1}
\put(5.12, 2.8){$\cdot$}
\put(0.1,2.7){$\vdots\kern 10pt \ddots\kern 18pt \vdots$\kern 12 pt 
$\vdots\kern 12pt \ddots\kern 7pt \vdots$ \kern 18pt \kern 10pt $\cdot$\kern 12pt $\cdot$ }
\put(4.3,2.9){$\vdots$}
\put(4.82, 2.5){$\cdot$}
\put(4.3,2.4){2\hskip 0.6cm $\cdot$}
\put(4.3,2.1){\hskip 0.6cm $\cdot$}
\put(0,2){2\kern 12pt $\cdots$\kern 18pt  2\kern 12pt 2\kern 8pt $\cdots$\kern 8pt  2}
\put(4.3,2){1}
\put(0, 1.5){2\kern 8pt $\cdots$\kern 8pt 2\kern 8pt 1}
\put(0.1, 1){$\vdots$ \kern 26pt $\cdot$ }
\put(1, 0.8){$\cdot$\kern 12pt $\cdot$ }
\put(0.8, 0.6){$\cdot$\kern 12pt $\cdot$ }
\put(0.05, 0.5){2}
\put(1, 0.4){$\cdot$}
\put(0.05, 0.1){1}
\end{picture}
\caption{\textit{Decomposition of a  2-modular diagram with 2 at 
the southeast corner of the Durfee square {\scriptsize \fbox{A}}}}
\end{figure}

It is clear that the g.f. of  ``A'' is
$xt^ry^rz^{r+(r-1)}q^{2r^2}$, 
that  of ``B'' is
$\frac{1}{1-zq^{2r}}$,
since  the number of distinct parts of a partition 
is equal to that of its conjugate (see \cite{All99} for other applications of this 
invariant), the g.f. of ``C'' can be obtained through the corresponding 
conjugate partitions  as follows~:
$$
q^{r}\prod_{i=1}^{r-1}\left(1+\frac{xzq^{2i}}{1-zq^{2i}}\right)=
q^r\frac{((1-x)zq^2;q^2)_{r-1}}{ (zq^2;q^2)_{r-1}}
$$
and that of ``D'' is 
$$
\prod_{i=0}^{r-1}\left(1+\frac{xyzq^{2i+1}}{ 1-yzq^{2i+1}}\right)=
\frac{((1-x)zyq;q^2)_{r}}{(zyq;q^2)_r}.
$$
Multiplying the four generating functions we get
$$
xy^rz^{2r}t^rq^{2r^2+r}
\frac{((1-x)zq^2;q^2)_{r-1}}{ (zq^2;q^2)_r}\cdot 
\frac{((1-x)zyq;q^2)_{r}}{(zyq;q^2)_r}.
$$

ii) If  $c(\la)=1$, then we decompose the 2-modular diagram of odd partitions as in Figure~2 where we move all the 1's at the end of $r$
 first rows to the last column of ``A'', 
which corresponds to the Durfee square,
and compute the corresponding generating functions 
of partitions corresponding to ``A'', ``B'' and ``C''. It is easy to see that 
the g.f. of ``A'' is $xt^ry^rz^{r+(r-1)}q^{2r^2-r}$, by enumerating its conjugate
partitions the g.f.  of ``B'' is
$$
\prod_{i=1}^{r-1}\left(1+\frac{xzq^{2i}}{1-zq^{2i}}\right)
=\frac{((1-x)zq^2;q^2)_{r-1}}{(zq^2;q^2)_{r-1}},
$$
that of ``C'' is
$$
\frac{1}{1-yzq^{2r-1}}
\prod_{i=0}^{r-2}\left(1+\frac{xyzq^{2i+1}}{ 1-yzq^{2i+1}}\right)=
\frac{((1-x)yzq;q^2)_{r-1}}{(yzq;q^2)_r}. 
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\unitlength=1cm
\begin{figure}
\begin{picture}(4,2)
\put(10,3.4){\line(1,0){1.2}}\put(11.5,3.4){\line(1,0){1.2}}
\put(10,3.4){\line(0,-1){1.2}}\put(10.4, 2.6){A}\put(11.2,3.4){\line(0,-1){1.2}}
\put(11.5,3.4){\line(0,-1){1.2}}\put(11.8, 3){B}
\put(10,2.2){\line(1,0){1.2}}\put(11.51, 2.21){\line(1,1){1.19}}
\put(10,2){\line(1,0){1.2}}
\put(10,2){\line(0,-1){1.2}}\put(10.4, 1.6){C}
\put(10, 0.81){\line(1,1){1.19}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\put(7, 1.6){\vector(1,0){1}}
\put(2,3.4){2\kern 8pt $\cdots$\kern 8pt 2\kern 8pt  1 \kern 8pt 
2\kern 2pt $\cdots$\kern 5pt  2 }
\put(5.2, 2.8){$\cdot$}
\put(2.1,2.7){$\vdots\kern 10pt \ddots\kern 18pt \vdots$\kern 12 pt 
$\vdots$ \kern 14pt $\cdot$}
\put(4.9, 2.5){$\cdot$}
\put(4.35,2.2){2}
\put(2,2){2\kern 8pt $\cdots$\kern 8pt 2\kern 8pt  1}
\put(2, 1.5){2\kern 8pt $\cdots$\kern 8pt 2\kern 8pt 1}
\put(2.1, 1){$\vdots$ \kern 26pt $\cdot$}
\put(3, 0.8){$\cdot$\kern 12pt $\cdot$ }
\put(2.8, 0.6){$\cdot$\kern 12pt $\cdot$ }
\put(2.1, 0.5){2}
\put(3, 0.4){$\cdot$}
\put(2.1, 0.1){1}
\end{picture}
\caption{\textit{Decomposition of a  2-modular diagram with 1 at 
the southeast corner of the Durfee square {\scriptsize \fbox{A}}}}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Multiplying the three generating functions  we get 
$$
xy^rz^{2r-1}t^rq^{2r^2-r}
\frac{((1-x)yzq;q^2)_{r-1}}{(yzq;q^2)_r}\cdot 
\frac{((1-x)zq^2;q^2)_{r-1}}{(zq^2;q^2)_{r-1}}
$$
Summing up the two generating functions of i) and ii) 
and replacing $r$ by $n+1$ we obtain the expression~(\ref{eq:gfsyl}).


