\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \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} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://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 Method For Examining Divisibility Properties Of Some Binomial Sums \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} \vskip .1in We introduce a notion that we call a ``$D$ sum'' and use it to examine the divisibility properties of some binomial sums. The method of $D$ sums consists of two theorems and their proofs. We present two applications of our method to non-negative sums with absolute values. The third application is for a known alternating binomial sum. In particular, our method of $D$ sums can be used to prove Dixon's formula. \end{abstract} \section{Introduction} Let $m$, $n$, and $k$ be non-negative integers such that $ m \geq 2 $. We consider the sum \begin{equation}\label{Eq:1.0.1}S(n,m)=\sum_{k=0}^{n}\binom{n}{k}^m F(n,k)\mbox{,} \end{equation} where $F(n,k) $ is an integer-valued function that depends only on $n$ and $k$. The aim is to examine some divisibility properties of sums of the form $S(n,m)$. To do this, we introduce the notion of ``$D$ sums''.\footnote{So-named in honor of professor Du\v{s}ko Joji\'{c}.} \begin{definition}\label{Def:1} Let $n$, $j$, and $t$ be non-negative integers such that $j \leq \lfloor \frac{n}{2}\rfloor $. Then the $D$ sums for $S(n,m) $ are as follows: \begin{equation}D_S(n,j,t)=\sum_{l=0}^{n-2j}\binom{n-j}{l}\binom{n-j}{j+l}\binom{n}{j+l}^tF(n,j+l)\mbox{.}\label{Eq:1.0.2} \end{equation} \end{definition} Obviously, for $m \geq 2$, the equation \begin{equation} S(n,m)=D_S(n,0,m-2)\label{Eq:1.0.3}\end{equation} holds. Hence, we can see the sum $D_S(n,j,m-2)$ as a generalization of the sum $S(n,m)$. We search for new theorems and facts about $D_S(n,j,t)$ sums, which may be useful for studying $S(n,m)$ sums. Let $n$, $j$, and $t$ be as in Definition \ref{Def:1}. The method of $D$ sums consists of the following two theorems: \begin{theorem} \[ D_S(n,j,t+1)=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor} \binom{n}{j+u}\binom{n-j}{u}D_S(n,j+u,t)\mbox{.} \]\label{Th:1}\end{theorem} \begin{theorem} \[D_S(n,j,0)=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n-j}{j+u}\binom{n-2j-u}{u}\sum_{v=0}^{n-2j-2u}\binom{n-2j-2u}{v}F(n,j+u+v)\mbox{.} \]\label{Th:2} \end{theorem} Together, Theorems \ref{Th:1} and \ref{Th:2} give a recursive definition of a $D_S$ sum. \section{Background} In $1891$, Dixon \cite{AD1} found the following identity \cite[Eq.\ (6.6), p.\ 51]{Gould}: \begin{equation} \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^3=(-1)^n\binom{2n}{n}\binom{3n}{2n}\mbox{,}\label{eq:2.0.1} \end{equation} where $n$ is a non-negative integer. The absolute value of the right-hand side is sequence \seqnum{A006480} in the {\it On-Line Encyclopedia of Integer Sequences}. Moreover, Dixon \cite{AD2} established the following generalization of Eq.~(\ref{eq:2.0.1}): \begin{equation} \sum_{k=-a}^{a}(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}=\frac{(a+b+c)!}{a!\cdot b!\cdot c!}\mbox{,}\label{eq:2.0.2} \end{equation} where $a$, $b$, and $c$ are non-negative integers. Both identities are known in the literature as Dixon's identities or Dixon's formulas. For $a$ = $b$ = $c$ = $n$, Eq.~(\ref{eq:2.0.2}) becomes Eq.~(\ref{eq:2.0.1}). There are many proofs of Eq.~(\ref{eq:2.0.2}): for example, see \cite{SE,DSIG,VG,JM}. However, there are not so many direct proofs of Eq.~(\ref{eq:2.0.1}). Our goal is to give an elementary proof of Eq.~(\ref{eq:2.0.1}) without using Eq.~(\ref{eq:2.0.2}). In order to find the desired proof, we discovered the method of $D$ sums. Unfortunately, the method of $D$ sums does not work on Eq.~(\ref{eq:2.0.2}). In $ 1998$, Calkin \cite[Thm.