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\begin{center}
\vskip 1cm{\LARGE\bf Two Properties of  \\
\vskip .1in
Catalan-Larcombe-French Numbers } 
\vskip 1cm \large
Xiao-Juan Ji \\
School of Mathematical Sciences \\
Soochow University \\
Suzhou \\
Jiangsu 215006 \\
P. R. China \\
\href{mailto:xiaojuanji2015@yahoo.com}{\tt xiaojuanji2015@yahoo.com}\\
\ \\
Zhi-Hong Sun\\
School of Mathematical Sciences\\
Huaiyin Normal University\\
Huaian\\
Jiangsu 223001\\
P. R. China\\
\href{mailto:zhihongsun@yahoo.com}{\tt zhihongsun@yahoo.com}
\end{center}

\vskip .2 in


\begin{abstract}
Let $(P_n)$ be the Catalan-Larcombe-French numbers. The numbers
$P_n$ occur in the theory of elliptic integrals, and are related  to
the arithmetic-geometric-mean. In this paper we investigate the
properties of the related sequence
$S_n = P_n/2^n$ instead, since $S_n$ is an Ap\'ery-like
sequence. We prove a congruence and an inequality for $P_n$.
\end{abstract}

\section{Introduction}\label{Section1}

Let $(P_n)$ be the sequence given by
\begin{equation}\label{eqdold1}P_0=1,\ P_1=8\qtq {and}
(n+1)^2P_{n+1}=8(3n^2+3n+1)P_n-128n^2P_{n-1}\ (n\geq
1).\end{equation} The numbers $P_n$ are called
Catalan-Larcombe-French numbers since Catalan first defined $P_n$ in
\cite{C}, and Larcombe and French \cite{LF1} proved that
$$P_n=2^n\sum_{k=0}^{\lfloor {n/2}\rfloor}(-4)^k
\binom{2n-2k}{n-k}^2\binom{n-k}k
=\sum_{k=0}^n\frac{\binom{2k}k^2\binom{2n-2k}{n-k}^2}{\binom nk},$$
where $ \lfloor x \rfloor$ is the greatest integer not exceeding
$x$. The numbers $P_n$ are related  to the
arithmetic-geometric-mean. See \cite{LF1} and 
\seqnum{A053175} in Sloane's
``On-Line Encyclopedia of Integer Sequences".

Let $(S_n)$ be defined by
\begin{equation}\label{eqdold2}
S_0=1,\ S_1=4\qtq {and} (n+1)^2S_{n+1}=4(3n^2+3n+1)S_n-32n^2S_{n-1}\
(n\geq 1).\end{equation}
 Comparing \eqref{eqdold2} with \eqref{eqdold1}, we see that
$$S_n=\frac{P_n}{2^n}.$$
Zagier noted that
$$S_n=\sum_{k=0}^{\lfloor {n/2}\rfloor}\binom{2k}k^2\binom{n}{2k}4^{n-2k}.$$ As
observed by Jovovi\'c \cite{LF2} in 2003 ,
$$ S_n=\sum_{k=0}^n\binom nk\binom
{2k}k\binom{2n-2k}{n-k}\quad (n=0,1,2,\ldots). $$ Recently Z. W. Sun
stated that
$$S_n=\sum_{k=0}^n\binom{2k}k^2\binom k{n-k}(-4)^{n-k}
=\frac 1{(-2)^n} \sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}
\binom k{n-k}(-4)^k.$$
 The first few values of $S_n$ are shown
below:
\begin{align*}&S_0=1,\ S_1=4,\ S_2=20,\ S_3=112,\ S_4=676,\
S_5=4304,\ S_6=28496,
\\&S_7=194240,\ S_8=1353508,\ S_9=9593104,\ S_{10}=68906320,
\\&S_{11}=500281280,\ S_{12}=3664176400, \
S_{13}=27033720640.\end{align*}


Let $p$ be an odd prime. Jarvis, Larcombe, and French \cite{JLF}
proved that if $n=a_rp^r+\cdots+a_1p+a_0$ with $a_0,a_1,
\ldots,a_r\in\{0,1,\ldots,p-1\}$, then
$$P_n\equiv P_{a_r}\cdots P_{a_1}P_{a_0}\pmod p.$$
 Jarvis and Verrill \cite{JV}  showed that
$$P_n\equiv (-1)^{\frac{p-1}2}128^nP_{p-1-n}\pmod p\qtq{for}
n=0,1,\ldots,p-1$$ and
$$P_{mp^r}\equiv P_{mp^{r-1}}\pmod{p^r}\qtq{for}m,r\in\mathbb{Z}^+,$$
 where $\mathbb{Z}^+$ is the set of positive integers.


