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\begin{document}

\noindent
{\bf Theorem 1}: For any prime $p$ and positive integer $r$, define

$$
A(r;p):=\left(\sum_{n=0}^{rp-1} { 2n \choose n}\right)  \text{mod} \, p \,\,.
$$


\vskip .2in
\noindent 
Then,  $A(1;p)=\left\{\begin{array}{l}
1 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  3\\
-1 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 2 \mod 3\,\,\,.\\
\end{array}\right.
$

\vskip .2in
\noindent 
{\bf Proof}:  Observe that  

$$
{ 2n \choose n} = CT_x  \left[ \frac{(1+x)^{2n}}{x^n} \right]
$$

Therefore,

\begin{eqnarray*}
\sum_{n=0}^{p-1} {2n \choose n} &=& \sum_{n=0}^{p-1} CT_x \left[ \left(\frac{(1+x)^{2n}}{x^n}\right)\right] \\
&=& \sum_{n=0}^{p-1} CT_x \left[ \left(2+x+\frac{1}{x}\right)^n\right] \\
&=& CT_x \left[ \frac{\left(2+x+\frac{1}{x}\right)^p-1}{2+x+\frac{1}{x}-1}\right] \\
&\equiv_p& CT_x \left[ \frac{1+x^{p}+\frac{1}{x^p}}{1+x+\frac{1}{x}}\right] \\
&\equiv_p& CT_x \left[ \frac{1+x^p+x^{2p}}{(1+x+x^2)x^{p-1}}\right] \\
&\equiv_p& COEFF_{x^{p-1}} \left[ \frac{1}{1+x+x^2}\right] \\
&\equiv_p& COEFF_{x^{p-1}} \left[ \frac{1-x}{1-x^3}\right] \\
\end{eqnarray*}


\noindent
The result folllows from series expansion of the last expression.

\newpage

\noindent
{\bf Corollary 1}:   $A(2;p)=\left\{\begin{array}{l}
3 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  3\\
-3 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 2 \mod 3\,\,\,.\\
\end{array}\right.
$



\noindent
{\bf Proof}: 

\begin{eqnarray*}
\sum_{n=0}^{2p-1} {2n \choose n} &=& \sum_{n=0}^{2p-1} CT_x \left[ \left(\frac{(1+x)^{2n}}{x^n}\right)\right] \\
&=&CT_x \left[ \frac{\left(2+x+\frac{1}{x}\right)^{2p}-1}{2+x+\frac{1}{x}-1}\right] \\
&=&CT_x \left[ \frac{\left(6+4x+\frac{4}{x}+x^2+\frac{1}{x^2}\right)^{p}-1}{2+x+\frac{1}{x}-1}\right] \\
&\equiv_p&CT_x \left[ \frac{\left(6+4x^p+\frac{4}{x^p}+x^{2p}+\frac{1}{x^{2p}}\right)-1}{2+x+\frac{1}{x}-1}\right] \\
&\equiv_p& COEFF_{x^{2p-1}} \left[  \frac{1+4x^p}{1+x+x^2}\right] \\
&\equiv_p& COEFF_{x^{2p-1}} \left[ \frac{1}{1+x+x^2}\right]  +4 COEFF_{x^{p-1}} \left[ \frac{1}{1+x+x^2}\right] \\
\end{eqnarray*}

\noindent The result folllows from Theorem 1 and the last congruence. 