\ 1]{nc98} proved that the alternating binomial sum $\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^m $ is divisible by $\binom{2n}{n} $ for all non-negative integers $n$ and all positive integers $m$. Calkin used arithmetical techniques in his proof. There are several generalizations of Calkin's result. In $2007$, Guo, Jouhet, and Zeng proved, among other things, two generalizations of Calkin's result \cite[Thm.\ 1.2, Thm.\ 1.3]{vg07}. In the same paper they proposed several conjectures including Conjecture $5$.$3$ which is a refinement of Calkin's result. In $2010$, Cao and Pan \cite[Thm.\ 1.1]{CaoPan} proved, among other results, Conjecture $5$.$3$. The rest of the paper is structured as follows. In Sections \ref{sec:3} and \ref{sec:4}, we prove our main Theorems \ref{Th:1} and \ref{Th:2}, respectively. In Section \ref{sec:5}, we give a brief explanation of our method of $D$ sums. In Section \ref{sec:6}, we consider two non-negative sums with absolute values: $S_1(2n,m)=\sum_{k=0}^{2n}\binom{2n}{k}^m|n-k|$ and $S_2(2n+1,m)=\sum_{k=0}^{2n+1}\binom{2n+1}{k}^m \left |\frac{2n+1}{2}-k \right |$. We assert, among other, that the first sum $S_1(2n,m)$ is divisible by $n\binom{2n}{n}$ for all positive integers $n$ and all positive integers $m$ . Similarly, we assert, among other, that the second sum $S_2(2n+1,m)$ is divisible by $(2n+1)\binom{2n}{n}$ for all non-negative integers $n$ and all positive integers $m$. In Section \ref{sec:7}, we begin with motivation for studying $S_1(2n,m)$ and $S_2(2n+1,m)$ sums. Then we apply the method of $D$ sums to prove our assertions for the sum $S_1(2n,m)$. Proofs of our assertions for the second sum $S_2(2n+1,m)$ are omitted because they are similar to the proofs for the first sum $S_1(2n,m)$. In Section \ref{sec:8}, we consider the alternating binomial sum $S_3(2n,m)=\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^m$. We give a sketch of the proof of Calkin's result \cite[Thm.\ 1]{nc98} by using our method of $D$ sums. Also, we give a sketch of the proof of Eq.~(\ref{eq:2.0.1}). \section{Proof of Theorem \ref{Th:1}}\label{sec:3} We use three well-known binomial identities. The first one is the Chu-Vandermonde convolution formula: \begin{equation} \sum_{k=0}^{c}\binom{a}{k}\binom{b}{c-k}=\binom{a+b}{c}\mbox{,}\label{Eq:4.0.1} \end{equation} where $a$, $b$, and $c$ are non-negative integers. Let $a$, $b$, and $c$ be non-negative integers such that $a\geq b \geq c $. The second identity is \begin{equation} \binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}\mbox{.}\label{Eq:4.0.2} \end{equation} The third identity is symmetry of binomial coefficients. \begin{proof} Let $ n$, $j$, and $t$ be as in Definition \ref{Def:1}. By Definition \ref{Def:1}, we know that \begin{equation} D_S(n,j,t+1)=\sum_{l=0}^{n-2j}\binom{n-j}{l}\binom{n-j}{j+l}\binom{n}{j+l}^{t+1}F(n,j+l)\mbox{.}\label{Eq:4.0.3} \end{equation} According to Eq.~(\ref{Eq:4.0.1}), we have \begin{equation}\label{Eq:4.0.4} \binom{n-j}{l}=\sum_{u=0}^{l}\binom{n-2j-l}{u}\binom{j+l}{l-u}\mbox{.} \end{equation} Obviously, the following inequalities \begin{align} u &\leq l \mbox{,}\label{Inq:4.0.5}\\ u&\leq n-2j-l\mbox{.}\label{Inq:4.0.6} \end{align} hold. From Inequalities (\ref{Inq:4.0.5}) and (\ref{Inq:4.0.6}), we obtain the following inequalities: \begin{alignat}{2} 0 &\leq u &\leq \left \lfloor \frac{n-2j}{2} \right \rfloor \mbox{,}\label{Inq:4.0.7}\\ u&\leq l &\leq n-2j-u \mbox{.}\label{Inq:4.0.8} \end{alignat} We use Eq.~(\ref{Eq:4.0.4}), the symmetry $\binom{n-j}{j+l}=\binom{n-j}{n-2j-l}$, and the symmetry $\binom{j+l}{l-u}= \binom{j+l}{j+u}$, and then permute terms in Eq.~(\ref{Eq:4.0.3}). Thus Eq.~(\ref{Eq:4.0.3}) becomes as follows: \begin{multline} D_S(n,j,t+1)=\sum_{l=0}^{n-2j} \sum_{u=0}^{l}\binom{n-2j-l}{u}\binom{j+l}{l-u}\binom{n-j}{j+l}\binom{n}{j+l}^{t+1}F(n,j+l)\mbox{ (Eq.~(\ref{Eq:4.0.4}))}\\ =\sum_{l=0}^{n-2j} \sum_{u=0}^{l}\binom{n-2j-l}{u}\binom{j+l}{j+u}\binom{n-j}{n-2j-l}\binom{n}{j+l}\binom{n}{j+l}^tF(n,j+l)\mbox{ ( by symmetry)}\\ =\sum_{l=0}^{n-2j} \sum_{u=0}^{l}\binom{n-j}{n-2j-l}\binom{n-2j-l}{u}\binom{n}{j+l}\binom{j+l}{j+u}\binom{n}{j+l}^tF(n,j+l)\mbox{ (perm.)