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\title{A combinatorial proof of an identity for the divisor generating function
}
\author{Masanori Ando}
\date{}
\address{Wakhok University, Hokkaido 097-0013, 
Japan}
\email{m-ando@wakhok.ac.jp}
\begin{document}
\pagestyle{empty}
\subjclass{05A15, 05A17}
\maketitle\thispagestyle{empty}
\begin{abstract}
We give combinatorial proofs and new generalizations of  $q$-series identities of Dilcher and Uchimura related to divisor function. 
Some interesting combinatorial results related to partition and arm length are also presented. 
\end{abstract}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In \cite{U}, 
Uchimura proved the following $q$-series identity
\begin{equation}
\label{eq;U}
\displaystyle \sum_{k = 1}^{\infty }{(-1)^{k-1} \frac {q^{\frac {k(k+1) }{2 }
}
 }{(q;q )_k (1-q^k) } }
=\sum_{k = 1}^{\infty }{\frac {q^k }{1-q^k } },  
\end{equation}
where $(a;b)_k=(1-a)(1-ab)(1-ab^2)\cdots (1-ab^{k-1})$. 
In fact the identity \eqref{eq;U} has been known 100 years before (\cite{K}). 
Moreover, it is obtained also as specialization of basic hypergeometric series (see \cite{F} section 12 on page 34). 
And, this identity is an infinite version of the following $q$-series identity called {\it {Problem 6407}} in American Mathematical Monthly \cite{vh}. 
\[
\sum_{k=1}^{m}{(-1)^{k-1}\left [ 
\begin{array}{l} 
m \\
k 
\end{array}
\right ] 
\frac{q^{\frac {k(k+1)}{2} }}{1-q^k} 
}
=\sum _{k=1}^{m} {\frac {q^k}{1-q^k}}. 
\]
Here $\left [ 
\begin{array}{l} 
m \\
k 
\end{array}
\right ]$ is a $q-$binomial coefficient. 
Many authors have generalized these identities (see e.g. \cite{vgcz}). 
In this paper, we translate these identities and Dilcher's generalization \cite{D} into combinatorics of partitions, and give a combinatorial proof of them. 
For example, we transform \eqref{eq;U} as 
\[
\sum_{\lambda \in {\mathcal {SP} } }{(-1)^{\ell (\lambda)-1 }\lambda _{\ell (\lambda) } q^{|\lambda | } } =\sum_{n = 1}^{\infty }{\sigma _0 (n)q^n }. 
\]
It is a $q$-series identity about {\it {strict partitions}} and a {\it {divisor function}}. 

\bigskip
The generalizations of \eqref{eq;U} we give in this paper are the following. 
\[
\displaystyle \sum_{k= 1}^{\infty }{(-1)^{k-1} \frac {b^kq^{\frac {k(k-1) }{2 }
+mk }
 }{(bq;q )_k (1-q^k)^m } }
=\sum_{j_1= 1 }^{\infty }{\frac {b^{j_1 }q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
}, 
\]
\[
\displaystyle \sum_{k= 1}^{t }{(-1)^{k-1} \frac {b^kq^{\frac {k(k-1) }{2 }
+mk }
 }{(1-q^k)^m } 
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] _{q,b}
}
=\sum_{j_1= 1 }^{t }{\frac {b^{j_1 }q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
}. 
\]
As a by-product of their proofs, we obtain some combinatorial results. 
%These translations are connected with combinatorial understanding of another generalizations. 
%And, new generalizations may be obtained by generalizing combinatorial results. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Young diagrams}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{Def}
Let $n$ be a positive integer. A partition $\lambda $ of $n$ is an integer sequence 
\[
\lambda =(\lambda_1,\lambda_2,\ldots,\lambda_\ell) 
\]
satisfying $\lambda _1 \geq \lambda _2 \geq \ldots \geq \lambda _\ell >0$ and $\displaystyle \sum _{i=1}^{\ell }{\lambda _i} =n$. 
We call $\ell (\lambda):= \ell$ the length of $\lambda $, 
and each $\lambda _i$ a part of $\lambda $. 
We let ${\mathcal {P}} $ and ${\mathcal {P}}(n) $ denote the set of partitions and the set of partitions of $n$. 
\end{Def}
\begin{Def}
A partition $\lambda $ is said to be strict if $\lambda _1 > \lambda _2 > \ldots > \lambda _\ell >0$. 
We let ${\mathcal {SP}} $ and ${\mathcal {SP}}(n) $ denote the set of partitions and the set of partitions of $n$. 
