\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} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf Sums of Products of Bernoulli Numbers, \\ \vskip .1in Including Poly-Bernoulli Numbers } \vskip 1cm \large Ken Kamano \\ Department of General Education\\ Salesian Polytechnic \\ 4-6-8, Oyamagaoka, Machida-city, Tokyo 194-0215\\ Japan\\ \href{mailto:kamano@salesio-sp.ac.jp}{\tt kamano@salesio-sp.ac.jp} \end{center} \vskip .2 in \begin{abstract} We investigate sums of products of Bernoulli numbers including poly-Bernoulli numbers. A relation among these sums and explicit expressions of sums of two and three products are given. As a corollary, we obtain fractional parts of sums of two and three products for negative indices. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{thm}[theorem]{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{cor}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{prop}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \def\Li{{\rm Li}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction and main results} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Bernoulli numbers $B_n$ ($n=0,1,2,\ldots$) are defined by the following generating function: \[ \frac{t}{e^t-1} = \sum_{n=0}^{\infty} \frac{B_n}{n!}t^n. \] The following identity on sums of two products of Bernoulli numbers is known as Euler's formula: \begin{equation}\label{eq:Euler1} \sum_{i=0}^n \binom{n}{i} B_iB_{n-i} = -nB_{n-1} -(n-1)B_{n} \ \ (n\ge 1). \end{equation} When $n$ is an even integer, the identity \eqref{eq:Euler1} can be written as \begin{equation}\label{eq:Euler2} \sum_{i=1}^{n-1} \binom{2n}{2i} B_{2i}B_{2n-2i} = -(2n+1)B_{2n}\ \ (n\ge 2), \end{equation} because $B_n=0$ for any odd integer $n\ge 3$. Many generalizations of \eqref{eq:Euler1} and \eqref{eq:Euler2} have been considered. As a generalization of \eqref{eq:Euler2}, Dilcher \cite{D} gave closed formulas of sums of $N$ products of Bernoulli numbers for any positive integer $N$. Chen \cite{Chen} gave generalizations of \eqref{eq:Euler1} for sums of $N$ products of Bernoulli polynomials, generalized Bernoulli numbers and Euler polynomials by using special values of certain zeta functions at non-positive integers. Other types of sums of products have been also studied; see, for example, \cite{AD2007,AD2009,E Trans,Ma,Pe,PS}. The reason why these formulas are valid is that the generating function of Bernoulli numbers satisfies simple differential equations. For example, Euler's formula \eqref{eq:Euler1} is derived by comparing the coefficients of the following identity: \begin{equation}\label{eq:diff.eq.of F_1} F(t)^2 = -tF'(t) +(1-t)F(t), \end{equation} where $F(t)=t/(e^t-1)$. For any integer $k$, Kaneko \cite{K} introduced poly-Bernoulli numbers of index $k$ (denoted by $B_n^{(k)}$) by the following generating function: \begin{equation}\label{eq:poly-Bernoulli defi} \frac{\Li_k(1-e^{-t})}{1-e^{-t}} = \sum_{n=0}^{\infty} \frac{B_n^{(k)}}{n!