\newpage

\noindent
{\bf Corollary 2} (Catalan Numbers):  Let $a(n)$ be the constant term of $\displaystyle{(1-x)\left(2+x+\frac{1}{x}\right)^n}$ and let

$$A(r;p):=\left(\sum_{n=0}^{rp-1}  a(n)\right) \text{mod} \,p $$
 
\noindent 
Then,  $A(1;p)=\left\{\begin{array}{l}
1 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  3\\
-2 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 2 \mod 3\,\,\,.\\
\end{array}\right.
$

\noindent
{\bf Proof}: 

\begin{eqnarray*}
\sum_{n=0}^{p-1} a(n) &=& \sum_{n=0}^{p-1} CT_x \left[(1-x)\left(2+x+\frac{1}{x}\right)^n\right] \\
&=& CT_x \left[ \frac{(1-x)\left(\left(2+x+\frac{1}{x}\right)^p-1\right)}{2+x+\frac{1}{x}-1}\right]\\
&\equiv_p& CT_x \left[ \frac{(1-x)\left(\left(2+x^p+\frac{1}{x^p}\right)-1\right)}{2+x+\frac{1}{x}-1}\right]\\
&\equiv_p& COEFF_{x^{p-1}} \left[  \frac{1-x}{1+x+x^2}\right] \\
&\equiv_p& COEFF_{x^{p-1}} \left[ \frac{1}{1+x+x^2}\right]  -  COEFF_{x^{p-2}} \left[ \frac{1}{1+x+x^2}\right] \\
\end{eqnarray*}

\noindent The result follows from Theorem 1 and the last congruence.

\newpage


\noindent
{\bf Theorem 2} (Motzkin Numbers):  Let $a(n)$ be the constant term of $\displaystyle{(1-x^2)\left(1+x+\frac{1}{x}\right)^n}$ and let

$$A(r;p):=\sum_{n=0}^{rp-1}  a(n)\,\,. $$
 
\noindent 
Then,  $A(1;p)=\left\{\begin{array}{l}
2 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  4\\
-2 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 3 \mod 4\,\,\,.\\
\end{array}\right.
$

\noindent
{\bf Proof}: 

\begin{eqnarray*}
\sum_{n=0}^{p-1} a(n) &=& \sum_{n=0}^{p-1} CT \left[(1-x^2)\left(1+x+\frac{1}{x}\right)^n\right] \\
&=& CT \left[ \frac{(1-x^2)\left(\left(1+x+\frac{1}{x}\right)^p-1\right)}{1+x+\frac{1}{x}-1}\right]\\
&\equiv_p& COEFF_{x^{p-1}} \left[  \frac{1-x^2}{1+x^2}\right] \\
&\equiv_p& COEFF_{x^{p-1}} \left[ \frac{1}{1+x^2}\right]  -  COEFF_{x^{p-3}} \left[ \frac{1}{1+x^2}\right] \\
\end{eqnarray*}

\noindent The result follows from series expansion of $\displaystyle{\frac{1}{1+x^2}}$ and the last congruence.

\newpage

\noindent
{\bf Corollary 3}:  Let $a(n)$ be the constant term of $\displaystyle{(1-x^2)\left(1+x+\frac{1}{x}\right)^n}$ and let

$$A(r;p):=\sum_{n=0}^{rp-1}  a(n)\,\,. $$
 
\noindent 
Then,  $A(1;p)=\left\{\begin{array}{l}
4 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  4\\
-4 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 3 \mod 4\,\,\,.\\
\end{array}\right.
$

\noindent
{\bf Proof}: 

\begin{eqnarray*}
\sum_{n=0}^{2p-1} a(n) &=& \sum_{n=0}^{2p-1} CT_x \left[(1-x^2)\left(1+x+\frac{1}{x}\right)^n\right] \\
&=& CT_x \left[ \frac{(1-x^2)\left(\left(1+x+\frac{1}{x}\right)^{2p}-1\right)}{1+x+\frac{1}{x}-1}\right]\\
&=& CT_x \left[ \frac{(1-x^2)\left(\left(3+2x+x^2+\frac{2}{x}+\frac{1}{x^2}\right)^{p}-1\right)}{1+x+\frac{1}{x}-1}\right]\\
&=& CT_x \left[ \frac{(1-x^2)\left(\left(3+2x^p+x^{2p}+\frac{2}{x^p}+\frac{1}{x^{2p}}\right)-1\right)}{1+x+\frac{1}{x}-1}\right]\\
&\equiv_p& COEFF_{x^{2p-1}} \left[  \frac{(1-x^2)(1+2x^p)}{1+x^2}\right] \\
&\equiv_p& COEFF_{x^{2p-1}} \left[ \frac{1}{1+x^2}\right]  + 2 COEFF_{x^{p-1}} \left[ \frac{1}{1+x^2}\right] \\& -&COEFF_{x^{2p-3}} \left[ \frac{1}{1+x^2}\right] -2COEFF_{x^{p-3}} \left[ \frac{1}{1+x^2}\right]\,. \\
\end{eqnarray*}