\end{Def}
\begin{Def}
Let $\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_\ell) $ be a partition. The Young diagram of $\lambda $ is defined by
\[
Y(\lambda):=\{ (i,j)\in \mathbb{N}\times \mathbb{N} \ |\  1\leq i\leq \ell ,1\leq j \leq \lambda_i \}. 
\]
We call $(i,j)\in Y(\lambda )$ the $(i,j)$-cell of $\lambda$. 
And the set of the corners of $\lambda $ is defined by 
\[
C(\lambda ):= 
\{ (i,j)\in Y(\lambda )\ |\ (i+1,j ), (i, j+1 )\not \in Y(\lambda ) 
\}. 
\]
We put $c(\lambda ):=\sharp C(\lambda )$, the number of the corners of $\lambda $. 
\end{Def}
\begin{Def}
Let $(i,j)\in Y(\lambda)$, The $(i,j)$-hook length of $\lambda $ is defined by
\[
h_{ij}(\lambda):=\sharp \{ (i',j')\in Y(\lambda)\  |\  i'=i,j'\geq j\ \  \textrm{or}\ \  j'=j,i'\geq i \ \}
\]
And we put $a_{ij}(\lambda):= \lambda_i - j+1$, the $(i,j)$-arm length of $\lambda $. 
We remark that our definition of arm length $a_{ij}$ is different by 1 from the usual definition \cite{Mac}. 
\end{Def}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$q$-series identity for the divisor function}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}$($Uchimura's identity$)$
\[
\displaystyle \sum_{k = 1}^{\infty }{(-1)^{k-1} \frac {q^{\frac {k(k+1) }{2 }
}
 }{(q;q )_k (1-q^k) } }
=\sum_{k = 1}^{\infty}{\frac {q^k }{1-q^k } },  
\]
where $(a;b)_k=(1-a)(1-ab)(1-ab^2)\cdots (1-ab^{k-1})$. 
\end{thm}
Remark that the right-hand side is computed as 
\[
\sum _{k = 1}^{\infty }{\frac {q^k}{1-q^k}} = \sum _{k=1}^{\infty }{(q^k+q^{2k}+q^{3k}+ \cdots )}=\sum _{n = 1}^{\infty}{\sigma _0 (n)q^n}, 
\]
where $\sigma _0 (n)$ is the number of positive divisors of $n$. 
We now translate this identity into a language of Young diagrams. Then we are able to prove this identity combinatorially. \\


\noindent
\textbf{Figure 1.}\\
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\\
Looking at each term of the left-hand side, 
$\displaystyle q^{\frac {k(k+1)}{2}}$ is translated into the stairs $B$ in Figure $1$. 
Since $\displaystyle \frac {1}{(q;q)_k}$ is the generating function of partitions whose lengths are at most $k$, 
this term corresponds to $C$ in Figure $1$. 
The leftover $\displaystyle \frac{1}{1-q^k}$ is the generating function of rectangular Young diagrams whose vertical lengths are equal to $k$. 
This part corresponds to $A$ in Figure $1$. 
Therefore the left-hand side of the identity is an alternating sum over $k \geq 1$ of $A+B+C$. 
As is noted in Figure $1$, the ``sum" $A+B+C$ is a strict partition.
For a strict partition $\lambda $, we count the number of tuples $(A,B,C)$ such that $A+B+C = \lambda$. 
Let $\lambda $ be a fixed strict partition of length $k$. 
One can embed the stairs $B$ into $\lambda $ in $\lambda _k$ ways. 
For each embedding the rectangle $A$ and the partition $C$ are uniquely determined, respectively. 
Therefore there are $\lambda _k$ tuples $(A,B,C)$, such that $A+B+C = \lambda $. 
Summing up over $k$, Theorem 3.1 reads 
\begin{eqnarray}
\label{id;div}
\sum_{\lambda \in {\mathcal {SP} } }{(-1)^{\ell (\lambda)-1 }\lambda _{\ell (\lambda) } q^{|\lambda | } } =\sum_{n = 1}^{\infty }{\sigma _0 (n)q^n }. 
\end{eqnarray}
The proof of this identity will follow from the next combinatorial theorem. 
\begin{thm}
For any positive integers $n$ and $k$, 
\begin{eqnarray*}
&&\sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda)},  \ell (\lambda ); \textrm {odd} \} \\
&-&\sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda)}, \ell (\lambda ); \textrm {even} \}\\
&=&\left \{ \begin{array}{ll}
1  &(k\mid n) \\
0  &(k\nmid n) .
