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\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 &amp;\leq l \mbox{,}\label{Inq:4.0.5}\\
   u&amp;\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 &amp;\leq u &amp;\leq \left \lfloor \frac{n-2j}{2} \right \rfloor \mbox{,}\label{Inq:4.0.7}\\
  u&amp;\leq l &amp;\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}&amp;=\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}&amp;=\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)&amp;=\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\\
&amp;=\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\\
&amp;=\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)&amp;=\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\\
&amp;=\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)&amp;=n\binom{2n}{n}\binom{2n-1}{n}\mbox{,}\label{Eq:1.0.5}\\
S_1(2n,3)&amp;=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)&amp;=(2n+1)\binom{2n}{n}^2\mbox{,}\label{Eq:1.0.8}\\
S_2(2n+1,3)&amp;=(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}&amp;=\binom{2n-2j-u}{n-j-u}\binom{n-j}{n-j-u}\mbox{ (by Eq.~(\ref{Eq:4.0.2}))}\\
                                                &amp;=\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)&amp;=\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\\
&amp;=\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)&amp;=\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\\
&amp;=\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)&amp;=\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\\
&amp;=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\\
&amp;=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)&amp;=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\\
&amp;=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, &amp; \text{if } n&gt;0;\\
1, &amp; \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)&amp;=(-1)^n\binom{2n}{n}\mbox{,}\quad &amp;&amp;(\text{Kummer's formula})\\
S_3(2n,3)&amp;=(-1)^n\binom{2n}{n}\binom{3n}{2n}\mbox{.}\quad &amp;&amp;(\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.

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\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.

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