On the other hand,
if $\mu$ is a strict partition,
then  
each integer less than $l(\mu)$ appears as a part in its conjugate $\mu'$ 
and there is a gap in the parts of $\mu$, i.e. $\mu_i-\mu_{i+1}\geq 2$, 
iff the corresponding part in $\mu'$ is repeated. 
Moreover, in view of the weight function (\ref{eq:B}) each odd (resp. even) 
part of 
$\mu'$ should be weighted by $yz$ (resp. $z$) and each 
reapeted odd part should be weighted by $x$.
Hence the relevant generating function of strict partitions of length 
$2n+2$ is
\begin{eqnarray*}
&& \frac{xzt^{n+1}q^{2n+2}}{ 1-zq^{2n+2}}
\prod_{i=0}^{n}yzq^{2i+1}\left(1+\frac{xyzq^{2i+1}}{ 1-yzq^{2i+1}}\right)
\prod_{i=1}^{n}zq^{2i}\left(1+\frac{xzq^{2i}}{1-zq^{2i}}\right)\\
&=& x(qty)^{n+1}z^{2n+2}q^{2(n+1)^2}
\frac{((1-x)yzq;q^2)_{n+1}((1-x)zq^2;q^2)_{n}}
{(yzq;q^2)_{n+1}(zq^2;q^2)_{n+1}};
\end{eqnarray*}
and that of strict partitions of length  $2n+1$ is
\begin{eqnarray*}
&&\frac{xyzt^{n+1}q^{2n+1}}{ 1-yzq^{2n+1}}
\prod_{i=0}^{n-1}yzq^{2i+1}\left(1+\frac{xyzq^{2i+1}}{1-yzq^{2i+1}}\right)
\prod_{i=1}^{n}zq^{2i}\left(1+\frac{xzq^{2i}}{1-zq^{2i}}\right)\\
&=& x(ty)^{n+1}z^{2n+1}q^{2n^2+3n+1}
 \frac{((1-x)yzq;q^2)_{n}((1-x)zq^2;q^2)_{n}}{(yzq;q^2)_{n+1}(zq^2;q^2)_{n}}.
\end{eqnarray*}
Summing up the above two cases we obtain the expression~(\ref{eq:gfsyl}).

\medskip

{\parindent=0pt\bf Remark}: One can also derive 
Theorem~2 by combining the formula for $A$ or $B$ with  Theorem~1.
The above proof shows  that the generating function of odd partitions obtained through
Durfee square dissection in the 2-modular diagram 
corresponds to that of strict partitions obtained through Sylvester's bijection. 





