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\title{\bf Products and sums divisible by\\ central binomial coefficients}

% input author, affilliation, address and support information as follows;
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% the street address

\author{Zhi-Wei Sun\thanks{Supported by the National Natural Science
Foundation (grant 11171140) of China and the PAPD of Jiangsu Higher
Education Institutions.}\\
\small Department of Mathematics\\[-0.8ex]
\small Nanjing University\\[-0.8ex]
\small Nanjing 210093, People's Republic of China\\
\small\tt zwsun@nju.edu.cn
}

%\date{\dateline{May 20, 2011}{Oct. 18, 2012}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{May 20, 2011}{Jan 5, 2013}{Jan 18, 2013}\\
\small Mathematics Subject Classifications: 05A10, 11B65, 11A07.}

\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
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\begin{abstract}
   In this paper we study products and sums divisible by
central binomial coefficients. We show that
$$2(2n+1)\bi{2n}n\ \bigg|\ \bi{6n}{3n}\bi{3n}n\ \ \t{for all}\ n=1,2,3,\ldots.$$
Also, for any nonnegative integers $k$ and $n$ we have
$$\bi {2k}k\ \bigg|\ \bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n-k+1}n$$
and
$$\bi{2k}k\ \bigg|\ (2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k},$$
where $C_m$ denotes the Catalan number
$\f1{m+1}\bi{2m}m=\bi{2m}m-\bi{2m}{m+1}$. On the basis of these
results, we obtain certain sums divisible by central binomial
coefficients.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:}  central binomial coefficients; divisibility; congruences
\end{abstract}

\section{Introduction}

Central binomial coefficients are given by $\bi{2n} n$ with
$n\in\N=\{0,1,2,\ldots\}$. The Catalan numbers
$$C_n=\f1{n+1}\bi{2n}n=\bi{2n}n-\bi{2n}{n+1}\ \ (n=0,1,2,\ldots)$$
play important roles in combinatorics (cf. R. P. Stanley~\cite[pp.\,219-229]{St}).
There are many sophisticated congruences involving
central binomial coefficients and Catalan numbers (see, e.g.,
\cite{S10a, ST1, ST2}).

 In 1998 N. J. Calkin \cite{C} proved that $\bi{2n}n\mid\sum_{k=-n}^n(-1)^k\bi{2n}{n+k}^m$
for any $m,n\in\Z^+$. See also V.J.W. Guo, F. Jouhet and J. Zeng
\cite{GJZ}, and H. Q. Cao and H. Pan \cite{CP} for further
extensions of Calkin's result.


In this paper we investigate a new kind of divisibility problems
involving central binomial coefficients.

Our first theorem is as follows.

\begin{theorem}\label{Th1.1}
{\rm (i)} For any positive integer $n$ we have

\begin{equation}\label{1.1}2(2n+1)\bi{2n}n\ \bigg|\ \bi{6n}{3n}\bi{3n}n.\end{equation}

{\rm (ii)} Let $k$ and $n$ be nonnegative integers. Then
\begin{equation}\label{1.2}\bi {2k}k\
\bigg|\ \bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n-k+1}n\end{equation}
 and
\begin{equation}\label{1.3}\bi{2k}k\ \bigg|\ (2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k}.\end{equation}
\end{theorem}

In view of (\ref{1.1}) it is worth introducing the sequence
$$S_n=\f{\bi{6n}{3n}\bi{3n}n}{2(2n+1)\bi{2n}n}\ \ (n=1,2,3,\ldots).$$
Here we list the values of $S_1,\ldots,S_8$:
\begin{gather*} 5,\
231,\ 14568,\ 1062347,\ 84021990,
\\7012604550,\ 607892634420,\ 54200780036595.
\end{gather*}
The author generated this sequence as A176898 at N.J.A Sloane's
OEIS (cf. \cite{S10b}). By Stirling's formula, $S_n\sim
108^n/(8n\sqrt{n\pi})$ as $n\to+\infty$.
 Set $S_0=1/2$.  Using {\tt Mathematica} we find that
  $$\sum_{k=0}^\infty S_kx^k=\f{\sin(\f23\arcsin(6\sqrt{3x}))}{8\sqrt{3x}}\ \ \left(0<x<\f1{108}\right)$$
  and in particular
  $$\sum_{k=0}^\infty\f{S_k}{108^k}=\f{3\sqrt3}8.$$
 {\tt Mathematica} also yields that
 $$\sum_{k=0}^\infty\f{S_k}{(2k+3)108^k}=\f{27\sqrt3}{256}.$$
 It would be interesting to find a combinatorial interpretation or recursion for the sequence $\{S_n\}_{n\gs1}$.

