<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}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.9in}

\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}

\begin{center}
\vskip 1cm{\LARGE\bf Bernoulli Numbers and a New Binomial \\
\vskip .1in
Transform Identity}
\vskip 1cm
\large
H.  W. Gould\\
Department of Mathematics\\
West Virginia University\\
Morgantown, WV 26506\\
USA\\
\href{mailto:gould@math.wvu.edu}{\tt gould@math.wvu.edu} \\
\ \\
Jocelyn Quaintance\\
Department of Mathematics\\
Rutgers University\\
Piscataway, NJ 08854\\
USA\\
\href{mailto:quaintan@math.rutgers.edu}{\tt quaintan@math.rutgers.edu}
\end{center}

\vskip .2 in

\begin{abstract} 
Let $(b_n)_{n \geq 0}$ be the binomial transform of
$(a_n)_{n \geq 0}$.
We show how a binomial transformation identity of Chen proves  a symmetrical Bernoulli number identity attributed to Carlitz.  We then modify Chen's identity to prove a new binomial transformation identity.
\end{abstract}

\vskip .2 in

Carlitz \cite{lc} posed as a problem the remarkable symmetric Bernoulli number identity
\begin{align}\label{eq1}
(-1)^m\sum_{k=0}^m{m\choose k}B_{n+k} &amp;= (-1)^n\sum_{k=0}^n{n\choose k}B_{m+k},
\end{align}
valid for arbitrary $m, n\geq 0$.  The published solution by Shannon \cite{ags} used mathematical induction on $m$ and $n$.  The identity was rediscovered recently by Vassilev and Vassilev-Missana \cite{vm}, but stated in the form
\begin{align}\label{eq2}
(-1)^m\sum_{k=0}^{m-1}{m\choose k}B_{n+k} &amp;= (-1)^n\sum_{k=0}^{n-1}{n\choose k}B_{m+k},
\end{align}
valid for arbitrary positive integers $m$ and $n$.  Identity (\ref{eq2}) is equivalent to Identity (\ref{eq1}) since $\left[(-1)^m - (-1)^n\right]B_{m+n} = 0$.  Their proof used the symmetry of a function $f_k(x,y)$ involving Bernoulli numbers introduced in a separate paper \cite{vm1}.  They give no reference to Carlitz's or to Shannon's proof.

An alternative proof of Equation (\ref{eq1}) is derived through an application of a binomial transformation identity discovered by Chen \cite{kwc}.  
Let $(a_n)$ be any sequence of numbers,
and define the binomial transform of $(a_n)$ to be the sequence $(b_n)$,
where $b_n = \sum_{k=0}^n{n\choose k}a_k$.
A corollary of  \cite[Thm.\ 2.1] {kwc} is
\begin{align}\label{eq3}
\sum_{k=0}^m{m\choose k}a_{n+k} &amp;= \sum_{k=0}^n(-1)^{n-k}{n\choose k}b_{m+k}.
\end{align}
The Bernoulli numbers satisfy the recurrence $\sum_{k=0}^n{n\choose k}B_k = (-1)^nB_n$ for $n\geq 0$.  Setting $a_k = B_k$, we then have $b_n = (-1)^nB_n$, so that Equation (\ref{eq3}) becomes
\begin{align*}
\sum_{k=0}^m{m\choose k}B_{n+k} = \sum_{k=0}^n(-1)^{n-k}{n\choose k}(-1)^{m+k}B_{m+k},
\end{align*}
which is precisely Identity (\ref{eq1}) of Carlitz.

Chen's proof of Equation (\ref{eq3}) relies on certain properties of Seidel matrices.  We present a direct proof which relies on the hypergeometric identity
\begin{align}\label{eq4}
\sum_{k=0}^m(-1)^k{m\choose k}{x+k\choose r} &amp;= (-1)^m{x\choose r-m};
\end{align}
see \cite[Identity 3.47, p.\ 27]{hwg}.  In Equation (\ref{eq4}) we require that $m$ and $r$ be nonnegative integers and $x$ be a complex number.