\end{array} \right.
\end{eqnarray*}
\end{thm}
\noindent
\textbf{Example}. Let $n=5$. 
We draw Young diagrams $Y(\lambda)$ of all strict partitions of $5$, and write arm length $a_{1j}(\lambda )$ in $(1,j)$-cell for $1\leq j \leq \lambda _{\ell (\lambda )}$. \\
\\
\textbf{Figure 2.}\\
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\\
\\
%Where the + and - means union and setminus in multiset. 
Here the numbers are regarded as variables. 
Numbers $5$ and $1$ are the positive divisors of $5$. 
We remark that it is check about Theorem 3.2 in all positive integer $k$ at the same time. 

\bigskip
\noindent
\textbf{Proof of Theorem 3.2.}\\
We consider the set of strict partitions of $n$ such that $a_{1,j}=k $ for some $1\leq j \leq \lambda _{\ell (\lambda )}$:
\[
{\mathfrak{D}}(n,k):= \{ \lambda \in {\mathcal {SP}}(n) \ |\  \lambda _1 - \lambda _{\ell (\lambda )} <  k \leq \lambda _1 \}. 
\]
We divide these strict partitions into two classes $A$ and $B$: 
\[
A=\{ \lambda \in {\mathfrak{D}}(n,k)\ |\ k\nmid \lambda_i \textrm{for any }i \} , B=\{ \lambda \in {\mathfrak{D}}(n,k)\ |\ k\mid \lambda_i \textrm{for some }i\}. 
\]
We consider a map between them that changes the length by $1$. 
\[
\alpha_k :\ A \rightarrow B , \ \ \ \ \ \alpha _k(\lambda )= \lambda ' , 
\]
%(recipe for $\lambda '$. )
where $\lambda ' \in B$ is defined in the following steps: 

\bigskip
\begin{enumerate}[Step 1.]
\item Append $0$ in the tail of $\lambda $ to get $\{ \lambda _1 , \ldots , \lambda _{\ell +1}\}$. 
\item Subtract $k$ from $\textrm {max} \{ \lambda _1 , \ldots , \lambda _{\ell+1}\}$, and add $k$ to $\lambda _{\ell+1}$. 
\item Repeat Step 2 till $\textrm {max} \{ \lambda _1 , \ldots , \lambda _{\ell+1}\} - \textrm {min} \{ \lambda _1 , \ldots , \lambda _{\ell+1}\}$ 
gets less than $k$. 
\item From the resulting composition we have the partition $\lambda '= (\lambda _1, \ldots , \lambda _{\ell +1})$ by arranging parts. 
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
\textbf{Example}. Let $n=13, k=4$. 
\[
\mathfrak {D} (13, 4)=
\{
(13), (8, 5), (7, 6), (6, 4, 3)
\}. 
\]
And we divide these partitions
\[
A=
\{ 
(13), (7, 6)
\}
, 
B=
\{
(8, 5), (6, 4, 3)
\}. 
\]
Then the map $\alpha _4 $ looks
\begin{eqnarray*}
\lambda &=&(13)\rightarrow \{13, 0\} \rightarrow \{9, 4\} \rightarrow \{ 5, 8\} \rightarrow (8, 5) =\alpha _4 (\lambda ), \\
\mu &=&(7, 6)\rightarrow \{7, 6, 0\} \rightarrow \{ 3, 6, 4\} \rightarrow (6, 4, 3)=\alpha _4 (\mu ). 
\end{eqnarray*}
When $k=4$, the pair of $\lambda = (8, 4, 1)$ seems not to exist. However, the partition $\lambda $ is not an element of  $\mathfrak {D}(13, 4)$ primarily. \\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bigskip
By the above construction, we have $\ell (\lambda ')=\ell (\lambda )+1$, and
\[
\sharp \{i \ |\ \lambda _i \equiv j  ({\textrm {mod}} \ k), 1\leq i \leq \ell \}
=\sharp \{i \ |\ \lambda '_i \equiv  j  ({\textrm {mod}} \ k), 1\leq i \leq \ell \}
\]
for $1\leq j \leq k-1$. 
The partition that $\alpha _k$ can not pair up is $(\lambda_1 , \ldots , \lambda _\ell )\equiv (0) \ (\textrm {mod} \ k)$. 
Therefore $\lambda = (n) $ is left when $n$ is a multiple of $k$. 
\qed\\
\\
We add in the proof that the identity and proof equivalent to Theorem 3.2 appears in \cite{P}. 