\noindent The result follows from Theorem 2 and the last congruence. 




\newpage

\noindent 
{\bf Theorem 3}:  Let $P(x)$  be a Laurent polynomial in $x$ and let $p$ be a prime.  Let $R(x)$ the denominator after simplifying the expression 
$$
\frac{P(x^p)-1}{P(x)-1}. 
$$

\noindent  Then, for any positive integer $r$ and Laurent polynomial $Q(x)$,

$$
\left( \sum_{n=0}^{rp-1} CT_x \left[P(x)^nQ(x)\right] \right) \text{mod} \,\, p
$$

\noindent is congruent to a $p$-finite sequence and can be expressed as linear combinations of finite terms of the coefficients of $\displaystyle{\frac{1}{R(x)}}$.


\newpage



\noindent 
{\bf {Theorem 4}}:  Let $p$ be a prime number and let $r$ be a positive integer.  Let 

$$
A(r,s;p) := \sum_{n=0}^{rp-1}\sum_{m=0}^{sp-1}  { m+n \choose m}  \mod p\,\,.
$$

\noindent 
Then, 
$$
A(r,s;p^{\alpha}) = {r+s\choose s}-1\,\,,\text{where} \,\,\,\alpha = 1,2, 3 \,\,.
$$


\noindent 
{\bf{Proof}}: Let  $\displaystyle{P(x)=1+\frac{1}{x}}$ and $\displaystyle{Q(x) = 1+x}$. Then,

\begin{eqnarray*}
 \sum_{n=0}^{rp-1}\sum_{m=0}^{sp-1}  { m+n \choose m}  &=& \sum_{n=0}^{rp-1}\sum_{m=0}^{sp-1}  CT_x\left[ P(x) ^mQ(x)^n\right]\\
&=&\sum_{n=0}^{sp-1} CT_x\left[\frac{P(x)^{rp}-1)Q(x)^n}{P(x)-1}\right] \\
&=&CT_x\left[\left(\frac{P(x)^{rp}-1}{P(x)-1}\right ) \left(\frac{Q(x)^{sp}-1}{Q(x)-1}\right)\right]\\
 &=& CT_x\left [\left(\left(1+\frac{1}{x}\right)^{rp}-1\right)\left(\left(1+x\right)^{sp}-1\right)\right]\\
&= & CT_x\left[\left(\left(1+\frac{1}{x}\right)^{rp}-1\right) \left(\left(1+x\right)^{sp}-1 \right)\right]\\
&= &  CT_X \left[\left(1+\frac{1}{x}\right)^{rp} \left(1+x\right)^{sp}-\left(1+\frac{1}{x}\right)^{rp}-\left(1+x\right)^{sp}+1\right]\\
&= & CT_X \left[\frac{\left(1+x\right)^{rp+sp}}{x^{sp}}-1-1+1\right]\\
&= &  { p(r+s) \choose ps}-1\\
&\equiv_{p^{\alpha}} & {r+s \choose s}-1, \text{where} \,\,\,\alpha = 1, 2,3\,\,.
\end{eqnarray*}

\noindent
The last equality follows from the beautiful result that  $\displaystyle{{ pa \choose pb} \equiv_{p^\alpha} { a \choose b}}$ for $\alpha = 1, 2, 3$.

