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\title{\bf Proof of a conjecture of Amdeberhan and Moll on\\ a divisibility property of binomial coefficients}

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\author{Quan-Hui Yang\thanks{Supported by the National Natural Science
Foundation of China, Grant No. 11371195 and the Startup Foundation
for Introducing Talent of NUIST, Grant No. 2014r029.}\\
\small School of Mathematics and Statistics\\[-0.8ex]
\small  Nanjing University of Information Science and Technology\\[-0.8ex]
\small Nanjing, P.R.~China\\
\small\tt yangquanhui01@163.com }

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% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Feb 10, 2014}{Dec 21, 2014}{Jan 9, 2015}\\
\small Mathematics Subject Classifications: 05A10, 11B65}

\begin{document}

\maketitle

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\begin{abstract}
  Let $a,b$ and $n$ be positive integers with $a>b$. In this note,
  we prove that
  $$(2bn+1)(2bn+3){2bn \choose bn}\bigg|3(a-b)(3a-b){2an \choose an}{an\choose bn}.$$
  This confirms a recent conjecture of Amdeberhan and Moll.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} binomial coefficients;
   $p$-adic order; divisibility properties
\end{abstract}

\section{Introduction}

In 2009, Bober \cite{bober} determined all cases such that
$$\frac{(a_1n)!\cdots (a_kn)!}{(b_1n)!\cdots (b_{k+1}n)!}\in \mathbb{Z},$$
where $a_s\not=b_t$ for all $s,t$, $\sum{a_s}=\sum{b_t}$ and
$\gcd(a_1,\ldots,a_k,b_1,\ldots,b_{k+1})=1$.

Recently, Z.-W. Sun \cite{sun12,sun13} studied divisibility
properties of binomial coefficients and obtained some interesting
results. For example,
$$2(2n+1){2n \choose n}\bigg| {6n \choose 3n}{3n \choose n},$$
$$(10n+1){3n \choose n}\bigg| {15n \choose 5n}{5n-1 \choose n-1}.$$
Later, Guo and Krattenthaler (see \cite{guo,{guokra}}) obtained
some similar divisibility results. Related results appear in
\cite{Calkin}-\cite{Fine} and \cite{Guo07}-\cite{Razpet}.


Introduce the notation $$S_n=\frac{{6n \choose 3n} {3n \choose
n}}{2(2n+1){2n \choose n}}\quad \text{and} \quad t_n=\frac{{15n
\choose 5n}{5n-1 \choose n-1}}{(10n+1){3n \choose n}}.$$

In \cite{guonew}, Guo proved the conjectures due to Z.W. Sun
\cite{sun12,sun13}.

\noindent{\bf  Theorem A.} (\cite[Conjecture 3(i)]{sun12}.){\em~
Let $n$ be a positive integer. Then
$$3S_n\equiv 0 \pmod {2n+3}.$$}
\noindent{\bf  Theorem B.} (\cite[Conjecture 1.3]{sun13}.){\em ~
Let $n$ be a positive integer. Then
$$21t_n\equiv 0 \pmod {10n+3}.$$}

Recently, T. Amdeberhan and  V. H. Moll proposed a conjecture
related to Theorems A and B, which was only presented as
Conjecture 7.1 in Guo's paper \cite{guonew} by private
communication.

This notes provides a proof of this conjecture.

\begin{theorem}\label{thm1} Let $a,b$ and $n$ be positive integers with $a>b$. Then
$$(2bn+1)(2bn+3){2bn \choose bn}\bigg|3(a-b)(3a-b){2an \choose an}{an\choose bn}.$$
\end{theorem}

\begin{remark} Theorem A is the special case $a=3$, $b=1$ of Theorem \ref{thm1}.
\end{remark}

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\section{Proofs}

For a real number $z$, denote the greatest integer not exceeding
$z$ by $\left\lfloor z\right\rfloor$ and  $\{z\}$ denotes the
fractional part of $z$. For an integer $n$ and a prime $p$, write
$p^{k}\|n$ if $p^k|n$ and $p^{k+1}\nmid n$. The integer $k$ above
is denoted by $\nu_p(n)$.

