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\begin{document}

%\title[Convolutions of Central Binomial Coefficients]{On Divisibility of Convolutions of Central Binomial Coefficients}
\title{\bf On Divisibility of Convolutions of Central Binomial Coefficients}
\author{Mark R. Sepanski\\
\small Department of Mathematics, Baylor University\\[-0.8ex]
\small One Bear Blace \#97328\\[-0.8ex]
\small Waco, TX \ 76798-7328, U.S.A.\\
\small\texttt{Mark\_Sepanski@baylor.edu}}

\date{\dateline{May 5, 2013}{Jan 11, 2014}{Feb 13, 2014}\\
\small Mathematics Subject Classifications: 11B65, 05A10}

\maketitle

\begin{abstract}
Recently, Z. Sun proved that%
\[
2\left(  2m+1\right)  \binom{2m}{m}\mid\binom{6m}{3m}\binom{3m}{m}%
\]
for $m\in\mathbb{Z}_{>0}$. In this paper, we consider a generalization of this
result by defining%
\[
b_{n,k}=\frac{2^{k}\,\left(  n+2k-2\right)  !!}{\left(  n-2\right)  !!\,k!}.
\]
In this notation, Sun's result may be expressed as $2\left(  2m+1\right)  \mid
b_{\left(  2m+1\right)  ,\left(  2m+1\right)  -1}$ for $m\in\mathbb{Z}_{>0}$.
In this paper, we prove that%
\[
2n\mid b_{n,un\pm2^{r}}%
\]
for $n\in\mathbb{Z}_{>0}$ and $u,r\in\mathbb{Z}_{\geq0}$ with $un\pm2^{r}%
>0$. In addition, we prove a type of converse. Namely, fix $k\in\mathbb{Z}$ and
$u\in\mathbb{Z}_{\geq0}$ with $u>0$ if $k<0$. If%
\[
2n\mid b_{n,un+k}%
\]
for all $n\in\mathbb{Z}_{>0}$ with $un+k>0$, then there exists a unique
$r\in\mathbb{Z}_{\geq0}$ so that either
\[
k=2^{r}\text{ \ or \ }k=-2^{r}\text{.}%
\]

\bigskip\noindent \textbf{Keywords:} central binomial coefficients
\end{abstract}
\maketitle


\section{Introduction}

There has been much recent work on topics relating to central binomial
coefficients (\cite{B2012,CZ2009,F2010,GS2008,S2006,SMA2009,ZWOCRCBC2011,ZWPSDCBC2013,ZWSPTBC2013,ZWNCCBC2010,T2010}). In particular,
Z. Sun in \cite{ZWPSDCBC2013} proved interesting results on congruences of
sums of products of central binomial coefficients. One such result is that
$2\left(  2m+1\right)  \binom{2m}{m}\mid\binom{6m}{3m}\binom{3m}{m}$ for
$m\in\mathbb{Z}_{>0}$. The new integer sequence $m\rightarrow\binom{6m}%
{3m}\binom{3m}{m}/[2\left(  2m+1\right)  \binom{2m}{m}]$ is given by A176898
in the OIES. Using this sequence, Sun proposed a number of open conjectures
(\cite[Conjectures 2, 4, 6 and 8]{ZWPSDCBC2013}) on certain divisibility
properties of this and related sequences.

In this paper, we consider a generalization of Sun's new sequence. Recognizing
$\binom{6m}{3m}\binom{3m}{m}/\binom{2m}{m}$ as the coefficient of $x^{2m}$ in
the $\left(  2m+1\right)  $-fold convolution of the central binomial sequence
with itself, we define $b_{n,k}$ to be the $k^{\text{th}}$ term of the
$n$-fold convolution of the central binomial sequence with itself--which turns
out to be%
\[
b_{n,k}=\frac{2^{k}\,\left(  n+2k-2\right)  !!}{\left(  n-2\right)  !!\,k!}.
\]
In this notation, Sun's result is that $2\left(  2m+1\right)  \mid
b_{2m+1,\left(  2m+1\right)  -1}$ for $m\in\mathbb{Z}_{>0}$.