 One can easily show that $S_p\eq15-30p+60p^2\ (\mo\ p^3)$ for any odd prime $p$.
Below we present a conjecture concerning congruence properties of
the sequence $\{S_n\}_{n\gs1}$.

\begin{conjecture}\label{Conj1.1} {\rm (i)} Let
$n\in\Z^+=\{1,2,3,\ldots\}$. Then
 $S_n$ is odd if and only if $n$ is a power of two. Also, $3S_n\eq0\ (\mo\ 2n+3)$.

 {\rm (ii)} For any prime $p>3$ we have
 $$\sum_{k=1}^{p-1}\f{S_k}{108^k}\eq\begin{cases} 0\ (\mo\ p)&\t{if}\ p\eq\pm1\ (\mo\ 12),
 \\-1\ (\mo\ p)&\t{if}\ p\eq\pm5\ (\mo\ 12).\end{cases}$$
 \end{conjecture}
 \begin{remark} Part (i) of Conjecture \ref{Conj1.1} might be
 shown by our method for proving Theorem \ref{Th1.1}(i).
\end{remark}
 \medskip

Our following conjecture is concerned with a companion sequence of
$\{S_n\}_{n\gs0}$.

\begin{conjecture}\label{Conj1.2} There are positive integers $T_1,T_2,T_3,\ldots$ such that
$$\sum_{k=0}^\infty S_kx^{2k+1}+\f1{24}-\sum_{k=1}^\infty T_kx^{2k}
=\f{\cos(\f23\arccos(6\sqrt 3x))}{12}$$
for all real $x$ with $|x|\ls 1/(6\sqrt3)$. Also, $T_p\eq-2\ (\mo\
p)$ for any prime $p$.
\end{conjecture}

Here we list the values of $T_1,\ldots,T_8$:
\begin{gather*} 1,\ 32,\ 1792,\
122880,\ 9371648,
\\763363328,\ 65028489216,\ 5722507051008.
\end{gather*}

In 1914 Ramanujan \cite{R} obtained that
$$\sum_{k=0}^\infty\f{4k+1}{(-64)^k}\bi{2k}k^3=\f2{\pi}$$
and
$$\sum_{k=0}^\infty(20k+3)\f{\bi{2k}k^2\bi{4k}{2k}}{(-2^{10})^k}=\f 8{\pi}.$$
(See also \cite{BB,BBC,Be} for such series.) Actually the first
identity was originally proved by G. Bauer in 1859.
 Both identities can be
proved via the WZ (Wilf-Zeilberger) method (cf. M. Petkov\v sek, H.
S. Wilf and D. Zeilberger \cite{PWZ}, and Zeilberger \cite{Z} for
this method). For WZ proofs of the two identities, see S.~B. Ekhad and D. Zeilberger \cite{EZ} and Guillera \cite{G}.
van Hamme \cite{vH} conjectured that the first identity has a
$p$-adic analogue. This conjecture was first proved by E. Mortenson
\cite{M}, and recently re-proved in \cite{Zu} via the WZ method.

On the basis of Theorem \ref{Th1.1}, we deduce the following new
result.

\begin{theorem}\label{Th1.2} For any positive integer $n$ we have
\begin{equation}\label{1.4}4(2n+1)\bi{2n}n\ \bigg|\
\sum_{k=0}^{n}(4k+1)\bi{2k}k^3(-64)^{n-k}\end{equation} and
\begin{equation}\label{1.5}4(2n+1)\bi{2n}n\ \bigg|\
\sum_{k=0}^{n}(20k+3)\bi{2k}k^2\bi{4k}{2k}(-2^{10})^{n-k}.\end{equation}
\end{theorem}

\medskip

Now we pose two more conjectures.

\begin{conjecture}\label{Conj1.3} {\rm (i)} For any $n\in\Z^+$ we have
$$a_n:=\f1{8n^2\bi{2n}n^2}\sum_{k=0}^{n-1}(205k^2+160k+32)(-1)^{n-1-k}\bi{2k}k^5\in\Z^+.$$

