\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}{-.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} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Note on the Convolution of \\ \vskip .12in Binomial Coefficients } \vskip 1cm \large Rui Duarte \\ Center for Research and Development in Mathematics and Applications\\ Department of Mathematics\\ University of Aveiro \\ Portugal \\ \href{mailto:rduarte@ua.pt}{\tt rduarte@ua.pt} \\ \ \\ Ant\'onio Guedes de Oliveira \\ CMUP and Department of Mathematics\\ Faculty of Sciences\\ University of Porto\\ Portugal\\ \href{mailto:agoliv@fc.up.pt}{\tt agoliv@fc.up.pt} \end{center} \vskip .2 in \newcommand\N{\mathbb{N}} \newcommand\R{\mathbb{R}} \begin{abstract} We prove that, for every integer $a$, real numbers $k$ and $\ell$, and nonnegative integers $n$, $i$ and $j$, $$ \sum_{i+j=n} {a\,i+k-\ell\choose i} {a\,j+\ell\choose j} = \sum_{i+j=n} {a\,i+k\choose i} {a\,j\choose j}, $$ by presenting explicit expressions for its value. We use the identity to generalize a recent result of Chang and Xu, and end the paper by presenting, in explicit form, the ordinary generating function of the sequence $\left\{{2n+k\choose n}\right\}_{n=0}^\infty$, where $k\in\R$. \end{abstract} \section{Introduction}\label{int} We consider the sequence $\big\{{a\,n+k\choose n}\big\}_{n=0}^\infty$, where, following the notation of \cite{RS}, for every real $\ell$ and every nonnegative integer $i$, $\big(\ell\big)_i=\ell(\ell-1)\cdots (\ell-i+1)$ is the falling factorial and ${\ell\choose i}=\frac{(\ell)_i}{ i!}$. Temporarily, we consider $k=0$ and take the convolution of this sequence with itself, defined by $C_a(n)=\sum_{i+j=n}{a\,i\choose i} {a\,j\choose j}$. When $a=2$, the former is sequence \seqnum{A000984} of \cite{oeis}, of the central binomial coefficients, and the latter is sequence \seqnum{A000302}, of the powers of $4$. In other words, \begin{equation}\label{mainId} C_2(n) \; = \; \sum_{i+j=n}{2\,i\choose i} {2\,j\choose j} \; = \; 4^n \, . \end{equation} In fact, we can prove directly \eqref{mainId} using, similarly to what we do here (see \cite{part} for details), the inclusion-exclusion principle after using identity \eqref{eqnova} for $a=2$, that is, the fact that $$\sum_{i+j=n} {2i\choose i} {2j\choose j} =\sum_{i+j=n} {2i-\ell\choose i} {2j+\ell\choose j}\,.$$ % In fact, we can prove directly \eqref{mainId} using \eqref{eqnova} % for $a=2$, and then the inclusion-exclusion principle, similarly to % what we do here (see \cite{part} for details). Note that \begin{equation} 2 \, C_2(n) = 2^{2n+1} = \sum_{i=0}^{2n+1} {2n+1\choose i} = 2 \sum_{i=0}^n {2n+1\choose i} \, . \label{eq3} \end{equation} For another identity, define as usual $[n]=\{1,\ldots,n\}$ for any natural number $n$ and consider the collection of the subsets of $[2n]$ with more than $n$ elements and with the same $(n+1)$-th element, $p$, say. Note that $p=n+1+i$ for some $i=0,\ldots,n-1$ and that there are ${n+i\choose n}\,2^{n-i-1}$ subsets in the collection. It follows that the number of all subsets of $[2n]$ is \begin{equation} C_2(n) = 2^{2n} = 2 \sum_{i=0}^{n-1} 2^{n-i-1} {n+i\choose i} + {2n\choose n} = \sum_{i=0}^n 2^{n-i} {n+i\choose i} \, . \label{eq4} \end{equation} We generalize these