} t^n, \end{equation} where $\Li_k(x)$ is the $k$-th polylogarithm defined by $\Li_k(x) = \sum_{n=1}^{\infty} x^n/n^k$. The list of poly-Bernoulli numbers $B_n^{(k)}$ with $-5\le k \le 5$ and $0\le n \le 7$ are given by Arakawa and Kaneko \cite{AK}. The numbers $B_n^{(k)}$ are rational numbers, in particular, are positive integers for $k\le 0$ (e.g., \cite[Section 1]{K}). When $k=1$, the left-hand side of \eqref{eq:poly-Bernoulli defi} is equal to $te^t/(e^t-1)$ because of $\Li_1(x)=-\log(1-x)$. Since \begin{equation}\label{eq:alternate Bernoulli} \frac{te^t}{e^t-1} = \sum_{n=0}^{\infty} \frac{(-1)^nB_n}{n!} t^n, \end{equation} we have $B_n^{(1)} =(-1)^nB_n$ for $n\ge 0$ (actually $B_n^{(1)}=B_n$ except for $n=1$). Poly-Bernoulli numbers of positive index are related to multiple zeta functions. To be more precise, special values of certain multiple zeta functions at non-positive integers are described in terms of poly-Bernoulli numbers (cf.\,\cite{AK_Nagoya}). For combinatorial interpretations of poly-Bernoulli numbers of negative index, see Brewbaker \cite{B} and Launois \cite{L}. In this paper we investigate the following type of sums of products of Bernoulli numbers including poly-Bernoulli numbers: \begin{equation}\label{eq:sop invol. poly} S_m^{(k)}(n) := \sum_{\substack{i_1+\cdots +i_{m}=n \\ i_1,\ldots ,i_{m}\ge 0}} \binom{n}{i_1,\ldots ,i_{m}} B_{i_1}\cdots B_{i_{m-1}} B_{i_{m}}^{(k)}\ \ (m\ge 1, n\ge 0), \end{equation} where $\binom{n}{i_1,\ldots ,i_{m}}$ are multinomial coefficients defined by \begin{equation*} \binom{n}{i_1,\ldots,i_{m}} = \frac{n!}{i_1! \cdots i_m!}. \end{equation*} Clearly, it holds that $S_1^{(k)}(n) = B_n^{(k)}$. We list $S_2^{(k)}(n)$ and $S_3^{(k)}(n)$ with $-4\le k \le 4$ and $0\le n \le 6$ in Tables \ref{tbl001} and \ref{tbl002}. We note that the numbers $S_m^{(k)}(n)$ appear as the coefficients of the following generating function: \begin{equation}\label{SP-generating} \left(\frac{t}{e^t-1}\right)^{m-1} \frac{\Li_k(1-e^{-t})}{1-e^{-t}} = \sum_{n=0}^{\infty} S_m^{(k)}(n) \frac{t^n}{n!}. \end{equation} The left-hand side of \eqref{SP-generating} satisfies a certain differential equation like \eqref{eq:diff.eq.of F_1} (see Proposition \ref{prop:dm_equation}), thus this type of sums \eqref{eq:sop invol. poly} is one of natural extensions of the classical sums of products of Bernoulli numbers. \begin{table}[t] \caption{$S_2^{(k)}(n)$}\label{tbl001}\vspace{-10pt} \begin{center} {\renewcommand\arraystretch{1.3} $\begin{array}{|c||c|c|c|c|c|c|c|} \hline k \backslash n &0& 1 & 2&3& 4 &5 & 6 \\ \hline \hline -4 & 1 & \frac{31}{2} &\frac{781}{6} & 855 &\frac{147479}{30}& 26025 & \frac{5474701}{42} \\ \hline -3 & 1 & \frac{15}{2} &\frac{229}{6} & 165 & \frac{19559}{30} & 2435 & \frac{367669}{42}\\ \hline -2 & 1 & \frac{7}{2} &\frac{61}{6} & 27 & \frac{2039}{30} &165 & \frac{16381}{42} \\ \hline -1 & 1 & \frac{3}{2} & \frac{13}{6} & 3 & \frac{119}{30} & 5 & \frac{253}{42} \\ \hline 0 & 1 & \frac{1}{2} &\frac{1}{6} & 0 & -\frac{1}{30} & 0 & \frac{1}{42} \\ \hline 1 & 1 & 0 &-\frac{1}{6} & 0 & \frac{1}{10} & 0 & -\frac{5}{42} \\ \hline 2 & 1 & -\frac{1}{4} & -\frac{1}{9} & \frac{1}{8} & \frac{17}{450} & -\frac{1}{8} & -\frac{23}{1470} \\ \hline 3 & 1 & -\frac{3}{8} & -\frac{1}{108} & \frac{13}{96} & -\frac{733}{13500} &-\frac{131}{1440} & \frac{65953}{617400}\\ \hline 4 & 1 & -\frac{7}{16} & \frac{43}{648} & \frac{115}{1152} & -\frac{70271}{810000} & -\frac{233}{9600} & \frac{26855027}{259308000} \\ \hline \end{array}$} \end{center} \end{table} \begin{table}[t] \caption{$S_3^{(k)}(n)$}\label{tbl002}\vspace{-10pt} \begin{center} {\renewcommand\arraystretch{1.3} $\begin{array}{|c||c|c|c|c|c|c|c|} \hline k \backslash n &0& 1 & 2&3& 4 &5 & 6 \\ \hline \hline -4 & 1 & 15 &\frac{689}{6} &\frac{1335}{2} & \frac{33361}{10} & \frac{30315}{2} & \frac{2708995}{42}\\ \hline -3 & 1 & 7 &\frac{185}{6} & \frac{223}{2} & \frac{3601}{10} & \frac{6473}{6} & \frac{128515}{42}\\ \hline -2 & 1 & 3 &\frac{41}{6} & \frac{27}{2} & \frac{241}{10} & \frac{79}{2} & \frac{2515}{42}\\ \hline -1 & 1 & 1 & \frac{5}{6} & \frac{1}{2} & \frac{1}{10} & -\frac{1}{6} & -\frac{5}{42}\\ \hline 0 & 1 & 0 &-\frac{1}{6} & 0 & \frac{1}{10} & 0 & -\frac{5}{42}\\ \hline 1 & 1 & -\frac{1}{2} & 0 & \frac{1}{4} & -\frac{1}{10} & -\frac{1}{4} & \frac{5}{21}\\ \hline 2 & 1 & -\frac{3}{4} & \frac{11}{36} & \frac{1}{6} & -\frac{107}{300} & \frac{11}{360} & \frac{209}{392}\\ \hline 3 & 1 & -\frac{7}{8} & \frac{115}{216} & -\frac{11}{288} & -\frac{6619}{18000} & \frac{899}{2700} & \frac{134563}{493920}\\ \hline 4 & 1 & -\frac{15}{16} &\frac{869}{1296} &-\frac{755}{3456}&-\frac{273653}{1080000}& \frac{279877}{648000} & -\frac{10347133}{207446400}\\ \hline \end{array}$} \end{center} \end{table} Now we state our main results of this paper. \begin{thm}\label{thm:maintheorem} For $k\in\Z$ and $m\ge 1$, we have \begin{equation}\label{eq:maintheorem} \begin{split} &\sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} S_{m+1}^{(k-l)}(n)\\ &= \begin{cases} n(n-1)\cdots (n-m+1) \displaystyle \sum_{l=1}^m {m \atopwithdelims[] l} B_{n-m+l}^{(k)} \, , & \text{\ \ if } n\ge m;\\ 0 , & \text{\ \ if } 0\le n \le m-1, \end{cases} \end{split} \end{equation} where ${m \atopwithdelims[] l}$ are (unsigned) Stirling numbers of the first kind. \end{thm} The definition of Stirling numbers of the first kind ${m \atopwithdelims[] l}$ will be given in Section 2. Although Theorem \ref{thm:maintheorem} only gives relations among sums of products $S_m^{(k)}(n)$, explicit formulas of $S_m^{(k)}(n)$ can be obtained for $m=2$ and $3$. \begin{thm}\label{thm:m=1 theorem} For $k\ge 1$ and $n\ge 0$, it holds that \begin{align} S_2^{(0)}(n) &= B_n^{(1)}, \label{eq:m=1theorem_0}\\ S_2^{(k)}(n) &= B_n^{(1)} -n \sum_{j=1}^k B_n^{(j)}, \label{eq:m=1theorem}\\ S_2^{(-k)}(n) &= B_n^{(1)} +n \sum_{j=0}^{k-1} B_n^{(-j)}. \label{eq:m=1theorem_negative} \end{align} \end{thm} When $k=1$ in \eqref{eq:m=1theorem}, we have \begin{equation}\label{eq:cor transform 1 to -1} \sum_{i=0}^n \binom{n}{i}(-1)^{n-i} B_{i} B_{n-i} = -(n-1)B_{n}. \end{equation} The identity \eqref{eq:cor transform 1 to -1} is equivalent to \eqref{eq:Euler1} because $B_n=0$ for any odd integer $n\ge 3$. Therefore Theorem \ref{thm:m=1 theorem} can be regarded as a generalization of Euler's formula \eqref{eq:Euler1}. \begin{thm}\label{thm:m=2 theorem} For $k\ge 1$ and $n\ge 1$, it holds that \begin{align} S_3^{(0)}(n) =& -(n-1)B_n, \label{eq:m=2theorem_0} \\ \begin{split} S_3^{(k)}(n) =& (-1)^n ( 1-2^{-k}) B_{n-1} -(n-1)B_n\\ &+n(n-1) \sum_{j=1}^k (1-2^{j-k-1}) \left(B_n^{(j)}+B_{n-1}^{(j)}\right), \end{split} \label{eq:m=2 theorem} \\ \begin{split} S_3^{(-k)}(n) =& n (2^k-1)(-1)^{n-1} B_{n-1} -(n-1)B_n \\ & +n(n-1) \sum_{j=0}^{k-2} (2^{k-1-j}-1) \left(B_n^{(-j)} + B_{n-1}^{(-j)}\right). \end{split} \label{eq:m=2 