Since the binomial transform inverts to give $a_n = \sum_{k=0}^n(-1)^{n-k}{n\choose k}b_k$ we find that
\begin{align*}
\sum_{k=0}^m{m\choose k}a_{n+k} &amp;= \sum_{k=0}^m{m\choose k}\sum_{j=0}^{n+k}(-1)^{n+k-j}{n+k\choose j}b_j\\
&amp;= \sum_{j=0}^{n+m}(-1)^{-j}b_j\sum_{k=j-n}^m(-1)^{n+k}{m\choose k}{n+k\choose j}\\
&amp;= \sum_{j=0}^{n+m}(-1)^{-j}b_j\sum_{k=0}^m(-1)^{n+k}{m\choose k}{n+k\choose j}\\
&amp;= \sum_{j=0}^{n+m}(-1)^{n+m-j}{n\choose j-m}b_j = \sum_{j=0}^n(-1)^{n-j}{n\choose j}b_{j+m}.
\end{align*}
A careful analysis of this preceding proof yields a short proof of 
\cite[Thm.\ 3.2]{kwc}, where Chen relies on lengthy induction arguments.  We will instead use Equation (\ref{eq4}).
\begin{theorem}\cite[Thm.\ 3.2]{kwc} Let $b_n$ be the binomial transform of $a_n$.  Then 
\begin{align}\label{eq5}
\sum_{k=0}^m{m\choose k}{n+k\choose s}a_{n+k-s} &amp;= \sum_{k=0}^n(-1)^{n-k}{n\choose k}{m+k\choose s}b_{m+k-s},
\end{align}
for arbitrary nonnegative $m, n,$ and $s$. 
\end{theorem}
\begin{proof}
 By definition $b_n = \sum_{k=0}^n{n\choose k}a_k$.  This implies that $a_n = \sum_{k=0}^n(-1)^{n-k}{n\choose k}b_k$.  Hence
\begin{align*}
\sum_{k=0}^m{m\choose k}{n+k\choose s}a_{n+k-s} &amp;= \sum_{k=0}^m{m\choose k}{n+k\choose s}\sum_{j=0}^{n+k-s}(-1)^{n+k-s-j}{n+k-s\choose j}b_j\\
&amp;= \sum_{j=0}^{n+m-s}(-1)^{n-s-j}b_j\sum_{k=0}^m(-1)^k{m\choose k}{n+k\choose s}{n+k-s\choose j}\\
&amp;= \sum_{j=0}^{n+m-s}(-1)^{n-s-j}{s+j\choose s}b_j\sum_{k=0}^m(-1)^k{m\choose k}{n+k\choose s+j}\\
&amp;= \sum_{j=m-s}^{n+m-s}(-1)^{m+n-j-s}{s+j\choose s}{n\choose j+s-m}b_j\\
&amp;= \sum_{j=0}^n(-1)^{n-j}{n\choose j}{m+j\choose s}b_{m+j-s},
\end{align*}
where the fourth equality follows by Equation (\ref{eq4}).
\end{proof}

Equation (\ref{eq5}) allows us to establish a generalization of the curious formula
\begin{align}\label{eq6}
\sum_{k=0}^n{n\choose k}{n+k\choose n}x^k &amp;= \sum_{k=0}^n(-1)^{n-k}{n\choose k}{n+k\choose n}(1+x)^k,
\end{align}
discovered by Simons \cite{ss}.  A quick proof of this was given
by Gould \cite{hwg1} using elementary properties of Legendre polynomials.  Instead, choose $a_n = x^n$ for all $n\geq 0$.  Then  $b_n = (1+x)^n$ and Identity (\ref{eq5}) tells us that
\begin{align*}
\sum_{k=0}^m{m\choose k}{n+k\choose s}x^{n+k-s} &amp;= \sum_{k=0}^n(-1)^{n-k}{n\choose k}{m+k\choose s}(1+x)^{m+k-s}.
\end{align*}
Letting $m = s = n$ recovers Identity (\ref{eq6}).

