<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}


\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\DeclareMathOperator{\Li}{Li}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\newcommand{\A}{\mathcal A}
\newcommand{\Q}{\mathbb Q}
\newcommand{\Z}{\mathbb Z}

\newcommand{\mpB}{{B}_n^{(k_1,\ldots,k_r)}}
\newcommand{\mpC}{{C}_n^{(k_1,\ldots,k_r)}}

\newcommand{\cycle}[2]{\left[{#1}\atop{#2}\right]}
\newcommand{\subs}[2]{{{#1}\brace{#2}}}

\def\={\;=\;} \def\+{\,+\,} \def\m{\,-\,}



\begin{center}
\vskip 1cm{\LARGE\bf 
Multi-Poly-Bernoulli Numbers and  \\
\vskip .1in
Finite Multiple Zeta Values
}
\vskip 1cm
\large
Kohtaro Imatomi, 
Masanobu Kaneko,\footnote{The second author is supported by the Japan Society for the Promotion of Science, Grant-in-Aid for Scientific Research (B) 23340010.}
and Erika Takeda\\
Graduate School of Mathematics \\
Kyushu University\\
Motooka 744, Nishi-ku \\
Fukuoka 819-0395 \\
Japan\\
\href{mailto:k-imatomi@math.kyushu-u.ac.jp}{\tt k-imatomi@math.kyushu-u.ac.jp} \\
\href{mailto:mkaneko@math.kyushu-u.ac.jp}{\tt mkaneko@math.kyushu-u.ac.jp} \\
\href{mailto:eririntake@yahoo.co.jp}{\tt eririntake@yahoo.co.jp}\\
\end{center}

\vskip .2 in

\begin{abstract} 
We define the multi-poly-Bernoulli numbers slightly differently from
similar numbers
given in earlier papers by Bayad, Hamahata, and Masubuchi, and study 
their basic properties.  Our motivation for 
the new definition is the connection to
``finite multiple zeta values'',
which have been studied by Hoffman and Zhao, among
others, and are recast in a recent work by Zagier and the second author.
We write the finite multiple zeta values in terms
of our new multi-poly-Bernoulli numbers.
\end{abstract}


\section{Introduction}

For any index set $(k_1,\ldots,k_r)$ with $k_i \in \mathbb{Z}$ and $r\ge1$, 
we define two kinds of {\it multi-poly-Bernoulli numbers}  $B_{n}^{(k_1,\ldots,k_r)}$ and $C_{n}^{(k_1,\ldots,k_r)}$ 
by the following generating series:
\begin{eqnarray}
\dfrac{\Li_{k_1, \ldots, k_r} (1-e^{-t})}{1-e^{-t}} &amp;\=&amp;
\sum^\infty_{n=0} B_{n}^{(k_1,\ldots,k_r)} \dfrac{t^n}{n!}, \label{defB}\\
\dfrac{\Li_{k_1, \ldots, k_r}(1-e^{-t})}{e^t-1} &amp;\=&amp;
\sum^\infty_{n=0} C_{n}^{(k_1,\ldots,k_r)} \dfrac{t^n}{n!},\label{defC}
\end{eqnarray} 
where $\Li_{k_1, \ldots, k_r} (z)$ is the  multiple polylogarithm series given by
\begin{equation}
\Li_{k_1 ,\ldots, k_r}  (z) \= \sum_{m_1 &gt;\cdots &gt; m_r &gt;0}
\dfrac{z^{m_1}}{m_1^{k_1} \cdots m_r^{k_r}}.
\end{equation}
When $r=1$, the numbers $B_n^{(k)}$ and $C_n^{(k)}$ are ``poly-Bernoulli numbers'' defined and studied in our previous work \cite{Ka, AK}.
When $r=1$ and $k_1=1$,
the multiple polylogarithm is just the usual logarithm $-\log(1-z)$,
and thus both $B_n^{(1)}$ and $C_n^{(1)}$ are the classical Bernoulli numbers
(with $B_1^{(1)}=1/2$ and $C_1^{(1)}=-1/2$).

The {\it finite multiple zeta value} is the collection of truncated sums
$$\sum_{p&gt;m_1 &gt;\cdots &gt; m_r &gt;0}
\frac{1}{m_1^{k_1} \cdots m_r^{k_r}}$$
modulo $p$ for all primes $p$,
considered in the quotient ring $\prod_p \Z/p\Z/\bigoplus_p \Z/p\Z$.
(For more details on finite multiple zeta values, we refer the reader to papers by Hoffman \cite{H}, Zhao \cite{Z}, and 
Kaneko-Zagier \cite{KZ}, among others.)
In \S\ref{fMZV} we show the congruence  
$$\sum_{p&gt;m_1 &gt;\cdots &gt; m_r &gt;0}
\frac{1}{m_1^{k_1} \cdots m_r^{k_r}}\,\equiv\, -C_{p-2}^{(k_1-1,k_2,\dots,k_r)} \pmod p,$$
which reveals that each component of a finite multiple zeta value is a
multi-poly-Bernoulli number.

