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\title{\bf On the Positive Moments of  Ranks \\of Partitions}

\author{William Y.C. Chen$^{1}$, \
Kathy Q. Ji$^{2}$, \  and \  Erin  Y.Y. Shen$^{3}$\\[5pt]
\small Center for Combinatorics, LPMC-TJKLC\\[-0.8ex]
\small Nankai University\\[-0.8ex]
\small Tianjin 300071, P.R. China\\
\small\tt  $^1$chen@nankai.edu.cn, $^2$ji@nankai.edu.cn,  $^3$shenyiying@mail.nankai.edu.cn
}


\date{\dateline{Nov 1, 2013}{Jan 12, 2014}{Feb 7, 2014}\\
\small Mathematics Subject Classifications: 05A17, 11P83, 05A30}

\begin{document}

\maketitle


\begin{abstract}
 By introducing
$k$-marked Durfee symbols, Andrews found a
 combinatorial interpretation of the  $2k$-th symmetrized  moment $\eta_{2k}(n)$ of ranks of partitions of $n$ in terms of $(k+1)$-marked Durfee symbols of $n$.      In this paper, we consider the $k$-th symmetrized positive moment $\bar{\eta}_k(n)$ of ranks of partitions of $n$ which is
 defined as the truncated sum over positive ranks of partitions of $n$.   As combinatorial interpretations of  $\bar{\eta}_{2k}(n)$ and $\bar{\eta}_{2k-1}(n)$, we show that for given $k$ and $i$ with  $1\leq i\leq k+1$,  $\bar{\eta}_{2k-1}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$  with the   $i$-th rank  being zero  and $\bar{\eta}_{2k}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$ with
  the $i$-th rank being positive.    The interpretations of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$ are independent of $i$, and they imply
  the interpretation of $\eta_{2k}(n)$ given by Andrews
  since $\eta_{2k}(n)$ equals $\bar{\eta}_{2k-1}(n)$ plus twice of $\bar{\eta}_{2k}(n)$.   Moreover,  we obtain the generating functions of $\bar{\eta}_{2k}(n)$ and $\bar{\eta}_{2k-1}(n)$.


  \bigskip\noindent \textbf{Keywords:}  rank of a partition; $k$-marked Durfee symbol;  moment of ranks
\end{abstract}

\section{Introduction}


This paper is concerned with a combinatorial study of the symmetrized positive moments of ranks of partitions.
The notion of symmetrized moments was introduced
by Andrews \cite{Andrews-07-a}. Any  odd   symmetrized moment  is  zero because of the symmetry of ranks. For an even symmetrized moment, Andrews found a combinatorial interpretation  by introducing $k$-marked Durfee symbols.
It is natural to
 investigate the combinatorial interpretation of   an odd symmetrized moment
 which is defined as a  truncated sum over positive ranks of partitions of $n$.  We give  combinatorial interpretations of both the even and the odd  positive moments   in terms of $k$-marked Durfee symbols, which also lead to the combinatorial interpretation of an even symmetrized moment of ranks given by Andrews.




The rank of a partition $\lambda$ introduced by Dyson \cite{Dyson-1944} is defined as the largest part minus the number of parts.  Let $N(m,n)$ denote the number of partitions of $n$ with rank $m$. The following generating function of $N(m,n)$
was conjectured by Dyson \cite{Dyson-1944} in 1944 and proved by Atkin and Swinnerton-Dyer \cite{Atki54} in 1954. A combinatorial proof was found by Dyson  \cite{Dyson-1969} in 1969.

\begin{thm} \label{gf-r} For given integer $m$, we have
\begin{equation}\label{gf-c-e}
\sum_{n=0}^{+\infty}
N(m,n)q^n=\frac{1}{(q;q)_{\infty}}\sum_{n=1}^{+\infty}(-1)^{n-1}
q^{n(3n-1)/2+|m|n}(1-q^n).
\end{equation}
\end{thm}