{\rm (ii)} Let $p$ be an odd prime. If $p\not=3$ then
$$\sum_{k=0}^{(p-1)/2}(205k^2+160k+32)(-1)^k\bi{2k}k^5\eq 32p^2+\f{896}3p^5B_{p-3}\ (\mo\ p^6),$$
where $B_0,B_1,B_2,\ldots$ are Bernoulli numbers. If $p\not=5$ then
$$\sum_{k=0}^{p-1}(205k^2+160k+32)(-1)^k\bi{2k}k^5\eq 32p^2+64p^3H_{p-1}\ (\mo\ p^7),$$
where $H_{p-1}=\sum_{k=1}^{p-1}1/k$.
\end{conjecture}
\begin{remark}. Note that $a_1=1$ and
$$4(2n+1)^2a_{n+1}+n^2a_n=(205n^2+160n+32)\bi{2n-1}n^3\ \ \ \t{for}\ n=1,2,\ldots.$$
The author generated the sequence $\{a_n\}_{n>0}$ at OEIS as A176285
(cf. \cite{S10b}). In 1997 T. Amdeberhan and D. Zeilberger \cite{AZ} used the
WZ method to obtain
$$\sum_{k=1}^\infty\f{(-1)^k(205k^2-160k+32)}{k^5\bi{2k}k^5}=-2\zeta(3).$$
\end{remark}

\begin{conjecture}\label{Conj1.4} {\rm (i)} For any odd prime $p$, we
have
$$\sum_{k=0}^{p-1}\f{28k^2+18k+3}{(-64)^k}\bi{2k}k^4\bi{3k}k\eq 3p^2-\f 72p^5B_{p-3}\ (\mo\ p^6),$$
and
$$\sum_{k=0}^{(p-1)/2}\f{28k^2+18k+3}{(-64)^k}\bi{2k}k^4\bi{3k}k\eq 3p^2+6\left(\f{-1}p\right)p^4E_{p-3}\ (\mo\ p^5),$$
where $E_0,E_1,E_2,\ldots$ are Euler numbers.

{\rm (ii)} For any integer $n>1$, we have
$$\sum_{k=0}^{n-1}(28k^2+18k+3)\bi{2k}k^4\bi{3k}k(-64)^{n-1-k}\eq0\ \ \(\mo\ (2n+1)n^2\bi{2n}n^2\).$$
Also,
$$\sum_{k=1}^\infty\f{(28k^2-18k+3)(-64)^k}{k^5\bi{2k}k^4\bi{3k}k}=-14\zeta(3).$$
\end{conjecture}
\begin{remark} The conjectural series for
$\zeta(3)=\sum_{n=1}^\infty1/n^3$ was first announced by the author
in a message to Number Theory Mailing List (cf. \cite{S10c}) on
April 4, 2010.
\end{remark}

\medskip

For more conjectures similar to Conjectures \ref{Conj1.3} and
\ref{Conj1.4} the reader may consult \cite{S09} and \cite{S10b}.

In the next section we will establish three auxiliary inequalities
involving the floor function. Sections 3 and 4 are devoted to the
proofs of Theorem \ref{Th1.1} and Theorem \ref{Th1.2} respectively.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Three auxiliary inequalities}


In this section, for a rational number $x$ we let $\{x\}=x-\lfloor
x\rfloor$ be the fractional part of $x$, and set $\{x\}_m=m\{x/m\}$
for any $m\in\Z^+$.

\begin{theorem}\label{Th2.1}  Let $m>1$ be an integer. Then for any $n\in\Z$ we have
\begin{equation}\label{2.1}\left\lfloor\f nm\right\rfloor+\left\lfloor\f{6n}m\right\rfloor\gs\left\lfloor\f{2n}m\right\rfloor+\left\lfloor\f{2n+1}m\right\rfloor
+\left\lfloor\f{3n}m\right\rfloor.\end{equation}
\end{theorem}
\begin{proof} Let $A_m(n)$ denote the left-hand side of (\ref{2.1}) minus the
right-hand side. Then
$$A_m(n)=\left\{\f{2n}m\right\}+\left\{\f{2n+1}m\right\}+\left\{\f{3n}m\right\}-\f1m-\left\{\f nm\right\}-\left\{\f{6n}m\right\},$$
which only depends on $n$ modulo $m$. So, without any loss of generality we may simply assume that $n\in\{0,\ldots,m-1\}$.
Hence $A_m(n)\gs0$ if and only if
\begin{equation}\label{2.2}\left\{\f{2n}m\right\}+\left\{\f{2n+1}m\right\}+\left\{\f{3n}m\right\}\gs \f{n+1}m.\end{equation}
(Note that $2n+(2n+1)+3n-(n+1)=6n$.)

(\ref{2.1}) is obvious when $n=0$. If $1\ls n<m/2$, then $\{2n/m\}=2n/m\gs(n+1)/m$ and hence (\ref{2.2}) holds.
In the case $n\gs m/2$, (\ref{2.2}) can be simplified as
$$\f{3n}m+\left\{\f{3n}m\right\}\gs2,$$
which holds since $3n\gs m+m/2$.