theorem negative} \end{align} \end{thm} As a corollary, for a negative index $-k$, we obtain the following formulas on fractional parts of $S_2^{(-k)}(n)$ and $S_3^{(-k)}(n)$. \begin{cor}\label{cor:denominator} For $k\ge 1$ and $n\ge 0$, we have \begin{align} S_2^{(-k)}(n) &\equiv B_n \pmod{1}, \label{eq:denominator S_2}\\ S_3^{(-k)}(n) &\equiv \begin{cases} n(2^k-1)B_{n-1} \pmod{1}, & $if $ n $ is odd$; \\ -(n-1)B_{n} \pmod{1}, & $if $ n $ is even$. \end{cases} \label{eq:denominator S_3} \end{align} Here $\alpha \equiv \beta \pmod{1}$ means $\alpha - \beta \in\Z $ for rational numbers $\alpha$ and $\beta$. \end{cor} The classical von Staudt-Clausen theorem (e.g., \cite[Section 7.9]{H-W}) states that \begin{equation*} B_n \equiv -\sum_{\substack{p : \text{prime} \\ (p-1)\mid n}} \frac{1}{p} \pmod{1} \end{equation*} for any even integer $n\ge 2$. Therefore, by using Corollary \ref{cor:denominator}, we can determine the fractional parts of $S_2^{(-k)}(n)$ and $S_3^{(-k)}(n)$ if $k\ge 1$ and $n\ge 0$ are given. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of Theorem \ref{thm:maintheorem}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% We first recall (unsigned) Stirling numbers of the first kind. Let $m$ be a positive integer. For $0\le l\le m$, Stirling numbers of the first kind ${m \atopwithdelims[] l}$ are defined as \begin{equation} x(x+1)\cdots (x+m-1) = \sum_{l=0}^m{m \atopwithdelims[] l} x^l. \end{equation} It follows immediately that ${m \atopwithdelims[] 0}=0$ and ${m \atopwithdelims[] m} = 1$ for all $m\ge 1$. For $l\ge m+1$ and $l\le -1$, we define ${m \atopwithdelims[] l}=0$. Then the recurrence relation \begin{equation}\label{eq:Stirling recurrence} {m+1 \atopwithdelims[] l} = {m \atopwithdelims[] l-1} + m{m \atopwithdelims[] l}\end{equation} holds for all $m\ge 1$ and $l\in\Z$. We set the generating function of poly-Bernoulli numbers of index $k$ as $F_k(t)$, i.e., \[ F_k(t) := \frac{\Li_k(1-e^{-t})}{1-e^{-t}}.\] For $k=1$, $0$ and $-1$, they have simple expressions; $ F_1(t)=te^t/(e^t-1)$, $F_0(t) = e^t$ and $F_{-1}(t)= e^{2t}$. Now let us prove Theorem \ref{thm:maintheorem}. The $n$-th coefficient of $t^m \frac{d^l}{dt^l}F_k(t)$ is equal to \[ \begin{cases} \dfrac{n(n-1)\cdots (n-m+1) B_{n-m+l}^{(k)}}{n!}, & \text{ if } n\ge m; \\ 0 , & \text{ if } 0\le n\le m-1. \end{cases} \] Therefore it suffices to show the following proposition to get Theorem \ref{thm:maintheorem}. \begin{prop}\label{prop:dm_equation} For $k\in \Z$ and $m\ge 1$ we have \begin{equation}\label{eq:dm_equation} \begin{split} & \left( {m \atopwithdelims[] m} \frac{d^{m}}{dt^m} + {m \atopwithdelims[] m-1} \frac{d^{m-1}}{dt^{m-1}} + \cdots + {m \atopwithdelims[] 1} \frac{d}{dt} \right) F_k(t) \\ &= \frac{1}{(e^t-1)^m} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t). \end{split} \end{equation} \end{prop} \vspace{10pt} \begin{proof} We prove the proposition by induction on $m$. Since $\frac{d}{dt} F_k(t) = F_{k-1}(t)/t$, we can easily prove that \begin{equation}\label{eq:diffeq_F_k} \frac{d}{dt}F_k(t)= \frac{1}{e^t-1} (F_{k-1}(t) - F_k(t))\ \ \ \ (k\in\Z). \end{equation} Hence the case $m=1$ holds. We assume that \eqref{eq:dm_equation} holds for a certain $m$. By \eqref{eq:Stirling recurrence}, we