\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}

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

\begin{center}
\vskip 1cm{\LARGE\bf A Proof of a Famous Identity Concerning the Convolution of the Central Binomial Coefficients \\
\vskip 1cm}
\large
Jovan Miki\'{c}\\
J. U. S\v{S}C ``Jovan Cviji\'{c}''\\
74480 Modri\v{c}a\\
Bosnia and Herzegovina\\
\href{mailto:jnmikic@gmail.com}{\tt jnmikic@gmail.com} \\
\end{center}

\vskip .2 in

\begin{abstract}
We give an elementary proof of a famous binomial identity by using
recurrence relations and telescoping. We also prove the alternating
version of that identity and one of its generalizations.
\end{abstract}

\section{Introduction}
We consider a famous identity concerning
the convolution of the central binomial coefficients:
\begin{equation}
\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}= 4^n\text{,}\label{I:1}
\end{equation}
where $n$ is a nonnegative integer.

The identity (\ref{I:1}) can be easily proved with help of generating
functions (see \cite[Exercises 1.2.c, 1.4.a]{RS} and
\cite[Example 4.3]{TC}). Also, there are many combinatorial
proofs of Eq.~(\ref{I:1}). For example, Sved \cite{MASV} gave
a proof using a path-counting argument, and De Angelis \cite{VA}
gave another by considering signed permutations.
Petkov\v{s}ek, Wilf, and Zeilberger gave a proof using
the WZ method \cite{MPHWDZ}. Chang and Xu
\cite{GCX} gave a probabilistic proof of a generalization of
Eq.~(\ref{I:1}). Duarte and Guedes de Oliveira \cite{RDAG1}
gave an elementary proof of a different generalization of 
Eq.~(\ref{I:1}); their proof relies on the inclusion-exclusion principle.
The same authors \cite{RDAG3} also gave a combinatorial proof.

In this paper we give an elementary proof of Eq.~(\ref{I:1})
by using recurrence relations and telescoping.  We consider the
main sum $ \sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k} $ (note
that this sum is the left-hand side of (\ref{I:1})), the
auxiliary sum
$\sum_{k=0}^{n}(n-k)\binom{2k}{k}\binom{2(n-k)}{n-k}$, and we
show that there is  a close relationship between these two sums.

More precisely, we derive two
simple recurrence relations for these two sums, and we obtain a
recurrence relation for the main sums. We can evaluate the main sum
by telescoping. In this construction, we only use two well-known
identities; namely, $\binom{2n}{n}=2\binom{2n-1}{n}$ and
$k\binom{n}{k}=n\binom{n-1}{k-1}$.

We also prove the alternating version of Eq.~(\ref{I:1}):
\begin{equation}\sum_{k=0}^{n}(-1)^k\binom{2k}{k}\binom{2(n-k)}{n-k}=\begin{cases} 2^n\binom{n}{\frac{n}{2}},&\text{if $n$ is even;}\\
 0,&\text{if $n$ is odd.} \end{cases}\label{I:2}
\end{equation}
Finally, we will use this idea to prove a generalization by
Chang and Xu \cite{GCX}:
 \begin{equation}
 \sum_{\underset{i_1\geq 0,\ldots, i_t \geq 0}{i_1+i_2+\cdots+i_t=n}}^{}\binom{2i_1}{i_1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t}=\begin{cases}4^n\binom{n+\frac{t}{2}-1}{n},&\text{if $t$ is even;}\\\\ \frac{\binom{2n+t-1}{2n}}{\binom{n+\frac{1}{2}(t-1)}{n}}\binom{2n}{n},&\text{if $t$ is odd.}\end{cases}\label{I:3}
 \end{equation}
In the above identity, $t$ is a positive integer.

The identities (\ref{I:2}) and (\ref{I:3}) can be proved by using
generating functions (see~\cite{MZS} for a combinatorial proof of
(\ref{I:2}) and \cite{RDAG2} for a proof of (\ref{I:3})).

We end this paper with some interesting
 identities which can be proved with our formulas for auxiliary sums.

Throughout the paper, $\mathbb{N} $ denotes the set of
positive integers, and $\mathbb{N}_0 $ denotes the set of
nonnegative integers. For $n\in \mathbb{N}$ we let $[n] $ denote
the set $\{1,2,\ldots,n\} $.