For a prime $p$ let $\mathbb{Z}_p$ denote the set of those rational
numbers whose denominator is not divisible by $p$.  Let $p$ be an
odd prime, $n\in\mathbb{Z}_p$ and $n\not\equiv 0,-16\pmod p$. The
second author \cite{S2} proved that
$$\sum_{k=0}^{p-1}\binom{2k}k\frac{S_k}{(n+16)^k}\equiv
\Ls {n(n+16)}p\sum_{k=0}^{p-1}
\frac{\binom{2k}k^2\binom{4k}{2k}}{n^{2k}}\pmod p,$$ where $(\frac
ap)$ is the Legendre symbol.

In 1894 Franel \cite{F} introduced the following Franel numbers
$(f_n)$:
$$f_n=\sum_{k=0}^n\binom nk^3\quad(n=0,1,2,\ldots).$$
The first few Franel numbers are as below:
$$f_0=1,\ f_1=2,\ f_3=10,\ f_4=56,\
f_5=346,\ f_6=2252,
 \ f_7=15184.$$
Franel \cite{F} noted that the sequence $(f_n)$ satisfies the
recurrence relation:
 $$(n+1)^2f_{n+1}=(7n^2+7n+2)f_n+8n^2f_{n-1}\ (n\geq 1).$$

Let $r\in\mathbb{Z}^+$ and $ p$ be a prime with $p\equiv 5,7\pmod
8$. The second author \cite{S2} conjectured that
\begin{equation}\label{eqdold3}
S_{\frac{p^r-1}2}\equiv0\pmod
{p^r}\qtq{and}f_{\frac{p^r-1}2}\equiv0\pmod {p^r}.\end{equation} In
this paper we prove \eqref{eqdold3} in the case $r=2$. We also prove
the second author's conjecture \cite{S2}:

$$\big(1+\frac 1{m(m-1)}\big)S_m^2>S_{m+1}S_{m-1}\qtq{for}m=2,3,\ldots.$$






\section{Basic lemmas}
  \begin{lemma}[{Lucas' theorem \cite{M}}]\label{Lemma1}
  Let $p$ be an odd prime.
 Suppose $a=a_rp^r+\cdots+a_1p+a_0$ and $b=b_rp^r+\cdots+b_1p+b_0$,
 where $a_r,\ldots,a_0,b_r,\ldots,b_0\in\{0,1,\ldots,p-1\}$. Then
 $$\binom ab\equiv\binom{a_r}{b_r}\cdots \binom{a_0}{b_0}\pmod p.$$
 \end{lemma}
 Lucas' theorem is often formulated as follows.
 \begin{lemma}[{\cite{M}}]\label{Lemma2}
 Let $p$ be an odd prime and $a,b\in\mathbb{Z}^+$.
  Suppose $a_0,b_0\in\{0,1,\ldots,p-1\}$. Then
  $$\binom{ap+a_0}{bp+b_0}\equiv\binom ab\binom{a_0}{b_0}\pmod p.$$
  \end{lemma}


 \begin{lemma}[{\cite[Lemma 2.7]{JS}}]\label{Lemma3}
  For any positive integer $n$
we have
$$S_n=2\sum_{k=1}^n\binom{n-1}{k-1}\binom{2k}k\binom{2n-2k}{n-k}.$$
\end{lemma}

\begin{lemma}[{\cite{S}}]\label{Lemma4}
Let $p$ be an odd prime.
  Suppose $n=n_1p+n_0$ and $k=k_1p+k_0$ with $k_1,n_1\in\mathbb{Z}^+$ and
  $k_0,n_0\in\{0,1,\ldots,p-1\}$. Then
  $$\binom nk\equiv\binom {n_1}{k_1}\Big((1+n_1)\binom{n_0}{k_0}-
  (n_1+k_1)\binom{n_0-p}{k_0}-k_1\binom{n_0-p}{k_0+p}\Big) \pmod {p^2}.$$
\end{lemma}