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some special cases}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We first prove that the $t=1$ case of Theorem~2 is equivalent to 
the so-called 
Rogers-Fine identity~\cite{And72}~:
\begin{equation}\label{eq:RF}
(1-t)\sum_{n\geq 0}\frac{(aq)_n}{ (bq)_n}\, t^n=
\sum_{n\geq 0}\frac{(aq)_n\, (atq/b)_n\, (1-atq^{2n+1})}{ 
(bq)_n\, (tq)_n}\, (bt)^nq^{n^2}.
\end{equation}
Indeed, when $t=1$, the corresponding generating function of 
odd partitions (\ref{eq:A})
with largest part $2j+1$ ($j\geq 0$) is
$$
\sum_{\lambda\in {\cal P}_O\atop \lambda_1=2j+1}x^{n_d(\lambda)}y^{l(\lambda)}
z^{l(\lambda)+(\lambda_1-1)/2}q^{|\lambda|}= \frac{xyz^{j+1}q^{2j+1}}{1-yzq^{2j+1}}
\prod_{i=0}^{j-1}\left (1+\frac{xyzq^{2i+1}}{1-yzq^{2i+1}}\right)
$$
therefore,
\begin{equation}\label{lhsRF}
\sum_{\lambda\in {\cal 
P}_O}x^{n_d(\lambda)}y^{l(\lambda)}
z^{l(\lambda)+(\lambda_1-1)/2}q^{|\lambda|}= 
\sum_{j\geq 0} \frac{xyz^{j+1}q^{2j+1}((1-x)yzq;q^2)_j}{ (yzq;q^2)_{j+1}}.
\end{equation}
It follows from (\ref{lhsRF}) and Theorem~2 with $t=1$
that~:
\begin{eqnarray}\label{zen'}
 &&\sum_{j\geq 0} \frac{xyz^{j+1}q^{2j+1}\,((1-x)yzq;\,q^2)_j}{ (yzq;\,q^2)_{j+1}}
=x\,\sum_{n\geq 0}
y^{n+1}z^{2n+1}q^{2n^2+3n+1} \\
&& \times \,\left(1-(1-x)yz^2\,q^{4n+3}\right)\,
\frac{((1-x)zq^2;\,q^2)_{n}((1-x)yzq;\,q^2)_{n}}{ (zq^2;\,q^2)_{n+1}(yzq;\,q^2)_{n+1}}.\nonumber
\end{eqnarray}
To see that (\ref{zen'}) is equivalent to Rogers-Fine 
identity~(\ref{eq:RF}) we need to do some manipulations.
In (\ref{zen'}), substituting $y$ by $qy$,  $q^2$ by $q$ and 
simplifying  $xyzq$, we obtain~:
\begin{eqnarray}\label{zen''}
&&\sum_{j\geq 0} \frac{((1-x)yzq)_{j}}{(zyq)_{j+1}}(zq)^{j}\\
&&=\sum_{n\geq 0}\frac{((1-x)tq)_{n}\,((1-x)yzq)_{n}\,
(1-(1-x)yz^2q^{2n+2})}{ 
(zq)_{n+1}(yzq)_{n+1}}z^{2n}y^{n}q^{n^2+2n}.\nonumber
\end{eqnarray}
Finally, multiplying (\ref{zen''}) by $(1-zq)(1-z\,y\,q)$ we get
Rogers-Fine identity~(\ref{eq:RF}) 
with $a=(1-x)yz$, $b=yzq$ and $t=zq$.

Now, setting 
$z=1$ in  (\ref{zen'}), the left-hand side ofbecomes
\begin{eqnarray}
&&\sum_{j\geq 0}\frac{xyq^{2j+1}}{ 1-yq^{2j+1}}
\frac{((1-x)yq; q^2)_j}{(yq;q^2)_j}\nonumber \\
&=& \sum_{j\geq 0}\left[\frac{((1-x)yq;q^2)_{j+1}}{(yq;q^2)_{j+1}}-
\frac{((1-x)yq;q^2)_{j}}{(yq;q^2)_{j}}\right]\nonumber \\
&=& \frac{((1-x)yq;q^2)_{\infty}}{(yq;q^2)_{\infty}}-1,\label{eq:z=1}
\end{eqnarray}
while the right-hand side plus 1 reduces to
\begin{eqnarray*}
&&1+\sum_{n\geq 0}y^{n+1}q^{(n+1)(2n+1)}
\frac{(1-x;q^2)_{n+1}((1-x)yq;q^2)_{n+1}}{(q^2;q^2)_n(yq;q^2)_{n+1}}\\
&&+\sum_{n\geq 0}y^{n+1}q^{(n+1)(2n+3)}
\frac{(1-x;q^2)_{n+1}((1-x)yq;q^2)_{n+1}}{(q^2;q^2)_{n+1}(yq;q^2)_{n+1}}\\
&&=\sum_{n\geq 0}\frac{(1-x;q^2)_{n}((1-x)yq;q^2)_{n}y^nq^{n(2n+1)}}
{(q^2;q^2)_{n}(yq;q^2)_{n+1}}\left[1-(1-x)yq^{4n+1}\right].
\end{eqnarray*}
So, substituting $x$ by $1-x$, (\ref{zen'}) with $z=1$ reduces to
\begin{eqnarray}\label{eq:4phi5}
\sum_{n\geq 0}y^{n}q^{n(2n+1)}\frac{(x;q^2)_{n}(xyq;q^2)_n}
{(q^2;q^2)_{n}(yq;q^2)_{n+1}}
\left(1-xyq^{4n+1}\right)
=\frac{(xyq;q^2)_{\infty}}{(yq;q^2)_{\infty}}.
\label{z=1}
\end{eqnarray}
Furthertmore, setting $y=1$ 
 the left-hand side
of (\ref{z=1})
reduces to
\begin{eqnarray}
&&\sum_{n\geq 0}q^{n(2n+1)}\left[1-q^{2n+1}+q^{2n+1}(1-xq^{2n})\right]
\frac{(x)_{2n}}{ (q)_{2n+1}}\nonumber\\
&=&\sum_{n\geq 0}q^{n(2n+1)}\frac{(x)_{2n}}{(q)_{2n}}+
\sum_{n\geq 0}q^{(n+1)(2n+1)}\frac{(x)_{2n+1}}{(q)_{2n+1}}\nonumber\\
&=& \sum_{n\geq 1}q^{n(n+1)/2}\frac{(x)_n}{(q)_n}.\label{eq:lebl}
\end{eqnarray}
Hence, in view of  
Euler's formula~(1),  (\ref{eq:4phi5}) reduces to
 Lebesgue's formula (\ref{eq:lebesgue})~:
\begin{equation}\label{eq:lebesgue}
\sum_{n\geq 0}\frac{(x)_nq^{n(n+1)/2}}{ (q)_n}=\prod_{m=1}^{\infty}
\left(1-xq^{2m-1}\right)(1+q^m).
\end{equation}
Originally,Sylvester  derived from his bijection the following
partition theorem, which is known (see \cite[p. 24-25]{And81}) to be
equivalent  to Lebesgue's identity.
\begin{cor}[Sylvester]
The number of partitions in ${\cal P}_{\cal O}(n)$
with $k$  distinct odd parts is equal to the number 
of  partitions in ${\cal P}_{\cal D}(n)$  with $k$ runs.
\end{cor}