identities, namely \eqref{eq3} and \eqref{eq4}. When $a=3$ and $a=4$, we have sequences \seqnum{A006256} and \seqnum{A078995} of \cite{oeis}, and no such simple formulas for $C_3(n)$ and $C_4(n)$ are known as in case $a=2$. For these sequences, we obtain, for every real $\ell$, \begin{align*} \sum_{i+j=n} {3i\choose i} {3j\choose j} & = \sum_{i+j=n} 2^i {3n+1\choose j} = \sum_{i+j=n} 3^i {2n+j\choose j} = \sum_{i+j=n} {3i-\ell\choose i} {3j+\ell\choose j} \, , \\ \sum_{i+j=n} {4i\choose i} {4j\choose j} & = \sum_{i+j=n} 3^i {4n+1\choose j} = \sum_{i+j=n} 4^i {3n+j\choose j} = \sum_{i+j=n} {4i-\ell\choose i} {4j+\ell\choose j} \, . \end{align*} More generally, we obtain the following theorem. \begin{theorem}\label{thm} For every nonnegative integers $i$, $j$ and $n$, and for every real numbers $k$ and $\ell$, \begin{align} \sum_{i+j=n} {a\,i+k-\ell\choose i} {a\,j+\ell\choose j} & = \sum_{i+j=n} {a\,i+k\choose i} {a\,j\choose j} \label{eqnova} \\ & = \sum_{i=0}^n (a-1)^{n-i} \, {a\,n+k+1\choose i} \label{eq6} \\ & = \sum_{i=0}^n a^{n-i} \, {(a-1)n+k+i\choose i} \label{eq7} \end{align} where we take $0^0=1$. \end{theorem} Gould \cite{gould} proves \eqref{eqnova} (together with the Rothe-Hagen identity) using generating functions, and asks for proofs ``using finite series, the method of finite differences, or otherwise''. So, part of this paper can be seen as an extension of a recent paper by Chu \cite{chu}. Coming back to the case where $a=2$, we consider the convolution of more than two copies of the sequence of central binomial coefficients: given integers $n\geq0$ and $t>0$, define $$P_t(n) = \sum_{i_1 + \cdots + i_t = n} {2i_1\choose i_1}{2i_2\choose i_2} \cdots {2i_t\choose i_t}.$$ where $i_1,\ldots,i_t$ are nonnegative integers. Recently, Chang and Xu showed \cite{ChX}, with a probabilistic proof, that $P_t(n)$ depends only on $n$ and $t$. We give here a proof of combinatorial nature of this fact and obtain a generalization that also includes \eqref{mainId} as a special case. Namely, we prove the following theorem, that is based on the results of Chang and Xu \cite{ChX}. \begin{theorem}\label{newgen} Let %$i_1,\ldots,i_n$ be any nonnegative integers and let $\ell_1, \ldots, \ell_t$ be any real numbers such that $\ell_1 + \cdots + \ell_t = 0$. Then \begin{equation}\label{neq1} \sum_{i_1 + \cdots + i_t = n} {2i_1 + \ell_1\choose i_1}{2i_2 + \ell_2\choose i_2} \cdots {2i_t + \ell_t\choose i_t} = 4^n {n+\frac{t}{2}-1\choose n}\,, \end{equation} where $i_1,\ldots,i_t$ are nonnegative integers. \end{theorem} Finally, in Section~\ref{sec3} we obtain formulas for the generating functions of the sequences involved in these identities. \section{General case} For the proof of Theorem~\ref{thm} we need some technical results. \begin{lemma}\label{lemma1} Let, for any real $\ell$ and integers $a$ and $n$ such that $n\geq0$, \begin{align*} & S_{a,\ell}(n) \; = \; \sum_{i=0}^n (-1)^i {\ell-(a-1)i\choose i} {\ell-a\,i\choose n-i} \, . \\ \intertext{Then} \\ & \sum_{p=0}^n {n\choose p} \, S_{a,\ell}(p) \; = \; S_{a+1,\ell+n}(n) \, . \end{align*} \end{lemma} \begin{proof} \begin{align*} \sum_{p=0}^n{n\choose p}\,S_{a,\ell}(p) & = \sum_{i=0}^n \left[ (-1)^i {\ell-(a-1)i\choose i} \sum_{p=i}^n {\ell-a\,i\choose