In this paper (Theorem \ref{Main Divisibility Theorem}), we prove that%
\[
2n\mid b_{n,un\pm2^{r}}%
\]
for $n\in\mathbb{Z}_{>0}$ and $u,r\in\mathbb{Z}_{\geq0}$ with $un\pm2^{r}>0$.
In particular, Sun's result is a special case of the above theorem in which
$n=2m+1$, $u=1$, $r=0$, and the $-$ sign is chosen. In addition, we prove a
type of converse (Theorem \ref{uniuqeness thm}). Namely, fix $k\in\mathbb{Z}$
and $u\in\mathbb{Z}_{\geq0}$ with $u>0$ if $k<0$. If%
\[
2n\mid b_{n,un+k}%
\]
for all $n\in\mathbb{Z}_{>0}$ with $un+k>0$, then there exists a unique
$r\in\mathbb{Z}_{\geq0}$ so that either
\[
k=2^{r}\text{ \ or \ }k=-2^{r}\text{.}%
\]


\section{Definition}

Write $b$ for the sequence of \emph{central binomial coefficients},
$b_{j}=\binom{2j}{j}$, with $j\in\mathbb{Z}_{\geq0}$. For $n,k\in
\mathbb{Z}_{\geq0}$ with $n\geq1$, we define the doubly indexed sequence
$b_{n,k}\in\mathbb{Z}_{>0}$ to be the $k^{\text{th}}$ term of the $n$-fold
convolution of $b$ with itself, $b_{n,k}=\left(  b^{\ast n}\right)  _{k}$. In
the degenerate case of $n=0$, we define $b_{0,0}=1$ and $b_{0,k}=0$ for $k>0$.
It follows that the generating function for the sequence $k\rightarrow
b_{n,k}$ is $\left(  1-4x\right)  ^{-n/2}$ and a trivial calculation shows
that%
\[
b_{n,k}=\frac{2^{k}\,\left(  n+2k-2\right)  !!}{\left(  n-2\right)  !!\,k!}%
\]
when $n\geq2$. For use in Theorem \ref{uniuqeness thm} and in order to compare
with \cite{ZWPSDCBC2013}, we note that it is straightforward to verify that%
\[
b_{2m,k}=2^{2k}\binom{m+k-1}{k}, \quad  b_{2m+1,k}=\frac{\binom{2m+2k}%
{m+k}\binom{m+k}{m}}{\binom{2m}{m}}%
\]
for $m\in\mathbb{Z}_{>0}$ in the first formula above and $m\in\mathbb{Z}%
_{\geq0}$ in the second.

\section{Divisibility}

Here we present the main result on the divisibility of the sequence $b_{n,k}$.

\begin{theorem}
\label{Main Divisibility Theorem}For $n\in\mathbb{Z}_{>0}$ and $u,r\in
\mathbb{Z}_{\geq0}$,%
\[
2n\mid b_{n,un+2^{r}}%
\]
and, if also $un-2^{r}>0$,%
\[
2n\mid b_{n,un-2^{r}}.
\]

\end{theorem}

\begin{proof}
Begin with the convolution definition%
\[
b_{n,un\pm2^{r}}=\sum_{\substack{a_{1},\ldots,a_{n}\in\mathbb{Z}_{\geq0}
\\a_{1}+\cdots+a_{n}=un\pm2^{r}}}%
%TCIMACRO{\dprod \limits_{j=1}^{n}}%
%BeginExpansion
{\displaystyle\prod\limits_{j=1}^{n}}
%EndExpansion
\binom{2a_{j}}{a_{j}}%
\]
and let $X=\left\{  \left(  a_{1},\ldots,a_{n}\right)  \in\mathbb{Z}_{\geq
0}^{n}\mid a_{1}+\cdots+a_{n}=un\pm2^{r}\right\}  $. Observe that $0\notin X $
since $un+2^{r},un-2^{r}\geq1$.