Finally, setting $x=0$ in (\ref{z=1}) we get
$$
\sum_{n=0}^\infty\frac{q^{n(2n+1)}y^n}{(q^2;q^2)_n(yq;q^2)_{n+1}}
=\frac{1}{(yq;q^2)_\infty},
$$
which is  clearly equivalent to Cauchy's formula~\cite[(2.2.8), p. 20]{And81}~:
\begin{equation}\label{cauchy}
\sum_{i=0}^\infty\frac{q^{i^2}y^i}{(q)_i(yq)_i}=\frac{1}{(yq)_\infty}.
\end{equation} 
The corresponding partition theorem is the following
\begin{cor} The number  of partitions
of ${\cal P}_{\cal O}(n)$ with length $l$ 
is equal to the number of 
partitions of ${\cal P}_{\cal D}(n)$ with alternative length $l$. 
\end{cor}
 
Note that the above result follows also from Sylvester's bijection 
seems to be published for the first time in \cite{Bes94} (see also
\cite{BME,KY99}).

\section{On Fine's two partition theorems}
In~\cite{Fine48,Fine88} Fine  gave two refinements of Euler's result, that we call Fine's first 
and second partition theorems respectively.  His first partition theorem reads as follows~:
\begin{cor}[Fine 1] 
The number of partitions of ${\cal P}_{\cal D}(n)$  
 with largest part $k$ is equal to  the number of 
partitions  in ${\cal P}_{\cal O}(n)$
 such that $2k+1$ equals the largest part plus twice the 
number  of parts.  
\end{cor}
This corresponds to the $x=y=1$ case of (\ref{zen'}), i.e.,
\begin{equation}\label{eq:And}
\sum_{j\geq 0}\frac{z^{j+1}q^{2j+1}}{(zq;q^2)_{j+1}}=
\sum_{n\geq 0}\frac{(zq^{n+1})^{2n+1}}{(zq)_{2n+2}}.
\end{equation}
 A proof of (\ref{eq:And}), based on Heine's transformation,
 can be found in \cite[p. 27]{And81}. In fact we can
 give a short proof of this result
by paraphrasing his  statement as follows.
\begin{cor}[Fine 1 bis]
The number of partitions of ${\cal P}_{\cal D}(n)$  with largest part 
$\leq k$ equals the number of partitions of ${\cal P}_{\cal O}(n)$ such that the largest part 
plus twice the number of parts is
$\leq 2k+1$.
\end{cor}
Since the $q$-binomial coefficient
$\qbi{n}{k}{q}=(q)_n/(q)_k\,(q)_{n-k}$
is  the generating function of partitions whose Ferrer diagrams are inscribed in the rectangle 
$k\times (n-k)$,
the above corollary is equivalent to the identity~:
\begin{equation}\label{rogers}
\sum_{l=0}^kq^l\left\lbrack{k\atop
l}\right\rbrack_{q^2}=(1+q)(1+q^2)\cdots (1+q^k),
\end{equation}
which is a special value of the Rogers-Szeg\"o 
polynomials~\cite[Example~5, p. 49]{And81} and can be calculated as follows~:
Set $f_k(t;q)=\sum_{l=0}^nt^l\left\lbrack{n\atop l}\right\rbrack$, then
\begin{equation}
\sum_{k\geq 0}\frac{f_k(t;q)}{(q;q)_n}x^k=
\sum_{l\geq 0}\frac{t^lx^l}{(q;q)_l}\sum_{k\geq
0}\frac{x^k}{(q;q)_k}
=\frac{1}{(tx;q)_\infty}\frac{1}{(x;q)_\infty}.
\end{equation}
Substituting $q$ by $q^2$ and $t$ by $q$ we get
$$
\sum_{n\geq
0}\frac{f_n(q;q^2)}{(q^2;q^2)_n}x^n=\frac{1}{(x;q)_\infty}
=\sum_{n\geq 0}\frac{x^n}{(q;q)_n}.
$$
Equating the coefficients of $x^n$ yields (\ref{rogers}).
Note that
(\ref{rogers}) can also be derived from Sylvester's bijection through
the following graphical decomposition, 
where $k=9$~:
$$
\leq l\overbrace{
\left\lbrace\begin{array}{ccccc}
2&2&2&2\\
2&2&2&2\\
2&2&&\\
2&&&
\end{array}\right.}^{\displaystyle \leq k-l}\quad 
\left.\begin{array}{c}
1\\
1\\
1\\
1\\
1
\end{array}\right\rbrace\ l\quad\to 
l\overbrace{
\left\lbrace\begin{array}{cccc}
2&2&2&2\\
2&2&2&2\\
2&2&1&\\
2&1&&\\
1&&&
\end{array}\right.}^{\displaystyle \leq k-l}\begin{array}{c}
1\\
1\\
\\
\\
\\
\end{array}
\to (9,\ 7,\ 6,\ 4,\ 1).
$$
Since $2(k-l)+1+2l=2k+1$, this completes the proof. 