p-i} {n\choose p} \right] \\ & = \sum_{i=0}^n \left[ (-1)^i {\ell-(a-1)i\choose i} \sum_{p=i}^n {\ell-a\,i\choose \ell-(a-1)i-p} {n\choose p} \right] \\ & = \sum_{i=0}^n (-1)^i {\ell-(a-1)i\choose i} {\ell+n-a\,i\choose \ell-(a-1)\,i} \\ & = \sum_{i=0}^n (-1)^i {(\ell+n)-a\,i\choose i \ , \ n-i \ , \ \ell-ai} \\ & = \sum_{i=0}^n (-1)^i {(\ell+n)-a\,i\choose i} {(\ell+n)-(a+1)\,i\choose n-\,i} \, , \end{align*} where we use Vandermonde's convolution in the third equality. \end{proof} \begin{lemma}\label{lemma2} With the notation of the previous lemma, $$S_{a,\ell}(n)\;=\;(a-1)^n.$$ \end{lemma} \begin{proof}[First proof] First note that we may assume that $\ell$ is a natural number, since $S_{a,\ell}(n)$ is a polynomial in $\ell$, and thus is constant. Now, suppose that fixed $a$, there exist $x$ such that for all $\ell$ and $p$, $S_{a,\ell}(p)=x^p$. Then, from Lemma~\ref{lemma1} it follows that $S_{a+1,\ell+n}(n)=(1+x)^n$. Hence, all we must prove is that $S_{a,\ell}(n)=0$ when $a=1$ and $\ell\in\N$. For this purpose, define $\mathcal{A}$ as the set of $n$-subsets of the set $[\ell]=\{1,2,\ldots,\ell\}$ and let $\mathcal{A}_i$ be the set of elements of $\mathcal{A}$ that do not contain $i$, for $i\in[\ell]$. Then, on the one hand, $|\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_\ell| = {\ell\choose p}$ and on the other hand, by the inclusion-exclusion principle, $$ |\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_\ell| = \sum_{i=1}^\ell (-1)^{i+1} \bigg( \sum_{1 \leq j_1 < \cdots < j_i \leq \ell} |\mathcal{A}_{j_1} \cap \cdots \cap \mathcal{A}_{j_i}| \bigg) \, . $$ The proof is completed by noting that, for every integers $j_1,\ldots, j_i$ such that $1 \leq j_1 < \cdots < j_i \leq p$, $|\mathcal{A}_{j_1} \cap \cdots \cap \mathcal{A}_{j_i}| = {\ell-i\choose n-i}$. \end{proof} \begin{proof}[Second proof] Notice first that, for any function $f$, we have by induction that \begin{equation} \label{formula1} (\Delta^n f)(x) = \sum_{i=0}^n (-1)^i {n\choose i} f(x+n-i) \end{equation} where $\Delta$ is the forward-difference operator defined by $(\Delta f)(x)=f(x+1)-f(x)$, and $\Delta^n$ denotes the operator $\Delta$ applied successively $n$ times. In addition, if $f$ is a polynomial, the degree reduces each time we apply $\Delta$ and, if the degree of $f$ is $n$, the left-hand side of \eqref{formula1} must be a constant which equals $(n!)a_n$, where $a_n$ is the coefficient of $x^n$ in $f$. In particular, $$\sum_{i=0}^n (-1)^i {n\choose i} f(n-i) = n!$$ if $f$ is a monic polynomial of degree $n$. Now, $S_{a,\ell}(n)$ can be rewritten as $\sum_{i=0}^n (-1)^i {n\choose i} {\ell-(a-1)i\choose n}$. If $a=1$, the identity is clearly satisfied. When $a \neq 1$, the identity can be written as $$ \sum_{i=0}^n (-1)^i {n\choose i} \left( \frac{\ell}{a-1} -i \right) \left( \frac{\ell-1}{a-1} -i \right) \cdots \left( \frac{\ell - (n-1)}{a-1} -i \right) = n! \, . $$ If $f$ is the monic polynomial of degree $n$ defined as $$ f(x)=\left( \frac{\ell}{a-1} -n+x \right) \left( \frac{\ell-1}{a-1} -n+x \right) \cdots \left( \frac{\ell - (n-1)}{a-1} -n+x \right) \, , $$ then clearly $$ \sum_{i=0}^n (-1)^i {n\choose i} f(n-i) \; = \; (\Delta^n f)(0) \; = \; n! \, , $$ which proves the identity. \end{proof} \begin{lemma}\label{lemma3} Let $n$ be a nonnegative integer and $s$ and $t$ be two real numbers. Then $${s+t+1\choose n} = \sum_{i=0}^n {s-i\choose n-i} {t+i\choose i}.