By the above we have proved (\ref{2.1}).
\end{proof}

\begin{theorem}\label{Th2.2} Let $m\in\Z^+$ and $k,n\in\Z$. Then we have
\begin{equation}\label{2.3}\left\lfloor\f{4n+2k+2}m\right\rfloor-\left\lfloor\f{2n+k+1}m\right\rfloor+2\left\lfloor \f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor
\gs\left\lfloor\f nm\right\rfloor+\left\lfloor\f{n-k+1}m\right\rfloor,\end{equation}
unless $2\mid m$ and $k\eq n+1\eq m/2\ (\mo\ m)$ in which case the
right-hand side of the inequality equals the left-hand side plus
one.
\end{theorem}
\begin{proof} Since
$$(4n+2k+2)-(2n+k+1)+2k-2(2k)=n+(n-k+1),$$
(\ref{2.3}) has the following equivalent form:
\begin{equation}\label{2.4}\left\{\f{4n+2k+2}m\right\}-\left\{\f{2n+k+1}m\right\}+2\left\{\f
km\right\}-2\left\{\f{2k}m\right\}\ls \left\{\f nm\right\}+\left\{\f{n-k+1}m\right\}.\end{equation}
Note that this only depends on $k$ and $n$ modulo $m$. So, without any loss of generality,
we may simply assume that $k,n\in\{0,\ldots,m-1\}$.

{\it Case} 1. $k<m/2$ and $\{2n+k+1\}_m<m/2$.

In this case, (\ref{2.4}) can be simplified as
$$\f{n+2k}m+\left\{\f{n-k+1}m\right\}\gs\left\{\f{2n+k+1}m\right\},$$
which is true since the left-hand side is nonnegative and $(n+2k)+(n-k+1)\eq2n+k+1\ (\mo\ m)$.

{\it Case} 2. $k<m/2$ and $\{2n+k+1\}_m\gs m/2$.

In this case, (\ref{2.4}) can be simplified as
$$\f{n+2k}m+\left\{\f{n-k+1}m\right\}\gs\left\{\f{2n+k+1}m\right\}-1,$$
which holds trivially since the right-hand side is negative.

{\it Case} 3. $k\gs m/2$ and $\{2n+k+1\}_m<m/2$.

In this case, (\ref{2.4}) can be simplified as
$$\f{n+2k}m+\left\{\f{n-k+1}m\right\}\gs 2+\left\{\f{2n+k+1}m\right\}.$$
Since $(n+2k)+(n-k+1)=2n+k+1$, this is equivalent to
$$n+2k+\{n-k+1\}_m\gs 2m.$$

If $k>n+1$, then
$$n+2k+\{n-k+1\}_m=n+2k+(n-k+1+m)=2n+k+1+m\gs 2m$$
since $2n+k+1>k\gs m/2$ and $\{2n+k+1\}_m<m/2$.

Now assume that $k\ls n+1$. Clearly
$$n+2k+\{n-k+1\}_m=n+2k+(n-k+1)=2n+k+1\gs 3k-1.$$
If $k>m/2$ then $3k-1\gs 3(m+1)/2-1>3m/2$.
If $k\ls n$ then $2n+k+1>3k\gs 3m/2$.
So, except the case $k=n+1=m/2$ we have
$$n+2k+\{n-k+1\}_m=2n+k+1\gs 3m/2$$ and
hence $n+2k+\{n-k+1\}_m=2n+k+1\gs 2m$ since $\{2n+k+1\}_m<m/2$.

When $k=n+1=m/2$, the left-hand side of (\ref{2.4}) minus the right-hand side equals
$$\f{m-2}m-\f{m/2-1}m+2\f{m/2}m-\f{m/2-1}m=1.$$


{\it Case} 4. $k\gs m/2$ and $\{2n+k+1\}_m\gs m/2$.

In this case, clearly $m\not=1$, and (\ref{2.4}) can be simplified as
$$\f{n+2k}m+\left\{\f{n-k+1}m\right\}\gs 1+\left\{\f{2n+k+1}m\right\}$$
which is equivalent to
$$n+2k+\{n-k+1\}_m\gs m.$$
If $k\ls n+1$, then
$$n+2k+\{n-k+1\}_m=n+2k+(n+1-k)=2n+k+1\gs 3k-1\gs \f {3m}2-1\gs m.$$
If $k>n+1$, then
$$n+2k+\{n-k+1\}_m=n+2k+(n+1-k)+m=2n+k+1+m>m.$$

