% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} \specs{P1.9}{22(1)}{2015} % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf Proof of a conjecture of Amdeberhan and Moll on\\ a divisibility property of binomial coefficients} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \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 } % \date{\dateline{submission date}{acceptance date}\\ % \small Mathematics Subject Classifications: comma separated list of % 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 % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract will be distributed without the rest of the % paper so it must be entirely self-contained. \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} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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