have \begin{equation}\label{eq:prop1} \begin{split} & \left( {m+1 \atopwithdelims[] m+1} \frac{d^{m+1}}{dt^{m+1}} + {m+1 \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + \cdots + {m+1 \atopwithdelims[] 1} \frac{d}{dt} \right) F_k(t) \\ & = \frac{d}{dt} \left( {m \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + \cdots + {m \atopwithdelims[] 1} \frac{d}{dt} \right) F_k(t) +m \left( {m \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + \cdots + {m \atopwithdelims[] 1} \frac{d}{dt} \right) F_k(t). \end{split} \end{equation} By the inductive assumption and \eqref{eq:diffeq_F_k}, the right-hand side of \eqref{eq:prop1} is equal to \begin{align*} & \frac{d}{dt} \left( \frac{1}{(e^t-1)^m} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t) \right) + \frac{m}{(e^t-1)^m} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t) \\ =& \frac{-me^t}{(e^t-1)^{m+1}} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t)\\ &+ \frac{1}{(e^t-1)^{m+1}} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} (F_{k-l-1} (t) - F_{k-l}(t)) \\ &+ \frac{m}{(e^t-1)^m} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t) \\ =& \frac{-m-1}{(e^t-1)^{m+1}} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l} (t)\\ &+ \frac{1}{(e^t-1)^{m+1}} \sum_{l=0}^{m} (-1)^{m-l} {m+1 \atopwithdelims[] l+1} F_{k-l-1} (t) \\ =& \frac{1}{(e^t-1)^{m+1}} \sum_{l=0}^{m+1} (-1)^{(m+1)-l} \left( (m+1){m+1 \atopwithdelims[] l+1} + {m+1 \atopwithdelims[] l} \right) F_{k-l} (t). \end{align*} As a consequence, by using the relation \eqref{eq:Stirling recurrence} again, we obtain \begin{align*} & \left( {m+1 \atopwithdelims[] m+1} \frac{d^{m+1}}{dt^{m+1}} + {m+1 \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + \cdots + {m+1 \atopwithdelims[] 1} \frac{d}{dt} \right) F_k(t) \\ =& \frac{1}{(e^t-1)^{m+1}} \sum_{l=0}^{m+1} (-1)^{(m+1)-l} {(m+1)+1 \atopwithdelims[] l+1} F_{k-l} (t). \end{align*} Therefore \eqref{eq:dm_equation} also holds for $m+1$ and this completes the proof. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Explicit formulas of $\boldsymbol{S_2^{(k)}(n)}$ and $\boldsymbol{S_3^{(k)}(n)}$} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section, we prove Theorem \ref{thm:m=1 theorem}, Theorem \ref{thm:m=2 theorem} and Corollary \ref{cor:denominator}. \begin{proof}[Proof of Theorem \ref{thm:m=1 theorem}] First we prove \eqref{eq:m=1theorem_0}. We recall $F_0(t)=e^t$. By setting $m=2$ and $k=0$ in \eqref{SP-generating}, we obtain that the generating function of $S_2^{(0)}(n)$ is equal to $te^t/(e^t-1)$. Then \eqref{eq:m=1theorem_0} follows from \eqref{eq:alternate Bernoulli}. Next we prove the positive index case \eqref{eq:m=1theorem}. Since the negative index case \eqref{eq:m=1theorem_negative} can be proved similarly, we omit its proof. By \eqref{eq:diffeq_F_k}, we have \[ \sum_{j=1}^k \frac{d}{dt} F_j(t) = \frac{1}{e^t-1} (F_{0}(t) -F_k(t) ). \] Since $F_0(t)=e^t$, it holds that \[\frac{t}{e^t-1}F_k(t) = \frac{te^t}{e^t-1} -t \sum_{j=1}^k \frac{d}{dt} F_j(t).