\section{Recurrences and explicit formulas for the multi-poly-Bernoulli numbers}

In this section, we first prove a simple relation between $\mpB$ and $\mpC$
and proceed to give recursion relations and explicit formulas for the multi-poly-Bernoulli numbers.
Since the two generating series \eqref{defB} and \eqref{defC} differ only by the factor $e^t$, two
obvious relations between  $\mpB$ and $\mpC$ are 
\[ \mpB\=\sum_{i=0}^n{n\choose{i}}C^{(k_1,\ldots,k_r)}_{i}\quad\ \text{and}\ \quad
\mpC\=\sum_{i=0}^n(-1)^{n-i}{n\choose i}B^{(k_1,\ldots,k_r)}_{i}.\] 
A simpler relation expressing $\mpB$ in terms of $\mpC$ is the following.

\begin{proposition}  For any $r\ge1, k_i\in\Z$ and $n\ge1$, we have 
\[ \mpB\,=\,\mpC + C^{(k_1-1,k_2,\cdots,k_r)}_{n-1}. \]
\end{proposition}
\begin{proof}  From the obvious identity
\[ \frac{d}{dz}\Li_{k_1, \ldots, k_r} (z)=\frac1z \Li_{k_1-1,k_2, \ldots, k_r} (z) \]
(note that we allow the first index to be $0$ or negative), we have
\[ \frac{d}{dt}\Li_{k_1, \ldots, k_r}(1-e^{-t}) =\frac{\Li_{k_1-1,k_2, \ldots, k_r} (1-e^{-t})}{1-e^{-t}}\cdot e^{-t}
=\frac{\Li_{k_1-1,k_2, \ldots, k_r} (1-e^{-t})}{e^t-1}, \]
and hence
\begin{eqnarray*}
\sum_{n=1}^\infty C^{(k_1-1,k_2,\ldots,k_r)}_{n-1}\frac{t^n}{n!}&amp;=&amp;\int_0^t  \sum_{n=0}^\infty C^{(k_1-1,k_2,\ldots,k_r)}_{n}\frac{t^n}{n!}\,dt
=\int_0^t \frac{\Li_{k_1-1,k_2, \ldots, k_r} (1-e^{-t})}{e^t-1}\,dt\\
&amp;=&amp;\Li_{k_1, \ldots, k_r}(1-e^{-t})
= \Li_{k_1, \ldots, k_r}(1-e^{-t})\left(\frac{e^t}{e^t-1}-\frac1{e^t-1}\right)\\
&amp;=&amp;\frac{\Li_{k_1, \ldots, k_r}(1-e^{-t})}{1-e^{-t}}-\frac{\Li_{k_1, \ldots, k_r}(1-e^{-t})}{e^t-1}\\
&amp;=&amp;\sum^\infty_{n=0} B_{n}^{(k_1,\ldots,k_r)} \dfrac{t^n}{n!}-\sum^\infty_{n=0} C_{n}^{(k_1,\ldots,k_r)} \dfrac{t^n}{n!}.
\end{eqnarray*}
This proves the proposition. \end{proof}

\begin{proposition}\label{zenka b}
For any $k_1,\ldots,k_r\in\Z$ and $n\ge0$, we have
\[ \mpB =
\frac{1}{n+1}\left( B^{(k_1-1,k_2,\ldots k_r)}_n - \sum^{n-1}_{m=1}{n \choose m-1}
 B^{(k_1,\ldots,k_r)}_m\right)
\]
and
\[ \mpC=
\frac{(-1)^n}{n+1}\left(\sum^{n}_{m=0}(-1)^{m}{n \choose m} C^{(k_1-1,k_2,\ldots,k_r)}_m
- \sum^{n-1}_{m=1}(-1)^{m}{n \choose m-1} C^{(k_1,\ldots,k_r)}_{m}\right),
\]
where an empty sum is understood to be $0$.
\end{proposition}
\begin{proof}
For $\mpB$, we multiply the defining equation (\ref{defB}) by $1-e^{-t}$ and differentiate
with respect to $t$ to obtain 
\[ \frac{e^{-t}}{1-e^{-t}}\,\Li_{k_1-1 ,k_2,\ldots,k_r}(1-e^{-t})=
e^{-t}\sum^\infty_{n=0}\mpB \frac{t^n}{n!} 
+(1-e^{-t})\sum^\infty_{n=1}\mpB \frac{t^{n-1}}{(n-1)!},
\]
and from this we have
\[  \sum^\infty_{n=0}\mpB \frac{t^n}{n!} =\sum^\infty_{n=0}B^{(k_1-1,k_2,\ldots,k_r)}_n \frac{t^n}{n!}
- (e^{t}-1)\sum^\infty_{n=1}\mpB \frac{t^{n-1}}{(n-1)!}.
\]
Comparing the coefficients of $t^n/n!$ on both sides, we obtain the
desired relation for $\mpB$.  The relation for $\mpC$ is obtained
similarly.
\end{proof}