The set $X$ carries a natural action of the symmetric group, $S_{n}$, acting
by permuting the coordinates. Write $\mathcal{O}_{1},\ldots,\mathcal{O}_{N}$
for the orbits of $X$ under the action of $S_{n}$. Clearly each $\mathcal{O}%
_{k}$, $1\leq k\leq N$, has a unique representative of the form%
\[
x_{k}=(\overset{d_{1k}}{\overbrace{c_{1k},\ldots c_{1k}}},\overset{d_{2k}%
}{\overbrace{c_{2k},\ldots,c_{2k}}},\ldots,\overset{d_{m_{k}k}}{\overbrace
{c_{m_{k}k},\ldots c_{m_{k}k}}})
\]
with $0\leq c_{1k}<c_{2k}<\cdots<c_{m_{k}k}$ and $d_{jk},m_{k}\in
\mathbb{Z}_{>0}$. Since the stabilizer of $x_{k}$ in $S_{n}$ is clearly
isomorphic to $S_{d_{1k}}\times S_{d_{2k}}\times\cdots\times S_{d_{m_{k}k}}$,
if follows that%
\[
\left\vert \mathcal{O}_{k}\right\vert =\frac{n!}{d_{1k}!d_{2k}!\cdots
d_{m_{k}k}!}=\binom{n}{d_{k}}%
\]
where we use multinomial notation above and write $d_{k}=\left(  d_{1k}%
,d_{2k},\ldots,d_{m_{k}k}\right)  $. Using this, we may rewrite the formula
for $b_{n,un\pm2^{r}}$ as%
\[
b_{n,un\pm2^{r}}=\sum_{k=1}^{N}\binom{n}{d_{k}}%
%TCIMACRO{\dprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\displaystyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\binom{2c_{jk}}{c_{jk}}^{d_{jk}}.
\]
We will prove the theorem by demonstrating that $2n\mid\binom{n}{d_{k}}%
%TCIMACRO{\tprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\textstyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\binom{2c_{jk}}{c_{jk}}^{d_{jk}}$ for each $k$.

As $x_{k}\in X$, it follows that%
\begin{align*}
\sum_{j=1}^{m_{k}}d_{jk}  & =n\\
\sum_{j=1}^{m_{k}}d_{jk}c_{jk}  & =un\pm2^{r}.
\end{align*}
Therefore,%
\[
\sum_{j=1}^{m_{k}}d_{jk}\left(  c_{jk}-u\right)  =\pm2^{r}%
\]
and we may write the greatest common divisor of $d_{1k},d_{2k},\ldots
,d_{m_{k}k}$ as $2^{q_{k}}$ for some $q_{k}$, $0\leq q_{k}\leq r$. Since it
follows that $2^{q_{k}}\mid n$, we also see that $2^{q_{k}}\leq n$.

Choose $w_{jk}\in\mathbb{Z}$, $1\leq j\leq m_{k}$, so that $\sum_{j=1}^{m_{k}%
}w_{jk}d_{jk}=2^{q_{k}}$. Write $e_{j}$ for the $j^{\text{th}}$ standard basis
vector, $e_{j}=(\overset{j}{\overbrace{0,\ldots0,1}},0,\ldots0)\in
\mathbb{Z}^{m_{k}}$ (suppressing the $m_{k}$ dependence). Then%
\begin{align*}
2^{q_{k}}\binom{n}{d_{k}}  & =\sum_{j=1}^{m_{k}}w_{jk}d_{jk}\binom{n}{d_{k}%
}=\sum_{\substack{j=1 \\d_{jk}\geq1}}^{m_{k}}w_{jk}d_{jk}\binom{n}{d_{k}}\\
& =\sum_{\substack{j=1 \\d_{jk}\geq1}}^{m_{k}}w_{jk}\binom{n-1}{d_{k}-e_{j}}n
\end{align*}
so that $n\mid2^{q_{k}}\binom{n}{d_{k}}$ and
\[
2n\mid\binom{n}{d_{k}}2^{q_{k}+1}.
\]


As $\frac{1}{2}\tbinom{2j}{j}=\tbinom{2j-1}{j}$ for $j\geq1$, it is well known
that $2\mid\tbinom{2c_{jk}}{c_{jk}}$ whenever $c_{jk}\neq0$. If $c_{1k}\neq0$,
then $2^{n}\mid%
%TCIMACRO{\dprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\displaystyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\tbinom{2c_{jk}}{c_{jk}}^{d_{jk}}$. As $2^{q_{k}+1}\leq2n\leq2^{n}$, it
follows that $2n\mid\binom{n}{d_{k}}%
%TCIMACRO{\tprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\textstyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\binom{2c_{jk}}{c_{jk}}^{d_{jk}}$ and we are done.