\] By comparing the coefficients of both sides, we obtain \eqref{eq:m=1theorem}. \end{proof} \begin{proof}[Proof of Theorem \ref{thm:m=2 theorem}] By setting $m=3$ and $k=0$ in \eqref{SP-generating}, we obtain that the generating function of $S_3^{(0)}(n)$ is $t^2e^t/(e^t-1)^2$. This is exactly the same as the generating function of $S_2^{(1)}(n)$, therefore \eqref{eq:m=2theorem_0} follows from the relation \eqref{eq:cor transform 1 to -1}. We prove the positive index case \eqref{eq:m=2 theorem}. We also omit the proof of the negative index case \eqref{eq:m=2 theorem negative} because it can be proved similarly. Setting $m=2$ in Proposition \ref{prop:dm_equation}, we have \begin{equation}\label{eq:pf_thm:m=2_first} \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_k(t)= \frac{1}{(e^t-1)^2} \left((2F_k(t)-F_{k-1}(t))- (2F_{k-1}(t)-F_{k-2}(t))\right). \end{equation} By this equation, we get \begin{equation}\label{eq:pf_thm:m=2_second} \frac{2F_l(t)}{(e^t-1)^2} - \frac{F_{l-1}(t)}{(e^t-1)^2} = \frac{2F_0(t)-F_{-1}(t)}{(e^t-1)^2} +\sum_{j=1}^l \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t). \end{equation} In fact, this can be proved by replacing $k$ with $j$ in \eqref{eq:pf_thm:m=2_first} and summing over $j$ from $1$ to $l$. Furthermore we multiply both sides of \eqref{eq:pf_thm:m=2_second} by $2^{l-1}$ and sum over $l$ from $1$ to $k$. Then we obtain \begin{align*} 2^k \frac{F_k(t)}{(e^t-1)^2} =& \frac{F_0(t)}{(e^t-1)^2} + \left(\sum_{l=1}^k 2^{l-1}\right) \frac{2F_0(t)-F_{-1}(t)}{(e^t-1)^2}\\ & + \sum_{l=1}^k 2^{l-1} \sum_{j=1}^l \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t) \\ =& \frac{(2^{k+1}-1)F_0(t) - (2^k-1)F_{-1}(t)}{(e^t-1)^2}\\ & + \sum_{j=1}^k \left(\sum_{l=j}^ k 2^{l-1}\right) \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t). \end{align*} Hence we have \begin{equation} \begin{split} 2^k \left(\frac{t}{e^t-1}\right)^2 F_k(t) =& (2^{k+1}-1) \frac{t}{e^t-1} \frac{te^t}{e^t-1} - (2^k-1)\frac{te^t}{e^t-1}\frac{te^t}{e^t-1}\\ & + t^2 \sum_{j=1}^k (2^k-2^{j-1}) \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t). \end{split} \end{equation} By comparing the coefficients of both sides, we obtain for $n\ge 1$ \begin{equation*} \begin{split} 2^k S_3^{(k)}(n) =& (2^{k+1}-1) \sum_{i=0}^n \binom{n}{i} (-1)^{n-i} B_{i} B_{n-i} - (2^k-1)\sum_{i=0}^n \binom{n}{i} (-1)^{n} B_{i} B_{n-i} \\ & +n(n-1) \sum_{j=1}^k (2^k-2^{j-1}) \left(B_n^{(j)} + B_{n-1}^{(j)} \right). \end{split} \end{equation*} By \eqref{eq:Euler1}, \eqref{eq:cor transform 1 to -1} and the fact $(-1)^n(n-1)B_n = (n-1)B_n$ for all $n\ge 1$, it holds that \begin{equation*} \begin{split} & (2^{k+1}-1) \sum_{i=0}^n \binom{n}{i} (-1)^{n-i} B_{i} B_{n-i} - (2^k-1)\sum_{i=0}^n \binom{n}{i} (-1)^{n} B_{i} B_{n-i} \\ &= -(2^{k+1}-1)(n-1)B_n - (-1)^n(2^k-1) (-nB_{n-1} - (n-1)B_n)\\ &=-n(2^k-1) (-1)^{n-1} B_{n-1} -(n-1)2^kB_n. \end{split} \end{equation*} Therefore we obtain \begin{equation}\label{eq: 2^k} \begin{split} 2^k S_3^{(k)}(n) =& -n\left(2^k- 1\right) (-1)^{n-1} B_{n-1} -(n-1)2^k B_n \\ & +n(n-1) \sum_{j=1}^k (2^k -2^{j-1}) (B_n^{(j)}+B_{n-1}^{(j)}). \end{split} \end{equation} Dividing both sides of \eqref{eq: 2^k} by $2^k$, we get \eqref{eq:m=2 theorem}. \end{proof} \begin{proof}[Proof of Corollary \ref{cor:denominator}] The congruence \eqref{eq:denominator S_2} immediately follows from \eqref{eq:m=1theorem_negative} and the fact $B_n^{(-k)}$ are integers for $k\ge 0$. The congruence \eqref{eq:denominator S_3} holds for $n=0$ because $S_3^{(-k)}(0)=1$ for