We make a remark on the following point: 
since we allow $k_1$ to be positive or negative, the recursions
above need never `stop'.  However, by the identity
$$\Li_{0,k_2,\ldots,k_r}(z)=\frac{z}{1-z}\Li_{k_2,\ldots,k_r}(z),$$
the $B^{(0,k_2,\ldots,k_r)}_n$ or $C^{(0,k_2,\ldots,k_r)}_n$ can be written as simple linear combinations of $B^{(k_2,\ldots,k_r)}_m$ or
$C^{(k_2,\ldots,k_r)}_m$ $(0\le m&lt;n)$. The proposition therefore gives a way to express $\mpB$ or $\mpC$
in terms of numbers with smaller $n$ and lower ``depth'' (length $r$ of
the index set).

Next, we write our multi-poly-Bernoulli numbers as finite sums involving Stirling numbers of the 
second kind.
Recall that
the Stirling numbers of the first kind $\displaystyle{\cycle{n}{m}}$ and the second kind $\displaystyle{\subs{n}{m}}$  are the integers uniquely determined by the following recursions for all integers $m,n$ (see Knuth \cite{K}):

\[ \cycle{0}{0}=1, \cycle{\overset{}{n}}{0}=\cycle{0}{m} = 0\quad (n,m \neq 0),\quad
\cycle{n+1}{m}= {\cycle{n}{m-1}} + n{\cycle{\overset{}{n}}{m}}\quad  (\forall n,m\in \Z),
\]
\[ \displaystyle \subs{0}{0}=1, \displaystyle \subs{n}{0}=\subs{0}{m} = 0, \quad (n,m \neq 0),\quad
\displaystyle \subs{n+1}{m} = \subs{n}{m-1} + m{\subs{n}{m}}\quad  (\forall n,m\in \Z).
\]



\begin{theorem}\label{explicit}
For any $r\ge1$, $k_1 ,k_2 ,\cdots,k_r \in  \Z$, and $n \ge 0 $, we have 
\[ \mpB \=
(-1)^{n}\sum_{n+1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1-1}({m_1}-1)!
{{n}\brace{m_1-1}}}{{m_1}^{k_1} \cdots {m_r}^{k_r}}\]
and
\[ \mpC \=
(-1)^{n}\sum_{n+1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1-1}({m_1}-1)!
{{n+1}\brace{m_1}}}{m^{k_1}_1 \cdots m^{k_r}_r}. \]
\end{theorem}
\begin{proof}
Using the well-known generating series (cf.\ Graham et al. \cite[\S7.4]{GKP})
\begin{equation}\label{st2formula} 
\frac{ (e^t - 1)^m}{m!} =  \sum_{n=m}^\infty \subs{n}{m}\frac{t^n}{n!}\:\ \    (m \geq 0 ),
\end{equation}
we have 
\begin{eqnarray*}
\sum^\infty_{n=0}\mpB \frac{t^n}{n!}
&amp;=&amp;\frac{\Li_{k_1,\cdots,k_r}(1-e^{-t})}{1-e^{-t}}\\
&amp;=&amp;\sum_{m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(1-e^{-t})^{m_1-1}}{{m_1}^{k_1}\cdots{m_r}^{k_r}}\\
&amp;=&amp;\sum_{m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1-1}(m_1-1)!}{{m_1}^{k_1}\cdots{m_r}^{k_r}}
\sum^\infty_{n=m_1-1}{{n}\brace{m_1-1}}\frac{(-t)^n}{n!}\\
&amp;=&amp;\sum^\infty_{n=r-1}(-1)^n \sum_{n+1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}
\frac{(-1)^{m_1-1}(m_1-1)!{{n}\brace{m_1-1}}}{{m_1}^{k_1}\cdots{m_r}^{k_r}}\frac{t^n}{n!}.
\end{eqnarray*}
Comparing the coefficients of $t^n$, we obtain the formula for $\mpB$.

