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\newtheorem{theorem}{Theorem}
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\begin{center}
\vskip 1cm{\LARGE\bf Special Multi-Poly-Bernoulli Numbers}

\vskip 1cm
\large
Y. Hamahata and H. Masubuchi \\
Department of Mathematics  \\
Tokyo University of Science \\
Noda, Chiba, 278-8510 \\
 Japan \\
{\tt hamahata\_yoshinori@ma.noda.tus.ac.jp}\\
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\begin{abstract}
In this paper we investigate
generalized poly-Bernoulli numbers.
We call them multi-poly-Bernoulli numbers, and
we establish a closed formula
and a duality property for them. 
\end{abstract}


\section{Introduction and background}

Kaneko \cite{Kaneko1} introduced
poly-Bernoulli numbers $B_n^{(k)}$ 
($k\in\mathbb{Z}$, $n=0, 1, 2, \ldots$)
which are
generalizations of Bernoulli numbers.
One knows that special values of certain
zeta functions at non-positive integers can be
described in terms of poly-Bernoulli numbers.
Kaneko \cite{Kaneko2} suggests to study
multi-poly-Bernoulli numbers,
which are generalizations of
poly-Bernoulli numbers, as an open problem.
Kim and Kim \cite{Kim-Kim} consider them and
give a relationship with special values
of certain zeta functions.
We consider special
multi-poly-Bernoulli numbers. 
It seems for the authors that they are more natural
than multi-poly-Bernoulli numbers when 
one tries to generalize the results of
poly-Bernoulli numbers.
The purpose of the present paper is to
establish some results for them.
To be more precise, we prove the closed formula
and the duality for them.


\par
We briefly recall poly-Bernoulli numbers.
For an integer $k\in\mathbb{Z}$, put
$$
Li_k(z)=\sum_{n=1}^{\infty}
\frac{z^n}{n^k}.
$$
The formal power series
$Li_k(z)$ is the $k$-th polylogarithm 
if $k\geq 1$, and a rational function
if $k\leq 0$.
When $k=1$, 
$Li_1(z)=-\log (1-z)$. 
Using $Li_k(z)$, one can introduce poly-Bernoulli
numbers.
The {\it poly-Bernoulli numbers} $B_n^{(k)}$
($n=0, 1, 2, \ldots$) 
are defined by the generating series
$$ 
\frac{Li_k(1-e^{-x})}{1-e^{-x}}
=\sum_{n=0}^{\infty}B_n^{(k)}
\frac{x^n}{n!} .
$$
We find that for any $n\geq 0$, 
$B_n^{(1)}=B_n$, the classical Bernoulli number.

\par
For nonnegative integers $n$, $m$, put
$$
\left\{\begin{array}{c}n\\ m\end{array}\right\}
=\frac{(-1)^m}{m!}
\sum_{l=0}^m(-1)^l
\left(\begin{array}{c}m\\ l\end{array}\right)
l^n .
$$
We call it the {\it Stirling number of the
second kind}. Kaneko obtained in \cite{Kaneko1}
an explicit formula for $B_n^{(k)}$:

\begin{theorem}[\cite{Kaneko1}]\label{1} For a nonnegative integer
$n$ and an integer $k$, we have
$$
B_n^{(k)}=(-1)^n\sum_{m=1}^{n+1}
\frac{(-1)^{m-1}(m-1)!
\left\{\begin{array}{c}n\\ m-1\end{array}\right\}}
{m^k} .
$$
\end{theorem}
 
Using it, the following formula can be shown:

\begin{theorem}[Closed formula \cite{Arakawa-Kaneko}]\label{2} 
For any $n, k\geq 0$, we have
$$
B_n^{(-k)}=\sum_{j=0}^{\min (n,k)}
(j!)^2
\left\{\begin{array}{c}n+1\\ j+1\end{array}
\right\}
\left\{\begin{array}{c}k+1\\ j+1\end{array}
\right\} .
$$
\end{theorem}

\par
By this theorem, we get

\begin{theorem}[Duality \cite{Arakawa-Kaneko}, \cite{Kaneko1}]\label{3}
For $n, k\geq 0$, 
$B_n^{(-k)}=B_k^{(-n)}$ holds.
\end{theorem}

\par
The last theorem can be proved in another way.
Namely, using Theorem~\ref{1}, we have

\begin{theorem}[Symmetric formula \cite{Kaneko1}]\label{4}
$$
\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}
B_n^{(-k)}\frac{x^n}{n!}\frac{y^k}{k!}
=\frac{e^{x+y}}{e^x+e^y-e^{x+y}} .
$$
\end{theorem}
As corollary to this theorem, we have the duality theorem.

\par
We would like to extend these results to our generalized poly-Bernoulli
numbers.