\section{The proof of Eq.~(\ref{I:1}) }

\begin{proof}
     Let $n \in \mathbb{N}_{0}$.
We let  $ T(n)$ denote  the left-hand side of Eq.~(\ref{I:1}), i.e.,\[
T(n)=\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}. \] We
introduce the auxiliary sum $P(n)$ as follows
\[ P(n)=\sum_{k=0}^{n}(n-k)\binom{2k}{k}\binom{2(n-k)}{n-k}. \]
Obviously, we have that
\begin{equation} P(n)=\frac{nT(n)}{2}.\label{E:1} 
\end{equation}
Let $n\in \mathbb{N}$.
 Since we know that the last term of $P(n)$ is equal to zero, we have
\begin{align*}
P(n)&=\sum_{k=0}^{n-1}(n-k)\binom{2k}{k}\binom{2(n-k)}{n-k}\\
&=2\sum_{k=0}^{n-1}\binom{2k}{k}(n-k)\binom{2(n-k)-1}{n-k}\\
&=2\sum_{k=0}^{n-1}\binom{2k}{k}(2(n-k)-1)\binom{2(n-1-k)}{n-1-k}\\
&=2\sum_{k=0}^{n-1}\binom{2k}{k}(2(n-1-k)+1)\binom{2(n-1-k)}{n-1-k}\\
&=4\sum_{k=0}^{n-1}(n-1-k)\binom{2k}{k}\binom{2(n-1-k)}{n-1-k}+
2\sum_{k=0}^{n-1}\binom{2k}{k}\binom{2(n-1-k)}{n-1-k}.
\end{align*}
From the last equation above, we  conclude that
\begin{equation} P(n)=4P(n-1)+2T(n-1)\text{,}\,n\in \mathbb{N}.\label{E:2} 
\end{equation}
From  Equations (\ref{E:1}) and (\ref{E:2}), it follows that
\begin{displaymath}
\frac{nT(n)}{2}=\frac{4(n-1)T(n-1)}{2}+2T(n-1) = 
2nT(n-1) .
\end{displaymath}
Therefore
\begin{equation} T(n)=4T(n-1);\,\,n \in \mathbb{N}.\label{E:3}
\end{equation}
Using telescoping and the fact $T(0)=1 $, Equation (\ref{E:3}) becomes
\[T(n)=4^n,\,\,n \in \mathbb{N}_{0}.\]
\end{proof}


\section{The proof of Eq.~(\ref{I:2}) }
    \begin{proof}
Let $n \in \mathbb{N}_{0}$. We let  $ Q(n)$ denote the left-hand
side of  Eq.~(\ref{I:2}), i.e.,\[
Q(n)=\sum_{k=0}^{n}(-1)^k\binom{2k}{k}\binom{2(n-k)}{n-k}. \]
 It is easy to note that \begin{equation}Q(2n+1)=0.\label{E:4}
 \end{equation}
We consider the auxiliary sum $H(n)$ as follows
\[ H(n)=\sum_{k=0}^{n}(-1)^k(n-k)\binom{2k}{k}\binom{2(n-k)}{n-k} .\]
Obviously,
\begin{equation} H(2n)=nQ(2n).\label{E:5} 
\end{equation}
Let $n\in \mathbb{N}$. If we continue in a similar manner, as we do in the second part of the previous proof, then it follows that
\begin{equation} H(n)=4H(n-1)+2Q(n-1)\text{,}\,n\in \mathbb{N}.\label{E:6}
\end{equation}
From relations (\ref{E:4}) and (\ref{E:6}), it follows that
    \begin{equation}
    H(2n)=4H(2n-1)+2Q(2n-1) = 4H(2n-1). \label{E:7}
\end{equation}
   
Also, from (\ref{E:6}), we obtain that
\begin{equation} H(2n-1)=4H(2n-2)+2Q(2n-2); \label{E:8}
\end{equation}
and relations (\ref{E:7}) and (\ref{E:8}) give us that
\begin{equation}H(2n)=16H(2n-2)+8Q(2n-2). \label{E:9} 
\end{equation}
Finally, from (\ref{E:5}) and (\ref{E:9}), it follows that
\begin{displaymath}
nQ(2n)=16(n-1)Q(2n-2)+8Q(2n-2) =8(2n-1)Q(2n-2),
\end{displaymath}
i.e.,
\begin{equation}
Q(2n)=\frac{8(2n-1)}{n}\cdot Q(2n-2),\,n\in \mathbb{N}. \label{E:10}
\end{equation}
By using telescoping and the fact $Q(0)=1$, from (\ref{E:10}) we
obtain that
\[
Q(2n)=\frac{8^n(2n-1)!!}{n!} .\] The equation above can be
transformed as follows:
\begin{displaymath}
Q(2n)=\frac{8^n(2n-1)!!}{n!}=\frac{8^n(2n-1)!!}{n!}\cdot \frac{2^nn!}{2^nn!}=\frac{4^n(2n-1)!!(2n)!!}{(n!)^2} 
=\frac{4^n(2n)!}{(n!)^2} =4^n\binom{2n}{n} .
\end{displaymath}