\newpage

\noindent 
{\bf {Theorem 5}}:  Let $p$ be a prime number and let $r$ be a positive integer.  Let 

$$
A(r,s;p) := \left(\sum_{n=0}^{rp-1}\sum_{m=0}^{sp-1} { n+m \choose m}^2\right) \mod p\,\,.
$$

\vskip .2in
\noindent 
Then,  $A(1,1;p)=\left\{\begin{array}{l}
0 ,\,\,\, \,\,\,\, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,0 \mod 3\\
1 ,\,\,\, \,\,\, \, \,\text{if} \,\,\,p\,\,\,\,\equiv \,\,1 \mod  3\\
-1 ,\,\,\,   \text{if} \,\,\,p\,\,\,\,\equiv\,\, 2 \mod 3\,\,\,.\\
\end{array}\right.
$

\vskip .2in
\noindent 
Furthermore,  $\displaystyle{A(2,2;p) \equiv_p COEFF_{[x^{2p-1}y^{2p-1}]} \frac{1+4x^p+4y^p+16x^py^p}{(1+x+xy)(1+y+xy)}}$.\\

\vskip .2in
\noindent 
{\bf{Proof}}: Let  $\displaystyle{P(x,y)=\left(1+y\right)\left(1+\frac{1}{x}\right)}$ and $\displaystyle{ Q(x,y) = \left(1+x\right)\left(1+\frac{1}{y}\right)}$. First observe that 
$$
 { n+m \choose m}^2 =  { n+m \choose m} { n+m \choose n}=CT_{xy} \left( P(x,y)^n Q(x,y)^m\right)\,\,.
$$

\noindent
Then,

\begin{eqnarray*}
 \sum_{m=0}^{p-1}\sum_{n=0}^{p-1}  { m+n \choose m}^2  &=& \sum_{m=0}^{p-1}\sum_{n=0}^{p-1}  CT_{xy}\left[ P(x,y) ^nQ(x,y)^m\right]\\
&=&\sum_{m=0}^{p-1} CT_{xy}\left[\frac{P(x,y)^{p}-1)Q(x,y)^m}{P(x,y)-1}\right] \\
&=&CT_{xy}\left[\left(\frac{P(x,y)^{p}-1}{P(x,y)-1}\right ) \left(\frac{Q(x,y)^{p}-1}{Q(x,y)-1}\right)\right]
\end{eqnarray*}

\noindent Using the fact that $(a+b)^p \equiv_p a^p+b^p$, we get 

\begin{eqnarray*}
\sum_{m=0}^{p-1}\sum_{n=0}^{p-1}  { m+n \choose m}^2  &\equiv_p&CT_{xy}\left[\left(\frac{P(x^p,y^p)-1}{P(x,y)-1}\right ) \left(\frac{Q(x^p,y^p)-1}{Q(x,y)-1}\right)\right]\\
&\equiv_p&CT_{xy}\left[\frac{(1+y^p+x^py^p)(1+x^p+x^py^p)}{(1+y+xy)(1+x+xy)x^{p-1}y^{p-1}}\right]\\
&\equiv_p&COEFF[x^{p-1}y^{p-1}]\left[\frac{(1+y^p+x^py^p)(1+x^p+x^py^p)}{(1+y+xy)(1+x+xy)}\right]\\
&\equiv_p&COEFF[x^{p-1}y^{p-1}]\left[\frac{1}{(1+y+xy)(1+x+xy)}\right]\\
\end{eqnarray*}


\noindent
IUsing the Almkvist-Apagodu-Zeilberger algorithm, the diagonal coefficients satisfy the recurrence $N^2+N+1=0$  with the initial conditions $a(0)=1,a(1)=1, a(2)=0$. The result now follows from the fact that the recurrence is equivalent to $N^3-1=0$.