\begin{lemma}\label{Lemma5}
Let $p$ be an odd prime. Then
$$\sum_{t=0}^{(p-1)/2}(-1)^t
\Big(\binom{\frac {p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}\Big)
  \binom{-\frac{1}2}t^2\equiv 0 \pmod {p^2}.$$
\end{lemma}
 \begin{proof}
 For $0\leq t\leq (p-1)/2$, from Lemma \ref{Lemma2} we have
 \begin{align*}\binom{\frac{p-1}2-p}{p+t}
 &=(-1)^{t+1}\binom{p+\frac{p+1}2+t-1}{p+t}
 \equiv(-1)^{t+1}\binom{\frac{p+1}2+t-1}{t}
 \\&=-\binom{\frac{p-1}2-p}t
 \pmod p
 \end{align*}
 and so $$\binom{\frac{p-1}2-p}t+\binom{\frac{p-1}2-p}{p+t}
 =(-1)^t\Big(\binom{\frac{p-1}2+t}t-
 \binom{p+\frac{p-1}2+t}{p+t}\Big)\equiv 0\pmod p.$$
We first assume $p\equiv1\pmod 4$. Applying Lemma \ref{Lemma4} we
get
\begin{align*}
&\binom{\frac{3(p-1)}4}{\frac{p-1}4}
-\binom{p+\frac{3(p-1)}4}{p+\frac{p-1}4}
\equiv\binom{\frac{3(p-1)}4}{\frac{p-1}2}-
\Big(2\binom{\frac{3(p-1)}4}{\frac{p-1}2}
-\binom{\frac{3(p-1)}4-p}{\frac{p-1}2}\Big)
 \\&=-\binom{\frac{3(p-1)}4}{\frac{p-1}2}+(-1)^{\frac{p-1}2}
 \binom{\frac{3(p-1)}4}{\frac{p-1}2}
 =0\pmod {p^2}\end{align*} and
 \begin{align*}&\binom{\frac {p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}
 +\binom{p-1-t}{\frac{p-1}2-t}-\binom{p+p-1-t}{p+\frac{p-1}2-t}
 \\&\equiv-\binom{\frac {p-1}2+t}{\frac{p-1}2}
 +\binom{\frac{p-1}2-p+t}{\frac{p-1}2}
 -\binom{p-1-t}{\frac{p-1}2}
 +\binom{-1-t}{\frac {p-1}2}
\\&=\Big((-1)^{\frac{p-1}2}-1\Big)\binom{\frac {p-1}2+t}{\frac {p-1}2}
+\Big((-1)^{\frac{p-1}2}-1\Big)\binom{p-1-t}{\frac {p-1}2}
\\&=0\pmod {p^2}.\end{align*}
Also,
$$(-1)^t\binom{-\frac{1}2}t^2-(-1)^{\frac{p-1}2-t}
\binom{-\frac{1}2}{\frac{p-1}2-t}^2
 \equiv(-1)^t\Big(\binom{\frac{p-1}2}t^2-
 \binom{\frac{p-1}2}{\frac{p-1}2-t}^2\Big)
=0\pmod p.$$ By the above four congruences, we have
 \begin{align*}
 &\sum_{t=0}^{(p-1)/2}(-1)^t
 \Big(\binom{\frac {p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}\Big)
  \binom{-\frac{1}2}t^2
  \\&=\sum_{t=0}^{(p-5)/4}(-1)^t\binom{-\frac{1}2}t^2
  \Big(\binom{\frac{p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}\Big)
  \\&\quad+
 (-1)^{\frac{p-1}4}\binom{-\frac {1}2}{\frac {p-1}4}^2
 \Big(\binom{\frac{3(p-1)}4}{\frac{p-1}4}-
 \binom{p+\frac{3(p-1)}4}{p+\frac{p-1}4}\Big)
 \\&\quad+\sum_{t=0}^{(p-5)/4}(-1)^{\frac{p-1}2-t}
 \binom{-\frac{1}2}{\frac{p-1}2-t}^2
 \Big(\binom{p-1-t}{\frac{p-1}2-t}-
 \binom{p+p-1-t}{p+\frac{p-1}2-t}\Big)
 \\&\equiv\sum_{t=0}^{(p-5)/4}\Big((-1)^t\binom{-\frac{1}2}t^2
 -(-1)^{\frac{p-1}2-t}\binom{-\frac{1}2}{\frac{p-1}2-t}^2\Big)
 \Big(\binom{\frac{p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}\Big)
\\&\quad+ (-1)^{\frac{p-1}4}\binom{-\frac {1}2}{\frac{p-1}4}^2
\Big(\binom{\frac{3(p-1)}4}{\frac{p-1}4}
-\binom{p+\frac{3(p-1)}4}{p+\frac{p-1}4}\Big)
  \equiv 0 \pmod {p^2}.\end{align*}
Thus the result is true for $p \equiv1\pmod 4$.