}\label{Eq:4.0.9} \end{multline} We apply Eq.~(\ref{Eq:4.0.2}) twice. It follows that \begin{align} \binom{n-j}{n-2j-l}\binom{n-2j-l}{u}&=\binom{n-j}{u}\binom{n-j-u}{n-2j-l-u}\mbox{ and }\label{Eq:4.0.10}\\ \binom{n}{j+l}\binom{j+l}{j+u}&=\binom{n}{j+u}\binom{n-j-u}{l-u}\mbox{.}\label{Eq:4.0.11} \end{align} If we use Eqns.~(\ref{Eq:4.0.10}) and (\ref{Eq:4.0.11}), Eq.~(\ref{Eq:4.0.9}) becomes \begin{multline}\label{Eq:4.0.12} D_S(n,j,t+1)\\=\sum_{l=0}^{n-2j} \sum_{u=0}^{l} \binom{n-j}{u}\binom{n-j-u}{n-2j-l-u}\binom{n}{j+u}\binom{n-j-u}{l-u}\binom{n}{j+l}^tF(n,j+l)\mbox{.} \end{multline} We now exchange the order of summation. If we take Inequalities (\ref{Inq:4.0.7}) and (\ref{Inq:4.0.8}) into consideration along with the symmetry $\binom{n-j-u}{n-2j-l-u}=\binom{n-j-u}{j+l} $, Eq.~(\ref{Eq:4.0.12}) becomes \begin{multline}\label{Eq:4.0.13} D_S(n,j,t+1)\\ =\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\sum_{l=u}^{n-2j-u}\binom{n-j}{u}\binom{n-j-u}{n-2j-l-u}\binom{n}{j+u}\binom{n-j-u}{l-u}\binom{n}{j+l}^tF(n,j+l)\mbox{ (Ineq.)}\\ =\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\sum_{l=u}^{n-2j-u}\binom{n-j}{u}\binom{n-j-u}{j+l}\binom{n}{j+u}\binom{n-j-u}{l-u}\binom{n}{j+l}^tF(n,j+l)\mbox{ (symmetry)}\\ =\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n}{j+u}\binom{n-j}{u}\sum_{l=u}^{n-2j-u}\binom{n-j-u}{l-u}\binom{n-j-u}{j+l}\binom{n}{j+l}^tF(n,j+l)\mbox{.} \end{multline} We substitute $u+v$ for $l$. From the Inequality (\ref{Inq:4.0.7}), we know that $0\leq j+u \leq \lfloor \frac{n}{2} \rfloor $. Then Eq.~(\ref{Eq:4.0.13}) becomes as follows: \begin{gather} \sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n}{j+u}\binom{n-j}{u}\sum_{v=0}^{n-2j-2u}\binom{n-j-u}{v}\binom{n-j-u}{j+u+v}\binom{n}{j+u+v}^tF(n,j+u+v)\notag\\ =\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n}{j+u}\binom{n-j}{u}D_S(n,j+u,t)\mbox{ (by Eq.~(\ref{Eq:1.0.2})).}\label{Eq:4.0.14} \end{gather} From Eqns.~(\ref{Eq:4.0.13}) and (\ref{Eq:4.0.14}), we obtain \[D_S(n,j,t+1)=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n}{j+u}\binom{n-j}{u}D_S(n,j+u,t)\mbox{.} \] This completes the proof of Theorem \ref{Th:1}. \end{proof} \section{Proof of Theorem \ref{Th:2}}\label{sec:4} The proof of Theorem \ref{Th:2} is similar to the proof of Theorem \ref{Th:1}. We use the same three binomial identities as in the previous proof. \begin{proof} By Definition \ref{Def:1} and Eq.~(\ref{Eq:1.0.2}), we know that \begin{equation} D_S(n,j,0)=\sum_{l=0}^{n-2j}\binom{n-j}{l}\binom{n-j}{j+l}F(n,j+l)\mbox{.}\label{Eq:5.0.1} \end{equation} We use Eq.~(\ref{Eq:4.0.4}) and the symmetry $\binom{j+l}{l-u}=\binom{j+l}{j+u}$. Again, Inequalities (\ref{Inq:4.0.5}), (\ref{Inq:4.0.6}), (\ref{Inq:4.0.7}), and (\ref{Inq:4.0.8}) hold. Then Eq.~(\ref{Eq:5.0.1}) becomes as follows: \begin{align} D_S(n,j,0)&=\sum_{l=0}^{n-2j}\left( \sum_{u=0}^{l}\binom{n-2j-l}{u}\binom{j+l}{l-u} \right)\binom{n-j}{j+l}F(n,j+l)\mbox{ (by Eq.~(\ref{Eq:4.0.4})) }\notag\\ &=\sum_{l=0}^{n-2j}\sum_{u=0}^{l}\binom{n-j}{j+l}\binom{j+l}{l-u}\binom{n-2j-l}{u}F(n,j+l)\mbox{ (permutation)}\notag\\ &=\sum_{l=0}^{n-2j}\sum_{u=0}^{l}\binom{n-j}{j+l}\binom{j+l}{j+u}\binom{n-2j-l}{u}F(n,j+l)\mbox{ (by symmetry).}\label{Eq:5.0.2} \end{align} By Eq.~(\ref{Eq:4.0.2}) and the symmetry $\binom{n-2j-u}{l-u}=\binom{n-2j-u}{n-2j-l}$ , it follows that \begin{equation} \binom{n-j}{j+l}\binom{j+l}{j+u}=\binom{n-j}{j+u}\binom{n-2j-u}{n-2j-l}\mbox{.}\label{Eq:5.0.3} \end{equation} If we use Eq.~(\ref{Eq:5.0.3}), Eq.~(\ref{Eq:5.0.2}) becomes \begin{equation} D_S(n,j,0)=\sum_{l=0}^{n-2j}\sum_{u=0}^{l}\binom{n-j}{j+u}\binom{n-2j-u}{n-2j-l}\binom{n-2j-l}{u}F(n,j+l)\mbox{.}\label{Eq:5.0.4} \end{equation} By Eq.