Moreover, the identity which generalized $\sigma_0$ to $\sigma _n$ in Theorem 3.1 appears there. \\
\noindent
\textbf{Proof of theorem 3.1.}\\
Sum of the left-hand side of Theorem 3.2 over $k$ is 
\begin{eqnarray*}
&&\sum_{k = 1}^{\infty } \sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda )}, \ell (\lambda ); \textrm {odd} \} \\
&-&\sum _{k = 1}^{\infty }\sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda )}, \ell (\lambda ); \textrm {even} \}\\
&=&\sum _{\lambda \in {\mathcal {SP} }(n) }{(-1)^{\ell (\lambda )-1}\lambda _{\ell (\lambda )} }. 
\end{eqnarray*}
And sum of the right-hand side is 
\begin{eqnarray*}
\sum _{k\mid n}{1}
=\sigma _0 (n). 
\end{eqnarray*}
They are the coefficients of $q^n$ in \eqref{id;div}. 
\qed
\bigskip

Theorem 3.1 is the generating function for the total sum of Theorem 3.2. 
Taking the sum over $k$ from $1$ to $m$ for Theorem 3.2, 
we have the following identity of the generating function. 
\begin{thm}$($Problem 6407$)$
\[
\sum_{k=1}^{m}{(-1)^{k-1}\left [ 
\begin{array}{l} 
m \\
k 
\end{array}
\right ] 
\frac{q^{\frac {k(k+1)}{2} }}{1-q^k} 
}
=\sum _{k=1}^{m} {\frac {q^k}{1-q^k}}, 
\]
where the $q$-binomial coefficient is defined as
\[
\left [ 
\begin{array}{l} 
m \\
k 
\end{array}
\right ]
= 
\sum _{
\begin{array}{c}
\lambda \in {\mathcal {P }}\\
\lambda _1 \leq m-k , \ell (\lambda )\leq k
\end{array}
}
{q^{|\lambda |}}. 
\]
\end{thm}
We draw the same figure as Figure 1. %The difference of Theorem 3.3 and Theorem 3.1 is a $q$-binomial coefficient. 
The Young diagram $C$ in Figure is restricted that the first row is at most $m-k$. 
Then, there are new restriction $a_{1j}\leq m$ on the left-hand side of identity. 
Here Theorem 3.3 reads 
\begin{eqnarray}
\sum _{\lambda \in {\mathcal {SP} } }\sum_{j\leq \lambda _\ell , a_{1,j} \leq m}{(-1)^{\ell (\lambda )-1} }q^{|\lambda |}
=\sum _{n= 1}^{\infty}{
\sum _{k \mid n, k\leq m}
{q^{n}
}
}.
\end{eqnarray}
\begin{proof}
Sum of the left-hand side of Theorem 3.2 over $1\leq k\leq m$ is
\begin{eqnarray*}
&&\sum_{k=1}^{m} \sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda )}, \ell (\lambda ): \textrm {odd} \} \\
&-&\sum _{k=1}^{m} \sharp \{ \lambda \in {\mathcal {SP} }(n) \ |\  \lambda_1 \geq k > \lambda _1 -\lambda _{\ell (\lambda )}, \ell (\lambda ): \textrm {even} \}\\
=&&\sum _{\lambda \in {\mathcal {SP} }(n) }\sum_{j\leq \lambda _\ell , a_{1,j} \leq m}{(-1)^{\ell (\lambda )-1} }. 
\end{eqnarray*}

\end{proof}
\begin{cor}
For $n=2(2m+1)$, 
\begin{eqnarray*}
&&\sum_{\lambda \in {\mathcal {SP} }(n)}\sharp \{ h_{1,j}(\lambda )\ | \ 1\leq j\leq \lambda_{\ell (\lambda)}, h_{1,j}(\lambda ) \textrm{is odd} \} \\
&=&\sum_{\lambda \in {\mathcal {SP} }(n)}\sharp \{h_{1,j}(\lambda )\ | \ 1\leq j\leq \lambda_{\ell (\lambda)}, h_{1,j}(\lambda ) \textrm{is even} \}. 
\end{eqnarray*}
\end{cor}
\noindent
\textbf{Example.} Let $n=6$. 
We draw Young diagrams $Y(\lambda)$ of all strict partitions of $6$, and write hook length $h_{1,j}(\lambda )$ in $(1,j)$-cell for $1\leq j \leq \lambda _{\ell (\lambda)}$. \\
\\
\noindent
\textbf{Figure 3.}
\\
\\
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\\
The number of odd numbers equals the number of even numbers. 
\begin{proof}
In Theorem 3.2, the strict partitions they have same arm length and different parity length are pair. 