Now we assume $p\equiv3\pmod 4$. By Lemma \ref{Lemma4},
$$\binom{\frac {p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}
\equiv
-\Big(\binom{\frac{p-1}2+t}{\frac{p-1}2}
+\binom{p-1-t}{\frac{p-1}2}\Big)
\pmod{p^2}.$$ As \begin{align*}
&\binom{\frac{p-1}2+t}{\frac{p-1}2}+\binom{p-1-t}{\frac{p-1}2}
\equiv\binom{\frac{p-1}2+t}{\frac{p-1}2}+\binom{-1-t}{\frac{p-1}2}
\\&=\binom{\frac{p-1}2+t}{\frac{p-1}2}+(-1)^{\frac{p-1}2}
\binom{t+\frac{p-1}2}{\frac{p-1}2} =0\pmod p\end{align*} and
$$(-1)^t\binom{-\frac{1}2}t^2
 +(-1)^{\frac{p-1}2-t}\binom{-\frac{1}2}{\frac{p-1}2-t}^2
 \equiv(-1)^t\Big(\binom{\frac{p-1}2}t^2-
 \binom{\frac{p-1}2}{\frac{p-1}2-t}^2\Big)
=0\pmod p,
$$
 we obtain
\begin{align*}&\sum_{t=0}^{(p-1)/2} (-1)^t\Big(\binom{\frac
{p-1}2+t}t-\binom{p+\frac{p-1}2+t}{p+t}\Big) \binom{-\frac{1}2}t^2
\\&\equiv-\sum_{t=0}^{(p-1)/2}
(-1)^t\Big(\binom{\frac{p-1}2+t}{\frac{p-1}2}+
\binom{p-1-t}{\frac{p-1}2}\Big) \binom{-\frac{1}2}t^2
\\&=-\sum_{t=0}^{(p-3)/4}
 \Big(\binom{\frac{p-1}2+t}{\frac{p-1}2}+
 \binom{p-1-t}{\frac{p-1}2}\Big)
  \Big((-1)^t\binom{-\frac{1}2}t^2+(-1)^{\frac{p-1}2-t}
  \binom{-\frac{1}2}{\frac{p-1}2-t}^2\Big)
  \\&\equiv 0 \pmod {p^2}.
  \end{align*}
Hence the result is also true in this case. The proof is now
complete.
\end{proof}

\begin{lemma}[{\cite[Theorem 3.3]{S1}}]\label{Lemma6}
Let $p$ be a prime with $p\equiv5,7\pmod 8$. Then
$$\sum_{k=0}^{\frac{p-1}2}\frac{\binom{2k}k^3}{(-64)^k}\equiv 0\pmod {p^2}.$$
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.8]{JS}}]\label{Lemma7}
Let $m\in \mathbb{Z}$ and $k,p\in \mathbb{Z}^+$. Then
$$\binom {m{p^r}-1}{k}=(-1)^{k-\lfloor {\frac kp}\rfloor}
\binom {m{p^{r-1}}-1}{\lfloor {k/p}\rfloor} \prod_{ \atop{i=1,\
p\nmid i}}^k\Big(1-\frac{mp^r}i\Big).$$
\end{lemma}