Recently,  Andrews \cite{Andrews-07-a} introduced the $k$-th symmetrized moment $\eta_{k}(n)$ of ranks  of partitions of $n$ as given by
\begin{align}\label{symRankMom}
\eta_{k}(n)=\sum_{m=-\infty}^{+\infty}{m+\lfloor\frac{k-1}{2}\rfloor \choose k}N(m,n).
\end{align}
 It can be easily seen that for any $k$, $\eta_{k}(n)$ is  a linear combination of the moments $N_j(n)$ of ranks given by Atkin and Garvan \cite{Atkin-Garvan-03}
\begin{align*}
N_j(n)=\sum_{m=-\infty}^{\infty}m^jN(m,n).
\end{align*}
For example,
\[\eta_6(n)=\frac{1}{720}N_6(n)-\frac{1}{144}N_4(n)+\frac{1}{180}N_2(n).\]

In view of the symmetry $N(-m,n)=N(m,n)$,  we have $\eta_{2k+1}(n)=0$.  As for an even symmetrized
moment  $\eta_{2k}(n)$, Andrews  gave the following combinatorial interpretation by introducing $k$-marked Durfee symbols. For the definition of $k$-marked Durfee symbols, see Section 2.

\begin{thm}[Andrews \cite{Andrews-07-a}]\label{Andrews} For any $k\geq 1$,
  $\eta_{2k}(n)$  is equal to the number of $(k+1)$-marked Durfee symbols of $n$.
\end{thm}

 Andrews  \cite{Andrews-07-a}  proved the above theorem by using the $k$-fold generalization of Watson's $q$-analog of Whipple's theorem. Ji \cite{Ji-2011} found a combinatorial proof of Theorem \ref{Andrews} by  establishing a map from  $k$-marked Durfee symbols to ordinary partitions.  Kursungoz \cite{Kursungoz-2011} gave another proof of Theorem \ref{Andrews}  by using an alternative representation of $k$-marked Durfee symbols.

In this paper,  we introduce the $k$-th symmetrized positive moment $\bar{\eta}_{k}(n)$ of ranks as given by
\begin{align*}
\overline{\eta}_k(n)=\sum_{m=1}^{\infty}
{m+\lfloor\frac{k-1}{2}\rfloor\choose k}N(m,n),
\end{align*}
or equivalently,
\begin{equation}
\overline{\eta}_{2k-1}(n)=\sum_{m=1}^{\infty}
{m+k-1\choose 2k-1}N(m,n)
\end{equation}
and
\begin{equation}
\overline{\eta}_{2k}(n)=\sum_{m=1}^{\infty}
{m+k-1\choose 2k}N(m,n).
\end{equation}
 Furthermore, it is easy to see that for any $k$, $\bar{\eta}_k(n)$ is a linear combination of the positive moments $\overline{N}_j(n)$ of ranks introduced by  Andrews, Chan and Kim \cite{Andrews-Chan-Kim} as given by
\begin{align*}
\overline{N}_j(n)=\sum_{m=1}^{\infty}m^jN(m,n).
\end{align*}
For example,
\begin{align*}
\bar{\eta}_4(n)&=
\frac{1}{24}\overline N_4(n)-\frac{1}{12}\overline N_3(n)-\frac{1}{24}\overline N_2(n)
+\frac{1}{12}\overline N_1(n),\\[5pt]
\bar{\eta}_5(n)&=\frac{1}{120}\overline{N}_5(n)-\frac{1}{24}\overline{N}_3(n)
+\frac{1}{30}\overline{N}_1(n).
\end{align*}
By the symmetry $N(-m,n)=N(m,n)$, it is readily seen that
 \begin{equation}\label{re}
 \eta _{2k}(n)=2 \overline{\eta}_{2k}(n)+ \overline{\eta} _{2k-1}(n).
\end{equation}

 The main objective of this paper is to give combinatorial interpretations of  $\bar{\eta}_{2k}(n)$ and $\bar{\eta}_{2k-1}(n)$. We show that for given $k$ and $i$ with  $1\leq i\leq k+1$,  $\bar{\eta}_{2k-1}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$  with the   $i$-th rank  being zero  and $\bar{\eta}_{2k}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$ with
  the $i$-th rank being positive. It should be noted that
  $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$ are independent of $i$
  since the ranks of $k$-marked Durfee symbols are symmetric, see Andrews \cite[Corollary 12]{Andrews-07-a}.