~(\ref{Eq:4.0.2}) and the symmetry $\binom{n-2j-2u}{n-2j-l-u}=\binom{n-2j-2u}{l-u}$, it follows that \begin{equation} \binom{n-2j-u}{n-2j-l}\binom{n-2j-l}{u}=\binom{n-2j-u}{u}\binom{n-2j-2u}{l-u}\mbox{.}\label{Eq:5.0.5} \end{equation} If we use Eq.~(\ref{Eq:5.0.5}), Eq.~(\ref{Eq:5.0.4}) becomes \begin{equation} D_S(n,j,0)=\sum_{l=0}^{n-2j}\sum_{u=0}^{l}\binom{n-j}{j+u}\binom{n-2j-u}{u}\binom{n-2j-2u}{l-u}F(n,j+l)\mbox{.}\label{Eq:5.0.6} \end{equation} We exchange the order of summation. If we take Inequalities (\ref{Inq:4.0.7}) and (\ref{Inq:4.0.8}) into consideration, Eq.~(\ref{Eq:5.0.6}) becomes as follows: \begin{align} D_S(n,j,0)&=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\sum_{l=u}^{n-2j-u}\binom{n-j}{j+u}\binom{n-2j-u}{u}\binom{n-2j-2u}{l-u}F(n,j+l)\mbox{ (Ineq.)}\notag\\ &=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n-j}{j+u}\binom{n-2j-u}{u}\sum_{l=u}^{n-2j-u}\binom{n-2j-2u}{l-u}F(n,j+l)\mbox{.}\label{Eq:5.0.7} \end{align} We substitute $l$ by $u+v$. Then Eq.~(\ref{Eq:5.0.7}) becomes \begin{equation} D_S(n,j,0)=\sum_{u=0}^{\lfloor \frac{n-2j}{2} \rfloor}\binom{n-j}{j+u}\binom{n-2j-u}{u}\sum_{v=0}^{n-2j-2u}\binom{n-2j-2u}{v}F(n,j+u+v)\mbox{.}\label{Eq:5.0.8} \end{equation} Eq.~(\ref{Eq:5.0.8}) completes the proof of Theorem \ref{Th:2}. \end{proof} \section{How Does This Method Work?}\label{sec:5} In one particular situation, Theorem \ref{Th:1} implies a simple consequence which is important for us. Let $n$ be a fixed non-negative integer. Let $t_0$ be a non-negative integer, and let $j$ be an arbitrary integer in the range $0\leq j \leq \lfloor \frac{n}{2} \rfloor$. Suppose that $q=q(n)$ is a positive integer which divides $D_S(n,j,t_0)$ sums for all $j$ in the given range. We want $q$ to be as large as possible. Then it can be shown, by Theorem \ref{Th:1}, that $q$ divides $D_S(n,j,t_0+1)$ for all $j$ in the given range. By induction, it follows that $q$ divides $D_S(n,j,t)$ for all $t$ such that $t \geq t_0 $ and for all $j$ in the given range. By Eq.~(\ref{Eq:1.0.3}), it follows that $q$ divides $S(n,t+2)$ for all $t$ such that $t \geq t_0 $. This is exactly how this method works. \section{Two applications for non-negative sums}\label{sec:6} We give two applications of our method of $D$ sums for non-negative sums with absolute values. Let $n$ and $j$ be non-negative integers such that $j \leq n $, and let $t$ and $m$ be positive integers. First, we consider the sum $S_1(2n,m)=\sum_{k=0}^{2n}\binom{2n}{k}^m|n-k|$. It is known \cite[p.\ 3]{rpb259} that \begin{equation}\label{Eq:1.0.4} \sum_{k=0}^{2n}\binom{2n}{k}|n-k|=n\binom{2n}{n}\mbox{.} \end{equation} Therefore, we have \begin{equation} S_1(2n,1)=n\binom{2n}{n}\mbox{.}\notag \end{equation} We establish the following two lemmas: \begin{lemma}\label{lm:1} \begin{equation} D_{S_1}(2n,j,0)=n\binom{2n-j}{n}\binom{2n-j-1}{n}\mbox{\quad for}\quad 0 \leq j\leq n\mbox{.}\notag \end{equation} \end{lemma} \begin{lemma}\label{lm:2} \begin{equation} D_{S_1}(2n,j,1)=n\binom{2n}{n}\sum_{u=0}^{n-1-j}\binom{n}{j+u}\binom{2n-j}{u}\binom{2n-j-u-1}{n}\mbox{\quad for}\quad 0 \leq j\leq n\mbox{.}\notag \end{equation} \end{lemma} In particular, Lemmas \ref{lm:1} and \ref{lm:2} imply the following formulas: \begin{align} S_1(2n,2)&=n\binom{2n}{n}\binom{2n-1}{n}\mbox{,}\label{Eq:1.0.5}\\ S_1(2n,3)&=n\binom{2n}{n}\sum_{u=0}^{n-1}\binom{n}{u}\binom{2n}{u}\binom{2n-u-1}{n}\mbox{.}\label{Eq:1.0.6} \end{align} If $n$ is a positive integer, Lemma \ref{lm:2} suggests setting $ q_1(2n)=n\binom{2n}{n}$. We establish the following theorem: \begin{theorem} Let $n$ be a positive integer. The sum $D_{S_1}(2n,j,t)$ is divisible by $n\binom{2n}{n}$ for all positive integers $t$ and all integers $j$ such that $0\leq j \leq n $.\label{Th:3} \end{theorem} We conclude that \begin{corollary}\label{Cor:1} Let $n$ be a positive integer. The sum $S_1(2n,m)$ is divisible by $n\binom{2n}{n}$ for all positive integers $m$ . \end{corollary} Next, we consider the sum $S_2(2n+1,m)=\sum_{k=0}^{2n+1}\binom{2n+1}{k}^m \left |\frac{2n+1}{2}-k \right |$. It is known \cite[p.