\section{Congruences for
$S_{\frac{p^2-1}2}$ and $f_{\frac{p^2-1}2}\pmod{p^2}$}

\begin{theorem}
 Let $p$ be a prime with $p\equiv 5,7\pmod 8$. Then
$$S_{\frac{p^2-1}2}\equiv f_{\frac{p^2-1}2}\equiv 0\pmod{p^2}.$$
Moreover,
$$S_{\frac{p^2-1}2}\equiv f_{\frac{p^2-1}2}\pmod{p^3}.$$
\end{theorem}
 \begin{proof}
  For $\frac{p-1}2<t< p$ and $0\leq s\leq \frac{p-1}2$,
  from Lemma \ref{Lemma2} we see that
 $$\binom{p^2-1}{\frac{p^2-1}2}\equiv
 \binom{p-1}{\frac{p-1}2}^2\equiv 1\pmod p,$$
 $$\binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}
 \equiv\binom{\frac{p-1}2}s\binom{\frac{p-1}2}t=0\pmod p,$$
 $$\binom{2sp+2t}{sp+t}=\binom{(2s+1)p+2t-p}{sp+t}
 \equiv\binom{2s+1}s\binom{2t-p}{t}=0\pmod p$$
 and
  \begin{align*}\binom{p^2-1-2sp-2t}{\frac{p^2-1-2sp-2t}2}
 &=\binom{(p-2s-2)p+2p-2t-1}{(\frac{p-1}2-s-1)p+p+\frac{p-1}2-t}
 \\&\equiv\binom{p-2s-2}{\frac{p-1}2-s-1}\binom{2p-2t-1}{p+\frac{p-1}2-t}=0
 \pmod p.\end{align*}