To obtain the formula for $\mpC$, we proceed similarly by using the equation
\[ \frac{ e^t(e^t - 1)^{m-1}}{(m-1)!} =  \sum_{n=m-1}^\infty \subs{n+1}{m}\frac{t^n}{n!}\:\ \    (m \geq 1 ),
\]
which follows from equation \eqref{st2formula} by differentiation:
\begin{eqnarray*}
\sum^\infty_{n=0}\mpC \frac{t^n}{n!}
&amp;=&amp;\frac{\Li_{k_1 ,\ldots,k_r}(1-e^{-t})}{e^{t}-1}\\
&amp;=&amp;\sum_{m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{e^{-t}(1-e^{-t})^{m_1-1}}{{m_1}^{k_1}\cdots{m_r}^{k_r}}\\
&amp;=&amp;\sum_{m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1-1}(m_1-1)!}{{m_1}^{k_1}\cdots{m_r}^{k_r}}
\sum^\infty_{n=m_1-1}{{n+1}\brace{m_1}}\frac{(-t)^n}{n!}\\
&amp;=&amp;\sum^\infty_{n=r-1}(-1)^{n}\sum_{n+1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}
\frac{(-1)^{m_1-1}(m_1-1)!{{n+1}\brace{m_1}}}{{m_1}^{k_1}\cdots{m_r}^{k_r}}\frac{t^n}{n!}.
\end{eqnarray*}
\end{proof}

We end this section by giving formulas for the special case $k_1=\cdots =k_r=1$.

\begin{proposition}\label{111} For any $r\ge1$ and $n\ge r-1$, we have
\[ 
B^{(\overbrace{\scriptstyle{1,1,\ldots,1}}^r)}_n \=\frac{1}{n+1}{n+1 \choose r}B_{n-r+1}^{(1)},\quad
C^{(\overbrace{\scriptstyle{1,1,\ldots,1}}^r)}_n \=\frac{1}{n+1}{n+1 \choose r}C_{n-r+1}^{(1)}.
\]
\end{proposition}
\begin{proof}
Use the identity $\Li_{\underbrace{\scriptstyle{1,\ldots,1}}_{r}}(z)=(-\log(1-z))^r/r!$ to
obtain
\[ \sum_{n=0}^\infty B_n^{(\overbrace{\scriptstyle{1,\ldots,1}}^{r})}\frac{t^n}{n!}
\= \frac{t^{r-1}}{r!}\frac{t}{1-e^{-t}}\=\frac{t^{r-1}}{r!}\sum_{n=0}^\infty B_n^{(1)}\frac{t^n}{n!}\]
and
\[ \sum_{n=0}^\infty C_n^{(\overbrace{\scriptstyle{1,\ldots,1}}^{r})}\frac{t^n}{n!}
\= \frac{t^{r-1}}{r!}\frac{t}{e^t-1}\=\frac{t^{r-1}}{r!}\sum_{n=0}^\infty C_n^{(1)}\frac{t^n}{n!}.\] 
The proposition follows.  \end{proof}


\section{Multi-poly-Bernoulli numbers with negative indices}

\begin{proposition}\footnote{This proposition and the corollary below have
also been pointed out by Genki Shibukawa.}\label{negative} 
We have the following generating series identities for multi-poly-Bernoulli numbers
with non-positive upper indices:

1)  \begin{eqnarray*}
&amp;&amp;\sum_{k_1,\ldots,k_r=0}^\infty\sum_{n=0}^\infty B_n^{(-k_1,\ldots,-k_r)}\frac{t^n}{n!}\frac{x_1^{k_1}}{k_1!}\cdots \frac{x_r^{k_r}}{k_r!}\\
&amp;=&amp;\frac{(1-e^{-t})^{r-1}}{(e^{-x_1}+e^{-t}-1)(e^{-x_1-x_2}+e^{-t}-1)\cdots (e^{-x_1-\cdots -x_r}+e^{-t}-1)},
\end{eqnarray*}

2) \begin{eqnarray*}
&amp;&amp;\sum_{k_1,\ldots,k_r=0}^\infty\sum_{n=0}^\infty C_n^{(-k_1,\ldots,-k_r)}\frac{t^n}{n!}\frac{x_1^{k_1}}{k_1!}\cdots \frac{x_r^{k_r}}{k_r!}\\
&amp;=&amp;\frac{e^{-t}(1-e^{-t})^{r-1}}{(e^{-x_1}+e^{-t}-1)(e^{-x_1-x_2}+e^{-t}-1)\cdots (e^{-x_1-\cdots -x_r}+e^{-t}-1)}.
\end{eqnarray*}
\end{proposition}