Recall that $h_{1j}(\lambda ) = a_{1j}(\lambda ) + \ell (\lambda )-1$. 
Therefore the parity of their hook length are different. 
And the leftovers are divisors of $n$. 
When $n$ equals $2(2m+1)$, the number of odd divisors of $n$ equals the number of even divisors of $n$. 
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Generalizations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm} For any positive integers $k,m$ and $n$, we have 
\begin{eqnarray*}
&&\sharp \bigg\{ (\lambda ,  i_1 ,  \ldots , i_{m})\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {odd}
\end{array}
\bigg\} \\
&-&\sharp \bigg\{ (\lambda , i_1 , \ldots , i_{m} )\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {even}
\end{array}
\bigg\} \\
&=&
\sharp \bigg\{ (\lambda , t_1, \ldots , t_{m- c(\lambda )} ) \ \Big|
\begin{array}{c}
\lambda \in {\mathcal {P}}(n), c(\lambda )\leq m, \lambda _1 = k, \lambda_{t_i}= \lambda_{t_i +1} \\
1\leq t_1 < \ldots <t_{m-c(\lambda )}<\ell (\lambda )
\end{array}
\bigg\}. 
\end{eqnarray*}
\end{thm}
\noindent
When $m=1$, this identity is specialized as 
\begin{eqnarray*}
&&\sharp \{ (\lambda ,  i)\ |
\ \lambda \in {\mathcal {SP} }(n), 1\leq i \leq \lambda_{\ell (\lambda )}, 
a_{1,i}(\lambda )=k , \ell (\lambda ):\textrm {odd}
\} \\
&-&\sharp \{ (\lambda , i )\ |
\ \lambda \in {\mathcal {SP} }(n), 1\leq i \leq \lambda_{\ell (\lambda )}, 
a_{1,i}(\lambda )=k , \ell (\lambda ):\textrm {even}
\} \\
&=&
\sharp \{ \lambda  \ |
\ \lambda \in {\mathcal {P}}(n), c(\lambda )=1, \lambda _1 = k
\}. 
\end{eqnarray*}
When $c(\lambda )=1$, the shape of $Y(\lambda )$ is rectangle. 
Therefore this identity is equivalent to Theorem 3.2. \\
\noindent
\textbf{Example. }�@For $n=5, m=2$, 
we draw the same figure as Figure 2. \\
\\
\textbf{Figure 4.}\\
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\\
We count the pairs of arm lengths in a partition $\lambda $ with sign $(-1)^{\ell (\lambda )}$. 
The pairs with positive sign are $(5,1)$, $(4,1)$, $(3,1)$, $(2,1)$, $(5,2)$, $(4,2)$, $(3,2)$, $(5,3)$, $(4,3)$, $(5,4)$. 
The pair with negative sign is $(3,2)$. 
On the other hand, all Young diagrams of partition of $5$ that made by concatenating $2$ rectangles are followings. \\
\\
\noindent
\textbf{Figure 5.}\\
\\

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\\

\bigskip

\noindent
There are $9$ such Young diagrams. 
Here we count the ways of concatenating rectangles. 
Hence, for example, the first $4$ diagrams must be thought of as the different ones. 
The number 9 equals the number of the pairs with positive sign minus the number of the pairs with negative sign. 
%Therefore the number of the pairs with positive sign minus the number of the pairs with negative sign equals the number of the Young diagrams made by concatenating $2$ rectangles. 
And more fix smaller number $b$, the number of pair $(a,b)$ with positive sign minus the number of pair $(a,b)$ with negative sign also equals 
the number of Young diagrams $\lambda$ made by concatenating $2$ rectangles that $\lambda _1$ equals $b$. 
For example, the number of the pairs that smaller number is $1$ and the number of Young diagrams that size of first line is $1$ are both $4$. 

\bigskip
\noindent
\textbf{Proof of Theorem 4.1.}
When $\lambda \in {\mathcal {SP}}(n) $, $i_1 <\ldots <i_m \leq \lambda _\ell $ are given, 
we consider the strict partitions $\lambda ^{(1)}, \ldots , \lambda ^{(m)}$ in the following procedure. 
First, we put $\lambda ^{(1)} :=\lambda $. 
When $\lambda ^{(h)}$ is determined, we put $\mu:=\lambda ^{(h)} $ and $j_{h} :=a_{1,i_{m+1-h } }(\lambda ^{(h ) } ) $. 