Now we assert that
 \begin{equation}\label{eqdold4}\sum\limits_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3\equiv 0
  \pmod {p^2} \qtq {for} s=0,1,2,\ldots.\end{equation}
  We prove the result by induction on s. For $0\leq t\leq (p-1)/2$ we see that
   $$\binom{\frac{p^2-1}2}t\equiv\binom{-\frac 12}t
   =\frac{\binom{2t}t}{(-4)^t}\pmod {p^2}.$$
  From Lemma \ref{Lemma6} we know that the result is true for $s=0$.
 Suppose that \eqref{eqdold4} holds for $s=k$.
  For $s=k+1$, applying  Lemma \ref{Lemma4} we have
 \begin{align*}&\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{(k+1)p+t}^3
 \\&\equiv\binom {\frac{p-1}2}{k+1}^3\ \sum_{t=0}^{\frac{p-1}2}
 \Big(\frac{p+1}2\binom {\frac{p-1}2}t-\big(\frac{p-1}2+k\big)
 \binom {\frac{p-1}2-p}t
 \\&\quad-k\binom {\frac{p-1}2-p}{t+p}
 -\Big(\binom {\frac{p-1}2-p}t+\binom {\frac{p-1}2-p}{t+p}\Big)\Big)^3
\pmod {p^2}.\end{align*}
 Hence $\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{(k+1)p+t}^3\equiv 0\pmod{p^2}$
 for $k\geq \frac{p-1}2$.
 For $k<\frac{p-1}2$,  by the inductive hypothesis and Lemma \ref{Lemma4} we have
 $$\sum\limits_{t=0}^{(p-1)/2}
\Big(\frac{p+1}2\binom {\frac{p-1}2}t-\big(\frac{p-1}2+k\big)
 \binom {\frac{p-1}2-p}t-k\binom {\frac{p-1}2-p}{t+p}\Big)^3
 \equiv 0\pmod {p^2}.$$
Also,
 $\binom {\frac{p-1}2}t\equiv\binom{\frac{p-1}2-p}t
 \equiv\binom {-\frac{1}2}t\pmod p$
 and
 $\binom{\frac{p-1}2-p}t+\binom{\frac{p-1}2-p}{t+p}
 =(-1)^t\Big(\binom{\frac{p-1}2+t}t-\binom{p+\frac{p-1}2+t}{t+p}\Big)
 \equiv 0\pmod p$ for $t\in\{0,1,\ldots,\frac{p-1}2\}$.
 By Lemma \ref{Lemma5},
 \begin{align*}&\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{(k+1)p+t}^3
\\&\equiv\binom {\frac{p-1}2}{k+1}^3\Big(\sum_{t=0}^{(p-1)/2}
 \Big(\frac{p+1}2\binom {\frac{p-1}2}t-\big(\frac{p-1}2+k\big)
 \binom {\frac{p-1}2-p}t-
 k\binom {\frac{p-1}2-p}{t+p}\Big)^3
\\&\quad+3\sum\limits_{t=0}^{(p-1)/2}
 \Big(\frac{p+1}2\binom{\frac{p-1}2}t-\big(\frac{p-1}2+k\big)
 \binom {\frac{p-1}2-p}t-
 k\binom{\frac{p-1}2-p}{t+p}\Big)
 \\&\quad\quad\quad\quad\quad\times\Big(\binom{\frac{p-1}2-p}t+
 \binom{\frac{p-1}2-p}{t+p}\Big)^2
\\&\quad -3\sum_{t=0}^{(p-1)/2}
 \Big(\frac{p+1}2\binom {\frac{p-1}2}t-\big(\frac{p-1}2+k\big)
 \binom {\frac{p-1}2-p}t-
 k\binom{\frac{p-1}2-p}{t+p}\Big)^2
 \\&\quad\quad\quad\quad\quad\times\Big(\binom{\frac{p-1}2-p}t+
 \binom{\frac{p-1}2-p}{t+p}\Big)
 \\&\quad-\sum_{t=0}^{(p-1)/2}
 \Big(\binom{\frac{p-1}2-p}t+\binom{\frac{p-1}2-p}{t+p}\Big)^3\Big)
 \\&\equiv-3\binom{\frac{p-1}2}{k+1}^3\ \sum_{t=0}^{(p-1)/2}
 \binom{-\frac{1}2}t^2
 \Big(\binom{\frac{p-1}2-p}t+\binom{\frac{p-1}2-p}{t+p}\Big)
 \\&=-3\binom{\frac{p-1}2}{k+1}^3\ \sum_{t=0}^{(p-1)/2}
 (-1)^t\binom{-\frac{1}2}t^2
 \Big(\binom{\frac{p-1}2+t}t-\binom{p+\frac{p-1}2+t}{t+p}\Big)
 \\&\equiv 0\pmod {p^2}.
 \end{align*} Hence
 $$f_{\frac{p^2-1}2}\equiv\sum\limits_{s=0}^{(p-1)/2}
 \sum\limits_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3\equiv0\pmod{p^2}.$$
 \par Set $H_0=H_0(1,1)=0$, $H_k=\sum_{i=1}^k \frac 1k$
  and $H_k(1,1)=\sum_{1\leq i<j\leq k}\frac 1{ij}$