\begin{proof} Use the defining identity
\[ \sum_{n=0}^\infty B_n^{(-k_1,\ldots,-k_r)}\frac{t^n}{n!}=(1-e^{-t})^{-1}\sum_{m_1 &gt;\cdots &gt; m_r &gt;0} m_1^{k_1}\cdots m_r^{k_r}(1-e^{-t})^{m_1} \]
to obtain 
\begin{eqnarray*}
&amp;&amp;\sum_{k_1,\ldots,k_r=0}^\infty\sum_{n=0}^\infty B_n^{(-k_1,\ldots,-k_r)}\frac{t^n}{n!}\frac{x_1^{k_1}}{k_1!}\cdots \frac{x_r^{k_r}}{k_r!}
=(1-e^{-t})^{-1} \sum_{m_1 &gt;\cdots &gt; m_r &gt;0}(1-e^{-t})^{m_1}e^{m_1x_1+\cdots +m_rx_r} \\
&amp;=&amp; (1-e^{-t})^{-1}\sum_{n_1,\ldots,n_r=1}^\infty (1-e^{-t})^{n_1+\cdots+n_r}e^{(n_1+\cdots+n_r)x_1+(n_2+\cdots+n_r)x_2+\cdots +n_rx_r} \\
&amp;=&amp; (1-e^{-t})^{-1}\sum_{n_1,\ldots,n_r=1}^\infty \bigl((1-e^{-t})e^{x_1}\bigr)^{n_1}\bigl((1-e^{-t})e^{x_1+x_2}\bigr)^{n_2}\cdots
\bigl((1-e^{-t})e^{x_1+\cdots+x_r}\bigr)^{n_r}\\
&amp;=&amp;(1-e^{-t})^{r-1}\frac{e^{x_1}}{1-\bigl((1-e^{-t})e^{x_1}\bigr)}\frac{e^{x_1+x_2}}{1-\bigl((1-e^{-t})e^{x_1+x_2}\bigr)}\cdots
\frac{e^{x_1+\cdots+x_r}}{1-\bigl((1-e^{-t})e^{x_1+\cdots+x_r}\bigr)}\\
&amp;=&amp;\frac{(1-e^{-t})^{r-1}}{(e^{-x_1}+e^{-t}-1)(e^{-x_1-x_2}+e^{-t}-1)\cdots (e^{-x_1-\cdots -x_r}+e^{-t}-1)}.
\end{eqnarray*}
The identity for $C_n^{(-k_1,\ldots,-k_r)}$ readily follows from this because the defining generating series differ only by 
the factor $e^{-t}$. \end{proof}

\begin{corollary} For any integers $k_1,\ldots,k_r\ge0$ and $n\ge0$, the multi-poly-Bernoulli numbers
$B_n^{(-k_1,\ldots,-k_r)}$ and $C_n^{(-k_1,\ldots,-k_r)}$ are positive integers.
\end{corollary}
\begin{proof}
The right-hand sides of 1) and 2) of the proposition can be rewritten as
\[ e^t(e^t-1)^{r-1}\cdot\frac{e^{x_1}}{1-(e^t-1)(e^{x_1}-1)}\cdot\frac{e^{x_1+x_2}}{1-(e^t-1)(e^{x_1+x_2}-1)}\cdots  \frac{e^{x_1+\cdots+x_r}}{1-(e^t-1)(e^{x_1+\cdots+x_r}-1)}\]
and
\[ (e^t-1)^{r-1}\cdot\frac{e^{x_1}}{1-(e^t-1)(e^{x_1}-1)}\cdot\frac{e^{x_1+x_2}}{1-(e^t-1)(e^{x_1+x_2}-1)}\cdots  \frac{e^{x_1+\cdots+x_r}}{1-(e^t-1)(e^{x_1+\cdots+x_r}-1)}\]
respectively, from which the positivity of $B_n^{(-k_1,\ldots,-k_r)}$ and $C_n^{(-k_1,\ldots,-k_r)}$ is obvious. 
That both are integers follows from the explicit formulas in Theorem \ref{explicit}. \end{proof}

We can give some explicit formulas different from Theorem
\ref{explicit} for special indices.

\begin{proposition} For $r\ge1,\,k\ge0$ and $n\ge0$, we have

1) \begin{eqnarray*}
 B_n^{(-k,\overbrace{\scriptstyle{0,\ldots,0}}^{r-1})}&amp;=&amp;(-1)^n\sum_{j=0}^n(-1)^j(j+1)^kj!{j \choose r-1}\subs{n}{j},\\
 C_n^{(-k,\overbrace{\scriptstyle{0,\ldots,0}}^{r-1})}&amp;=&amp;(-1)^n\sum_{j=0}^n(-1)^j(j+1)^kj!{j \choose r-1}\subs{n+1}{j+1}.
\end{eqnarray*}