Now, we come to Fine's second partition theorem.
\begin{cor}[Fine 2]
The number of  partitions of $n$ into odd parts 
 with largest part equal to $2r+1$ ($r\geq 0$ ) 
is equal to the number of 
partitions of $n$ into distinct parts with rank $2r$ or $2r+1$.
\end{cor}
This theorem of Fine does not follow from Sylvester's bijection
directly.
For example, for $r=0$, the partition $n=1+1+\cdots +1$ maps to $n$ through 
Sylvester's bijection, but $rank(n)=n-1\geq 2$ if $n\geq 3$.
Actually no direct combinatorial proof of this result is known. The analytical
 proof was based on an ``elegant and difficult'' $q$-identity 
(see \cite{And66,And83,Fine88}). 
However, as we will see, Fine's second theorem follows 
from Lebesgue's identity, which is
the generating function version of Sylvester's partition theorem.
Clearly Fine's second theorem is equivalent to the following statement
\begin{cor}[Fine 2 bis] 
The number of partitions of $n$ into odd
parts each $\leq 2r+1$ ($r\geq 0$) is equal to the number of partitions of $n$ into
distinct parts with rank $\leq 2r+1$. 
\end{cor}
Now, the above corollary is clearly  equivalent to the 
following $q$-series identity~:
\begin{equation}
\sum_{n=0}^{\infty}\left\lbrack{n+2r+1\atop n}
\right\rbrack q^{n(n+1)/2}=\frac{1}{ (1-q)(1-q^3)\cdots (1-q^{2r+1})},
\label{fine2}
\end{equation}
which  is Lebesgues's identity (\ref{eq:lebesgue}) with $x=q^{2r+2}$.


{\parindent=0pt\bf Remark}: It is interesting to note that (\ref{rogers}) and
(\ref{fine2}) are two finite extensions of Euler's identity (1).
Other finite extensions of (1) are given in \cite{BME} in the framework of 
\emph{Lecture Hall partitions}.

We end this section with an even analogue of Fine's second theorem.
Let $D(n,2r)$ denote the number of partitions of $n$ with rank less than 
or equal to $2r$, and $E(n, 2r)$ the number of 
partitions of $n$ with even parts each of them is less than or equal to $2r$.
Then we have the following related result.
\begin{thm} We have
$$
D(n,\, 2r)=\sum_{m\geq 0}E(n-m(m+1)/2,\, 2r).
$$
\end{thm}
\begin{proof} First of all, setting  $x=q$ in Lebesgue's formula 
we obtain the following result of Gauss:
$$
\sum_{n=0}^{\infty} q^{n(n+1)/2}=
\prod_{m=1}^{\infty} (1-q^{2m})(1+q^m).
$$
Now, applying Lebesgue's formula with $x=q^{2r+1}$ ($r\geq 0$) and 
the Gauss formula we have
\begin{eqnarray*}
\sum_{n=0}^{\infty}\left\lbrack{n+2r\atop n}\right\rbrack q^{n(n+1)/2}=
{\prod_{m\geq 1}(1+q^m)(1-q^{2m})\over (q^2;q^2)_{r}}=
{\sum_{n=0}^{\infty} q^{n(n+1)/2}\over (q^2;q^2)_{r}}.
\end{eqnarray*}

This is clearly equivalent to our result.
\end{proof}


{\parindent=0pt\bf Remark}: Hirschhorn~\cite{Hir74} obtained a similar result related to 
Sylvester's partition theorem.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{On (m, c)-analogue of Sylvester's theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $c$ and $m$ be two integers such that $1\leq c<m$.
We can replace the sequence of odd integers $\{1, 3, 5, \ldots \}$
by any arithmetical sequence of form 
$\{c, m+c, \ldots, km+c, \ldots\}$.
Instead of 2-modular diagram
we can draw an $m$-modular diagram as follows: if $\la_i=mk_i+c$ then
we draw a Ferrer diagram of the partition $(k_1+1, k_2+1,\ldots)$
and interpret each dot except the rightmost at each row as m dots
and put $c$ at the rightmost dot.
If we read this diagram by ``hook''
then we obtain some special type of partitions $\mu=(l_1^c, l_2^{m-c},
l_3^c, l_4^{m-c},\ldots)$. For example, if $\lambda=(( 4m+c)^2,\,
2m+c,\, m+c,\, c)$ then we draw the $m$-modular diagram
$$
\begin{array}{ccccccccc}
m&-&m&-&m&-&m&-&c\\
|&&&&&&&&\\
m&&m&-&m&-&m&-&c\\
|&&|&&&&&&\\
m&&m&&{\mathbf c}&&\\
|&&|&&&&&&\\
m&&c&&&\\
|&&&&&&&&\\
c&&&&
\end{array}
$$
reading the above diagram by ``hooks'' we obtain
$$ 
\mu=(9^{c},\, 7^{m-c},\, 6^{c},  4^{m-c},\, 1^{c}).
$$
Denote by $\varphi_{m,c}$ the $(m,c)$-analog of 
Sylvester's bijection~:$\la\mapsto \mu$. In what follows we shall state 
the similar results for this bijection and leave the proofs to the interested reader.