\ 3]{rpb259} that \begin{equation}\label{Eq:1.0.7} \sum_{k=0}^{2n+1}\binom{2n+1}{k}\left|\frac{2n+1}{2}-k\right|=(2n+1)\binom{2n}{n}\mbox{.} \end{equation} Therefore, we have \begin{equation} S_2(2n+1,1)=(2n+1)\binom{2n}{n} \mbox{.}\notag \end{equation} We establish the following two lemmas: \begin{lemma}\label{lm:3} \begin{equation} D_{S_2}(2n+1,j,0)=(2n+1-j)\binom{2n-j}{n}^2\mbox{\quad for}\quad 0 \leq j\leq n\mbox{.}\notag \end{equation} \end{lemma} \begin{lemma}\label{lm:4} \begin{equation} D_{S_2}(2n+1,j,1)=(2n+1)\binom{2n}{n}\sum_{u=0}^{n-j}\binom{n}{j+u}\binom{2n+1-j}{u}\binom{2n-j-u}{n}\mbox{\quad for}\quad 0 \leq j\leq n \mbox{.}\notag \end{equation} \end{lemma} In particular, Lemmas \ref{lm:3} and \ref{lm:4} imply the following formulas: \begin{align} S_2(2n+1,2)&=(2n+1)\binom{2n}{n}^2\mbox{,}\label{Eq:1.0.8}\\ S_2(2n+1,3)&=(2n+1)\binom{2n}{n}\sum_{u=0}^{n}\binom{n}{u}\binom{2n+1}{u}\binom{2n-u}{n}\mbox{.}\label{Eq:1.0.9} \end{align} Lemma \ref{lm:4} suggests setting $ q_2(2n+1)=(2n+1)\binom{2n}{n}$. We establish the following theorem: \begin{theorem} Let $n$ be a non-negative integer. The sum $D_{S_2}(2n+1,j,t)$ is divisible by $(2n+1)\binom{2n}{n}$ for all positive integers $t$ and all integers $j$ such that $0\leq j \leq n $.\label{Th:4} \end{theorem} We conclude that \begin{corollary}\label{Cor:2} Let $n$ be a non-negative integer. Then the sum $S_2(2n+1,m)$ is divisible by $(2n+1)\binom{2n}{n}$ for all positive integers $m$. \end{corollary} Due to clarity and brevity of this paper, we prove only Lemma \ref{lm:1}, Lemma \ref{lm:2}, Theorem \ref{Th:3}, and Corollary \ref{Cor:1}. Proofs of Lemma \ref{lm:3}, Lemma \ref{lm:4}, Theorem \ref{Th:4}, and Corollary \ref{Cor:2} are similar to proofs of Lemma \ref{lm:1}, Lemma \ref{lm:2}, Theorem \ref{Th:3}, and Corollary \ref{Cor:1}, respectively. Therefore, proofs of Lemma \ref{lm:3}, Lemma \ref{lm:4}, Theorem \ref{Th:4}, and Corollary \ref{Cor:2} are omitted. \section{Details of Theorem \ref{Th:3}}\label{sec:7} \subsection{Motivation} The Identity (\ref{Eq:1.0.4}) has a long history \cite[Introduction]{rpb255}. It was a problem in the $1974$ Putnam competition \cite[Problem A4]{Putnam}. Best \cite [Thm.\ 3]{Best} considered this identity in an application to Hadamard matrices. See also \cite[Thm.\ 15.2]{ErdSpe}, \cite[Chapter 2.5]{AlSpe}, and \cite{BrSpe}. Tuenter \cite{Tuenter} considered centered binomial sums of the form \[S_r(n)=\sum_{k=0}^{2n}\binom{2n}{k}|n-k|^r\mbox{,}\] which are a generalization of the Identity (\ref{Eq:1.0.4}). Brent \cite{rpb259} considered binomial sums \[U_r(n)=\sum_{k=0}^{n}\binom{n}{k}|n/2-k|^r\mbox{,}\] which are a generalization of Tuenter's sums. Brent gave recurrence relations for $U_r(n)$, where he used facts $U_1(2n)=n\binom{2n}{n}$ and $U_1(2n+1)=(2n+1)\binom{2n}{n}$. Brent cited a solution by Hillman \cite{Hillman} of the Putnam problem 35-A4. Hillman gave a closed form for $U_1(n)$, i.e., $U_1(n)=n\binom{n-1}{\lfloor \frac{n}{2}\rfloor} $. Identities (\ref{Eq:1.0.4}) and (\ref{Eq:1.0.7}) are connected with the following binomial identity \cite{rpb260,rpb255}: \begin{equation}\label{Eq:2.0.1} \sum_{i=-n}^{n}\sum_{j=-n}^{n}\binom{2n}{n+i}\binom{2n}{n+j}|i^2-j^2|=2n^2\binom{2n}{n}^2 \mbox{.} \end{equation} The Identity (\ref{Eq:2.0.1}) can be used in proofs of lower bounds for the Hadamard maximal determinant problem. We consider the sum $S_1(2n,m)$ as another generalization of the Identity (\ref{Eq:1.0.4}). Similarly, we consider the sum $S_2(2n+1,m)$ as another generalization of the Identity (\ref{Eq:1.0.7}). \subsection{Proof of Lemma \ref{lm:1}} Let $n$ be a non-negative integer, and let $m$ be a positive integer. We defined the sum $ S_1(2n,m)$ as \begin{equation} S_1(2n,m)=\sum_{k=0}^{2n}\binom{2n}{k}^m|n-k|\mbox{.