\section{Multi-poly-Bernoulli numbers}

In this section, we investigate generalized
poly-Bernoulli numbers.
First, we define a generalization of
$Li_k(z)$.

\begin{definition}\label{5}{\em
For $k_1, k_2, \ldots , k_r\in\mathbb{Z}$,
define
$$
Li_{k_1,k_2,\ldots ,k_r}(z)=
\sum_{m_1,\ldots ,m_r\in\mathbb{Z}\atop 0<m_1<m_2<\cdots <m_r}
\frac{z^{m_r}}{m_1^{k_1}\cdots m_r^{k_r}} .
$$
}
\end{definition}

\par
Now let us introduce a generalization of
poly-Bernoulli numbers with the use of $L_{k_1,k_2,\ldots ,k_r}(z)$.

\begin{definition}\label{6} Multi-poly-Bernoulli numbers
$B_n^{(k_1,k_2,\ldots ,k_r)}$ ($n=0, 1, 2, \ldots$)
{\em are defined for each integer $k_1, k_2, \ldots , k_r$ by
the generating series
$$
\frac{Li_{k_1,k_2,\ldots ,k_r}(1-e^{-t})}
{(1-e^{-t})^r}=
\sum_{n=0}^{\infty}B_n^{(k_1,k_2,\ldots ,k_r)}
\frac{t^n}{n!}.
$$
}
\end{definition}

\par
We generalize Theorem~\ref{1}:

\begin{theorem}\label{7} For a nonnegative
integer $n$ and integers $k_1, \ldots , k_r$,
we have
$$
B_n^{(k_1,k_2,\ldots ,k_r)}
=(-1)^n\sum_{m_r=r}^{n+r}
\sum_{0<m_1<\cdots <m_r}
\frac{(-1)^{m_r-r}(m_r-r)!\left\{\begin{array}{c}
n\\ m_r-r\end{array}\right\}}
{m_1^{k_1}\cdots m_r^{k_r}} .
$$
\end{theorem}

\begin{proof}
By definition,  
\begin{eqnarray*}
\sum_{n=0}^{\infty}B_n^{(k_1,k_2,\ldots ,k_r)}
\frac{t^n}{n!}
&=&
\sum_{0<m_1<\cdots <m_r}
\frac{(-1)^{m_r-r}(e^{-t}-1)^{m_r-r}}
{m_1^{k_1}\cdots m_r^{k_r}} .
\end{eqnarray*}
Here we apply the formula
$$
\frac{(e^t-1)^m}{m!}=
\sum_{n=m}^{\infty}
\left\{\begin{array}{c}n\\ m\end{array}\right\}
\frac{t^n}{n!}\quad (n\geq m\geq 0)
$$
(see (4) in \cite{Kaneko1}) to the 
right hand side of the above equation.
\begin{eqnarray*}
\text{R.H.S.}&=&
\sum_{0<m_1<\cdots <m_r}
\frac{(-1)^{m_r-r}(m_r-r)!}
{m_1^{k_1}\cdots m_r^{k_r}}
\sum_{n=m_r-r}^{\infty}
\left\{\begin{array}{c}n\\ m_r-r\end{array}
\right\}\frac{(-t)^n}{n!} \\
&=&
\sum_{0<m_1<\cdots <m_r}
\sum_{n=m_r-r}^{\infty}
\frac{(-1)^{m_r-r}(m_r-r)!\left\{\begin{smallmatrix}
n\\ m_r-r\end{smallmatrix}\right\}}
{m_1^{k_1}\cdots m_r^{k_r}}\cdot
\frac{(-1)^nt^n}{n!} \\
&=&
\sum_{n=0}^{\infty}\sum_{m_r=r}^{n+r}
\sum_{0<m_1<\cdots <m_r}
\frac{(-1)^{m_r-r}(m_r-r)!\left\{\begin{smallmatrix}
n\\ m_r-r\end{smallmatrix}\right\}}
{m_1^{k_1}\cdots m_r^{k_r}}\cdot
\frac{(-1)^nt^n}{n!} \\
&=&
\sum_{n=0}^{\infty}
\left(
(-1)^n\sum_{m_r=r}^{n+r}\sum_{0<m_1<\cdots <m_r}
\frac{(-1)^{m_r-r}(m_r-r)!\left\{\begin{smallmatrix}
n\\ m_r-r\end{smallmatrix}\right\}}
{m_1^{k_1}\cdots m_r^{k_r}}\right)
\frac{t^n}{n!} .
\end{eqnarray*}
This shows the claim.
\end{proof}