And we make a new strict partition [$\mu $] by replacing $\mu_1$ by $\mu_1 - j_h$ in $\mu$. 
This operation keeps strictness, since $\mu $ is strict to modulus $k$. 
We repeat this operation until we get $a_{1, i_{m-h } }(\mu )\leq j_h $. 
We put $\lambda ^{(h+1)}$ with $\mu $ obtained in this way. 
And let $t_h$ be the number of times of the operations. 
The length of $\lambda ^{(i )}$ is same as the length of $\lambda $. 
We %consider that $j_m $ is $k$ of Theorem 3.2, and 
pair up $\lambda ^{(m)}$ by $\alpha _{j_m}$ the map defined in the proof of Theorem 3.2. 
Then there is only $1$ difference between the lengths of partitions of each pair. 
Leftovers are $\lambda ^{(m)}=(\lambda^{(m)}_1 )$ that $\lambda^{(m)}_1 $ is multiple of $j_m$. 
Since $j_1 \geq \ldots \geq j_m$, it corresponds with Young diagram that is made by concatenating rectangles $j_h\times {t_h} $, 
where $\displaystyle t_m=\frac{\lambda^{(m)}_1 }{j_m}$. 
\qed
\\
\\
\noindent
\textbf{Example}. Let $k=13$, $m=3$, $n=70$, $\lambda =(23, 19, 16, 12 )$, $i_1=4$, $i_2=5$, $i_3=11$.\\
First, we put $\lambda ^{(1)} =(23, 19, 16, 12), j_1=a_{1, 11}(\lambda ^{(1)})=13$. 
\[
\lambda ^{(1)}=(23, 19, 16, 12) \stackrel{-13}{\rightarrow} (19, 16, 12, 10)\stackrel{-13}{\rightarrow} (16, 12, 10, 6). 
\]
Because $a_{1, 5}(16, 12, 10, 6)=12\leq j_1$, operation stops. 
And we put $\lambda ^{(2)} =(16, 12, 10, 6), j_2 =a_{1, 5}(\lambda ^{(2)})=12$. 
\[
\lambda ^{(2)}=(16, 12, 10, 6) \stackrel{-12}{\rightarrow} (12, 10, 6, 4). 
\]
We put $\lambda ^{(3)}=(12, 10, 6, 4), j_3=a_{1, 4}(\lambda ^{(3)})=9$. Then, 
\begin{eqnarray*}
\lambda ^{(3)}=(12, 10, 6, 4) &\rightarrow &\{ 12, 10, 6, 4, 0\} \rightarrow \{ 3, 10, 6, 4, 9\} \\
&\rightarrow &\alpha_{9}(\lambda ^{(3)})=(10, 9, 6, 4, 3). 
\end{eqnarray*}
And we perform inverse operation, 
\begin{eqnarray*}
a_{9}(\lambda ^{(3)})=(10, 9, 6, 4, 3) &\stackrel{12}{\rightarrow}& (15, 10, 9, 6, 4) \\
&\stackrel{13}{\rightarrow}& (17, 15, 10, 9, 6) \stackrel{13}{\rightarrow} (19, 17, 15, 10, 9). 
\end{eqnarray*}
Here, $a_{1, 2}(10, 9, 6, 4, 3)$, $a_{1, 4}(15, 10, 9, 6, 4)$ and
$a_{1, 7}(19, 17, 15, 10, 9)$ are equal to $j_3$, $j_2$ and $j_1$ respectively. 
Then, we checked $((23, 19, 16, 12), 4, 5, 11)$ and $((19, 17, 15, 10, 9), 2, 4, 7)$ are pairs. \\
As another example, let $\lambda =(70), i_1=13, i_2=19, i_3=58$. \\
We put $\lambda ^{(1)} = (70), j_{1}=a_{1, 58}(\lambda ^{(1)})=13$. 
\[ 
\lambda ^{(1)}=(70)\stackrel{-13}{\rightarrow} (57) \stackrel{-13}{\rightarrow}(44) \stackrel{-13}{\rightarrow} (31). 
\]
We put $\lambda ^{(2)}=(31), j_2 =a_{1, 19}(\lambda ^{(2)})=13$. 
\[
\lambda ^{(2)}=(31)\stackrel{-13}{\rightarrow} (18). 
\]
We put $\lambda ^{(3)} =(18), j_3 =a_{1, 13}(\lambda ^{(3)})=6$. 
Then, $\lambda ^{(3)}_1=18$ is multiple of $j_3=6$. 