}\label{Eq:6.0.1} \end{equation} Obviously, the sum $S_1(2n,m)$ is an instance of the sum (\ref{Eq:1.0.1}), where \begin{equation} F_1(2n,k)=|n-k|\mbox{.}\label{Eq:6.0.2} \end{equation} Let $j$ be a non-negative integer such that $j \leq n $. The proof of Lemma \ref{lm:1} is based on Theorem \ref{Th:2} and Eq.~(\ref{Eq:1.0.4}). \begin{proof} By Theorem \ref{Th:2}, we have \begin{gather} D_{S_1}(2n,j,0)=\sum_{u=0}^{\lfloor \frac{2n-2j}{2} \rfloor}\binom{2n-j}{j+u}\binom{2n-2j-u}{u}\sum_{v=0}^{2n-2j-2u}\binom{2n-2j-2u}{v}F_1(2n,j+u+v)\notag\\ =\sum_{u=0}^{n-j}\binom{2n-j}{j+u}\binom{2n-2j-u}{u}\sum_{v=0}^{2n-2j-2u}\binom{2n-2j-2u}{v}F_1(2n,j+u+v)\label{Eq:6.1.1}\\ =\sum_{u=0}^{n-j}\binom{2n-j}{j+u}\binom{2n-2j-u}{u}\sum_{v=0}^{2n-2j-2u}\binom{2n-2j-2u}{v}|n-j-u-v|\mbox{.}\label{Eq:6.1.2} \end{gather} Note that in Eq.~(\ref{Eq:6.1.1}), we used Eq.~(\ref{Eq:6.0.2}). By Eq.~(\ref{Eq:1.0.4}), we obtain \begin{equation} \sum_{v=0}^{2n-2j-2u}\binom{2n-2j-2u}{v}|n-j-u-v|=(n-j-u)\binom{2(n-j-u)}{n-j-u}\mbox{.}\label{Eq:6.1.3} \end{equation} If we use Eq.~(\ref{Eq:6.1.3}), the symmetry $\binom{2n-j}{j+u}=\binom{2n-j}{2n-2j-u}$, and the symmetry $\binom{2n-2j-u}{u}=\binom{2n-2j-u}{2n-2j-2u}$, Eq.~(\ref{Eq:6.1.2}) becomes as follows: \begin{gather} \sum_{u=0}^{n-j}\binom{2n-j}{j+u}\binom{2n-2j-u}{u}(n-j-u)\binom{2(n-j-u)}{n-j-u}\mbox{ (by Eq.~(\ref{Eq:6.1.3}))}\label{Eq:6.1.4}\\ =\sum_{u=0}^{n-1-j}\binom{2n-j}{2n-2j-u}\binom{2n-2j-u}{2n-2j-2u}\binom{2n-2j-2u}{n-j-u}(n-j-u)\mbox{ (symmetry).}\label{Eq:6.1.5} \end{gather} Note that the last term of the sum in Eq.~(\ref{Eq:6.1.4}) is equal to zero. Furthermore, we have \begin{align*} \binom{2n-2j-u}{2n-2j-2u}\binom{2n-2j-2u}{n-j-u}&=\binom{2n-2j-u}{n-j-u}\binom{n-j}{n-j-u}\mbox{ (by Eq.~(\ref{Eq:4.0.2}))}\\ &=\binom{2n-2j-u}{n-j}\binom{n-j}{u}\mbox{ (by symmetry).} \end{align*} Therefore, we obtain \begin{equation} \binom{2n-2j-u}{2n-2j-2u}\binom{2n-2j-2u}{n-j-u}=\binom{2n-2j-u}{n-j}\binom{n-j}{u}\mbox{.}\label{Eq:6.1.6} \end{equation} If we use Eq.~(\ref{Eq:6.1.6}), Eq.~(\ref{Eq:6.1.5}) becomes \begin{equation} D_{S_1}(2n,j,0)=\sum_{u=0}^{n-1-j}\binom{2n-j}{2n-2j-u}\binom{2n-2j-u}{n-j}\binom{n-j}{u}(n-j-u)\mbox{.}\label{Eq:6.1.7} \end{equation} By Eq.~(\ref{Eq:4.0.2}) and the symmetry $\binom{2n-j}{n-j}=\binom{2n-j}{n} $, it follows that \begin{equation} \binom{2n-j}{2n-2j-u}\binom{2n-2j-u}{n-j}=\binom{2n-j}{n}\binom{n}{n-j-u}\mbox{.}\label{Eq:6.1.8} \end{equation} If we use Eq.~(\ref{Eq:6.1.8}), Eq.~(\ref{Eq:6.1.7}) becomes as follows: \begin{align} D_{S_1}(2n,j,0)&=\sum_{u=0}^{n-1-j}\binom{2n-j}{n}\binom{n}{n-j-u}\binom{n-j}{u}(n-j-u)\mbox{ (by Eq.~(\ref{Eq:6.1.8}))}\notag\\ &=\binom{2n-j}{n}\sum_{u=0}^{n-1-j}\binom{n-j}{u}(n-j-u)\binom{n}{n-j-u}\mbox{ (permutation)}\label{Eq:6.1.9} \end{align} It is well-known that \begin{equation} k\binom{n}{k}=n\binom{n-1}{k-1}\mbox{.}\label{Eq:6.1.10} \end{equation} By Eq.~(\ref{Eq:6.1.10}), we get \begin{equation} (n-j-u)\binom{n}{n-j-u}=n\binom{n-1}{n-1-j-u}\mbox{.}\label{Eq:6.1.11} \end{equation} If we use Eq.~(\ref{Eq:6.1.11}), Eq.~(\ref{Eq:6.1.9}) becomes \begin{equation} D_{S_1}(2n,j,0)=n\binom{2n-j}{n}\sum_{u=0}^{n-1-j}\binom{n-j}{u}\binom{n-1}{n-1-j-u}\mbox{.}\label{Eq:6.1.12} \end{equation} By Eq.~(\ref{Eq:4.0.1}) and the symmetry $\binom{2n-1-j}{n-1-j}=\binom{2n-1-j}{n}$, we get \begin{equation} \sum_{u=0}^{n-1-j}\binom{n-j}{u}\binom{n-1}{n-1-j-u}=\binom{2n-1-j}{n}\mbox{.}\label{Eq:6.1.13} \end{equation} Finally, if we put Eq.~(\ref{Eq:6.1.13}) in Eq.~(\ref{Eq:6.1.12}), we obtain \begin{equation} D_{S_1}(2n,j,0)=n\binom{2n-j}{n}\binom{2n-1-j}{n}\mbox{.}\notag \end{equation} This completes the proof of Lemma \ref{lm:1}. \end{proof} \subsection{Proof of Lemma \ref{lm:2}} This proof is based on Theorem \ref{Th:1} and Lemma \ref{lm:1}. \begin{proof} By Theorem \ref{Th:1}, we have \begin{align} D_{S_1}(2n,j,1)&=\sum_{u=0}^{\lfloor \frac{2n-2j}{2} \rfloor}\binom{2n}{j+u}\binom{2n-j}{u}D_{S_1}(2n,j+u,0)\mbox{\quad (by Theorem \ref{Th:1})}\notag\\ &=\sum_{u=0}^{n-j}\binom{2n}{j+u}\binom{2n-j}{u}D_{S_1}(2n,j+u,0)\mbox{.