   With the aid of  Theorem \ref{main-1} and Theorem \ref{main-2} together with the   generating function \eqref{gf-c-e} of $N(m,n)$,  we  obtain the generating functions of  $\bar{\eta}_{2k}(n)$ and  $\bar{\eta}_{2k-1}(n)$.



\section{Combinatorial interpretations}

 In this section, we   give  combinatorial interpretations of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$ in terms of   $k$-marked Durfee symbols.
For a partition $\lambda$ of $n$, we write $\lambda=(\lambda_1, \lambda_2, \ldots, \lambda_s)$ with the entries $\lambda_i$ in nonincreasing order such that
 $\lambda_1+\lambda_2+\cdots +\lambda_s=n$. We assume that all the parts of $\lambda$ are positive. The number of parts of $\lambda$ is called the length of $\lambda$, denoted by $\ell(\lambda)$. The weight of $\lambda$ is the sum of parts, denoted $|\lambda|$.

 Recall that a $k$-marked Durfee symbol  of $n$ introduced by Andrews \cite{Andrews-07-a} is a two-line array composed of $k$ pairs of
partitions  $(\alpha^1,\beta^1),\, (\alpha^2,\beta^2),\ldots, (\alpha^k,\beta^k)$ along with a positive integer $D$ which
 is represented in the following form:
  \[\tau=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D,\] where  the partitions $\alpha^i=(\alpha_1^i,\alpha_2^i,\ldots,\alpha^i_s)$ and $\beta^i=(\beta_1^i,\beta_2^i,\ldots,\beta^i_s)$  satisfy
the following four conditions:
\begin{itemize}

\item[{\rm (1)}]The partitions $\alpha^i$ ($1\leq i<k$) are nonempty, while $\alpha^k$ and  $\beta^i$ ($1\leq i\leq k$) are allowed to be empty{\rm ;}

\item[{\rm (2)}] $\beta^{i-1}_1\leq \alpha^{i-1}_1 \leq
\min\{\alpha^i_s,\beta^{i}_{s}\}$ for $2\leq i\leq k${\rm ;}



\item[{\rm (3)}]$\alpha_1^k$, $\beta_1^k \leq D${\rm ;}

\item[{\rm (4)}]$\sum_{i=1}^{k}(|\alpha^i|+|\beta^i|)+D^2=n$.
\end{itemize}









Let
\[\tau=\left(\begin{array}{cccc}
\alpha^k,&\alpha^{k-1},&\ldots,&\alpha^1\\[2pt]
\beta^k,&\beta^{k-1},&\ldots,&\beta^1
\end{array}
\right)_D\]
be a $k$-marked Durfee symbol. The pair
$(\alpha^i,\beta^i)$ of partitions
is called the $i$-th vector of $\tau$. Andrews defined
 the $i$-th rank $\rho_i(\tau)$ of $\tau$ as follows
\[\rho_i(\tau)=\left\{
\begin{array}{ll}
\ell(\alpha^i)-\ell(\beta^i)-1, \ \ &\text{ for }\ 1\leq i<k,\\[5pt]
\ell(\alpha^k)-\ell(\beta^k), \ \ &\text{ for }\ i=k.
\end{array}\right.
\]


For example, consider  the following   $3$-marked Durfee symbol
\[\tau=\left(\begin{array}{cccccccc}
\overbrace{5_3,4_3}^{\alpha^3},&
\overbrace{4_2,3_2,3_2,2_2}^{\alpha^2},&
\overbrace{2_1}^{\alpha^1}\\[3pt]
\underbrace{4_3}_{\beta^3},&
\underbrace{3_2,2_2,2_2}_{\beta^2},&
\underbrace{2_1,2_1}_{\beta^1}
\end{array}\right)_5.\]
We have  $\rho_1(\tau)=-2,\,\rho_2(\tau)=0,$ and $\rho_3(\tau)=1$.