It is well known that
\begin{equation}\label{eq1}\nu_p{(n!)}=\sum_{i=1}^{\infty}\left\lfloor \frac {n}{p^i}\right\rfloor.
\end{equation}


The proof of Theorem \ref{thm1} begins with a preliminary result.

\begin{lemma}\label{lem1} Let $x$ and $y$ be two real numbers. Then
$$\left\lfloor 2x\right\rfloor +\left\lfloor y\right\rfloor \ge \left\lfloor x\right\rfloor +\left\lfloor x-y\right\rfloor +\left\lfloor
2y\right\rfloor.$$
\end{lemma}
\begin{proof} The identity $2x+y=x+(x-y)+2y$ shows that it
suffices to $\{2x\}+\{y\}\le \{x\}+\{x-y\}+\{2y\}$. The proof now
follows by comparing $\{x\}$ and $\{y\}$ to 1/2. The details are
left to the reader.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1}] Let
$$T(a,b,n):={2an \choose an}{an \choose bn}\bigg/{2bn \choose bn}=\frac{(2an)!(bn)!}{(an)!(an-bn)!(2bn)!}.$$
By \eqref{eq1}, for any prime $p$,
$$\nu_p(T(a,b,n))=\sum_{i=1}^{\infty}\left( \left\lfloor \frac{2an}{p^i}\right\rfloor
+\left\lfloor\frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\right).$$
Lemma \ref{lem1} shows that each term of $\nu_p(T(a,b,n))$ is
nonnegative. Hence $\nu_p(T(a,b,n))\ge 0$. Therefore, $T(a,b,n)\in
\mathbb{Z}$.

Since $\gcd(2bn+1,2bn+3)=1$, it suffices to prove that
$$2bn+1|3(a-b)(3a-b)T(a,b,n)$$ and $$2bn+3|3(a-b)(3a-b)T(a,b,n).$$

The second statement is established here. The proof of the first
statement is similar and the details are omitted. Suppose that
$p^{\alpha}\| 2bn+3$ with $\alpha\ge 1$. It is shown that
\begin{equation}\label{eq5}p^{\alpha}| 3(a-b)(3a-b)T(a,b,n).\end{equation}

Let $p^{\beta}\|a-b$ and $p^{\gamma}\|3a-b$ with $\beta\ge 0$ and
$\gamma\ge 0$. Write $\tau=\max\{\beta,\gamma\}$. If $\alpha \le
\tau$, then \eqref{eq5} clearly holds. Now we assume $\alpha >
\tau$.

Suppose that $p\ge 5$. The statement
$$\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor=1$$ is
established for $i=\tau+1,\tau+2,\ldots,\alpha$. This is proven
next. Noting that $p|2bn+3$ and $p\ge 5$, it follows that
$\gcd(p,n)=1$.

Observe that $p^{\alpha}\| 2bn+3$, it follows that $2bn \equiv
p^{\alpha}-3 \pmod {p^{\alpha}}$ and $bn\equiv (p^{\alpha}-3)/2
\pmod {p^{\alpha}}$.

Take $i\in \{\tau+1,\tau+2,\ldots,\alpha\}$. Then $2bn \equiv
p^{i}-3 \pmod {p^{i}}$ and $bn\equiv (p^{i}-3)/2 \pmod {p^{i}}$.
Now we divide into several cases according to the value of $an
\pmod {p^i}$.