}\label{Eq:6.2.1} \end{align} By Lemma \ref{lm:1}, we get \begin{equation} D_{S_1}(2n,j+u,0)=n\binom{2n-j-u}{n}\binom{2n-1-j-u}{n}\mbox{.}\label{Eq:6.2.2} \end{equation} Then Eq.~(\ref{Eq:6.2.1}) becomes as follows: \begin{align} D_{S_1}(2n,j,1)&=\sum_{u=0}^{n-j}\binom{2n}{j+u}\binom{2n-j}{u}n\binom{2n-j-u}{n}\binom{2n-1-j-u}{n}\mbox{ (by Eq.~(\ref{Eq:6.2.2}))}\notag\\ &=n\sum_{u=0}^{n-j}\binom{2n}{j+u}\binom{2n-j-u}{n}\binom{2n-j}{u}\binom{2n-1-j-u}{n}\mbox{ (permutation)}\notag\\ &=n\sum_{u=0}^{n-j}\binom{2n}{2n-j-u}\binom{2n-j-u}{n}\binom{2n-j}{u}\binom{2n-1-j-u}{n}\mbox{.}\label{Eq:6.2.3} \end{align} Note that we used the symmetry $\binom{2n}{j+u}=\binom{2n}{2n-j-u} $ in Eq.~(\ref{Eq:6.2.3}). Furthermore, the last term in Eq.~(\ref{Eq:6.2.3}) equals zero. It is readily verified that \begin{equation} \binom{2n}{2n-j-u}\binom{2n-j-u}{n}=\binom{2n}{n}\binom{n}{j+u}\mbox{.}\label{Eq:6.2.4} \end{equation} Then Eq.~(\ref{Eq:6.2.3}) becomes as follows: \begin{align} D_{S_1}(2n,j,1)&=n\sum_{u=0}^{n-j}\binom{2n}{n}\binom{n}{j+u}\binom{2n-j}{u}\binom{2n-1-j-u}{n}\mbox{ (by Eq.~(\ref{Eq:6.2.4}))}\notag\\ &=n\binom{2n}{n}\sum_{u=0}^{n-1-j}\binom{n}{j+u}\binom{2n-j}{u}\binom{2n-1-j-u}{n}\mbox{.}\label{Eq:6.2.5} \end{align} Eq.~(\ref{Eq:6.2.5}) implies that \[D_{S_1}(2n,j,1)=n\binom{2n}{n}\sum_{u=0}^{n-1-j}\binom{n}{j+u}\binom{2n-j}{u}\binom{2n-1-j-u}{n}\mbox{.} \] This completes the proof of Lemma \ref{lm:2}. \end{proof} \subsection{Proof of Theorem \ref{Th:3}} We assume that $n$ is a fixed positive integer and $j$ is a fixed non-negative integer such that $j \leq n$. We use induction on $t$. \begin{proof} When $t=1$, $D_{S_1}(2n,j,t)$ is divisible by $n\binom{2n}{n}$. This follows from Lemma \ref{lm:2}. The base case of induction is confirmed. We now assume that $D_{S_1}(2n,k,t)$ is divisible by $n\binom{2n}{n}$ for all non-negative integers $k$ such that $k \leq n$. What happens with $D_{S_1}(2n,j,t+1)$? By Theorem \ref{Th:1}, we have \begin{equation} D_{S_1}(2n,j,t+1)=\sum_{u=0}^{n-j}\binom{2n}{j+u}\binom{2n-j}{u}D_{S_1}(2n,j+u,t) \mbox{.}\label{Eq:6.3.1} \end{equation} Obviously, $0 \leq u+j \leq n$ for $ 0 \leq u \leq n-j$. By induction hypothesis, $D_{S_1}(2n,j+u,t)$ is divisible by $n\binom{2n}{n}$. By Eq.~(\ref{Eq:6.3.1}), it follows that $D_{S_1}(2n,j,t+1)$ is divisible by $n\binom{2n}{n}$. By induction, Theorem \ref{Th:3} follows. \end{proof} \subsection{Proof of Corollary \ref{Cor:1}} \begin{proof} By Eq.~(\ref{Eq:1.0.4}), it follows that $S_1(2n,1)$ is divisible by $n\binom{2n}{n}$. Setting $j=0$ in Lemma \ref{lm:1}, we obtain Eq.~(\ref{Eq:1.0.5}). By Eq.~(\ref{Eq:1.0.5}), it follows that $S_1(2n,2)$ is divisible by $n\binom{2n}{n}$. Let $m\geq 3$. By Eq.~(\ref{Eq:1.0.3}), we know that $S_1(2n,m)=D_{S_1}(2n,0,m-2)$. Since $m-2 \geq 1$, we can apply Theorem \ref{Th:3}. By Theorem \ref{Th:3}, $D_{S_1}(2n,0,m-2)$ is divisible by $n\binom{2n}{n}$. By Eq.~(\ref{Eq:1.0.3}), $S_1(2n,m)$ is divisible by $n\binom{2n}{n}$ for $m \geq 3 $. This completes the proof of Corollary \ref{Cor:1}. \end{proof} \begin{remark}\label{rem:1} Note that \begin{equation} S_1(2n,2)=\frac{n}{2}\binom{2n}{n}^2\mbox{.}\label{Eq:6.4.1} \end{equation} Setting $j=0$ in Lemma \ref{lm:2} and using Eq.~\textup{(}\ref{Eq:1.0.3}\textup{)}, we obtain Eq.~\textup{(}\ref{Eq:1.0.6}\textup{)}. \end{remark} \section{The third application for the alternating sum}\label{sec:8} We give a sketch of the proof of Calkin's result \cite[Thm.\ 1]{nc98} by using our method of $D$ sums. We use the well-known identity \cite [Eq.\ (1.25), p.\ 4]{Gould} \begin{equation}\label{Eq:8.0.1} \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}= \begin{cases}0, & \text{if } n>0;\\ 1, & \text{if } n=0, \end{cases} \end{equation} where $n$ is a non-negative integer. Eq.~(\ref{Eq:8.0.1}) plays the same role as Eq.