For an odd symmetrized moment  $\bar{\eta}_{2k-1}(n)$, we have the following combinatorial interpretation.

\begin{thm}\label{main-1} For given positive integers  $k$ and $i$ with $1\leq i\leq k+1$,
 $\bar{\eta}_{2k-1}(n)$ is equal to the number of $(k+1)$-marked Durfee symbols of $n$ with the $i$-th rank equal to zero.
\end{thm}

For the even case, we
have the following interpretation.

\begin{thm}\label{main-2}For given positive integers  $k$ and $i$ with $1\leq i\leq k+1$,
 $\bar{\eta}_{2k}(n)$  is equal to the number of $(k+1)$-marked Durfee symbols of $n$ with the $i$-th rank being positive.
\end{thm}

The proofs of the above two interpretations are based on the following partition identity obtained by Ji  \cite{Ji-2011}.
We shall adopt the notation $ {D}_k(m_1,m_2,\ldots,m_k;n)$ as used by
Andrews \cite{Andrews-07-a} to denote the   number of
$k$-marked Durfee symbols of $n$ with the $i$-th rank equal to
$m_i$ for $1\leq i\leq k$.

 \begin{thm}\label{r-thm} For $k\geq 2$ and $n\geq 1$, we have
\begin{align}\label{r-thm-e}
 {D}_{k}(m_1,\ldots,m_k;n)
=\sum_{t_1,\ldots,t_{k-1}=0}^{\infty}
N\left(\sum_{i=1}^{k}|m_i|+2\sum_{i=1}^{k-1}t_i+k-1,n\right).
\end{align}
\end{thm}

To derive the above  interpretations of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$, we also need
 the following symmetric property given by Andrews \cite{Andrews-07-a}.  Boulet and Kursungoz \cite{Boulet-2011} found a combinatorial proof of this fact.

\begin{thm} \label{r-sym-2}For $k\geq 2$ and $n\geq 1$,
$ {D}_k(m_1,\ldots,m_k;n)$ is symmetric in $m_1,\ m_2,\ldots,m_k$.
\end{thm}





We are now in a position to prove Theorem \ref{main-1} and Theorem \ref{main-2}.



\noindent {\it Proof of Theorem \ref{main-1}.}
By Theorem \ref{r-sym-2}, it suffices to show that
\begin{equation}\label{step-1}
\sum_{m_2,m_3,\ldots,m_{k+1}=-\infty}^{\infty}
 {D}_{k+1}(0,m_2,m_3,\ldots,m_{k+1};n)=\bar{\eta}_{2k-1}(n).
\end{equation}
Using Theorem \ref{r-thm}, we get
\begin{align}\label{main-1-e}
&\sum_{m_2,m_3,\ldots,m_{k+1}=-\infty}^{\infty}
 {D}_{k+1}(0,m_2,m_3,\ldots,m_{k+1};n)\nonumber \\
&\hskip 2cm =\sum_{m_2,m_3,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{t_1,\ldots,t_{k}=0}^{\infty}
N\left(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k,n\right).
\end{align}

For $k\geq 1$ and $m\geq k$, let $c_k(m)$ denote the number  of
 integer solutions to the equation
\[|m_2|+\cdots+|m_{k+1}|+2t_1+\cdots+2t_k=m-k,\]
where   $m_i$ are integers and   $t_i$ are nonnegative integers. It is easy to see that the generating function of $c_k(m)$ is equal to
\begin{align}
\sum_{m=k}^{\infty}c_k(m)q^{m-k}&=(1+2q+2q^2+2q^3+\cdots)^{k}
(1+q^2+q^4+q^6+\cdots)^{k}\nonumber\\[3pt]
&=\left(\frac{1+q}{1-q}\right)^k\left(\frac{1}{1-q^2}\right)^k
\nonumber\\[3pt]
&=\frac{1}{(1-q)^{2k}}\nonumber\\[3pt]
&=\sum_{m=k}^{\infty}{m+k-1 \choose 2k-1}q^{m-k}. \label{gf-ck}
\end{align}
Hence
\[c_k(m)={m+k-1 \choose 2k-1},\]
and \eqref{main-1-e} can be written as
\begin{align*}
&\sum_{m_2,m_3,\ldots,m_{k+1}=-\infty}^{\infty}
 {D}_{k+1}(0,m_2,m_3,\ldots,m_{k+1};n)\nonumber \\
&\hskip 2cm = \sum_{m=1}^{\infty}
{m+k-1 \choose 2k-1}N(m,n),
\end{align*}
which is the defining expression of $\bar{\eta}_{2k-1}(n)$. This completes the proof.  \qed