In view of the above, we have completed the proof of Theorem \ref{Th2.2}.
\end{proof}

\begin{theorem}\label{Th2.3} Let $m\in\Z^+$ and $k,n\in\Z$. Then we have
\begin{equation}\label{2.5}\begin{aligned}&\left\lfloor\f{2n+2k}m\right\rfloor
-\left\lfloor\f{n+k}m\right\rfloor+2\left\lfloor \f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor
\\&\ \ \gs2\left\lfloor\f
nm\right\rfloor-\left\lfloor\f{2n+1}m\right\rfloor+\left\lfloor\f{n-k+1}m\right\rfloor,
\end{aligned}\end{equation}
unless $2\mid m$ and $k\eq n+1\eq m/2\ (\mo\ m)$ in which case the
right-hand side of the inequality equals the left-hand side plus
one.
\end{theorem}
\begin{proof} Since
$$2n+2k-(n+k)+2k-2(2k)=2n-(2n+1)+(n-k+1),$$
(\ref{2.5}) is equivalent to the following inequality:
\begin{equation}\label{2.6}\begin{aligned}&\left\{\f{2n+2k}m\right\}-\left\{\f{n+k}m\right\}+2\left\{\f km\right\}-2\left\{\f{2k}m\right\}
\\ &\ \ \ls2\left\{\f nm\right\}-\left\{\f{2n+1}m\right\}+\left\{\f{n-k+1}m\right\}.
\end{aligned}\end{equation}
As (\ref{2.6}) only depends on $k$ and $n$ modulo $m$, without loss of generality we simply assume that $k,n\in\{0,\ldots,m-1\}$.

{\it Case} 1. $k<m/2$ and $\{n+k\}_m<m/2$.

In this case, (\ref{2.6}) can be simplified as
$$\f{2n+2k}m+\left\{\f{n-k+1}m\right\}\gs\left\{\f{2n+1}m\right\}+\left\{\f{n+k}m\right\}$$
which holds since
$$\f{2n+2k}m-\left\{\f{n+k}m\right\}+\left\{\f{n-k+1}m\right\}\gs0$$
and $2n+2k-(n+k)+(n-k+1)=2n+1.$

{\it Case} 2. $k<m/2$ and $\{n+k\}_m\gs m/2$.

In this case, (\ref{2.6}) can be simplified as
$$\f{2n+2k}m+\left\{\f{n-k+1}m\right\}\gs\left\{\f{2n+1}m\right\}+\left\{\f{n+k}m\right\}-1$$
which holds since
$$\f{2n+2k}m\gs\f{n+k}m\gs\left\{\f{n+k}m\right\}
\ \t{and}\ \left\{\f{n-k+1}m\right\}\gs0>\left\{\f{2n+1}m\right\}-1.$$


{\it Case} 3. $k\gs m/2$ and $\{n+k\}_m<m/2$.

In this case, we must have $n+k\gs m$ and hence $\{n+k\}_m=n+k-m$. Thus (\ref{2.6}) can be simplified as
$$\f{n+k-m}m+\left\{\f{n-k+1}m\right\}\gs\left\{\f{2n+1}m\right\}$$
which holds trivially since $n+k-m+(n-k+1)\eq 2n+1\ (\mo\ m)$.

{\it Case} 4. $k\gs m/2$ and $\{n+k\}_m\gs m/2$.

In this case, (\ref{2.6}) can be simplified as
$$\f{2n+2k}m-\left\{\f{n+k}m\right\}+\left\{\f{n-k+1}m\right\}\gs 1+\left\{\f{2n+1}m\right\}$$
which is equivalent to
\begin{equation}\label{2.7}\f{2(n+k)}m-\left\{\f{n+k}m\right\}+\left\{\f{n-k+1}m\right\}\gs1\end{equation}
since $2n+2k-(n+k)+(n-k+1)=2n+1$.

Clearly (\ref{2.7}) holds if $n+k\gs m$.
If $n+k<m$ and $k>n+1$, then the left-hand side of the inequality (\ref{2.7}) is
$$\f{n+k}m+\f{n+1-k}m+1=\f{2n+1}m+1>1.$$

Now assume that $n+k<m$ and $k\ls n+1$. Then (\ref{2.7}) is equivalent to $2n+1\gs m$.
If $k\ls n$ then $2n+1>2k\gs m$. If $k=n+1\not= m/2$, then
$k=n+1\gs(m+1)/2$ and hence $2n+1=2(n+1)-1\gs m$.