Through an induction argument Chen proves 
\begin{theorem}\cite[Thm.\ 3.1]{kwc} Let $b_n$ be the binomial transform of $a_n$.  Then
\begin{align}\label{eq7}
\sum_{k=0}^m\frac{{m\choose k}}{{n+k+s\choose s}}a_{n+k+s} &amp;= \sum_{k=0}^n(-1)^{n-k}\frac{{n\choose k}}{{m+k+s\choose s}}b_{m+k+s}\notag\\
&amp;+ \sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}{s-1-j\choose i}{s-1\choose j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i){m+n+i\choose n}},
\end{align}
where $m$, $n$, and $s$ are nonnegative integers.  
\end{theorem}
If we use Equation (\ref{eq4}) and the following hypergeometric identity attributed to Frisch \cite{rf}, \cite[p.\ 337]{en},
\begin{align}\label{eq8}
\sum_{k=0}^n(-1)^k{n\choose k}\frac{1}{{b+k\choose c}} &amp;= \frac{c}{n+c}\frac{1}{{n+b\choose b-c}},\qquad b\geq c &gt; 0,
\end{align}
\cite[Identity 4.2, p.\ 46]{hwg}, we are able to prove the following new binomial transformation identity.
\begin{theorem} Let $b_n$ be the binomial transform of $a_n$.  Let $m$, $n$, and $s$ be nonnegative integers.  Then
\begin{align}\label{eqA}
\sum_{j=0}^s\frac{{s\choose j}a_j}{(m+n+s+1-j){m+n+s-j\choose m}} &amp;= \sum_{j=0}^s\frac{(-1)^{s-j}{s\choose j}b_j}{(m+n+s+1-j){m+n+s-j\choose n}}.
\end{align}
\end{theorem}
\begin{proof}
By definition $b_n = \sum_{k=0}^n{n\choose k}a_k$.  Hence $a_n = \sum_{k=0}^n(-1)^{n-k}{n\choose k}b_k$ and 
\begin{align*}
&amp;\sum_{k=0}^m\frac{{m\choose k}}{{n+k+s\choose s}}a_{n+k+s} = \sum_{k=0}^m\frac{{m\choose k}}{{n+k+s\choose s}}\sum_{j=0}^{n+k+s}(-1)^{n+k+s-j}{n+k+s\choose j}b_j\\
&amp;= \sum_{j=0}^{m+n+s}(-1)^{n+s-j}b_j\sum_{k=0}^m(-1)^k\frac{{m\choose k}}{{n+k+s\choose s}}{n+k+s\choose j}\\
&amp;= \sum_{j=s}^{m+n+s}(-1)^{n+s-j}\frac{b_j}{{j\choose s}}\sum_{k=0}^m(-1)^k{m\choose k}{n+k\choose j-s} +  \sum_{j=0}^{s-1}(-1)^{n+s-j}b_j\sum_{k=0}^m(-1)^k\frac{{m\choose k}}{{n+k+s\choose s}}{n+k+s\choose j}\\
&amp;= \sum_{j=s+m}^{m+n+s}(-1)^{m+n+s-j}\frac{{n\choose j-s-m}}{{j\choose s}}b_j +  \sum_{j=0}^{s-1}(-1)^{n+s-j}b_j\sum_{k=0}^m(-1)^k\frac{{m\choose k}}{{n+k+s\choose s}}{n+k+s\choose j}\\
&amp;= \sum_{j=0}^n(-1)^{n-j}\frac{{n\choose j}}{{m+j+s\choose s}}b_{m+j+s} +  \sum_{j=0}^{s-1}(-1)^{n+s-j}b_j\sum_{k=0}^m(-1)^k\frac{{m\choose k}}{{n+k+s\choose s}}{n+k+s\choose j}\\
&amp;= \sum_{j=0}^n(-1)^{n-j}\frac{{n\choose j}}{{m+j+s\choose s}}b_{m+j+s} + \sum_{j=0}^{s-1}(-1)^{n+s-j}{s\choose j}b_j\sum_{k=0}^m(-1)^k\frac{{m\choose k}}{{n+k+s-j\choose s-j}}\\
&amp;=  \sum_{j=0}^n(-1)^{n-j}\frac{{n\choose j}}{{m+j+s\choose s}}b_{m+j+s} + \sum_{j=0}^{s-1}(-1)^{n+s-j}\frac{(s-j){s\choose j}}{(m+s-j){m+n+s-j\choose n}}b_j\\
&amp;=  \sum_{j=0}^n(-1)^{n-j}\frac{{n\choose j}}{{m+j+s\choose s}}b_{m+j+s} + \sum_{j=0}^{s-1}(-1)^{n+s-j}\frac{s{s-1\choose j}}{(m+n+s-j){m+n+s-j-1\choose n}}b_j.
\end{align*}
The fourth line follows from Equation (\ref{eq4}) while the seventh follows from Equation (\ref{eq8}).