$$ \end{lemma} \begin{proof}[First proof] Let $\ F(n,i) = {s-i\choose s-n} {t+i\choose i} \ $. By Zeilberger's algorithm \cite{npwz,aeqb}, as implemented by Paule and Schorn \cite{pp} and by Krattenthaler \cite{krat}, we know that $T(n) = \sum_{i=0}^n F(n,i)$ verifies \begin{equation} \label{zeil} (s+t+1-n)\,T(n) - (n+1)\,T(n+1) = 0, \end{equation} which is also verified by $T(n) = {s+t+1\choose n}$, and for both it holds $T(0)=1$. In fact, we can see that for every $i$ with $0 \leq i \leq n+1$, \begin{equation*} (s+t+1-n)\,F(n,i) - (n+1)\,F(n+1,i) = G(n,i+1)-G(n,i) \end{equation*} with $G(n,i) = i\,{s+1-i\choose s-n} {t+i\choose i}$. Hence, \eqref{zeil} holds since $F(n,n+1)=G(n,n+2)=G(n,0)=0$. \end{proof} \begin{proof}[Second proof] Since ${t+i\choose i}=(-1)^i{-t-1\choose i}$ and by Vandermonde's convolution, \begin{align*} \sum_{i=0}^n {s-i\choose n-i} {t+i\choose i} & = (-1)^n \sum_{i=0}^n {-t-1\choose i} {-s-1+n\choose n-i} \\ & = (-1)^n {-t-s+n-2\choose n} \\ & = {t+s+1\choose n} \, . \qedhere \end{align*} \end{proof} \begin{lemma}\label{lemma4} For all nonnegative integers $i$ and $n$, and for all real numbers $a$ and $b$, $$ \sum_{i=0}^n a^{n-i} \, {(b-1)n+k+i\choose i} = \sum_{i=0}^n (a-1)^{n-i} \, {b\,n+k+1\choose i}.$$ \end{lemma} \begin{proof} \begin{align*} \sum_{i=0}^n a^{n-i} \, {(b-1)n+k+i\choose i} & = \sum_{i=0}^n \left[ \sum_{j=0}^{n-i} (a-1)^{n-j} {n-i\choose j} \right] {(b-1)n+k+i\choose i} \\ & = \sum_{j=0}^n (a-1)^{n-j} \left[ \sum_{i=0}^{n-j} {n-i\choose n-j-i} {\big((b-1)n+k\big)+i\choose i} \right] \, . \end{align*} The result now follows from Lemma~\ref{lemma3}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm}] Let $\mathfrak{S}=\sum_{i+j=n} {a\,i+k-\ell\choose i} {a\,j+\ell\choose j} = \sum_{i+j=n} (-1)^i {\ell-k'-(a-1)i\choose i} {a\,n+\ell-a\,i\choose j}$, with $k'=k+1$. Then, by Vandermonde's convolution, \begin{eqnarray*} \mathfrak{S} & = & \sum_{i+j=n} \left[ (-1)^i {\ell-k'-(a-1)i\choose i} \sum_{p+m=j} {a\,n+k'\choose p} {\ell-k'-a\,i\choose m} \right] \\ & = & \sum_{p=0}^n \left[ {a\,n+k'\choose p} \sum_{i+m=n-p} (-1)^i {\ell-k'-(a-1)i\choose i} {\ell-k'-a\,i\choose m} \right] \, . \end{eqnarray*} Now, \eqref{eq6} follows immediately from Lemma~\ref{lemma2} and \eqref{eq7} follows from \eqref{eq6} and Lemma~\ref{lemma4}. \end{proof} We end this section with a problem based on a new result that, when we represent by $\left(\!\!{n\choose k}\!\!\right)$ the number ${n+k-1\choose k}$ of $k$-multisets of elements of an $n$-set, can be formulated in the following elegant terms. \begin{theorem}\label{thm2} For every real $\ell$ and every nonnegative integer $n$, $$\sum_{i=0}^n (-1)^i \left(\!\!\!{\ell-a\,i\choose i}\!\!\!\right) {\ell-a\,i\choose n-i}= a(a-1)^{n-1}.