2) \begin{eqnarray*}
 B_n^{(\overbrace{\scriptstyle{0,\ldots,0}}^{r-1},-k)}&amp;=&amp;\sum_{j=0}^{\min\{n+1-r,k\}}(r+j-1)!j!\subs{n+1}{r+j}\subs{k+1}{j+1},\\
 C_n^{(\overbrace{\scriptstyle{0,\ldots,0}}^{r-1},-k)}&amp;=&amp;\sum_{j=0}^{\min\{n+1-r,k\}}(r+j-1)!j!\subs{n}{r+j-1}\subs{k+1}{j+1}.
\end{eqnarray*}
\end{proposition}
\begin{proof} Set $x_2=\cdots =x_r=0$ in 1) of Proposition \ref{negative} to obtain
\[ \sum_{k=0}^\infty \sum_{n=0}^\infty B_n^{(-k,\overbrace{\scriptstyle{0,\ldots,0}}^{r-1})}\frac{t^n}{n!}
\frac{x^k}{k!} = \frac{(1-e^{-t})^{r-1}}{(e^{-x}+e^{-t}-1)^r}. \]
From this we have 
\begin{eqnarray*}
\sum_{r=1}^\infty\sum_{k,n=0}^\infty  B_n^{(-k,\overbrace{\scriptstyle{0,\ldots,0}}^{r-1})}\frac{t^n}{n!}
\frac{x^k}{k!}y^{r-1}
&amp;=&amp;\frac1{(e^{-x}+e^{-t}-1)}\sum_{r=1}^\infty \left(\frac{1-e^{-t}}{e^{-x}+e^{-t}-1}\right)^{r-1}y^{r-1}\\
&amp;=&amp; \frac1{(e^{-x}+e^{-t}-1)}\frac1{1-\frac{1-e^{-t}}{e^{-x}+e^{-t}-1}y}\\
&amp;=&amp;\frac1{e^{-x}+e^{-t}-1-(1-e^{-t})y}\\
&amp;=&amp;\frac{e^x}{1-e^x(1+y)(1-e^{-t})}\\
&amp;=&amp;\sum_{j=0}^\infty e^{(j+1)x}(1+y)^j(1-e^{-t})^j\\
&amp;=&amp;\sum_{j=0}^\infty e^{(j+1)x}(1+y)^j(-1)^jj!\sum_{n=j}^\infty (-1)^n\subs{n}{j}\frac{t^n}{n!}\\
&amp;=&amp; \sum_{n=0}^\infty\sum_{j=0}^n e^{(j+1)x}(1+y)^j(-1)^{n+j}j!\subs{n}{j}\frac{t^n}{n!}.
\end{eqnarray*}
Comparing the coefficients of $\frac{t^n}{n!}\frac{x^k}{k!}y^{r-1}$ on both sides, we obtain 1) for
$B_n^{(-k,0,\ldots,0)}$.

Similarly, we compute 
\begin{eqnarray*}
\sum_{r=1}^\infty\sum_{k,n=0}^\infty  C_n^{(-k,\overbrace{\scriptstyle{0,\ldots,0}}^{r-1})}\frac{t^n}{n!}
\frac{x^k}{k!}y^{r-1}
&amp;=&amp;e^{-t}\cdot\frac{e^x}{1-e^x(1+y)(1-e^{-t})}\\
&amp;=&amp;\sum_{j=0}^\infty e^{(j+1)x}(1+y)^je^{-t}(1-e^{-t})^j\\
&amp;=&amp;\sum_{j=0}^\infty e^{(j+1)x}(1+y)^j(-1)^jj!\sum_{n=j}^\infty (-1)^n\subs{n+1}{j+1}\frac{t^n}{n!}\\
&amp;=&amp; \sum_{n=0}^\infty\sum_{j=0}^n e^{(j+1)x}(1+y)^j(-1)^{n+j}j!\subs{n+1}{j+1}\frac{t^n}{n!}
\end{eqnarray*}
and obtain the formula for $C_n^{(-k,0,\ldots,0)}$.

For 2), we set $x_1=\cdots =x_{r-1}=0$ in the formulas in Proposition \ref{negative} and have
\begin{eqnarray*}
\sum_{k=0}^\infty \sum_{n=0}^\infty B_n^{(\overbrace{\scriptstyle{0,\ldots,0}}^{r-1},-k)}\frac{t^n}{n!}
\frac{x^k}{k!} &amp;=&amp; \frac{(e^t-1)^{r-1}}{e^{-x}+e^{-t}-1}\\
&amp;=&amp;(e^t-1)^{r-1}\cdot\frac{e^{t+x}}{1-(e^t-1)(e^x-1)} \\
&amp;=&amp; \sum_{j=0}^\infty e^t(e^t-1)^{r+j-1}e^x(e^x-1)^j\\
&amp;=&amp; \sum_{j=0}^\infty (r+j-1)!\sum_{n=r+j-1}^\infty \subs{n+1}{r+j}\frac{t^n}{n!}\cdot 
j!\sum_{k=j}^\infty \subs{k+1}{j+1}\frac{x^k}{k!}
\end{eqnarray*}
and 
\[ \sum_{k=0}^\infty \sum_{n=0}^\infty C_n^{(\overbrace{\scriptstyle{0,\ldots,0}}^{r-1},-k)}\frac{t^n}{n!}
\frac{x^k}{k!} \= \sum_{j=0}^\infty (r+j-1)!\sum_{n=r+j-1}^\infty \subs{n}{r+j-1}\frac{t^n}{n!}\cdot 
j!\sum_{k=j}^\infty \subs{k+1}{j+1}\frac{x^k}{k!}.
\]
Comparing the coefficients of $\frac{t^n}{n!}\frac{x^k}{k!}$ on both sides, we obtain 2). 
 \end{proof}