\par
By this theorem, for the smaller $n$, 
we can compute $B_n^{(k_1,k_2,\ldots ,k_r)}$
more specifically. For example,
\begin{eqnarray*}
B_0^{(k_1,k_2,\ldots ,k_r)}&=&
\frac{1}{1^{k_1}2^{k_2}\cdots r^{k_r}}, \\
B_1^{(k_1,k_2,\ldots ,k_r)}&=&
\sum_{0<m_1<\cdots <m_r}
\frac{1}{m_1^{k_1}\cdots m_{r-1}^{k_{r-1}}
(r+1)^{k_r}} .
\end{eqnarray*}




\section{Closed formula and duality}

Let $n$ be a nonnegative integer, $r$ a positive
integer, and $k\in\mathbb{Z}$. We define
$$
B[r]_n^{(k)}=B_n^{(\overbrace{0,\ldots ,0}^{r-1},k)}.
$$
We say that $B[r]_n^{(k)}$ is a {\it special multi-Bernoulli
number} of order $r$.
By definition, it is clear that $B[1]_n^{(k)}=B_n^{(k)}$. Hence the notion of special
multi-poly-Bernoulli number is a generalization of
 that of poly-Bernoulli number.
For $B[r]_n^{(k)}$, we establish closed formula:

\begin{theorem}[Closed formula]\label{8}
For $n, k\geq 0$, we have
{\small
\begin{eqnarray*} 
&&B[r]_n^{(-k)} \\
&&= 
\sum_{n=n_1+\cdots +n_r\atop n_1,\ldots ,n_r\geq 0}
\sum_{k=k_1+\cdots +k_r\atop k_1,\ldots ,k_r\geq 0}
\frac{n!k!}{n_1!\cdots n_r!k_1!\cdots k_r!}  \\
&&\times\left(\sum_{j_1=0}^{\min (n_1,k_1)}
\cdots
\sum_{j_r=0}^{\min (n_r,k_r)}
(j_1!\cdots j_r!)^2
\left\{\begin{array}{c}n_1+1\\ j_1+1\end{array}\right\}
\cdots
\left\{\begin{array}{c}n_r+1\\ j_r+1\end{array}\right\}
\left\{\begin{array}{c}k_1+1\\ j_1+1\end{array}\right\}
\cdots
\left\{\begin{array}{c}k_r+1\\ j_r+1\end{array}\right\}\right) .
\end{eqnarray*}
}
\end{theorem}

\par
To prove the theorem, we need the following lemma:

\begin{lemma}\label{9}
$$
\sum_{n=m+1}^{\infty}
(e^y-e^{y-x})^n
=
\frac{e^{x+y}(1-e^{-x})}
{e^x+e^y-e^{x+y}}
(e^y-e^{y-x})^m.
$$
\end{lemma}

\begin{proof}
\begin{eqnarray*}
\text{L.H.S.}&=&
\frac{1}{1-(e^y-e^{y-x})}
(e^y-e^{y-x})^{m+1} \\
&=&
\frac{e^y-e^{y-x}}{1-(e^y-e^{y-x})}
(e^y-e^{y-x})^m \\
&=& \text{R.H.S.} .
\end{eqnarray*}
\end{proof}