Therefore $((70), 13, 19, 58)$ corresponds with Young diagram concatenating with rectangles $13\times 3$, $13\times 1$ and $6\times 3$. 
\begin{thm}$($Dilcher's identity 1$)$
For any positive integer $m$,  
\[
\displaystyle \sum_{k= 1}^{\infty }{(-1)^{k-1} \frac {q^{\frac {k(k-1) }{2 }
+mk }
 }{(q;q )_k (1-q^k)^m } }
=\sum_{j_1= 1 }^{\infty }{\frac {q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
}. 
\]
\end{thm}
\noindent
\textbf{Figure 6.}\\
\\
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\\
In the left-hand side of Dilcher's identity, $\displaystyle \Big(\frac{q^k}{1-q^k}\Big)^m$ is the generating function of 
$m$-tuples of rectangular Young diagrams whose vertical lengths are equal to $k$. 
Note the lower right angle of each rectangle. 
For a strict partition $\lambda $, there are $\displaystyle \binom {\lambda _{\ell (\lambda )}}{m}$ decompositions to the partitions of figure type. 
Therefore, in a language of Young diagrams, this identity is equivalent to the following. 
\begin{eqnarray}
\label{id;Dilch'}
\displaystyle
\sum_{\lambda \in {\mathcal {SP} } }{(-1)^{\ell (\lambda)-1 }\binom {\lambda _{\ell (\lambda) }}{m} q^{|\lambda | } } =
\sum_{
\lambda \in {\mathcal {P} }
}
{\binom {\ell (\lambda )-c(\lambda )}{m-c(\lambda ) }q^{|\lambda |}
}. 
\end{eqnarray}
\textbf{Proof of Theorem 4.2.}
Total sum of the left-hand side of Theorem 4.1 is
\begin{eqnarray*}
&&\sum_{k= 1}^{\infty }
\sharp \bigg\{ (\lambda ,  i_1 ,  \ldots , i_{m})\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {odd}
\end{array} 
\bigg\} \\
&-&\sum_{k= 1}^{\infty }
\sharp \bigg\{ (\lambda , i_1 , \ldots , i_{m} )\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {even} 
\end{array}
\bigg\} \\
&=&\sum_{\lambda \in {\mathcal {SP} }(n) }
{
(-1)^{\ell (\lambda )-1 }\binom {\lambda _{\ell (\lambda )}}{m}
}.
\end{eqnarray*}
And sum of right-hand side is
\begin{eqnarray*}
&&\sum_{k= 1}^{\infty }
\sharp \bigg\{ (\lambda , t_1, \ldots , t_{m- c(\lambda )} ) \ \Big|
\begin{array}{c}
\lambda \in {\mathcal {P}}(n), c(\lambda )\leq m, \lambda _1 = k, \lambda_{t_i}= \lambda_{t_i +1}\\
1\leq t_1 < \ldots <t_{m-c(\lambda )}<\ell (\lambda )
\end{array}
\bigg\}\\
&=&
\sum_{
\lambda \in {\mathcal {P}(n) }
}
{\binom {\ell (\lambda )-c(\lambda )}{m-c(\lambda ) }
}.
\end{eqnarray*}
They are the coefficients of $q^n$ in \eqref{id;Dilch'}
\qed

\bigskip
Analogously with Theorem 3.3, 
we have an identity by taking the sum over $k$ from $1$ to $t$ for Theorem 4.1. 
\begin{thm}$($Dilcher's identity 2$)$
For any positive integers $m$ and $t $, 
\[
\displaystyle \sum_{k= 1}^{t }{(-1)^{k-1} \frac {q^{\frac {k(k-1) }{2 }
+mk }
 }{(1-q^k)^m } 
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] 
}
=\sum_{j_1= 1 }^{t }{\frac {q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
}. 
\]
\end{thm}

\bigskip
\noindent
Theorem 4.1 is a generalization of Dilcher's identities. 
%Considering the generating functions of the both sides, we obtain the following. 
The both sides are the coefficients of $b^kq^n$ in the following generating functions
\begin{thm}
For any positive integer $m$, 
\[
\displaystyle \sum_{k= 1}^{\infty }{(-1)^{k-1} \frac {b^kq^{\frac {k(k-1) }{2 }
+mk }
 }{(bq;q )_k (1-q^k)^m } }
=\sum_{j_1= 1 }^{\infty }{\frac {b^{j_1 }q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
}. 