\newpage

\noindent 
Finally, if we apply the above method to $A(2,2;p)$ we get,  
$$\displaystyle{A(2,2;p) \equiv_p   COEFF_{[x^{2p-1}y^{2p-1}]} \frac{1+4x^p+4y^p+16x^py^p}{(1+x+xy)(1+y+xy)}}\,\,.$$

\noindent Expanding, we get 

 $$
COEFF_{[x^{2p-1}y^{2p-1}]} \frac{1}{(1+x+xy)(1+y+xy)} + 4COEFF_{[x^{p-1}y^{2p-1}]} \frac{1}{(1+x+xy)(1+y+xy)}
$$ 
$$
+4COEFF_{[x^{2p-1}y^{p-1}]} \frac{1}{(1+x+xy)(1+y+xy)}+16COEFF_{[x^{p-1}y^{p-1}]} \frac{1}{(1+x+xy)(1+y+xy)}
$$

\vskip .2in

\noindent By symmetry in the second and third term, we have 

$$
A(2,2;p) \equiv_p COEFF_{[x^{2p-1}y^{2p-1}]} \frac{1}{(1+x+xy)(1+y+xy)} +
$$ 
$$
8COEFF_{[x^{p-1}y^{2p-1}]} \frac{1}{(1+x+xy)(1+y+xy)}+ 16COEFF_{[x^{p-1}y^{p-1}]} \frac{1}{(1+x+xy)(1+y+xy)}
$$

\newpage


\noindent Let $p$ be a prime number.  Then,

$$
C(r,s;p):=\left(\sum_{n=0}^{rp-1}\sum_{m=0}^{sp-1} { n+m \choose m}^3\right) \mod p\,\,.
$$

\noindent 
Since 

$$
 { n+m \choose m}^3 =  CT_{xyz} \left[ \frac{(1+x)^{m+n}}{x^m}\frac{ (1+y)^{m+n}}{y^m}\frac{ (1+z)^{m+n}}{z^n}\right]\,\,.
$$

\noindent 
If we let  $\displaystyle{P(x,y,z)=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+z\right)}$ and $\displaystyle{ Q(x,y,z) = \left(1+x\right)\left(1+y\right)\left(1+\frac{1}{z}\right)}$, then we have 

$$
 { n+m \choose m}^3 =  CT_{xyz} \left(( P(x,y)^m Q(x,y)^n\right)\,\,.
$$


\noindent Repeating the above arguement verbatim, we arive at 

$$
C(1,1,;p) \equiv_p COEFF[x^{p-1}y^{p-1}z^{p-1}]\left[\frac{1}{p(x,y,z)q(x,y,z)}\right],
$$

\vskip .2in
\noindent where $p(x,y,z)=1+x+y+xy+xz+yz+xyz$ and $q(x,y,z)=1+y+z+xy+xz+yz+xyz$.  

\newpage

\noindent \noindent {\bf Theorem 6}:  Let $p$ be a prime number and let

$$
A(r,s,t;p):=\sum_{m_1=0}^{rp-1}\sum_{m_2=0}^{sp-1} \sum_{m_3=0}^{tp-1} { m_1+m_2+m_3 \choose m_1,m_2,m_3}   \,\,.
$$

\noindent  Then,   $\displaystyle{A(1,1,1;p) \equiv_p 1}$ for $p>2$.    \\


\noindent 
{\bf Proof ($\alpha =1$ case)}:  Observe that  

$$
 { m_1+m_2+m_3 \choose m_1,m_2,m_3}  = CT_{xyz} \left[\frac{ \left(1+x+y+z\right)^{m_1+m_2+m_3}}{x^{m_1}y^{m_2}z^{m_3}}\right]
$$