\medskip



\noindent {\it Proof of Theorem \ref{main-2}.}  Similarly, by Theorem \ref{r-sym-2},
it is sufficient  to show that
\begin{equation}\label{main-2-1}
\sum_{\stackrel{m_1>0}{m_2,m_3,\ldots,m_{k+1}=-\infty}}^{\infty}
 {D}_{k+1}(m_1,m_2,\ldots,m_{k+1};n)=\bar{\eta}_{2k}(n).
\end{equation}
Invoking Theorem \ref{r-thm}, we get
\begin{align}\label{last-main-2}
&\sum_{\stackrel{m_1>0}{m_2,m_3,\ldots,m_{k+1}=-\infty}}^{\infty}
 {D}_{k+1}(m_1,m_2,\ldots,m_{k+1};n)\nonumber \\[3pt]
&\hskip 1cm \qquad =\sum_{\stackrel{m_1>0}{m_2,m_3,\ldots,m_{k+1}=-\infty}}^{\infty}\sum_{t_1,\ldots,t_{k}=0}^{\infty}
N\left(m_1+\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k,n\right).
\end{align}
For $k\geq 1$ and $m\geq k+1$, let $\bar{c}_k(m)$ denote the number of integer solutions to the equation
\[m_1+|m_2|+\cdots+|m_{k+1}|+2t_1+\cdots+2t_k=m-k,\]
where  $m_1$ is a positive integer,  $m_i$ ($2\leq i\leq k+1$) are integers and     $t_i$ are  nonnegative integers. An easy computation shows that
\begin{align}
 \sum_{m=k+1}^{\infty}\bar{c}_k(m)q^{m-k} =
 \frac{q}{(1-q)^{2k+1}},
\end{align}
so that
\[\bar{c}_k(m)={m+k-1 \choose 2k}.\]
Thus, the sum on the right hand side of \eqref{last-main-2} becomes
\[
 \sum_{m=1}^{\infty}
{m+k-1 \choose 2k}N(m,n),
\]
which is in accordance with the definition of $\bar{\eta}_{2k}(n),$ and hence the proof is complete.   \qed

Note that   the number ${D}_k(m_1,\ldots,
m_k;n)$ has the mirror symmetry with
respect to each $m_i$, that is, for $1\leq i\leq k$, we have
\[
 {D}_k(m_1,\ldots,m_i,\ldots,
m_k;n)= {D}_k(m_1,\ldots,-m_i,\ldots, m_k;n).
\]
  Using this symmetry property, Theorem \ref{main-2} can be restated  as follows.

\begin{thm}\label{main-3}For given positive integers  $k$ and $i$ with $1\leq i\leq k+1$,
 $\bar{\eta}_{2k}(n)$ is also equal to the number of $(k+1)$-marked Durfee symbols of $n$ with the $i$-th rank being negative.
\end{thm}