\section{Connection to the finite multiple zeta values}\label{fMZV}

As recalled in the introduction, the finite multiple zeta value $\zeta_\A(k_1,\ldots,k_r)$
is an element in the ring $\A:=\prod_p \Z/p\Z/\bigoplus_p \Z/p\Z$ represented by
$\left(\zeta_\A(k_1,\ldots,k_r)_{(p)}\right)_p\in\prod_p \Z/p\Z$ where
\[ \zeta_\A(k_1,\ldots,k_r)_{(p)}\,:=\,\sum_{p&gt;m_1 &gt;\cdots &gt; m_r &gt;0}
\frac{1}{m_1^{k_1} \cdots m_r^{k_r}} \  \bmod p. \]

\begin{theorem} For any $r\ge1$ and $k_i\in\Z$, we have
\[ \zeta_\A(k_1,\ldots,k_r)_{(p)}\= -C_{p-2}^{(k_1-1,k_2,\dots,k_r)} \bmod p.\]
More generally, we have for any $r\ge1$, $j\ge0$, and $k_i\in\Z$
\[ \zeta_\A(\underbrace{1,\ldots,1}_j,k_1,\ldots,k_r)_{(p)}\= -C_{p-j-2}^{(k_1-1,k_2,\dots,k_r)} \bmod p.\]
\end{theorem}
\begin{proof}
By the explicit formula in Theorem \ref{explicit}, we have
\begin{eqnarray*} C_{p-2}^{(k_1-1,k_2,\dots,k_r)}&amp;=&amp;-\sum_{p-1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1-1}({m_1}-1)!
{{p-1}\brace{m_1}}}{m^{k_1-1}_1m_2^{k_2} \cdots m^{k_r}_r}\\
&amp;=&amp;\sum_{p-1\ge m_1&gt;m_2&gt;\cdots&gt;m_r&gt;0}\frac{(-1)^{m_1}m_1!
{{p-1}\brace{m_1}}}{m^{k_1}_1m_2^{k_2} \cdots m^{k_r}_r}. 
\end{eqnarray*}
By the well-known formula (cf.\  Graham et al. \cite[\S6.1]{GKP})
\[ (-1)^{m_1}m_1!{{p-1}\brace{m_1}}=\sum_{l=0}^{m_1}(-1)^l{m_1\choose l}l^{p-1},\]
we see 
\[ (-1)^{m_1}m_1!{{p-1}\brace{m_1}}\equiv \sum_{l=1}^{m_1}(-1)^l{m_1\choose l}\pmod p, \]
the sum on the right being equal to $(1-1)^{m_1}-1=-1$. This proves that 
\[ C_{p-2}^{(k_1-1,k_2,\dots,k_r)}\bmod p = -\zeta_\A(k_1,\ldots,k_r)_{(p)}.\]