\noindent 
As in the proof of Theorem 2 using the fact that $(x+y)^p\equiv_p x^p + y^p$, we get

\begin{eqnarray*}
\sum_{m1=0}^{p-1}\sum_{m2=0}^{p-1} \sum_{m_3=0}^{p-1} { m_1+m_2+m_3 \choose m_1,m_2,m_3}   &= &  \sum_{m_1=0}^{p-1}\sum_{m_2=0}^{p-1} \sum_{m_3=0}^{p-1} CT_{xyz} \left[\frac{ \left(1+x+y+z\right)^{m_1+m_1+m_3}}{x^{m_1}y^{m_2}z^{m_3}}\right]\\
&\equiv_p& CT_{xyz} \frac{(1+x^py^p+x^pz^p)(1+y^px^p+y^pz^p)(1+z^px^p+z^py^p)}{(xyz)^{p-1}(1+x+y)(1+x+z)(1+y+z)}\\
&\equiv_p& COEFF_{[x^{p-1}y^{p-1} z^{p-1}]}  \frac{1}{(1+x+y)(1+x+z)(1+y+z)}\\
\end{eqnarray*}

\vskip .1in

\noindent Using Almkvist-Apagodu-Zeilberger algorithm, the diagonal coefficients, $a(n):=a(n,n,n)$, satisfy the recurrence  

$$
24(5n+11)(3n+5)(3n+4)a(n)+(295n^3+1614n^2+2855n+1620)a(n+1)
$$

$$
+2(2n+5)(n+2)(5n+6)a(n+2)=0\,\,.
$$

\vskip .1in

\noindent Passing to mod p, the terms with powers of $p$ are zero and the recurrence reduces to the much simpler second order recurrence, 


$$
288a(p)+84a(p+1)+6a(p+2)=0
$$ 

\vskip .2in

\noindent with initial conditions  $a(0)=1$ and $a(1)=-14$, whose solution is given by 


$$
a(p)\equiv_p 4(-8)^p-3(-6)^p\,\,.
$$

\vskip .2in

\noindent From the above equation, we know that  $A(1,1,1;p) \equiv a(p-1)$ (mod p).  Thus, 


$$
A(1,1,1;p)=a(p-1) \equiv_p  4(-8)^{p-1} - 3 (-6)^{p-1} \equiv_p 2^{p-1} ,\,.
$$. 


\noindent  Now it suffices to show that $2^{p-1}-1 \equiv_p 0$.  But this follows fom 

$$
2^{p-1}-1 = \frac{1}{2}(2^p-2) = \frac{1}{2}((1+1)^p-2) \equiv_p \frac{1}{2}(1+1-2)=0\,\,.  
$$



\newpage

\noindent {\bf Observation 1}: $\displaystyle{A(1,1,1;p) \equiv_{p^{\alpha}} 1}$ for $p>2$, for $\alpha =2,3$.\\

\noindent {\bf Observation 2}: The above  computer data suggests that $\displaystyle{A(2,2,2;p) \equiv_{p^\alpha} 16}$  for $p\geq 17$,   for $\alpha = 1, 2, 3$ \\

\vskip .2in


\noindent In general, if we denote, 

$$
A(r_1,r_2,\ldots ,r_a;p):=\sum_{m_1=0}^{r_1p-1}\sum_{m_2=0}^{r_2p-1}\ldots  \sum_{m_a=0}^{r_ap-1} { m_1+m_2\ldots+m_a \choose m_1,m_2,\ldots,m_a}   \,\,.
$$

\noindent 

Then,  \\

\begin{enumerate}

\item

$\displaystyle{A(1,1,\ldots,1;p) \equiv_p Coeff_{[x_1^{p-1}x_2^{p-1} \ldots x_a^{p-1}]} \frac{1}{\prod_{j=1}^{a} (1+\sum_{i =1, i\neq j}^{a}x_i)} }$.\\

\item

{\bf Conjecture}: $\displaystyle{A(1,1,\ldots,1;p) \equiv_p 1}$ for $p>2$ and $A(1,1,\ldots,1;2) \equiv_p 0$. \\


\end{enumerate}


\noindent 
{\bf{Proof}} (of i) :  Observe that  

$$
 { m_1+m_2\ldots+m_a \choose m_1,m_2,\ldots,m_a}   = CT_{x_1x_2\ldots x_a} \left[\frac{ \left(1+x_1+x_2+\ldots +x_a\right)^{m_1+m_2+\ldots+m_a}}
 {\prod_{i=1}^{a}x_i^{m_i}}\right]
$$