any $k\ge 1$. We assume that $n\ge 1$. By \eqref{eq:m=2 theorem negative}, the fractional part of $S_3^{(-k)}(n)$ is \begin{equation}\label{eq:fractionalpart} n(2^k-1)(-1)^{n-1}B_{n-1} -(n-1)B_n. \end{equation} If $n$ is odd, then $(-1)^{n-1}B_{n-1}=B_{n-1}$ and $(n-1)B_n =0$. Thus we have $S_3^{(-k)}(n) \equiv n(2^k-1)B_{n-1}$ (mod $1$). If $n \ge 4$ is even, then $B_{n-1}=0$. Thus we have $S_3^{(-k)}(n) \equiv -(n-1)B_{n}$ (mod$1$) for even $n\ge 4$. This congruence also holds for $n=2$ because the first term of \eqref {eq:fractionalpart} becomes $2^k-1 \in \Z$, and this completes the proof of \eqref{eq:denominator S_3}. \end{proof} \begin{remark} For $m\ge 4$ we may give explicit formulas of $S_m^{(k)}(n)$ by the method similar to the proof of Theorem \ref{thm:m=1 theorem} and Theorem \ref{thm:m=2 theorem}. However, these formulas seem to be complicated to describe. \end{remark} \begin{thebibliography}{1} \bibitem{AD2007} T.~Agoh and K.~Dilcher, Convolution identities and lacunary recurrences for Bernoulli numbers, {\em J. Number Theory} {\bf 124} (2007), 105--122. \bibitem{AD2009} T.~Agoh and K.~Dilcher, Higher-order recurrences for Bernoulli numbers, {\em J. Number Theory} {\bf 129} (2009), 1837--1847. \bibitem{AK} T.~Arakawa and M.~Kaneko, On poly-Bernoulli numbers, {\em Comment. Math. Univ. St. Pauli} {\bf 48} (1999), 159--167. \bibitem{AK_Nagoya} 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{B} C.~Brewbaker, A combinatorial interpretation of the poly-Bernoulli numbers and two Fermat analogues, {\em Integers} {\bf 8} (2008), $\sharp$A02. \bibitem{Chen} K.-W.~Chen, Sums of products of generalized Bernoulli polynomials, {\em Pacific J. Math.} {\bf 208} (2003), 39--52. \bibitem{D} K.~Dilcher, Sums of products of Bernoulli numbers, {\em J. Number Theory} {\bf 60} (1996), 23--41. \bibitem{E Trans} M.~Eie, A note on Bernoulli numbers and Shintani generalized Bernoulli polynomials, {\em Trans. Amer. Math. Soc.} {\bf 348} (1996), 1117--1136. \bibitem{H-W} G.~H.~Hardy and E.~M.~Wright, {\em An Introduction to the Theory of Numbers}, 6th edition, Oxford Univ. Press, 2008. \bibitem{K} M.~Kaneko, Poly-Bernoulli numbers, {\em J. Th\'{e}or. Nombres Bordeaux} {\bf 9} (1997), 221--228. \bibitem{L} S.~Launois, Rank $t$ ${\cal H}$-primes in quantum matrices, {\em Comm. Algebra} {\bf 33} (2005), 837--854. \bibitem{Ma} T.~Machide, Sums of products of Kronecker's double series, {\em J. Number Theory} {\bf 128} (2008), 820--834. \bibitem{Pe} A.~Petojevi\'{c}, New sums of products of Bernoulli numbers, {\em Integral Transforms Spec. Funct.} {\bf 19} (2008), 105--114. \bibitem{PS} A.~Petojevi\'{c} and H.~M.~Srivastava, Computation of Euler's type sums of the products of Bernoulli numbers, {\em Appl. Math. Lett.} {\bf 22} (2009), 796--801. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B68; Secondary 11B73. \noindent \emph{Keywords: } poly-Bernoulli numbers, sums of products. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A027649}, \seqnum{A027650}, and \seqnum{A027651}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received September 27 2009; revised version received April 17 2010. Published in {\it Journal of Integer Sequences}, April 17 2010. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}