When $k=n+1=m/2$, the left-hand side of (\ref{2.6}) minus the right-hand side equals
$$\f{m-2}m-\f{m-1}m+2\f{m/2}m-2\f{m/2-1}m+\f{m-1}m=1.$$

Combining the discussion of the four cases we obtain the desired result.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem \ref{Th1.1}}

For a prime $p$,  the $p$-adic evaluation of an integer $m$ is given by
$$\nu_p(m)=\sup\{a\in\N:\ p^a\mid m\}.$$
For a rational number $x=m/n$ with $m\in\Z$ and $n\in\Z^+$, we set $\nu_p(x)=\nu_p(m)-\nu_p(n)$ for any prime $p$.
Note that a rational number $x$ is an integer if and only if $\nu_p(x)\gs0$ for all primes $p$.

\medskip\noindent
\begin{proof}[Proof of Theorem \ref{Th1.1}] (i) Fix $n\in\Z^+$, and define $A_m(n)$ for $m>1$ as in the proof of Theorem \ref{Th2.1}.
Observe that
$$Q:=\f{\bi{6n}{3n}\bi{3n}n}{(2n+1)\bi{2n}n}=\f{n!(6n)!}{(2n)!(2n+1)!(3n)!}.$$
So, for any prime $p$ we have
$$\nu_p(Q)=\sum_{i=1}^\infty A_{p^i}(n)\gs0$$
by Theorem \ref{Th2.1}. Therefore $Q$ is an integer.

Choose $j\in\Z^+$ such that $2^{j-1}\ls n<2^j$.
As $2n+1\ls 2(2^j-1)+1<2^{j+1}$, we have
\begin{align*} &\left\lfloor\f n{2^{j+1}}\right\rfloor+\left\lfloor\f{6n}{2^{j+1}}\right\rfloor
-\left\lfloor\f{2n}{2^{j+1}}\right\rfloor-\left\lfloor\f{2n+1}{2^{j+1}}\right\rfloor
-\left\lfloor\f{3n}{2^{j+1}}\right\rfloor
\\=&\left\lfloor\f{3n}{2^j}\right\rfloor-\left\lfloor\f{3n}{2^{j+1}}\right\rfloor
=\left\lfloor\f{3n+2^j}{2^{j+1}}\right\rfloor\gs\left\lfloor\f{2n+2^j}{2^{j+1}}\right\rfloor\gs1.
\end{align*}
Therefore
$$\nu_2(Q)=\sum_{i=1}^\infty A_{2^i}(n)\gs A_{2^{j+1}}(n)\gs1.$$
and hence $Q$ is even. This proves (\ref{1.1}).

(ii) (\ref{1.2}) and (\ref{1.3}) are obvious in the case $k=0$. If $k>n+1$, then
$$\bi{2n+k+1}{2k}=\bi{n+k+1}{2k}=0$$
and hence (\ref{1.2}) and (\ref{1.3}) hold trivially.
Below we assume that $1\ls k\ls n+1$.

 Recall that for any nonnegative integer $m$ and prime $p$ we have
 $$\nu_p(m!)=\sum_{i=1}^\infty\left\lfloor\f m{p^i}\right\rfloor.$$
 Since
 $$\f{\bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n+k+1}n}{\bi{2k}k}
 =\f{(4n+2k+2)!(k!)^2}{(2n+k+1)!((2k)!)^2n!(n-k+1)!}$$
 and
 $$\f{(2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k}}{\bi{2k}k}=\f{(2n+1)!(2n+2k)!(k!)^2}{(n!)^2(n+k)!((2k)!)^2(n-k+1)!},$$
 it suffices to show that for any prime $p$ we have
 $$\sum_{i=1}^\infty C_{p^i}(n,k)\gs0\ \ \t{and}\ \ \sum_{i=1}^\infty D_{p^i}(n,k)\gs0,$$
 where
 \begin{align*} C_m(n,k)=&\left\lfloor\f{4n+2k+2}m\right\rfloor-\left\lfloor\f{2n+k+1}m\right\rfloor
 +2\left\lfloor\f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor
 \\&-\left\lfloor\f nm\right\rfloor-\left\lfloor\f{n-k+1}m\right\rfloor
 \end{align*}
 and
 \begin{align*} D_m(n,k)=&\left\lfloor\f{2n+2k}m\right\rfloor-\left\lfloor\f{n+k}m\right\rfloor
 +2\left\lfloor\f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor
 \\&-2\left\lfloor\f nm\right\rfloor+\left\lfloor\f{2n+1}m\right\rfloor-\left\lfloor\f{n-k+1}m\right\rfloor.
\end{align*}

(a) By Theorem \ref{Th2.2}, $C_{p^i}(n,k)\gs0$ unless $p=2$ and $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$
in which case $C_{2^i}(n,k)=-1$. Suppose that $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$, $k=2^{i-1}k_0$ and $n+1=2^{i-1}n_0$, where
$1\ls k_0\ls n_0$ and $k_0$ and $n_0$ are odd. If $i\gs2$, then
$$C_{2^{i-1}}(n,k)=4n_0+2k_0-1-(2n_0+k_0-1)+2k_0-4k_0-(n_0-1)-(n_0-k_0)=1$$
and hence $C_{2^{i-1}}(n,k)+C_{2^i}(n,k)=1+(-1)=0$.
So it remains to consider the case $k\eq n+1\eq1\ (\mo\ 2)$.