{\bf Case 1.} $an\equiv t \pmod {p^i}$ with $0\le t<(p^i-3)/2$. It
follows that $2an\equiv 2t \pmod {p^i}$ and $0\le 2t< p^i-3$. Also
$$an-bn\equiv t-(p^i-3)/2+p^i \pmod {p^i},$$ where $0\le
t-(p^i-3)/2+p^i<p^i$. Hence
\begin{eqnarray*}&&\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\\
&=&\frac{2an-2t}{p^i}+\frac{bn-(p^i-3)/2}{p^i}-\frac{an-t}{p^i}\\
&&-
\left(\frac{an-bn-(t-(p^i-3)/2+p^i)}{p^i}\right)-\frac{2bn-(p^i-3)}{p^i}\\
&=&1.
\end{eqnarray*}

{\bf Case 2.} $an\equiv ({p^i}-3)/2 \pmod {p^i}$. Then,
$an-bn\equiv 0 \pmod {p^i}$. Since $\gcd(p,n)=1$, it follows that
$p^i|a-b$. However, $p^{\beta}\|a-b$ and $\beta\le \tau<i$. This
is a contradiction.

{\bf Case 3.} $an\equiv ({p^i}-1)/2 \pmod {p^i}$. It follows that
$$3an-bn\equiv \frac {3({p^i}-1)}{2}-\frac{({p^i}-3)}{2}\equiv  0 \pmod {p^i}.$$
The fact that $\gcd(p,n)=1$ implies $p^i|3a-b$. This contradicts
$p^{\gamma}\|3a-b$ and $\gamma<i$.

{\bf Case 4.} $an\equiv t \pmod {p^i}$ with $(p^i+1)/2\le t<p^i$.
Then
$$2an\equiv 2t-p^i \pmod {p^i}, \quad 0\le 2t-p^i< p^i,$$ and $$an-bn\equiv t-(p^i-3)/2
\pmod {p^i},\quad 0\le t-(p^i-3)/2<p^i.$$ Hence
\begin{eqnarray*}&&\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\\
&=&\frac{2an-(2t-p^i)}{p^i}+\frac{bn-(p^i-3)/2}{p^i}-\frac{an-t}{p^i}\\
&&-
\left(\frac{an-bn-(t-(p^i-3)/2)}{p^i}\right)-\frac{2bn-(p^i-3)}{p^i}\\
&=&1.
\end{eqnarray*}

Therefore, $\nu_p(T(a,b,n))\ge \alpha-\tau$, and this implies
$$\nu_p(3(a-b)(3a-b)T(a,b,n))\ge \alpha.$$ The proof of
\eqref{eq5}, for $p\ge 5$, is complete.

Now assume $p=3$. If $9|n$, then $3|2bn+3$ and $9\nmid 2bn+3$. It
follows that $\alpha=1$, and then \eqref{eq5} clearly holds. If
$9\nmid n$, then the proof of the case $p\ge 5$ applies to this
situation. In Case 2, $an-bn\equiv 0 \pmod {3^i}$ gives
$3^{i-1}|a-b$. In Case 3, $3^{i-1}|3a-b$. Thus, if $i\ge \tau+2$,
then $i-1\ge \tau+1$. It is a contradiction in both cases. Hence
$$\left\lfloor \frac{2an}{3^i}\right\rfloor+\left\lfloor \frac{bn}{3^i}\right\rfloor-\left\lfloor\frac{an}{3^i}\right\rfloor-\left\lfloor\frac{an-bn}
{3^i}\right\rfloor-\left\lfloor\frac{2bn}{3^i}\right\rfloor=1$$
for $i=\tau+2,\tau+3,\ldots,\alpha$. It follows that
$\nu_3(T(a,b,n))\ge \alpha-\tau-1$, and then
$$\nu_3(3(a-b)(3a-b)T(a,b,n))\ge \alpha.$$ That is, \eqref{eq5}
also holds. Hence, $2bn+3|3(a-b)(3a-b)T(a,b,n)$.

Therefore,
$$(2bn+1)(2bn+3){2bn \choose bn}\bigg|3(a-b)(3a-b){2an \choose an}{an\choose bn}.$$

 This completes the proof of Theorem \ref{thm1}.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
We are grateful to the anonymous referee for carefully reading our
original manuscript and giving many detailed comments.

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