 \begin{table}[h]
 \[
\begin{array}{c|c|c|c }
&\overline{\eta}_1(5)&\overline{\eta}_2(5)&\overline{\eta}_2(5)\\[5pt]
\hline
 &\left(\begin{array}{llll}
 1_2 &1_2 &1_2 &1_1\\
&&&\end{array}\right)_1&
\left(\begin{array}{llll}
 1_1 &1_1 &1_1 &1_1\\
 &&&\end{array}\right)_1&
\left(\begin{array}{llll}
  1_1&& \\
1_1 &1_1 &1_1\end{array}\right)_1\\[20pt]
& \left(\begin{array}{llll}
 1_2 &1_1 &1_1 \\
1_1&\end{array}\right)_1&\left(\begin{array}{llll}
 1_2 &1_1 &1_1&1_1 \\
&\end{array}\right)_1&\left(\begin{array}{llll}
 1_2 &1_1 \\
1_1&1_1 \end{array}\right)_1\\[20pt]
&\left(\begin{array}{llll}
1_2 &1_2&1_1\\
 1_2 &\end{array}\right)_1&\left(\begin{array}{llll}
1_2 &1_2&1_1&1_1\\
 &\end{array}\right)_1&\left(\begin{array}{llll}
1_2 &1_2&1_1 \\
1_1\end{array}\right)_1\\[20pt]
& \left(\begin{array}{llll}
1_1&\\
 1_2 &1_2&1_2\end{array}\right)_1&\left(\begin{array}{llll}
1_1&1_1&1_1\\
1_1\end{array}\right)_1&\left(\begin{array}{llll}
1_1&1_1\\
1_1&1_1\end{array}\right)_1\\[20pt]
&\left(\begin{array}{llll}
1_1 &1_1\\
1_2& 1_1\end{array}\right)_1&\left(\begin{array}{llll}
1_1 &1_1&1_1\\
1_2 \end{array}\right)_1&\left(\begin{array}{llll}
1_1 \\
1_2&1_1&1_1 \end{array}\right)_1\\[20pt]
&\left(\begin{array}{llll}
 1_2 &1_1 \\
1_2 &1_2\end{array}\right)_1&\left(\begin{array}{llll}
 1_2 &1_1&1_1 \\
1_2 \end{array}\right)_1&\left(\begin{array}{llll}
 1_2 &1_1 \\
1_2 &1_1\end{array}\right)_1\\[20pt]
&\left(\begin{array}{llll}
1_1\\
\ \end{array}\right)_2 &\left(\begin{array}{llll}
1_1&1_1\\
 1_2 &1_2\end{array}\right)_1&\left(\begin{array}{llll}
1_1 \\
 1_2 &1_2&1_1\end{array}\right)_1
 \end{array}
 \]
 \caption{$2$-Marked Durfee Symbols of $5$.}
\end{table}
For example,  for   $n=5$, $k=1$ and $i=1$,  there are twenty-one $2$-marked Durfee symbols of $5$ as listed in Table 2.1. The first column  in Table 2.1 gives seven  $2$-marked Durfee symbols $\tau$ with  $\rho_1(\tau)=0$,  the second column  contains seven $2$-marked Durfee symbols  $\tau$ with $\rho_1(\tau)>0$ and the third column contains seven $2$-marked Durfee symbols $\tau$ with  $\rho_1(\tau)<0$.  It can be verified that $\overline{\eta}_{1}(5)=7$, $\overline{\eta}_{2}(5)=7$ and $\eta_{2}(5)=\overline{\eta}_{1}(5)+2\overline{\eta}_{2}(5)=21.$

\section{The generating functions of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$}

In this section, we obtain the generating functions of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$ with the aid of Theorem \ref{main-1} and Theorem \ref{main-2}. In doing so, we use the generating function  of $N(m,n)$ to derive the  generating functions  of  $D_{k+1}(0,m_2,\ldots,m_{k+1};n)$ and $D_{k+1}(m_1,m_2,\ldots,m_{k+1};n)$.




\begin{thm}\label{GF-thm-1}
For $k \geq 1$, we have
\begin{align}\label{gf-thm-e}
&\sum_{m_2,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{n=0}^{\infty}D_{k+1}(0,m_2,\ldots,m_{k+1};n)
x_1^{m_2}\cdots x_{k}^{m_{k+1}}q^n \nonumber \\[3pt]
&\quad \quad =\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+kn}
\frac{(1-q^n)}{\prod_{j=1}^{k}(1-x_jq^n)(1-x_j^{-1}q^n)}.
\end{align}
\end{thm}

\pf  Let
\[G_{k}(x_1,\ldots,x_{k};q)
=\sum_{m_2,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{n=0}^{\infty}D_{k+1}(0,m_2,\ldots,m_{k+1};n)x_1^{m_2}\cdots x_{k}^{m_{k+1}}q^n.\]
By Theorem \ref{r-thm}, we have
\begin{align}
&G_{k}(x_1,\ldots,x_{k};q) \nonumber \\[3pt]
&\quad \quad=\sum_{m_2,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{t_1,\ldots,t_{k}=0}^{\infty}x_1^{m_2}\cdots x_{k}^{m_{k+1}}\sum_{n=0}^{\infty}
N\left(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k,n\right)
q^n. \label{gf-temp}
\end{align}
Using the generating function  \eqref{gf-c-e} of $N(m,n)$ with $m$ replaced by $\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k$, we find that
\begin{align}\label{thm3.1-gf}\sum_{n=0}^{\infty}
&N\left(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k,n\right)q^n \nonumber \\[3pt]
&\quad =\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+n(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k)}
(1-q^n).
\end{align}
Substituting \eqref{thm3.1-gf} into \eqref{gf-temp}, we get