Assume that $k$ is odd and $n$ is even. Write $k+1=2^jk_1$ and $n=2n_1$ with $k_1,n_1\in\Z^+$ and $2\nmid k_1$.
Then it is easy to see that
\begin{align*} 
C_{2^{j+1}}(n,k)
&=\left\lfloor\f{4n_1}{2^j}\right\rfloor+k_1-\left\lfloor\f{2n_1-2^{j-1}
+2^{j-1}(k_1-1)}{2^j}\right\rfloor
\\
&\quad+2\left\lfloor\f{k_1}2\right\rfloor-2\left\lfloor\f{2^jk_1-1}{2^j}\right\rfloor
-\left\lfloor\f{n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1-2^{j-1}k_1}{2^j}\right\rfloor
\\
&=\left\lfloor\f{4n_1}{2^j}\right\rfloor+k_1-\left\lfloor\f{2n_1-2^{j-1}}{2^j}\right\rfloor-\f{k_1+1}2+k_1-1-2(k_1-1)
\\
&\quad-\left\lfloor\f{n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1+2^{j-1}}{2^j}\right\rfloor+\f{k_1+1}2
\\
&=1+\left\lfloor \f{n_1+(n_1+1+2^{j-1})+(2n_1-2^{j-1})}{2^j}\right\rfloor
\\
&\quad-\left\lfloor\f {n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1+2^{j-1}}{2^j}\right\rfloor-\left\lfloor\f{2n_1-2^{j-1}}{2^j}\right\rfloor
\\
&\gs 1
\end{align*}
and hence $C_2(n,k)+C_{2^{j+1}}(n,k)\gs0$.

By the above, we do have $\sum_{i=1}^\infty C_{p^i}(n,k)\gs0$ for any prime $p$. So (\ref{1.2}) holds.


(b) By Theorem \ref{Th2.2}, $D_{p^i}(n,k)\gs0$ unless $p=2$ and $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$
in which case $D_{2^i}(n,k)=-1$. So, to prove (\ref{1.2}) it suffices to find a positive integer $j$
such that $D_{2^j}(n,k)\gs1$.

Clearly there is a unique positive integer $j$ such that $2^{j-1}\ls n+k<2^j$. Note that $k\ls(n+k)/2<2^{j-1}$
and $$D_{2^j}(n,k)=1+\left\lfloor\f{2n+1}{2^j}\right\rfloor\gs1.$$
This concludes the proof of (\ref{1.3}).

  The proof of Theorem \ref{Th1.1} is now complete.
  \end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem \ref{Th1.2}}

\begin{proof}[Proof of Theorem \ref{Th1.2}] (i) We first prove (\ref{1.4}). For $k,n\in\N$ define
$$F(n,k)=\f{(-1)^{n+k}(4n+1)}{4^{3n-k}}\bi{2n}n^2\f{\bi{2n+2k}{n+k}\bi{n+k}{2k}}{\bi{2k}k}$$
and
$$G(n,k)=\f{(-1)^{n+k}(2n-1)^2\bi{2n-2}{n-1}^2}{2(n-k)4^{3(n-1)-k}}\bi{2(n-1+k)}{n-1+k}\f{\bi{n-1+k}{2k}}{\bi{2k}k}.$$
Clearly $F(n,k)=G(n,k)=0$ if $n<k$. By \cite{EZ},
$$F(n,k-1)-F(n,k)=G(n+1,k)-G(n,k)$$
for all $k\in\Z^+$ and $n\in\N$.