A  partition $\lambda$ is said to be of type $(h_1,h_2, \ldots)$ 
if $\lambda$ 
has exactly $h_1\geq 1$ parts of maximal length, $h_2\geq 1$ second by
length parts, etc. 
Let ${\cal A}(n)$  be the set of  partitions of $n$ with
parts $\equiv c\pmod m$ and  ${\cal B}(n)$
the set of partitions of $n$ of type $(c,m-c,
c,m-c,\ldots)$. Set
${\cal A}=\cup_{n\geq 0}{\cal A}(n)\quad \hbox{and}\quad
{\cal B}=\cup_{n\geq 0}{\cal B}(n)$.

In \cite{Bes94} Bessenrodt briefly alluded to  this generalisation at the
end of her paper. Later on, 
 using Fronenius's representation of partitions,
Pak and Postnikov~\cite{PP98} independently worked out some more
details of this generalisation, but they overlooked the parameter of $\mu$
 corresponding to the length of $\la$. 

 For any partition $\la\in {\cal A}$ let $d_m(\la)$ be 
the  size of Durfee square in the
$m$-modular
diagram of $\la$, i.e.,  $d_m(\lambda)=\#\{i:\lambda_i\geq m(i-1)+c\}$.
The following is the $(m,c)$-analog of Theorem~1 and merges the results of 
 \cite{Bes94} and \cite{PP98} into one single whole.

\begin{thm}
Let $\varphi_{m,c} : {\cal A}\longrightarrow {\cal B}$ 
be $(m,c)$-analog of Sylvester's bijection and $\mu=\varphi(\la)$, 
then $|\la|=|\mu|$, and 
\begin{itemize}
\item $d_m(\la)=[(l(\mu)+m-c)/m]$,
\item $n_d(\la)=n_c(\mu)$,
\item $l(\la)=l_a(\mu)$,
\item $l(\la)+(\la_1-c)/m=\mu_1$.
\end{itemize}  
\end{thm}

{\parindent=0pt\bf Remark}: Bessenrodt~\cite{Bes94} pointed out the equation  $l(\la)=l_a(\mu)$, while
the  three other equations were proved by Pak \& Postnikov~\cite{PP98}.

Clearly Theorem~4 implies immediately the following 
identity~:
\begin{eqnarray}\label{mcgf}
&&\sum_{\la\in {\cal A}}x^{n_d(\la)}y^{l(\la)}z^{l(\la)
+(\la_1-c)/m}t^{d_m(\la)}q^{|\la|}\nonumber\\
&&\qquad =\sum_{\mu\in {\cal B}}x^{n_c(\mu)}
y^{l_a(\mu)}z^{\mu_1}t^{\lfloor(l(\mu)+m-c)/m\rfloor}q^{|\mu|}
\end{eqnarray}
Similarly we have the following result. 
\begin{thm} \label{BPP}
The generating functions of (\ref{mcgf}) have the following
explicit formula~:
\begin{eqnarray}
&&x\sum_{n\geq 
0}(ty)^{n+1}z^{2n+1}q^{(n+1)(mn+c)}\,
\left[1-(1-x)yz^2\,q^{m(2n+1)+c}\right]\nonumber\\
&&\qquad \times \,
\frac{((1-x)zq^m;\,q^m)_{n}((1-x)yzq^c;\,q^m)_{n}}
{ (zq^m;\,q^m)_{n+1}(yzq^c;\,q^m)_{n+1}}.\label{mgf}
\end{eqnarray}
\end{thm}

In what follows we shall  derive some interesting $q$-identities from 
the above theorem by specializing the parameters. 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{cor}[$(m,c)$-analog of Rogers-Fine identity] \label{mrf}
The following identity holds true
\begin{eqnarray}\label{zen}
&&\sum_{j=0}^\infty
xyz^{j+1}q^{mj+c}\frac{((1-x)yzq^c;q^m)_{j}}{\left(yzq^c;q^m\right)_{j+1}}
\nonumber\\
&& =x\sum_{n\geq 0}y^{n+1}z^{2n+1}q^{(n+1)(mn+c)}
\left[1-(1-x)yz^2q^{m(2n+1)+c}\right]\nonumber\\
&& \times \frac{((1-x)zq^m;q^m)_{n}
((1-x)yzq^c;q^m)_{n}}{(zq^m;q^m)_{n+1}(yzq^c;q^m)_{n+1}}.
\end{eqnarray}
\end{cor}
\begin{proof}
Since the g.f. of partitions in ${\cal A}$ with one distinct part
$mj+c$ 
is 
$$
1+\frac{xyz^{j+1}q^{mj+c}}{1-yzq^{mj+c}},\qquad  j\geq 0, 
$$
so the g.f. of  partitions in ${\cal A}$ 
 with largest part $mj+c$ ($j\geq 0$) is 
\begin{eqnarray*}
\sum_{\la\in {\cal A}}x^{n_d(\la)}y^{l(\la)}z^{l(\la)
+\frac{\la_1-c}{m}}q^{|\la|}&=&
\sum_{j\geq 0}\frac{xyz^{j+1}q^{mj+c}}{1-yzq^{mj+c}}
\prod_{k=0}^{j-1}\left(1+\frac{xyzq^{mk+c}}{1-yzq^{mk+c}}\right)\\
&=&\sum_{j=0}^\infty
\frac{xyz^{j+1}q^{mj+c}((1-x)yzq^c;q^m)_{j}}{\left(yzq^c;q^m\right)_{j+1}}.
\end{eqnarray*}
The identity (\ref{zen}) follows then from Theorem~5 with $t=1$. 
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
However  the following result shows that 
the $(m,c)$-analog of Rogers-Fine identity (\ref{zen}) is 
not more general than the classical Rogers-Fine identity.
\begin{prop} 
The $(m,c)$-analog of Rogers-Fine identity (\ref{zen}) is equivalent to
the classical Rogers-Fine identity, i.e., the case where $m=2$ and $c=1$.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{proof} We shall prove that (\ref{zen}) is equivalent to (\ref{zen'}).
Indeed,
setting  $m=2$ and $c=1$ in (\ref{zen}) yields (\ref{zen'}).
Inversely, substituting $y$ by $yq^{2c/m-1}$ and then $q$ by $q^{m/2}$ in 
(\ref{zen'}) we recover (\ref{zen}). 
\end{proof}