Suppose, therefore, that we are in the case of $c_{1k}=0$ (so $m_{k}\geq2$).
Let $s_{k}=n-d_{1k}$. Now all that we get is that $2^{s_{k}}\mid%
%TCIMACRO{\dprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\displaystyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\tbinom{2c_{jk}}{c_{jk}}^{d_{jk}}$. If $q_{k}+1\leq s_{k}$, then a similar
argument as in the above paragraph shows $2n\mid\binom{n}{d_{k}}%
%TCIMACRO{\tprod \limits_{j=1}^{m_{k}}}%
%BeginExpansion
{\textstyle\prod\limits_{j=1}^{m_{k}}}
%EndExpansion
\binom{2c_{jk}}{c_{jk}}^{d_{jk}}$ and we are done. It remains only to show
that $q_{k}+1\leq s_{k}$. So suppose that $q_{k}+1>s_{k}$ and write
$d_{jk}=2^{q_{k}}t_{jk}$ for $t_{jk}\in\mathbb{Z}_{>0}$. Since we must have
$t_{jk}\geq j-1\geq1$ when $j\geq2$, we get
\[
q_{k}+1>s_{k}=\sum_{j=2}^{m_{k}}d_{jk}=\sum_{j=2}^{m_{k}}2^{q_{k}}t_{jk}%
\geq\sum_{j=2}^{m_{k}}2^{q_{k}}=2^{q_{k}}\left(  m_{k}-1\right)  \geq2^{q_{k}%
}\text{.}%
\]
Since it is impossible to obtain $q+1>2^{q}$, we arrive at the desired contradiction.
\end{proof}

For the case of $2n\mid b_{n,un+2^{r}}$ in the above theorem, $n=0$ is ruled
out in order to make sure that division by $2n$ is well defined (note
$b_{0,0}=1$ and $b_{0,k}=0$ for $k>0$). For the case of $2n\mid b_{n,un-2^{r}}
$, we require $un>2^{r}$ since $b_{n,0}=1$.

We also note that Sun's result
\[
2\left(  2m+1\right)  \binom{2m}{m}\mid\binom{6m}{3m}\binom{3m}{m}%
\]
for $m\in\mathbb{Z}_{>0}$ is equivalent to
\[
2\left(  2m+1\right)  \mid\frac{\binom{6m}{3m}\binom{3m}{m}}{\binom{2m}{m}}.
\]
This is then a special case of our equation $2n\mid b_{n,un-2^{r}}$ in which
$n=2m+1$, $u=1$, and $r=0$. This gives the same statement as the above
equation, but written as%
\[
2\left(  2m+1\right)  \mid b_{2m+1,2m}.
\]


\section{A Type of Converse}

The next result is a type of converse to Theorem
\ref{Main Divisibility Theorem}.

\begin{theorem}
\label{uniuqeness thm}Fix $k\in\mathbb{Z}$ and $u\in\mathbb{Z}_{\geq0}$ with
$u>0$ if $k<0$. If%
\[
2n\mid b_{n,un+k}%
\]
for all $n\in\mathbb{Z}_{>0}$ with $un+k>0$, then there exists a unique
$r\in\mathbb{Z}_{\geq0}$ so that either
\[
k=2^{r}\text{ \ or \ }k=-2^{r}\text{.}%
\]

\end{theorem}

\begin{proof}
First we show that $k\neq0$. For this choose any odd prime $p$ and consider
$b_{2p,2up}$. Using Theorem 4 of \cite{FF} and the Division Algorithm, work
$\operatorname{mod}p$ to see that
\[
b_{2p,2up}=2^{4up}\binom{\left(  1+2u\right)  p-1}{p-1}\equiv2^{4u}\tbinom
{2u}{0}\tbinom{p-1}{p-1}\equiv2^{4u}.
\]
Thus $p\nmid b_{2p,2up}$ and so $k\neq0$.