$$ \end{theorem} \begin{proof} By Pascal's rule, \begin{align*} \sum_{i=0}^n (-1)^i {\ell-1-(a-1)i\choose i} {\ell-a\,i\choose n-i} = & \sum_{i=0}^n (-1)^i {\ell-(a-1)i\choose i} {\ell-a\,i\choose n-i} \\ & - \sum_{i=1}^n (-1)^i {\ell-(a-1)i-1\choose i-1} {\ell-a\,i\choose n-i} \\ = & S_{a, \ell}(n)+S_{a, \ell-a}(n-1) \, . \qedhere \end{align*} \end{proof} \begin{problem} Give a combinatorial proof of Theorem~\ref{thm2}. \end{problem} \section{The case $a=2$: on Chang \& Xu generalization of the identity~\eqref{mainId}} The main result of the article of Chang and Xu \cite{ChX} is a generalization of \eqref{mainId} that we may write as \begin{equation} \label{CXa} \sum_{i_1 + \cdots + i_t=n} {2i_1\choose i_1} {2i_2\choose i_2} \cdots {2i_t\choose i_t} = 4^n {n+\frac{t}{2}-1\choose n}. \end{equation} where $i_1,\ldots,i_n$ are (nonnegative) integers. Let $P_t (n)$ be the left-hand side of \eqref{CXa}, as we defined before. We remark that $P_1 (n) = {2n\choose n}$ and, by \eqref{mainId}, $P_2(n) = 4^n$. Note also that \eqref{CXa} can be obtained using induction and the following lemma. Finally, we observe that $4^n {n+\frac{t}{2}-1\choose n}= \frac{{2n+2 k \choose 2n}}{{n+ k \choose n}} {2n\choose n}= \frac{{2n+2 k \choose n+ k}}{{2 k \choose k }} {n+ k \choose n}$ when $t = 2k+1$. \begin{lemma} For every positive integer $t$ and every nonnegative integer $n$, \begin{equation*} P_{t+2} (n+1) = P_t (n+1) + 4\,P_{t+2} (n) \end{equation*} \end{lemma} \begin{proof} In fact, $P_{t+2} (n+1) = \sum_{j=0}^{n+1} P_2 (n+1-j) P_t (j) = P_t (n+1) + 4 \sum_{j=0}^{n} 4^{n-j} P_t (j)$. \end{proof} Now, we consider again \eqref{eqnova}, the first identity in Theorem~\ref{thm}. From this identity, as a clear consequence, we obtain, for every integer $a$, for every nonnegative integers $t$, $n$ and $i_1, i_2, \ldots, i_t$ and for every real numbers $k, k_1, k_2, \ldots, k_t$ such that $k = k_1+k_2+\cdots+k_t$, the following identity, which is also a clear consequence of statement (2) of Theorem~\ref{GenFun}: $$ \sum_{i_1 + \cdots + i_t = n} {a\,i_1 + k\choose i_1} {a\,i_2\choose i_2} \cdots {a\,i_t\choose i_t} = \sum_{i_1 + \cdots + i_t = n} {a\,i_1+k_1\choose i_1} {a\,i_2 + k_2\choose i_2} \cdots {a\,i_t+k_t\choose i_t} \, . $$ Whence we obtain Theorem~\ref{newgen} as stated in Section~\ref{int}. \section{Generating functions}\label{sec3} In what follows, we denote by $f^{(n)}$ the $n$-th derivative of a function $f$ of one real variable, $g(x) = \sum_{n \geq 0} {2n\choose n} \, x^n$ is the generating function of the central binomial coefficients and $C(x) = \sum_{n \geq 0} \frac{1}{n+1} {2n\choose n} \, x^n$ is the generating function of the Catalan numbers. We remember that $g(x)=\frac{1}{\sqrt{1-4x}}$ and $C(x) = \frac{2}{1 + \sqrt{1-4x}}$. Note that $g' = 2g^3$ and $C' = g \, C^2$. The following lemma can be easily proved by induction on $n$. \begin{lemma} \label{deriv} For every real numbers $t$ and $\ell$ and nonnegative integer $n$, \begin{align*} &\frac{\left( g^t \right)^{(n)}}{n!} = 4^n {n + \frac{t}{2} -1\choose n} g^{t+2n}, \\ &\frac{\left( g \, C^\ell \right)^{(n)}}{n!} = \sum_{i=0}^n {2n-i\choose n-i} {\ell+i-1\choose i} g^{1+2n-i} C^{\ell+i}, \\ &\left( C^\ell \right)^{(n+1)} = \left( \ell \, g \, C^{\ell+1} \right)^{(n)} \, . \end{align*} \end{lemma} Now, by Lemma~\ref{lemma3}, we obtain immediately the following theorem. Note that the second statement is a particular case, but in explicit form, of an identity of Gould \cite[p.\ 86 (9)]{gould}, and that the third statement was proved by Catalan in 1876 \cite[p.