Note that setting 
$z=1$ in  (\ref{zen}) the left-hand side  becomes
\begin{eqnarray}
&&\sum_{j\geq 0}\frac{xyq^{mj+c}}{ 1-yq^{mj+c}}
\frac{((1-x)yq^c; q^m)_j}{(yq^c;q^m)_j}\nonumber \\
&=& \sum_{j\geq 0}\left[\frac{((1-x)yq^c;q^m)_{j+1}}{(yq^c;q^m)_{j+1}}-
\frac{((1-x)yq^c;q^m)_{j}}{(yq^c;q^m)_{j}}\right]\nonumber \\
&=& \frac{((1-x)yq^c;q^m)_{\infty}}{(yq^c;q^m)_{\infty}}-1.\label{Eq:z=1}
\end{eqnarray}
On the other hand, 
since $1-wyq^{m(2n+1)+c}=1-q^{m(n+1)}+q^{m(n+1)}(1-wyq^{mn+c})$, if $z=1$ 
we can rewrite
the right-hand side of (\ref{zen}) plus 1 as follows~:
\begin{eqnarray*}
&& 1+ \sum_{n\geq 0}y^{n+1}q^{(n+1)(mn+c)}
\frac{(1-x;q^m)_{n+1}((1-x)yq; q^m)_n}{(q^m;q^m)_n(yq;q^m)_{n+1}}\\
&&+\sum_{n\geq 0}y^{n+1}q^{(n+1)(mn+m+c)}
\frac{(1-x;q^m)_{n+1}((1-x)yq; q^m)_{n+1}}
{(q^m;q^m)_{n+1}(yq;q^m)_{n+1}}\\
&&=\sum_{n\geq 0}\frac{(1-x;q^m)_{n}((1-x)yq; q^m)_{n}}
{(q^m;q^m)_{n}(yq^c;q^m)_{n+1}}y^nq^{n(mn+c)}\left[1-(1-x)yq^{2mn+c}\right].
\end{eqnarray*}
Therefore substituting $x$ by $1-x$ we get the following
\begin{cor} There holds
\begin{equation}\label{master}
\sum_{n\geq 0}\frac{(x;q^m)_{n}(xyq; q^m)_{n}y^nq^{n(mn+c)}}
{(q^m;q^m)_{n}(yq^c;q^m)_{n+1}}\left(1-xyq^{2mn+c}\right)
= \frac{(xyq^c;q^m)_{\infty}}{(yq^c;q^m)_{\infty}}.
\end{equation}
\end{cor}
Furthermore, for $y=1$ (\ref{master}) reduces to 
\begin{cor}[$(m,c)$-analogue of Lebesgue's identity]~:
\begin{equation}\label{mleb}
\sum_{n\geq 0}\frac{(x;q^m)_n(xq^c;q^m)_n\,q^{n(c+mn)}}
{(q^m;q^m)_n(q^c;q^m)_{n+1}}(1-x\,q^{c+2mn})
=\frac{(xq^{c};q^m)_\infty}{(q^c;q^m)_\infty};
\end{equation}
\end{cor}
while for $x=0$ (\ref{master}) reduces to
\begin{cor}[$(m,c)$-analogue of Cauchy's identity]~:
\begin{equation}\label{mcleb}
\sum_{n\geq 0}\frac{ q^{n(m(n-1)+c)}y^n}
{(q^m;q^m)_n(yq^c;q^m)_n}                            
=\frac{1}{(yq^{c};q^m)_\infty}.
\end{equation}
\end{cor}
Evidently, when 
 $m=2$ and $c=1$, (\ref{mleb}) and (\ref{mcleb}) reduce to identities of 
Lebesgue and Cauchy, respecticely.