In summary, we have shown that
\begin{align}\label{eq9}
\sum_{k=0}^m\frac{{m\choose k}}{{n+k+s\choose s}}a_{n+k+s} &amp;=  \sum_{j=0}^n\frac{(-1)^{n-j}{n\choose j}}{{m+j+s\choose s}}b_{m+j+s} + \sum_{j=0}^{s-1}\frac{(-1)^{n+s-j}s{s-1\choose j}}{(m+n+s-j){m+n+s-j-1\choose n}}b_j,
\end{align}
If we compare Identity (\ref{eq7}) to Identity (\ref{eq9}), we conclude that
\begin{align}\label{eq10}
\sum_{j=0}^{s-1}&amp;\frac{(-1)^{n+s-j}s{s-1\choose j}}{(m+n+s-j){m+n+s-j-1\choose n}}b_j \notag\\
&amp;= \sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}{s-1-j\choose i}{s-1\choose j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i){m+n+i\choose n}}.
\end{align}
Equation (\ref{eq10}) can be furthered simplified by applying Equation (\ref{eq8}).  In particular,
\begin{align*}
 \sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}{s-1-j\choose i}&amp;{s-1\choose j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i){m+n+i\choose n}}\\
  &amp;= \sum_{j=0}^{s-1}(-1)^{n+1}\frac{s}{n+1}{s-1\choose j}a_j\sum_{i=0}^{s-1-j}(-1)^i\frac{{s-1-j\choose i}}{{m+n+1+i\choose n+1}}\\
&amp;= \sum_{j=0}^{s-1}(-1)^{n+1}\frac{s}{n+1}{s-1\choose j}a_j\frac{n+1}{n+s-j}\frac{1}{{m+n+s-j\choose m}}\\
&amp;= (-1)^{n+1}\sum_{j=0}^{s-1}\frac{s{s-1\choose j}}{(n+s-j){m+n+s-j\choose m}}a_j.
\end{align*}
These calculations show that Equation (\ref{eq10}) is equivalent to
\begin{align}
-\sum_{j=0}^{s-1}\frac{{s-1\choose j}}{(n+s-j){m+n+s-j\choose m}}a_j &amp;= \sum_{j=0}^{s-1}(-1)^{s-j}\frac{{s-1\choose j}}{(m+n+s-j){m+n+s-j-1\choose n}}b_j.
\end{align}
Set $s\rightarrow s+1$ to obtain 
\begin{align}\label{eqA1}
\sum_{j=0}^s\frac{{s\choose j}a_j}{(n+s+1-j){m+n+s+1-j\choose m}} &amp;= \sum_{j=0}^s\frac{(-1)^{s-j}{s\choose j}b_j}{(m+n+s+1-j){m+n+s-j\choose n}}.
\end{align}
Since $(n+s+1-j){m+n+s+1-j\choose m} = (m+n+s+1-j){m+n+s-j\choose m}$, we see that Equation (\ref{eqA1}) is equivalent to Equation (\ref{eqA}).
\end{proof}
\begin{thebibliography}{99}
\bibitem{lc} L. Carlitz, Problem 795, {\it Math. Mag.} {\bf 44} (1971), 107.
\bibitem{ags} A. G. Shannon, Solution of Problem 795, {\it Math. Mag.} {\bf 45} (1972), 55--56.
\bibitem{kwc} K. W. Chen, Identities from the binomial transform, {\it J. Number Theory} {\bf 124} (2007), 142--150.
\bibitem{rf} R. Frisch, Sur les semi-invariants et moments employ\'{e}s dans l'\'{e}tude des distributions
statistiques, {\it Skrifter utgitt av Det Norske Videnskaps-Akademi i Oslo, II}.
Historisk-Filosofisk Klasse, 1926, No. 3, 87 pp.  
\bibitem{en} Eugen Netto, {\it Lehrbuch der Combinatorik}, 2nd edition, 1927.
Reprinted by Chelsea, 1958.
\bibitem{hwg} H. W. Gould, {\it Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations}, revised edition.  Published by the author, Morgantown, WV, 1972.
\bibitem{hwg1} H. W. Gould, A curious identity which is not so curious. {\it Math. Gaz.} {\bf 88} (2004), 87.
\bibitem{ss} S. Simons, A curious identity, {\it Math. Gaz.} {\bf 85} (2001), 296--298.
\bibitem{vm1} P. Vassilev and M. Vassilev-Missana, On the sum of equal powers of the first $n$ terms of an arbitrary arithmetic progression, {\it Notes on Number Theory and Discrete Mathematics} {\bf 11} (2005), 15--21.
\bibitem{vm} P. Vassilev and M. Vassilev-Missana, On one remarkable identity involving Bernoulli numbers, {\it Notes on Number Theory and Discrete Mathematics} {\bf 11} (2005), 22--24.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 05A10, 11B65.

\noindent \emph{Keywords: } 
Bernoulli number, binomial transform.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A027641} and
\seqnum{A027642}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 2 2013;
revised version received  January 3 2014.
Published in {\it Journal of Integer Sequences}, January 3 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>