\par
\noindent
{\it Proof of Theorem~\ref{8}.}
\begin{eqnarray*}
&&\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}
B[r]_n^{(-k)}\frac{x^n}{n!}\frac{y^k}{k!}\\
&&=
\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\left(
(-1)^n\sum_{m_r=r}^{n+r}\sum_{0<m_1<\cdots <m_r}
(-1)^{m_r-r}(m_r-r)!\left\{\begin{array}{c}
n\\ m_r-r\end{array}\right\} m_r^k\right)
\frac{x^n}{n!}\frac{y^k}{k!} \\
&&\hspace{8cm} (\text{by\ Theorem~\ref{7}}) \\
&&=\sum_{n=0}^{\infty}
\sum_{m_r=r}^{n+r}\sum_{m_{r-1}=r-1}^{m_r-1}\cdots
\sum_{m_1=1}^{m_2-1}\left\{\begin{array}{c}n\\ m_r-r\end{array}
\right\}\frac{(-x)^n}{n!}(m_r-r)!
(-1)^{m_r-r}\sum_{k=0}^{\infty}\frac{(m_ry)^k}{k!} \\
&&=
\sum_{n=0}^{\infty}\sum_{m_1=1}^{n+1}\sum_{m_2=m_1+1}^{n+2}
\cdots\sum_{m_r=m_{r-1}+1}^{n+r}
\left\{\begin{array}{c}n\\ m_r-r\end{array}
\right\}\frac{(-x)^n}{n!}(m_r-r)!
(-1)^{m_r-r}\sum_{k=0}^{\infty}\frac{(m_ry)^k}{k!} \\
&&=
\sum_{m_1=1}^{\infty}\sum_{m_2=m_1+1}^{\infty}\cdots
\sum_{m_r=m_{r-1}+1}^{\infty}\sum_{n=m_r-r}^{\infty}
\left\{\begin{array}{c}n\\ m_r-r\end{array}
\right\}\frac{(-x)^n}{n!}(m_r-r)!
(-1)^{m_r-r}\sum_{k=0}^{\infty}\frac{(m_ry)^k}{k!} \\
&&=
\sum_{m_1=1}^{\infty}\sum_{m_2=m_1+1}^{\infty}\cdots
\sum_{m_r=m_{r-1}+1}^{\infty}
(1-e^{-x})^{m_r-r}e^{m_ry} \\
&&=
\frac{1}{(1-e^{-x})^r}
\sum_{m_1=1}^{\infty}\sum_{m_2=m_1+1}^{\infty}\cdots
\sum_{m_r=m_{r-1}+1}^{\infty}
(e^y-e^{y-x})^{m_r}. 
\end{eqnarray*}
We  use Lemma~\ref{9} repeatedly to the right hand side
of the last equation. Then
\begin{eqnarray*}
\text{R.H.S}&=&
\frac{1}{(1-e^{-x})^r}
\left(\frac{e^{x+y}(1-e^{-x})}{e^x+e^y-e^{x+y}}
\right)^{r-1}
\sum_{m_1=1}^{\infty}(e^y-e^{y-x})^{m_1} \\
&=&
\frac{1}{(1-e^{-x})^r}
\left(\frac{e^{x+y}(1-e^{-x})}{e^x+e^y-e^{x+y}}
\right)^r \\
&=&
\left(\frac{e^{x+y}}{e^x+e^y-e^{x+y}}\right)^r .
\end{eqnarray*}
By Theorem~\ref{2}, the right hand side of the last
equation is equal to
\begin{eqnarray*}
&&\left(\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}
B_n^{(-k)}\frac{x^n}{n!}\frac{y^k}{k!}\right)^r \\
&=&
\left(\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}
\sum_{j=0}^{\min (n,k)}
(j!)^2\left\{\begin{array}{c}n+1\\ j+1\end{array}\right\}
\left\{\begin{array}{c}k+1\\ j+1\end{array}\right\}
\frac{x^n}{n!}\frac{y^k}{k!}\right)^r .
\end{eqnarray*}
This implies the theorem.\hfill
$\Box$


\par
A generalization of Theorem~\ref{3} follows from the last theorem:

\begin{corollary}[Duality]\label{10} For $n, k\geq 0$, we have
$$
B[r]_n^{(-k)}=B[r]_k^{(-n)}.
$$
\end{corollary}

\par
We note that in the process of proof of the last theorem,
we have obtained two formulae:

\begin{proposition}[Symmetric formula]\label{11}
For $n, k\geq 0$,
$$
\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}
B[r]_n^{(-k)}\frac{x^n}{n!}\frac{y^k}{k!}
=\left(\frac{e^{x+y}}
{e^x+e^y-e^{x+y}}\right)^r .
$$
\end{proposition}

\begin{proposition}\label{12} For $n, k\geq 0$,
$$
B[r]_n^{(-k)}= 
\sum_{n=n_1+\cdots +n_r\atop n_1,\ldots ,n_r\geq 0}
\sum_{k=k_1+\cdots +k_r\atop k_1,\ldots ,k_r\geq 0}
\frac{n!k!}{n_1!\cdots n_r!k_1!\cdots k_r!} 
B_{n_1}^{(-k_1)}\cdots
B_{n_r}^{(-k_r)} .
$$
\end{proposition}
Proposition~\ref{11} is a generalization of Theorem~\ref{4}.


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{\bf 48} (1999), 159--167.

\bibitem{Carlitz}L. Carlitz,
\newblock Some theorems on Bernoulli numbers of higher
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127--139.

\bibitem{Graham, Knuth, Patashnik}R. Graham, D. Knuth, and
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\newblock Addison-Wesley, 1989.

\bibitem{Ireland-Rosen}K. Ireland and M. Rosen,
\newblock{\em A Classical Introduction to Modern
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{\bf 84}, Springer-Verlag, 1990.

\bibitem{Kaneko1}M. Kaneko,
\newblock Poly-Bernoulli numbers. 
\newblock{\em J. Th\'eorie de Nombres} {\bf 9} (1997),
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\bibitem{Kaneko2}M. Kaneko,
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\bibitem{Kim-Kim}M.-S. Kim and T. Kim,
\newblock An explicit formula on the generalized
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\bibitem{Sanchez}R. S\'anchez-Peregrino,
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362--364.


\end{thebibliography}

\bigskip
\hrule
\bigskip
\noindent
2000 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 11B73.\\

\noindent
{\it Keywords}: poly-Bernoulli numbers, Stirling numbers.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 24 2007;
revised version received April 13 2007.
Published in {\it Journal of Integer Sequences}, April 13 2007.

\bigskip
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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