As for the Cauchy formula,
we can prove the $(m,c)$-Cauchy formula~(\ref{mcleb})
using  $m$-module Ferrers diagrams. 
Clearly, the right-hand side of (\ref{mcleb}) is the g.f. of partitions 
in ${\cal A}$ with $y$ enumerating  the length of each partition.
Now consider the $m$-module Ferrers diagrams of partitions  in ${\cal A}$.
Let $n\geq 0$ be the size of Durfee square.
  There are three cases~:
\begin{itemize}
\item[(1)] if the diagram is empty, the g.f. is 1;
\item[(2)] if $m$ is at the southeast corner, the g.f. of such partitions is
$$
\frac{q^{mn^2}}{(q^m;q^m)_n}\cdot \frac{y^nq^{nc}}{(yq^c;q^m)_n}
\qquad n\geq 1;
$$
\item[(3)] if $c$ is at the southeast corner, the g.f. of such partitions is
$$
\frac{q^{mn(n-1)}}{(q^m;q^m)_{n-1}}\cdot
\frac{y^nq^{nc}}{(yq^c;q^m)_n}
\qquad n\geq 1.
$$
\end{itemize}
Summing up the above three cases 
we get the generating function for the partitions having a
Durfee square of size $n\geq 0$:
$$
\frac{ q^{mn(n-1)+nc}y^n}
{(q^m;q^m)_n(yq^c;q^m)_n}. 
$$
This establishes (\ref{mcleb}).


$$
\begin{array}{cccccc}
m&m&m&m&m&c\\
m&m&m&m&c&\\
m&m&{\mathbf  m}&m&c&\\
m&m&c&&&\\
m&c&&&&\\
c&&&&&
\end{array}
\qquad 
\begin{array}{ccccc}
m&m&m&m&c\\
m&m&m&m&c\\
m&m&{\mathbf   c}&\\
m&c&&&\\
c&&&&
\end{array}
$$

For instance,
\begin{eqnarray*}
\lambda&=&(5m+c,\,( 4m+c)^2,\, 2m+c,\, m+c,\, c)\\
&& \mapsto \lambda'=(11^c, \, 9^{m-c},\, 7^c,\, 5^{m-c},  4^c,\, 2^{m-c}).
\end{eqnarray*}
Note that $rank \lambda'=11-3m$, which is negative if $m\geq 4$.




{\parindent=0pt\bf Remark}: It's known~\cite[p. 571]{And72} that  Rogers-Fine identity
(\ref{eq:RF}) is a special case
of Watson's transformation formula for a terminating very-well-poised
 ${}_8\phi_7$ series as a multiple of a terminating balanced
${}_4\phi_3$ series~\cite[p. 35]{GR90}. In view of Proposition~1, 
all the specializations of (\ref{zen}) or (\ref{eq:RF}) would be limiting
 cases of Watson's formula. For example, as pointed out by 
Krattenthaler~\cite{Kra}, 
setting $a=xyq^{c/m}$ and $b=x$, (\ref{master}) can be 
written in standard basic hypergeometric function notation as follows~:
$$
{}_4\phi_5\left[\begin{array}{llll}
a,&b, &q\sqrt{a}, &-q\sqrt{a}\\
\sqrt{a},&-\sqrt{a},&aq/b,&0,\;0
		\end{array}
; q,\, aq/b
\right]=\frac{(aq;q)_\infty}{(aq/b;q)_\infty},
$$
which is a limiting case of Bailey's ${}_6\phi_5$ summation~\cite[Appendix (II.20)]{GR90}.


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\end{document}




$$
\begin{array}{cccccc}
2&2&2&2&2&1\\
2&2&2&2&1&\\
2&2&\mathbf{2}&2&1&\\
2&2&1&&&\\
2&1&&&&\\
1&&&&&
\end{array}
\qquad 
\begin{array}{ccccc}
2&2&2&2&1\\
2&2&2&2&1\\
2&2&{\mathbf 1}&\\
2&1&&&\\
1&&&&
\end{array}
$$


By ${\cal A}(n,k,l)$ denote the set of partitions 
$\lambda\in {\cal A}(n)$ with  $l$ parts and exactly $k$ different
parts. By ${\cal B}(n,k,l)$
denote the set of all partitions $\lambda\in {\cal B}(n)$
such that $\lambda$ has exactly $k$ maximal chains and $l_a(\lambda)=l$.
On the other hand, for partitions $\mu\in {\cal B}$ such that 
$\mu_1'=mN$  the corresponding generating function is
\begin{eqnarray*}
&&\frac{zxq^{mN}}{1-zq^{mN}}\prod_{j=1}^{N-1}zq^{mj}
\left(1+\frac{xzq^{mj}}{1-zq^{mj}}\right)\prod_{j=0}^{N-1}zyq^{mj+c}
\left(1+\frac{xzq^{mj+c}}{1-zyq^{mj+c}}\right)\\
&&=\frac{z^{2N-1}xy^Nq^{N(mN+c)}((1-x)zq^m;q^m)_{N-1}((1-x)zyq^c;q^m)_N}
{(zq^m;q^m)_N(zyq^c;q^m)_N},
\end{eqnarray*}
and for partitions $\mu\in {\cal B}$ such that 
 $\mu_1'=m(N-1)+c$ with $N\geq 1$, the corresponding generating function is
\begin{eqnarray*}
&&\frac{zxyq^{m(N-1)+c}}{1-zq^{m(N-1)+c}}
\prod_{j=1}^{N-1}zq^{mj}\left(1+\frac{xzq^{mj}}{1-zq^{mj}}\right)
\prod_{j=0}^{N-2}zyq^{mj+c}
\left(1+\frac{xzq^{mj+c}}{1-zyq^{mj+c}}\right)\\
&&=\frac{z^{2N-1}xy^{N}q^{N(m(N-1)+c)}((1-x)zq^m;q^m)_{N-1}
((1-x)zyq^c;q^m)_{N-1}}
{(zq^m;q^m)_{N-1}(zyq^c;q^m)_N}.
\end{eqnarray*}
Summing up we get the right hand side of (\ref{zen}).