   for $k\in \mathbb{Z}^+$.
   For $0\leq s\leq (p-1)/2$, it is easily seen that
   $H_{p-1}\equiv 0\pmod p$,
   $\binom{p-1}{2s}\equiv 1-pH_{2s}+p^2H_{2s}(1,1)\pmod{p^3}$  and so
   $\frac {1}{\binom{p-1}{2s}}
   \equiv 1+pH_{2s}+p^2\big(H_{2s}^2-H_{2s}(1,1)\big)\pmod {p^3}$.
By Lemma \ref{Lemma7}, for $0\leq t\leq (p-1)/2$ we see that
$$\binom{p^2-1}{2sp+2t}
=\binom{p-1}{2s}\prod\limits_{\atop{i=1, p\nmid i}}
^{2sp+2t}\Big(1-\frac{p^2}i\Big)
\equiv\binom{p-1}{2s}\Big(1-p^2\sum\limits_{\atop{i=1, p\nmid i}}
^{2sp+2t}\frac 1i\Big)\pmod {p^3}.
$$
 Applying \eqref{eqdold4}, Lemma \ref{Lemma6} and the identity
 $$\binom{a-b}{c-d}\binom bd={\binom ac\binom cd\binom {a-c}{b-d}}
 \Big/{\binom ab}$$
  we derive that
 \begin{align*} S_{\frac{p^2-1}2}
 &\equiv\sum_{s=0}^{(p-1)/2}\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p^2-1}2}{sp+t}\binom{2sp+2t}{sp+t}
 \binom{p^2-1-2sp-2t}{\frac{p^2-1}2-sp-t}
\\&=\binom{p^2-1}{\frac{p^2-1}2}\sum_{s=0}^{(p-1)/2}\sum_{t=0}^{(p-1)/2}
\frac{\binom{(p^2-1)/2}{sp+t}^3}{\binom{p^2-1}{2sp+2t}}
\\&\equiv\binom{p^2-1}{\frac{p^2-1}2}\sum_{s=0}^{(p-1)/2}
\frac1{\binom{p-1}{2s}}\sum_{t=0}^{(p-1)/2}
\binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3
\Big(1+p^2\sum\limits_{\atop{i=1, p\nmid i}} ^{2sp+2t}\frac 1i\Big)
\\&\equiv\binom{p^2-1}{\frac{p^2-1}2}\sum_{s=0}^{(p-1)/2}
\frac1{\binom{p-1}{2s}}\sum_{t=0}^{(p-1)/2}
\binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3
\\&\quad+p^2\binom{p^2-1}{\frac{p^2-1}2}
\sum_{s=0}^{(p-1)/2}\frac{\binom{2s}s^3}{(-64)^s}
\sum_{t=0}^{(p-1)/2}\frac{\binom{2t}t^3}{(-64)^t}H_{2t}
\\&\equiv\binom{p^2-1}{\frac{p^2-1}2}\Big(\sum_{s=0}^{(p-1)/2}
\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3
 \\&\quad+p\sum_{s=0}^{(p-1)/2}\big(H_{2s}+p(H_{2s}^2-H_{2s}(1,1))\big)
 \sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3\Big)
 \\&\equiv\sum_{s=0}^{(p-1)/2}\sum_{t=0}^{(p-1)/2}
 \binom{\frac{p-1}2p+\frac{p-1}2}{sp+t}^3
 \\&\equiv f_{\frac{p^2-1}2}\pmod{p^3}.\end{align*}
Summarizing the above proves the theorem.
\end{proof}
\section{An inequality involving $(S_m)$ }
 \begin{theorem}\label{Theorem9}
For $m=2,3,4,\ldots$ we have
$$\big(1+\frac 1{m(m-1)}\big)S_m^2>S_{m+1}S_{m-1}.$$
\end{theorem}
\begin{proof}
 It is easily seen that
  $$\big(1+\frac 1{(m-1)(m-2)}\big)S_{m-1}^2>S_{m}S_{m-2}\qtq {for}
  m=3,4,\ldots,13.$$ Now suppose $m\geq 14$ and
  $\big(1+\frac 1{(m-1)(m-2)}\big)S_{m-1}^2>S_{m}S_{m-2}$.
By \eqref{eqdold2}, Lemma \ref{Lemma3} and the inductive hypothesis
we have
  \begin{align*} &\big(1+\frac 1{m(m-1)}\big)S_m^2-S_{m+1}S_{m-1}
\\&=\big(1+\frac 1{m(m-1)}\big)S_m^2-\frac{4(3m^2+3m+1)}{(m+1)^2}S_mS_{m-1}
  +\frac{32m^2}{(m+1)^2}S_{m-1}^2
\\&>\Big(\frac{m^2-m+1}{m(m-1)}S_m-\frac{4(3m^2+3m+1)}{(m+1)^2}S_{m-1}
+\frac{32m^2(m-1)(m-2)}{(m+1)^2(m^2-3m+3)}S_{m-2}\Big)S_m
\\&=\Big((20m^5-60m^4+52m^3+28m^2-36m+12)S_{m-1}
\\&\quad+(-128m^5+320m^4-256m^3-32m^2+192m-96)S_{m-2}\Big)
\\&\quad\times\frac{S_m}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\\&=\frac{16S_m}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\sum_{k=0}^{m-2}\binom{m-2}k\binom{2k}k\binom{2m-4-2k}{m-2-k}F(m,k),