\vskip .2in

\noindent  As in the proof of Theorem 4, 

\begin{eqnarray*}
\sum_{m_1=0}^{p-1}\ldots  \sum_{m_a=0}^{p-1} { m_1+m_2\ldots+m_a \choose m_1,m_2,\ldots,m_a}    &= &  
\sum_{m_1=0}^{p-1}\ldots  \sum_{m_a=0}^{p-1} CT_{x_1x_2\ldots x_a} \left[\frac{ \left(1+x_1+x_2+\ldots +x_a\right)^{\sum_{j=1}^{a}m_j}} {\prod_{i=1}^{a}x_i^{m_i}}\right]\\
 &\equiv_p& COEFF_{[x_1^{p-1}x_2^{p-1} \ldots x_a^{p-1}]} \frac{1}{\prod_{j=1}^{a} \left(1+\sum_{i =1, i\neq j}^{a}x_i\right)} .\\
\end{eqnarray*}

\newpage

\noindent
{\bf Apery's Number}:\\


\noindent Apery's number is defined by 

$$
A(n):=\sum_{k=0}^{n} { n \choose k}^2 {n+k \choose k}^2
$$


\noindent and it is known that 

$$
A(n)= CT_{xyz} \left[ \left(\frac{(1+x)(1+y)(1+z)(1+y+z+yz+xyz)}{xyz}\right)^n\right] 
$$


\noindent Let $p$ be a prime number and define 

$$
a(p):=\sum_{n=0}^{p-1}A(n)\,\,.
$$ 



\noindent We can express  $a(p)$ as 


\begin{eqnarray*}
\sum_{n=0}^{p-1}\sum_{k=0}^{n} { n \choose k}^2 {n+k \choose k}^2 &=& \sum_{n=0}^{p-1} CT_{xyz} \left[ \left(\frac{(1+x)(1+y)(1+z)(1+y+z+yz+xyz)}{xyz}\right)^n\right] \\
&=& CT_{xyz} \left[R1(x,y,z)+R2(x,y,z)\right]\\
&=&CT_{xyz}\left[\frac{R1(x,y,z)}{x^{p-1}y^{p-1}z^{p-1}}\right]\\
&=&CT_{xyz}\left[\frac{P(x,y,z)}{x^{p-1}y^{p-1}z^{p-1}Q(x,y,z)}\right]\\
&\equiv_p&COEFF_{x^{p-1}y^{p-1}z^{p-1}} \frac{P(x^p,y^p,z^p)}{Q(x,y,z)}\\
&\equiv_p& COEFF_{x^{p-1}y^{p-1}z^{p-1}} \left[\frac{1}{Q(x,y,z)}\right]\\
\end{eqnarray*}

\noindent where,  $R2(x,y,z)=\frac{xyz}{Q(x,y,z)}$ and has zero constant term and  $R1(x,y,z)= \frac{P(x,y,z)}{Q(x,y,z)}$, where

$$
P(x,y,z) = (x+1)^p(y+1)^p(1+z)^p((1+y+z+yz+xyz)^p
$$


\noindent and 

$$
Q(x,y,z) = x^2y^2z^2+x^2y^2z+x^2yz^2+2xy^2z^2+x^2yz+3xy^2z+3xyz^2+y^2z^2+xy^2
$$
$$
+4xyz+xz^2+2y^2z+2yz^2+2xy+2xz+y^2+4yz+z^2+x+2y+2z+1\,\,.
$$