\ 62 and Errata]{cat}. \begin{theorem} For every real numbers $t$ and $\ell$, \label{GenFun} \hfill \begin{align*} & g(x)^t = \sum_{n \geq 0} 4^n {n + \frac{t}{2} -1\choose n} x^n, \\ & g(x)\, C(x)^\ell = \sum_{n \geq 0} {2n+\ell\choose n} x^n, \\ & C(x)^\ell = 1 + \sum_{n \geq 1} \frac{\ell}{2n+\ell} {2n+\ell\choose n} x^n \, . \end{align*} \end{theorem} \section{Acknowledgments} We warmly thank the referee for his careful reading of the text, and specially for offering us the second proof of Lemma \ref{lemma2} and of Lemma \ref{lemma3}. We also thank Christian Krattenthaler for drawing our attention to an article of Gould \cite{gould}, and the \emph{On-Line Encyclopedia of Integer Sequences} \cite{oeis}, that was very useful throughout all work. The work of both authors was supported in part by the European Regional Development Fund through the program COMPETE - Operational Program Factors of Competitiveness (``Programa Operacional Factores de Competitividade'') and by the Portuguese government through FCT - Funda\c{c}\~ao para a Ci\^encia e a Tecnologia, under the projects PEst-C/MAT/UI0144/2011 and PEst-C/MAT/UI4106/2011. \begin{thebibliography}{99} \bibitem{cat} E. C. Catalan, \textit{M\'elanges Math\'ematiques}, Vol.\ 2, F. Hayez, Bruxelles, 1887. Available at \url{http://archive.org/details/mlangesmathm02catauoft} . \bibitem{ChX} G. Chang and C. Xu, Generalization and probabilistic proof of a combinatorial identity. \textit{Amer. Math. Monthly} \textbf{118} (2011), 175--177. \bibitem{chu} W. Chu, Elementary proofs for convolution identities of Abel and Hagen-Rothe. \textit{Electron. J. Combin.} \textbf{17} (2010), Note 24. \bibitem{part} R. Duarte and A. Guedes de Oliveira, A short proof of a famous combinatorial identity, preprint, \url{http://arxiv.org/abs/1307.6693}. \bibitem{gould} H. W. Gould, Some generalizations of Vandermonde's convolution. \textit{Amer. Math. Monthly} \textbf{63} (1956), 84--91. \bibitem{krat} C. Krattenthaler, HYP/HYPQ, available at \url{http://tinyurl.com/ld8dre8} . \bibitem{npwz} I. Nemes, M. Petkov\v{s}ek, H. S. Wilf, and D. Zeilberger, How to do Monthly problems with your computer. \textit{Amer. Math. Monthly} \textbf{104} (1997), 505--519. \bibitem{pp} P. Paule and M. Schorn, The Paule/Schorn implementation of Gosper's and Zeilberger's algorithms, available at \url{http://tinyurl.com/kbluqo4} . \bibitem{aeqb} M. Petkov\v{s}ek, H. S. Wilf, and D. Zeilberger, {$\mathit{A=B}$}, A K Peters, 1996. \bibitem{oeis} N. J. A. Sloane, \textit{The On-Line Encyclopedia of Integer Sequences}, published electronically at \url{http://oeis.org}, 2011. \bibitem{RS} R. Stanley, \textit{Enumerative Combinatorics}, Volume 1, Cambridge University Press, 1997. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 05A19; Secondary 05A10,05A15. \noindent \emph{Keywords: } combinatorial identity, binomial coefficient, Vandermonde's convolution, WZ-method. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000302}, \seqnum{A000984}, \seqnum{A006256}, and \seqnum{A078995}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received February 8 2013; revised version received March 8 2013; July 26 2013; August 21 2013. Published in {\it Journal of Integer Sequences}, August 22 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}