\end{align*}
where \begin{align*} F(m,k)&=(5m^5-15m^4+13m^3+7m^2-9m+3)
\frac{2k+1}{k+1}\\&\quad-8m^5+20m^4-16m^3-2m^2+12m-6.\end{align*}
 For $m\geq14$  it is easily seen that $3<\frac{(2m-7)(2m-5)}{(m-3)(m-2)}<4$,
 $5m^5-15m^4+13m^3+7m^2-9m+3>0$,
 $-8m^5+20m^4-16m^3-2m^2+12m-6<0$,
 $6m^7-75m^6+223m^5-283m^4-61m^3+427m^2-87m-42>0,$
 and  $F(m,k+1)>F(m,k)$ for $k=0,1,\ldots,m-3$. Thus,
$F(m,m-3)+F(m,1)>F(m,5)+F(m,1)>0$ and
 \begin{align*} F(m,k)\geq
F(m,2)&= \frac 53(5m^5-15m^4+13m^3+7m^2-9m+3)
\\&\quad-8m^5+20m^4-16m^3-2m^2+12m-6>0\qtq{for}k\geq 2.\end{align*}
From the above we derive that
\begin{align*} &\quad\big(1+\frac
1{m(m-1)}\big)S_m^2-S_{m+1}S_{m-1}
\\&>\frac{16S_m}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\Big(\sum_{k=0}^{2}\binom{m-2}k\binom{2k}k\binom{2m-4-2k}{m-2-k}F(m,k)
\\&\quad+\sum_{k=m-4}^{m-2}\binom{m-2}k\binom{2k}k\binom{2m-4-2k}{m-2-k}F(m,k)\Big)
\\&=\frac{16S_m}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\Big(\binom{2m-4}{m-2}\big(F(m,m-2)+F(m,0)\big)
\\&\quad+3(m-2)(m-3)\binom{2m-8}{m-4}\big(F(m,m-4)+F(m,2)\big)
\\&\quad+2(m-2)\binom{2m-6}{m-3}\big(F(m,1)+F(m,m-3)\big)\Big)
\\&>\Big(3(m^2-5m+6)F(m,m-4) +\frac{4(2m-7)(2m-5)}{(m-3)(m-2)}F(m,0)\Big)
\\&\quad\times\frac{16S_m\binom{2m-8}{m-4}}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\\&=\frac{S_m\binom{2m-8}{m-4}}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\Big(\big(6(m-2)(2m-7)+\frac{8(2m-7)(2m-5)}{(m-3)(m-2)}\big)
\\&\quad\times(40m^5-120m^4+104m^3+56m^2-72m+24)
\\&\quad+(-128m^5+320m^4-256m^3-32m^2+192m-96)
\\&\quad\times\big(3(m-2)(m-3)+\frac{4(2m-7)(2m-5)}{(m-3)(m-2)}\big)\Big)
\\&>\Big(\big(6(m-2)(2m-7)+24\big)
(40m^5-120m^4+104m^3+56m^2-72m+24)
\\&\quad+\big(3(m-2)(m-3)+16\big)
(-128m^5+320m^4-256m^3-32m^2+192m-96)\Big)
\\&\quad\times\frac{S_m\binom{2m-8}{m-4}}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\\&=(6m^7-75m^6+223m^5-283m^4-61m^3+427m^2-87m-42)
\\&\quad\times\frac{16S_m\binom{2m-8}{m-4}}{(m+1)^2(m^2-3m+3)m^3(m-1)}
\\&>0.\end{align*}
 Hence the inequality is proved by induction.
\end{proof}
\begin{corollary}
For $m=2,3,4,\ldots$ we have
$$\big(1+\frac 1{m(m-1)}\big)P_m^2>P_{m+1}P_{m-1}.$$
\end{corollary}

\begin{proof}
Since $P_m=2^mS_m$, the result follows from Theorem \ref{Theorem9}.
\end{proof}


\bibliographystyle{amsplain}

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\bibitem {JV} F. Jarvis and H. A. Verrill,
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:  Primary
11A07; Secondary 05A10, 05A19, 05A20.

\noindent \emph{Keywords: } congruence, inequality,
Catalan-Larcombe-French number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A053175}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 21 2015;
revised versions received  November 24 2015; February 23 2016.
Published in {\it Journal of Integer Sequences}, April 6 2016.


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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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