Fix a positive integer $N$. Then
\begin{align*} \sum_{n=0}^NF(n,0)-F(N,N)
&=\sum_{n=0}^NF(n,0)-\sum_{n=0}^NF(n,N)
\\
&=\sum_{k=1}^N\(\sum_{n=0}^NF(n,k-1)-\sum_{n=0}^NF(n,k)\)
\\
&=\sum_{k=1}^N\sum_{n=0}^N(G(n+1,k)-G(n,k))
=\sum_{k=1}^NG(N+1,k).
\end{align*}
Note that
$$\sum_{n=0}^NF(n,0)=\sum_{n=0}^N\f{4n+1}{(-64)^n}\bi{2n}n^3$$
and
$$F(N,N)=\f{4N+1}{4^{2N}}\bi{2N}N\bi{4N}{2N}=\f{(4N+1)(2N+1)}{4^{2N}}\bi{2N}NC_{2N}.$$
Also,
\begin{align*}
\sum_{k=1}^NG(N+1,k)
&=\f{(2N+1)^2}2\sum_{k=1}^N\f{(-1)^{N+k+1}}{4^{3N-k}}\bi{2N}N^2C_{N+k}\f{\bi{N+k+1}{2k}}{\bi{2k}k}
\\
&=\f{2(2N+1)\bi{2N}N}{(-64)^N}\sum_{k=1}^N(-4)^{k-1}\f{(2N+1)\bi{2N}NC_{N+k}\bi{N+k+1}{2k}}{\bi{2k}k}.
\end{align*}
and
\begin{align*} \f{\bi{2N}NC_{N+1}\bi{N+2}2}{\bi 21}=&\bi{2N-1}{N-1}\bi{2N+2}{N+1}\f{N+1}2
\\&=\bi{2N-1}{N-1}\bi{2N+1}{N+1}(N+1)
\\&=\bi{2N-1}{N-1}(2N+1)\bi{2N}N
\\&=2(2N+1)\bi{2N-1}{N-1}^2\eq0\ (\mo\ 2).
\end{align*}
So, with the help of (\ref{1.3}) we see that $\sum_{n=0}^N(4n+1)\bi{2n}n^3(-64)^{N-n}$
is divisible by $4(2N+1)\bi{2N}N$.


(ii) Now we turn to the proof of (\ref{1.5}).

For $n,k\in\N$, define
$$F(n,k):=\f{(-1)^{n+k}(20n-2k+3)}{4^{5n-k}}\cdot\f{\bi{2n}n\bi{4n+2k}{2n+k}\bi{2n+k}{2k}\bi{2n-k}n}{\bi{2k}k}.
$$
and
$$G(n,k)
:=\f{(-1)^{n+k}}{4^{5n-4-k}}\cdot\f{n\bi{2n}n\bi{4n+2k-2}{2n+k-1}\bi{2n+k-1}{2k}\bi{2n-k-1}{n-1}}{\bi{2k}k}.$$
Clearly $F(n,k)=G(n,k)=0$ if $n<k$. By \cite{Zu},
$$F(n,k-1)-F(n,k)=G(n+1,k)-G(n,k)$$
for all $k\in\Z^+$ and $n\in\N$.



Fix a positive integer $N$. As in part (i) we have
$$\sum_{n=0}^NF(n,0)-F(N,N)=\sum_{k=1}^NG(N+1,k).$$
Observe that
$$\sum_{n=0}^NF(n,0)=\sum_{n=0}^N\f{20n+3}{(-2^{10})^n}\bi{2n}n^2\bi{4n}{2n}$$
and
$$F(N,N)=\f{18N+3}{2^{8N}}\bi{6N}{3N}\bi{3N}N.$$
Also,
$$\sum_{k=1}^NG(N+1,k)
=\f{2(2N+1)\bi{2N}N}{(-2^{10})^N}\sum_{k=1}^N(-4)^{k-1}\f{\bi{4N+2k+2}{2N+k+1}\bi{2N+k+1}{2k}\bi{2N-k+1}N}{\bi{2k}k}.$$
Note that
$$\f{\bi{4N+4}{2n+2}\bi{2N+2}2\bi{2N}N}{\bi 21}=2\bi{4N+3}{2N+1}\bi{2N+2}2\bi{2N-1}{N-1}\eq0\ (\mo\ 2).$$
Applying (\ref{1.2}) we see that $(-2^{10})^N\sum_{k=1}^NG(N+1,k)$ is a multiple of
$4(2N+1)\bi{2N}N$. By (\ref{1.1}),
$$(-2^{10})^N\f{18N+3}{2^{8N}}\bi{6N}{3N}\bi{3N}N$$
is divisible by $8(2N+1)\bi{2N}N$.
Therefore
$$\sum_{n=0}^N(20n+3)\bi{2n}n^2\bi{4n}{2n}(-2^{10})^{N-n}$$
is a multiple of $4(2N+1)\bi{2N}N$.
\medskip

Combining the above, we have completed the proof of Theorem \ref{Th1.2}.
 \end{proof}


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