\begin{align}\label{temp}
G_{k}(x_1,\ldots,x_{k};q)&=\sum_{m_2,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{t_1,\ldots,t_{k}=0}^{\infty}x_1^{m_2}\cdots x_{k}^{m_{k+1}} \nonumber\\[3pt]
& \hskip 2cm \times \frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+n(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i+k)}
(1-q^n).
\end{align}
Write (\ref{temp}) in the following form
\begin{align}
G_{k}(x_1,\ldots,x_{k};q)
& =\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+kn}(1-q^n) \nonumber\\[3pt]
&\hskip 1cm \times \sum_{m_2,\ldots,m_{k+1}=-\infty}^{\infty}
\sum_{t_1,\ldots,t_{k}=0}^{\infty}x_1^{m_2}\cdots x_{k}^{m_{k+1}}
q^{n(\sum_{i=2}^{k+1}|m_i|+2\sum_{i=1}^{k}t_i)}.\label{gf-t-2}
\end{align}
Notice that
\begin{align}\label{temp-1}
&\sum_{a=-\infty}^{{+\infty}}\sum_{b=0}^{{+\infty}}x^{a}q^{n(|a|+2b)}
=\frac{1}{(1-xq^n)(1-x^{-1}q^n)}.
\end{align}
Applying the above formula (\ref{temp-1}) repeatedly to \eqref{gf-t-2}, we deduce that
\[G_{k}(x_1,\ldots,x_{k};q)=\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+kn}
\frac{(1-q^n)}{\prod_{j=1}^{k}(1-x_jq^n)(1-x_j^{-1}q^n)},
\]
as required.   \qed





Setting $x_j=1$ for $1 \leq j\leq k$ in Theorem
\ref{GF-thm-1} and applying Theorem \ref{main-1},  we arrive at the following generating function of $\bar{\eta}_{2k-1}(n)$.

\begin{coro}
For $k\geq 1$, we have
\begin{align}\label{GF-eq-1}
\sum_{n=1}^{\infty}\bar{\eta}_{2k-1}(n)q^n
=\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n-1)/2+kn}\frac{1}{(1-q^n)^{2k-1}}.
\end{align}
\end{coro}

 Since $\bar{\eta}_1(n)=\overline{N}_1(n)$, when taking $k=1$ in \eqref{GF-eq-1},  we are led to the
generating function   for $\overline{N}_1(n)$ as given  by
Andrews, Chan and Kim in \cite[Theorem 1]{Andrews-Chan-Kim}.


The following generating function can be derived by using the same reasoning as in the proof of Theorem \ref{GF-thm-1}.


\begin{thm}\label{GF-thm-2}
For $k \geq 1$, we have
\begin{align}
&\sum_{\stackrel{m_1>0}{m_2,\ldots,m_{k+1}=-\infty}}^{\infty}
\sum_{n=1}^{\infty}D_{k+1}(m_1,m_2,\ldots,m_{k+1};n)
x_1^{m_1}\cdots x_{k+1}^{m_{k+1}}q^n \nonumber \\[3pt]
&\quad =\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n+1)/2+kn}\frac{x_1(1-q^n)}
{(1-x_1q^n)\prod_{j=2}^{k+1}(1-x_jq^n)(1-x_j^{-1}q^n)}.
\end{align}
\end{thm}






Setting   $x_j=1$ for $1 \leq j\leq k+1$ in Theorem
\ref{GF-thm-2} and using Theorem \ref{main-2},
we come to the following generating function of $\bar{\eta}_{2k}(n)$.

\begin{coro}
For $k\geq 1$, we have
\begin{align}\label{GF-eq-2}
\sum_{n=1}^{\infty}\bar{\eta}_{2k}(n)q^n
=\frac{1}{(q;q)_\infty}\sum_{n=1}^{\infty}(-1)^{n-1}
q^{n(3n+1)/2+kn}\frac{1}{(1-q^n)^{2k}}.
\end{align}
\end{coro}

\medskip

\noindent{\bf Acknowledgments.}  We wish to thank the referee for helpful suggestions. This work was supported by the 973 Project and the National Science Foundation of China.






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\end{document}