    \end{proof}
\section{The proof of Eq.~(\ref{I:3}) }
    \begin{proof}
Let $n \in \mathbb{N}_{0}$ and $t\in \mathbb{N}$. We let $ S(n,t)$
denote the left-hand side of (\ref{I:3}), i.e.,
\[S(n,t)=\sum_{\underset{i_1\geq 0,\ldots, i_t \geq 0}
{i_1+i_2+\cdots+i_t=n}}^{}\binom{2i_1}{i_1}
\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} . \] We introduce the
auxiliary sum $P(n,t)$ as follows:
\[ P(n,t)=\sum_{\underset{i_1\geq 0,\ldots, i_t \geq 0}
{i_1+i_2+\cdots+i_t=n}}^{}i_1\binom{2i_1}{i_1}
\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} . \] Because of 
symmetry, $P(n,t)$ is also equal to
\[ P(n,t)=\sum_{\underset{i_1\geq 0,\ldots,
i_t \geq 0}{i_1+i_2+\cdots+i_t=n}}
^{}i_k\binom{2i_1}{i_1}\binom{2i_2}{i_2}
\cdots\binom{2i_t}{i_t}, \, k \in [t] .\] It is easy to obtain that
\begin{equation}P(n,t)=\frac{nS(n,t)}{t} .\label{E:11}
\end{equation}
For arbitrary $n \in \mathbb{N}$ we have
\begin{align*}
P(n,t)&\,=\sum_{\underset{i_1 > 0,i_2 \geq 0,\ldots, i_t \ge
0}{i_1+i_2+\cdots+i_t=n}}^{}i_1\binom{2i_1}{i_1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} \\
&\,=2\cdot \sum_{\underset{i_1-1 \geq 0,i_2 \geq 0,\ldots, i_t
\ge
0}{(i_1-1)+i_2+\cdots+i_t=n-1}}^{}i_1\binom{2i_1-1}{i_1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} \\
&\,=2\cdot \sum_{\underset{i_1-1 \geq 0,i_2 \geq 0,\ldots, i_t
\ge
0}{(i_1-1)+i_2+\cdots+i_t=n-1}}^{}(2i_1-1)\binom{2(i_1-1)}{i_1-1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} \\
&\,=2\cdot \sum_{\underset{i_1-1 \geq 0,i_2 \geq 0,\ldots, i_t
\ge
0}{(i_1-1)+i_2+\cdots+i_t=n-1}}^{}(2(i_1-1)+1)\binom{2(i_1-1)}{i_1-1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t} \\
&\,=4\cdot \sum_{\underset{i_1-1 \geq 0,i_2 \geq 0,\ldots, i_t
\ge
0}{(i_1-1)+i_2+\cdots+i_t=n-1}}^{}(i_1-1)\binom{2(i_1-1)}{i_1-1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t}+\\
&\quad\quad\quad\quad 
\,2\cdot \sum_{\underset{i_1-1 \geq 0,i_2 \geq 0,\ldots, i_t \ge
0}{(i_1-1)+i_2+\cdots+i_t=n-1}}^{}\binom{2(i_1-1)}{i_1-1}\binom{2i_2}{i_2}\cdots\binom{2i_t}{i_t}.
\end{align*} 
We  conclude that \begin{equation}
P(n,t)=4P(n-1,t)+2S(n-1,t),\,n \in \mathbb{N} .\label{E:12}
\end{equation} From relations (\ref{E:11}) and (\ref{E:12}), it follows
that \begin{align*}
\frac{nS(n,t)}{t}&=4\cdot\frac{(n-1)S(n-1,t)}{t}+2S(n-1,t) \\
&=\frac{4(n-1)S(n-1,t)+2tS(n-1,t)}{t} \\
&=\frac{2(2(n-1)+t)S(n-1,t)}{t},
\end{align*}
and so
\begin{equation} S(n,t)=\frac{2(2(n-1)+t)}{n}\cdot S(n-1,t)\text{,}\,n
\in \mathbb{N} .\label{E:13}
  \end{equation}
    By using telescoping and the fact $ S(0,t) =1$, Equation
    (\ref{E:13}) becomes \begin{equation} S(n,t)=2^n\cdot
\frac{(2n-2+t)\cdot (2n-4+t)\cdots (2+t)\cdot t}{n!}\text{,}\,n \in
\mathbb{N} .\label{E:4.4} \end{equation}
Now we have two cases:

\medskip


\noindent \textbf{Case} (\textbf{a}) The number $t$ is even: $t=2t_1
\text{,\,where} \,\,t_1 \in \mathbb{N} $. In that case relation
(\ref{E:4.4}) becomes \begin{align*} S(n,2t_1)&=4^n\cdot
\frac{(n-1+t_1)\cdot(n-2+t_1) \cdots(1+t_1)\cdot t_1}{n!},\,n \in
\mathbb{N} \\
S(n,2t_1)&=4^n\cdot\frac{(n-1+t_1)!}{n!\cdot(t_1-1)!}.
\end{align*}
So
\begin{equation} S(n,2t_1)=4^n\cdot
\binom{n+t_1-1}{n}.\label{E:14} \end{equation}
This proves the first case of (\ref{I:3}).\\

\noindent \textbf{Case} (\textbf{b})
  The number $t$ is odd: $t=2t_1+1, \text{where} \,\,t_1 \in \mathbb{N}_0 $.
In this case, from (\ref{E:4.4}) we obtain
\begin{equation} S(n,2t_1+1)=2^n\cdot \frac{(2n-1+2t_1)(2n-3+2t_1)\cdots(1+2t_1)}{n!},\,n \in \mathbb{N} .\label{E:15}
\end{equation}
Equation (\ref{E:15}) can be transformed as follows
\begin{align*}
S(n,2t_1+1)&=2^n\cdot\frac{(2n+2t_1)(2n-1+2t_1)\cdots(1+2t_1)}{n!(2n+2t_1)(2n-2+2t_1)\cdots (2+2t_1)}\\
&=2^n\cdot\frac{(2n+2t_1)(2n-1+2t_1)\cdots(1+2t_1)}{n!\cdot 2^n\cdot(n+t_1)(n+t_1-1)\cdots(1+t_1)}\\
&=\frac{(2n+2t_1)(2n-1+2t_1)\cdots(1+2t_1)\cdot(2t_1)!\cdot t_1!}{n!\cdot (n+t_1)(n+t_1-1)\cdots(1+t_1)\cdot t_1!\cdot(2t_1)!}\\
&=\frac{(2n+2t_1)!\cdot t_1!}{n!\cdot (n+t_1)!\cdot(2t_1)!}\\
&=\frac{(2n+2t_1)!}{(2n)!\cdot(2t_1)!}\cdot\frac{(2n)!\cdot t_1!}{n!\cdot (n+t_1)!}\\
&=\binom{2n+2t_1}{2n}\cdot \frac{(2n)!}{(n!)^2}\cdot \frac{t_1!\cdot n!}{(n+t_1)!}\\
&=\binom{2n+2t_1}{2n}\cdot \binom{2n}{n}\cdot \frac{1}{\binom{n+t_1}{n}}.\\
\end{align*}
This proves the second case of (\ref{I:3}).\\
    \end{proof}
\section{Conclusions}
We notice that recurrences (\ref{E:6}) and (\ref{E:12}) are same as Equation (\ref{E:2}). This is not a coincidence.\\
Let $ f $ be a function of $ k $, which is independent of $ n$. We denote the main sum $ S(n)$ as \[ S(n)=\sum_{k=0}^{n}f(k)\binom{2k}{k}\binom{2(n-k)}{n-k},\]
and the auxiliary sum $P(n)$ as \[ P(n)=\sum_{k=0}^{n}(n-k)f(k)\binom{2k}{k}\binom{2(n-k)}{n-k}.\]
Then, for any $ n \in \mathbb{N}$, the
equation \[ P(n)=4P(n-1)+2S(n-1) \] holds.

By using  auxiliary sums, we can prove the important identity
\[\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}\frac{1}{k+r}=\frac{\binom{2n+2r}{n+r}}{\binom{2r}{r}\cdot r}\,\text{;} \,r \in \mathbb{N}. \]
Recall that $ C_k=\frac{1}{k+1}\binom{2k}{k} $ are the Catalan numbers. When $ r=1 $, the last identity above becomes
\[\sum_{k=0}^{n}C_{k}\binom{2(n-k)}{n-k}=\frac{1}{2}\binom{2n+2}{n+1}.\]
It is easy to note that
\begin{align*}
\sum_{k=0}^{n}C_{k}\binom{2(n-k)}{n-k}=\sum_{k=0}^{n}C_{n-k}\binom{2k}{k}.
\end{align*}
From the last two above identities, we have 
\begin{align*}
\sum_{k=0}^{n}C_{k}\binom{2(n-k)}{n-k}+\sum_{k=0}^{n}C_{n-k}\binom{2k}{k}=\binom{2n+2}{n+1}.\\
(n+2)\sum_{k=0}^{n}\frac{1}{k+1}\binom{2k}{k}\frac{1}{n-k+1}\binom{2(n-k)}{n-k}=\binom{2n+2}{n+1}
\end{align*} 
and so
\begin{align*}
\sum_{k=0}^{n}\frac{1}{k+1}\binom{2k}{k}\frac{1}{n-k+1}\binom{2(n-k)}{n-k}=\frac{1}{n+2}\binom{2n+2}{n+1}.
\end{align*}
From the last identity above, Segner's recurrence relation \cite{TC}
for $C_{n+1}$ follows:
\[\sum_{k=0}^{n}C_k\cdot C_{n-k}=C_{n+1}. \]
Therefore, we can prove Segner's
recurrence relation by using auxiliary sums.