Next we consider the case of $u=0$ (so $k>0$ here). Actually the following
argument works whenever $k>0$ so that is all we actually assume. If $k$ has an
odd prime divisor, $p$, write $k=pk^{\prime}$ for some $k^{\prime}%
\in\mathbb{Z}_{>0}$. Consider $b_{2p,2up+k}$. Then, working
$\operatorname{mod}p$ again,
\[
b_{2p,2up+k}=2^{2(2u+k^{\prime})p}\binom{(1+2u+k^{\prime})p-1}{p-1}%
\equiv2^{2(2u+k^{\prime})}\binom{2u+k^{\prime}}{0}\binom{p-1}{p-1}%
\equiv2^{2(2u+k^{\prime})}.
\]
Thus $p\nmid b_{2p,2up+k}$ and so $k$ must be a power of $2$. Note that the
only reason this argument may fail for $k<0$ is that we might have
$2up+k\leq0$.

Finally, consider the case of $u\neq0$. Suppose there exists an odd prime $p $
so $p\mid k$. Then write $k=pk^{\prime}$ and fix $m_{0}\in\mathbb{Z}_{\geq0}$
so $2^{m_{0}+1}u+k^{\prime}>0$ and $2^{m_{0}+1}u$ is congruent
$\operatorname{mod}p$ to either $0$ or $1$ (depending on whether $p\mid u$ or
$p\nmid u$). Now consider $n=2^{m_{0}+1}p\left(  p^{N}+1\right)  $ for any
sufficiently large $N$. We will show that $p\nmid b_{n,un+k}$. For this, write%
\begin{align*}
un+k  & =2^{m_{0}+1}up\left(  p^{N}+1\right)  +k^{\prime}p\\
& =2^{m_{0}+1}up^{N+1}+\left(  2^{m_{0}+1}u+k^{\prime}\right)  p.
\end{align*}
For sufficiently large $N$, $un+k$ can be expanded in base $p$ as%
\begin{align*}
un+k  & =a_{r}p^{r}+a_{r-1}p^{r}+\cdots+a_{N+1}p^{N+1}\\
& +b_{s}p^{s}+b_{s-1}p^{s-1}+\cdots+b_{1}p
\end{align*}
with $0\leq a_{i},b_{j}\leq p-1$, $a_{N+1}\leq1$, and $s\leq N$. Now we apply
Kummer's theorem to the binomial coefficient in%
\[
b_{n,un+k}=2^{p\left(  p^{N}+1\right)  }\tbinom{p^{N+1}+p-1+un+k}{p^{N+1}%
+p-1}.
\]
Clearly adding $un+k$ to $p^{N+1}+\left(  p-1\right)  $ in base $p$ results in
no carries so that $p\nmid b_{n,un+k}$. As a result, $k$ has no odd prime
divisors and we are done.
\end{proof}

\section{Relation to Known Sequences}

As a result of Theorem \ref{Main Divisibility Theorem}, we have the following
integer sequences%
\[
n\rightarrow B_{n,u,r,\pm}\equiv\frac{b_{n,un\pm2^{r}}}{2n}.
\]
For most choices of parameters $u,r,\pm$, this sequence seems to be new.
However, for a few special choices, the sequence is known. Up to a shift and a
few initial terms, the sequence $B_{n,0,2,+}$ is the OEIS integer sequence
A077415, $B_{n,1,0,+}$ is A085614, $B_{n,1,1,+}$ is A078531, and $B_{n,1,0,-}
$ appears as every other term in A089073. In addition, the odd terms of
$B_{n,0,2,+}$ are A162540, the even terms of $B_{n,0,2,+}$ are A102860 and the
negative of A136264, and (by construction) the odd terms of $B_{n,1,0,-}$\ are
Sun's A176898.

\section{Final Remarks}

It would be interesting to find a combinatorial interpretation for the
sequences $B_{n,u,r,\pm}$. For instance, one is given the case of
$B_{n,1,0,-}$ (A089073) or $B_{n,1,1,+}$ (A078531) as the number of symmetric
non-crossing connected graphs on equidistant nodes of a circle and
$B_{n,1,0,+}$ (A085614) is the number of elementary arches of size $n$.

In addition, information on corresponding generating functions would be of
interest. Some are known. For example $B_{n,1,0,+}$ (A085614) is the series
reversion of $x-3x^{2}+2x^{3}$.

\bigskip

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\end{document}