\noindent 
Therefore, 

$$
a(p) \equiv \alpha (p-1,p-1,p-1) (\text{mod}\,\, p)
$$

\noindent where 


$$
Q(x,y,z)=\sum_{m,n,k} \alpha(m,n,k) x^ny^mz^k\,\, .
$$


\newpage



\noindent
{\bf Toward Z-W Sun's Conjecture}:\\

\noindent Z-W Sun conjectures that 

$$
\sum_{n=0}^{p-1}\sum_{k=0}^{n} (-1)^{n+k}8^k {n \choose k}^3 \equiv
\left(\frac{p}{3}\right) (\text{mod}\,\,\,\, p^2)\,\,.
$$


\noindent Using the fact that   


$$
{ n \choose k}^3 = CT_{xyz} \left[\frac{(1+x)^n (1+y)^n(1+z)^n}{x^ky^kz^k}\right]\,\,.
$$


\noindent we get 

\begin{eqnarray*}
\sum_{n=0}^{p-1}\sum_{k=0}^{n} (-1)^{n+k}8^k {n \choose k}^3 &=&\sum_{n=0}^{p-1}\sum_{k=0}^{n} (-1)^{n+k}8^k CT_{xyz} \left[\frac{(1+x)^n (1+y)^n(1+z)^n}{x^ky^kz^k}\right]\\
&=& CT_{xyz} \left[R1(x,y,z)+R2(x,y,z)+R3(x,y,z)+R4(x,y,z)\right]\\
\end{eqnarray*}


\noindent where 


$$
R1(x,y,z)=-\frac{(-1)^p((x+1)(y+1)(1+z))^pxyz}{(xyz+xy+xz+yz+x+y+z+2)(xyz+8)}
$$

$$
R2(x,y,z)=\frac{(-8)^{p+1}((x+1)(y+1)(1+z))^p}{x^{p-1}y^{p-1}z^{p-1}(7xyz+8xy+8xz+8yz+8x+8y+8z+8)(xyz+8)}
$$

$$
R3(x,y,z)=\frac{xyz}{(xyz+xy+xz+yz+x+y+z+2)(xyz+8)}
$$

\noindent and
$$
R4(x,y,z)=-\frac{8xyz}{(7xyz+8xy+8xz+8yz+8x+8y+8z+8)(xyz+8)}\,\,.
$$


\noindent 
Since the constant terms of $R1(x,y,z),R3(x,y,z)$ and $R4(x,y,z)$ are zero, we get, 


\begin{eqnarray*}
\sum_{n=0}^{p-1}\sum_{k=0}^{n} (-1)^{n+k}8^k {n \choose k}^3&=&COEFF_{x^{p-1}y^{p-1}z^{p-1}} \left[\frac{(-8)^{p+1}((1+x)(1+y)(1+z))^p}{(7xyz+8(xy+xz+yz+x+y+z+1))(xyz+8)}\right]\\
&\equiv_p&COEFF_{x^{p-1}y^{p-1}z^{p-1}} \left[\frac{(-8)^{p+1}}{(7xyz+8(xy+xz+yz+x+y+z+1))(xyz+8)}\right]\\
\end{eqnarray*}


\newpage

\noindent 
{\bf {Theorem 7}}:  Let  $P(x,y,z)$ and $Q(x,y,z)$ be Laurent polynomials in $x,y$ and $z$.  Let  $T(x,y,z)$ be the denominator (after clearing!) of 

$$
\frac{(P(x^p,y^p,z^p)-1)(Q(x^p,y^p,z^p)-1)}{(P(x,y,z)-1)(Q(x,y,z)-1)}\,\,.
$$

\noindent Then, for a prime $p$,  Laurent polynomial $R(x,y,z)$,  and positive integers $r,s,t$,

$$
\left(\sum_{k=0}^{rp-1}\sum_{n=0}^{sp-1}\sum_{n=0}^{tp-1}  CT_{xyz} \left(P(x,y,z)^nQ(x,y,z)^kR(x,y,z)\right)\ \right)\,\,\,\text{mod} \,\, p
$$


\noindent is congruent to a $P$-finite sequence and can be expressed as a finite linear combination of the coefficients of  
$\displaystyle{\frac{1}{T(x,y,z)}}$.

\end{document} 