\]
\end{thm}
\begin{thm}
For any positive integers $m$ and $t$, 
\[
\displaystyle \sum_{k= 1}^{t }{(-1)^{k-1} \frac {b^kq^{\frac {k(k-1) }{2 }
+mk }
 }{(1-q^k)^m } 
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] _{q,b}
}
=\sum_{j_1= 1 }^{t }{\frac {b^{j_1 }q^{j_1} }{1-q^{j_1} } 
\sum_{j_2 =1 }^{j_1 }{\frac {q^{j_2} }{1-q^{j_2} } 
\cdots
\sum_{j_m =1 }^{j_{m-1 } }{\frac {q^{j_m} }{1-q^{j_m} } 
}
}
},  
\]
where $\displaystyle \left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] _{q,b}$
is defined by
\[
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] _{q,b}
= 
\sum _{
\lambda \in {\mathcal {P }}\atop
\lambda _1 \leq t-k , \ell (\lambda )\leq k
%\begin{array}{c}
%\lambda \in {\mathcal {P }}\\
%\lambda _1 \leq t-k , \ell (\lambda )\leq k
%\end{array}
}
{b^{\lambda _1}q^{|\lambda |}}. 
\] 
\end{thm}
\noindent
When $b=1$, they are Dilcher's identities. 


\bigskip
\noindent
\textbf{Proof of Theorem 4.4.} 
By analogous transform in proof of Theorem 4.2, this identity is equivalent to the following. 
\[
\sum_{\lambda \in \mathcal{SP} }
{\sum _{i=1}^{\lambda _{\ell (\lambda )}}
{(-1)^{\ell (\lambda )-1}}\binom {i-1}{m-1}b^{a_{1, i}(\lambda )}q^{|\lambda |}
}
= 
\sum _{\lambda \in \mathcal{P}}
{\binom {\ell (\lambda )-c(\lambda )}{m-c(\lambda )}b^{\lambda _1}q^{\lambda }
}. 
\]
On the left-hand side of this identity, $i$ is the column number which has the right most gray box in Figure 6. 
And, $\binom {i-1}{m-1}$ is the number of arrangement of other gray box.  
Then, we count the left-hand side of Theorem 4.1, taking care about $i_m$. 
\begin{eqnarray*}
&&\sharp \bigg\{ (\lambda ,  i_1 ,  \ldots , i_{m})\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {odd}
\end{array}
\bigg\} \\
&-&\sharp \bigg\{ (\lambda , i_1 , \ldots , i_{m} )\ \Big|
\begin{array}{c}
\lambda \in {\mathcal {SP} }(n), 1\leq i_1 < \ldots < i_m \leq \lambda_{\ell (\lambda )}\\
a_{1,i_m}(\lambda )=k , \ell (\lambda ):\textrm {even}
\end{array}
\bigg\}\\
=&&
\sum_{\lambda \in \mathcal{SP}(n)}
{
\sum_{i\leq \lambda _\ell(\lambda ), a_{1, i}(\lambda )=k}
{
(-1)^{\ell (\lambda )-1}\binom {i-1}{m-1}
}
}. 
\end{eqnarray*}
And right-hand side is 
\begin{eqnarray*}
&&\sharp \bigg\{ (\lambda , t_1, \ldots , t_{m- c(\lambda )} ) \ \Big|
\begin{array}{c}
\lambda \in {\mathcal {P}}(n), c(\lambda )\leq m, \lambda _1 = k, \lambda_{t_i}= \lambda_{t_i +1} \\
1\leq t_1 < \ldots <t_{m-c(\lambda )}<\ell (\lambda )
\end{array}
\bigg\} \\
=&&\sum_{\lambda \in \mathcal{P}(n), \lambda _1=k}{\binom {\ell (\lambda )-c(\lambda )}{m- c(\lambda )}}. 
\end{eqnarray*}
Therefore the coefficients of $b^kq^ n$ of Theorem 4.4 are equal to the both sides of Theorem 4.1. 
\qed

\bigskip
Theorem 4.5 is the finite version of Theorem 4.4. 
The author does not have the good display of this analogue of $q$-binomial coefficient. 
We say that 
$
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] _{q,b}
$ satisfy following recurrence formula. 
\[
(bq)^{t-k}
\left [ 
\begin{array}{l} 
t \\
k 
\end{array}
\right ] 
+
\left [ 
\begin{array}{l} 
t \\
k+1 
\end{array}
\right ] _{q,b}
=
\left [ 
\begin{array}{l} 
t+1 \\
k+1
\end{array}
\right ] _{q,b}. 
\]
We remark that the first term is ordinary $q$-binomial coefficient and we will finish. 
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\end{document}