For the second identity, we proceed as follows. First note
\begin{eqnarray*} &amp;&amp;\sum_{p-1\ge i_1&gt;\cdots&gt;i_j&gt;m_1&gt;\cdots&gt; m_r\ge1}\frac1{i_1\cdots i_jm_1^{k_1}\cdots m_r^{k_r}}\\
&amp;=&amp;\sum_{p-j&gt;m_1&gt;\cdots &gt;m_r\ge1}\frac1{m_1^{k_1}\cdots m_r^{k_r}}
\sum_{p-1\ge i_1&gt;\cdots&gt; i_j&gt;m_1}\frac1{i_1\cdots i_j}. 
\end{eqnarray*}
By changing $i_l\to p-i_l$, we see that 
\[ \sum_{p-1\ge i_1&gt;\cdots&gt; i_j&gt;m_1}\frac1{i_1\cdots i_j}\equiv
(-1)^j \sum_{p-m_1&gt; i_j&gt;\cdots&gt; i_1\ge1}\frac1{i_1\cdots i_j} \pmod p. \]
Using the formula
\[ \sum_{p-m_1&gt; i_j&gt;\cdots &gt;i_1\ge1}\frac1{i_1\cdots i_j}=\frac1{(p-m_1-1)!}\cycle{p-m_1}{j+1}, \]
and the congruences $$\cycle{\overset{}{n}}{m}\equiv\subs{p-m}{p-n} \pmod p\quad (1\le m\le n\le p-1)$$ 
(see Hoffman \cite[\S5]{H}) and 
$$\frac1{(p-m_1-1)!}\equiv (-1)^{m_1+1}m_1! \pmod p,$$ we obtain (for odd $p$)
\begin{eqnarray*}
\zeta_\A(\underbrace{1,\ldots,1}_j,k_1,\ldots,k_r)_{(p)}&amp;=&amp;
\sum_{p-j&gt;m_1&gt;\cdots&gt; m_r\ge1}\frac{(-1)^{j+m_1+1}m_1!\subs{p-j-1}{m_1}}{m_1^{k_1}\cdots m_r^{k_r}}\bmod p\\
&amp;=&amp; -C_{p-j-2}^{(k_1-1,k_2,\ldots,k_r)} \bmod p.
\end{eqnarray*}
This concludes the proof of the theorem.
\end{proof}

Combining the theorem with Proposition \ref{111}, we see that 
\begin{eqnarray*} \zeta_\A(2,\overbrace{1,\ldots,1}^{k-2})&amp;\=&amp;\bigl(-C_{p-2}^{(\overbrace{\scriptstyle{1,\ldots,1}}^{k-1})}\bmod p\bigr)_p\=
\left(-\frac1{p-1}{p-1 \choose k-1}B_{p-k}\bmod p\right)_p\\
&amp;\=&amp;(B_{p-k}\bmod p)_p. 
\end{eqnarray*}
(We used ${p-1\choose k-1}\equiv (-1)^{k-1}\pmod p$ and $(-1)^{p-k}B_{p-k}=B_{p-k}$ for large enough $p$.)
As is discussed in Kaneko-Zagier \cite{KZ}, the element $(B_{p-k}\bmod p)_p$ on the right is regarded as an analogue
of $k\zeta(k)$. (Note that the obvious analogue $\zeta_\A(k)$ of $\zeta(k)$ is 0.)

\begin{thebibliography}{6}

\bibitem{AK} T.~Arakawa and M.~Kaneko, Multiple zeta values,
  poly-Bernoulli numbers, and related zeta functions, {\em Nagoya
    Math. J.} {\bf 153} (1999), 189--209.

\bibitem{BH} A.~Bayad and Y.~Hamahata, Multiple polylogarithms and multi-poly-Bernoulli polynomials,
{\em  Funct. Approx. Comment. Math.} {\bf 46} (2012), 45--61.

\bibitem{GKP} R.~Graham, D.~Knuth, and O.~Patashnik,  {\em Concrete
  Mathematics}, Addison-Wesley, 1989.
  
\bibitem{HM} Y.~Hamahata and H.~Masubuchi, Special multi-poly-Bernoulli numbers, {\em J. Integer Seq.}
{\bf 10} (2007), \href{https://cs.uwaterloo.ca/journals/JIS/VOL10/Hamahata/hamahata3.html}{Article 07.4.1}.

\bibitem{HM2} Y.~Hamahata and H.~Masubuchi, Recurrence formulae for
multi-poly-Bernoulli numbers, {\em Integers} {\bf 7} (2007), \#A46.

\bibitem{H} M.~Hoffman,  Quasi-symmetric functions and  mod $p$
multiple harmonic sums, preprint, \url{http://arxiv.org/abs/math/0401319}.

\bibitem{Ka} M.~Kaneko, Poly-Bernoulli numbers,  {\em J. Th\'eor.
Nombres Bordeaux} {\bf 9} (1997), 199--206.

\bibitem{KZ} M.~Kaneko and D.~Zagier, Finite multiple zeta values, in
preparation.

\bibitem{K}  D.~Knuth, Two notes on notation, {\em Amer. Math.
    Monthly} {\bf 99} (1992), 403--422.

\bibitem{Z} J.~Zhao, Wolstenholme type theorem for multiple harmonic sums,
{\em Int. J. Number Theory} {\bf 4} (2008), 73--106.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 11M32.

\noindent \emph{Keywords: } 
multi-poly-Bernoulli number, Stirling number, finite multiple zeta value.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A027643},
\seqnum{A027644},
\seqnum{A027645},
\seqnum{A027646},
\seqnum{A027647},
\seqnum{A027648},
\seqnum{A027649},
\seqnum{A027650}, and
\seqnum{A027651}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 24 2013;
revised version received February 17 2014. 
Published in {\it Journal of Integer Sequences}, February 17 2014.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


</pre></body></html>