Using  auxiliary sums, we can prove the companion binomial  theorem
\[ \sum_{k=0}^{n}\binom{a+k}{k}\binom{b+n-k}{n-k}=\binom{a+b+n+1}{n}.\]
We also can derive new formulas (to my knowledge), such as
\[\sum_{k=0}^{n}k^2\binom{2k}{k}\binom{2(n-k)}{n-k}=\frac{4^{n-1}\cdot n \cdot (3n+1)}{2}; \]
or
\[\sum_{k=0}^{n}k^3\binom{2k}{k}\binom{2(n-k)}{n-k}=4^{n-2}\cdot n^2 \cdot (5n+3). \]
\medskip
\begin{remark}
Marko Petkov\v{s}ek, Herbert Wilf, and Doron Zeilberger also give  proofs of binomial identities   via recurrence relations and telescoping  (see~\cite{MPHWDZ}).
However, the technique of auxiliary sums is different from their techniques. 

We consider recurrence relations between auxiliary sums and main sums. \emph{We do not consider  recurrence relations between summands of the main sum}. This is the main difference from Sister Celine's method. 

Our telescoping is quite different, especially, from the telescoping in the WZ method. 
We use telescoping in order to calculate the main sum, not only to prove an identity.

\end{remark}

\section {Acknowledgments}
I wish to thank to my parents. I am very grateful to  my professor Du\v{s}ko Joji\'{c}. Also I thank the referee for valuable comments.

\begin{thebibliography}{HD}


\bibitem{VA}
V. De Angelis, Pairings and signed permutations, 
\textit{Amer. Math. Monthly} \textbf{113} (2006), 642–-644.

\bibitem{GCX}
G. Chang and C. Xu, Generalization and probabilistic proof
of a combinatorial identity,
\textit{Amer. Math. Monthly} \textbf{118} (2011), 175-–177.

\bibitem{RDAG3}
R.~Duarte and A. G. de Oliveira, A famous identity of Haj\'os in terms of sets, \textit{J. Integer Seq.} \textbf{17} (2014),\,\href{https://cs.uwaterloo.ca/journals/JIS/VOL17/Duarte/duarte7.html}{Article 14.9.1}.

\bibitem{RDAG1}
R.~Duarte and A. G. de Oliveira, A short proof of a famous
combinatorial identity, 2013. Preprint available at
\url{http://arxiv.org/abs/1307.6693}.

\bibitem{RDAG2}
R.~Duarte and A. G. de Oliveira, New developments of an old identity,
2013. Preprint available at \url{http://arxiv.org/abs/1203.5424}.

\bibitem{TC}
T. Koshy, \textit{Catalan Numbers with Applications}, Oxford University
Press, 2009.

\bibitem{MPHWDZ}
M. Petkov\v{s}ek, H. Wilf, and D. Zeilberger, $A = B$, A. K. Peters,
1996.

\bibitem{MZS}
M. Z. Spivey, A combinatorial
proof for the alternating convolution of the central binomial coefficients, \textit{Amer. Math. Monthly} \textbf{121} (2014),
537–-540.

\bibitem{RS}
R. Stanley, \textit{Enumerative Combinatorics}, Vol.~1,
Cambridge Studies in Advanced Mathematics, \textbf{49}, Cambridge
University Press, 1997.

\bibitem{MASV}
M. Sved, Counting and recounting: the aftermath, \textit{Math. Intelligencer} \textbf{6} (1984), 44--45.

\end{thebibliography}
\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary  05A10 ; Secondary 05A19.

\noindent \emph{Keywords:} central binomial coefficient, recurrence equation, telescoping, auxiliary sum.
\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 8 2016;
revised version received June 18 2016.
Published in {\it Journal of Integer Sequences}, July 6 2016.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                