~(\ref{Eq:1.0.4}) in the proof of Theorem \ref{Th:3}. Let $n$ be a non-negative integer and let $m$ be a positive integer. Let $S_3(2n,m)$ denote the sum $\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^m$. Obviously, the sum $S_3(2n,m)$ is an instance of the sum (\ref{Eq:1.0.1}), where $F_3(2n,k)=(-1)^k$. By Eq.~\eqref{Eq:8.0.1}, we conclude that $S_3(2n,1)$ is divisible by $\binom{2n}{n}$. Furthermore, it is known \begin{align*} S_3(2n,2)&=(-1)^n\binom{2n}{n}\mbox{,}\quad &&(\text{Kummer's formula})\\ S_3(2n,3)&=(-1)^n\binom{2n}{n}\binom{3n}{2n}\mbox{.}\quad &&(\text{Dixon's formula (\ref{eq:2.0.1})}) \end{align*} Therefore, it follows that $S_3(2n,m)$ is divisible by $\binom{2n}{n}$ for $1\leq m\leq 3$. By Eq.~(\ref{Eq:8.0.1}) and Theorem \ref{Th:2}, it can be proved that \begin{equation} D_{S_3}(2n,j,0)=(-1)^n\binom{2n-j}{n}\mbox{\quad for }\quad 0 \leq j \leq n\mbox{.}\label{eq:8.0.2} \end{equation} By Theorem \ref{Th:1} and Eq.~(\ref{eq:8.0.2}), it can be shown that \begin{equation} D_{S_3}(2n,j,1)=(-1)^n\binom{2n}{n}\binom{3n-j}{2n}\mbox{\quad for }\quad 0 \leq j \leq n\mbox{.}\label{eq:8.0.3} \end{equation} Setting $j=0$ in Eq.~(\ref{eq:8.0.3}) and using Eq.~(\ref{Eq:1.0.3}), we obtain Eq.~(\ref{eq:2.0.1}). Hence, the method of $D$ sums gives a proof of Eq.~(\ref{eq:2.0.1}). Furthermore, Eq.~(\ref{eq:8.0.3}) suggests setting $q_3(2n)=\binom{2n}{n}$. By Theorem \ref{Th:1}, it can be shown that $D_{S_3}(2n,j,t)$ is divisible by $\binom{2n}{n}$ for all positive integers $t$ and all integers $j$ such that $0\leq j \leq n$. Let $m\geq 4$. By Eq.~(\ref{Eq:1.0.3}), we know that $S_3(2n,m)=D_{S_3}(2n,0,m-2)$. Since $m-2 \geq 2$, the sum $D_{S_3}(2n,0,m-2)$ is divisible by $\binom{2n}{n}$. By Eq.~(\ref{Eq:1.0.3}), $S_3(2n,m)$ is divisible by $\binom{2n}{n}$ for $m \geq 4 $. Finally, we can conclude that $S_3(2n,m)$ is divisible by $\binom{2n}{n}$ for all non-negative integers $n$ and all positive integers $m$. This proves Calkin's result. \begin{remark}\label{rem:2} By using asymptotic methods, Bruijn \cite{Bruijn} has proved that no closed form exists for $S_3(n,m)$ when $m\geq4$. However, there are formulas for $S_3(2n,4)$ and $S_3(2n,5)$ \cite[Eq.\ \textup{(}5.13\textup{)}, Eq.\ \textup{(}5.12\textup{)}]{vq09}. Both formulas can be derived by using the method of $D$ sums. \end{remark} \section{Acknowledgments} I would like to thank Professor Du\v{s}ko Joji\'{c}. Also I would like to thank the referee for a very careful reading of this manuscript and useful suggestions. \begin{thebibliography}{22} \bibitem{AlSpe} N. Alon and J. H. Spencer,\textit{ The Probabilistic Method}, 3rd edn.\ , Wiley 2008. \bibitem{Putnam} Anonymous, Putnam Competition, 1974, available at \url{https://mks.mff.cuni.cz/kalva/putnam/putn74.html}. \bibitem{Best} M. R. Best, The excess of a Hadamard matrix, \textit{Indag.\ Math.\ \textup{(}N.S.\textup{)}} \textbf{39} (1977), 357--361. \bibitem{rpb259} R. P. Brent, Generalising Tuenter's binomial sums, \textit{J.\ Integer Sequences} \textbf{18} (2015), \href{https://cs.uwaterloo.ca/journals/JIS/VOL18/Brent/brent5.html}{Article 15.3.2}. \bibitem{rpb260} R. P. Brent, H. Ohtsuka, J. H. Osborn, and H. 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Tuenter, Walking into an absolute sum, \emph{Fibonacci Quart.\ }\textbf{40} (2002), 175--180. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 05A10 ; Secondary 05A19. \noindent \emph{Keywords:} Method of $D$ sums, binomial sum with absolute values, alternating binomial sum, recurrence equation, Dixon's formula, Hadamard maximal determinant problem. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A006480}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received May 30 2018; revised versions received July 5 2018; September 29 2018; October 1 2018; October 7 2018. Published in {\it Journal of Integer Sequences}, November 25 2018. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .