Sum of product of squared binomial coefficients

Similar to the sum of product of binomial coefficients—what happens when I square the binomial coefficients? Is there a nice closed formula for that?

More precisely, I'm interested in the special case where the $k_i$ are all equal to the sum of all $x_i$ (to use the notation of the question linked above), which I'll call $m$ (not necessarily equal to $n$ here).

So what I'm looking for is making the following sum tractable:

$$S_{n,m}=\sum_{x_1+\ldots+x_n=m}\prod_{i=1}^{n}{\binom{m}{x_i}^2}$$

My goal is to show that $S_{n,m}\cdot n^{-2m}$ decreases as the number of summands, $n$, grows. I've tried expanding $n^{2m}$:

$$n^{2m} = \sum_{y_1+\ldots+y_n=2m}{\binom{2m}{y_1,\ldots,y_n}} = \sum_{y_1+\ldots+y_n=2m} {\prod_{i=1}^{w} {\binom{\sum_{j=1}^{i}{y_j}}{y_i}}}$$

but I didn't get anywhere with that. I'd be grateful for any pointers on how to approach this.

Answer

We consider the generating function
\begin{align*}
S_m(x)=\sum_{j=0}^m\binom{m}{j}^2x^j\qquad\qquad m\geq 0
\end{align*}
Note $S_{n,m}$ is the coefficient of $x^m$ of $\left(S_m(x)\right)^n$.

\begin{align*}
S_{n,m}=\sum_{{x_1+\ldots+x_n=m}\atop{x_j\geq 0}}\prod_{i=1}^{n}{\binom{m}{x_i}^2}
=[x^m]\left(S_m(x)\right)^n
\end{align*}

We can represent $S_m(x)$ in terms of Legendre polynomials $P_m(x)$. From the representation
\begin{align*}
P_m(x)=\frac{1}{2^m}\sum_{j=0}^m\binom{m}{j}^2(x-1)^{m-j}(x+1)^j
\end{align*}

we obtain

\begin{align*}
P_m\left(\frac{x+1}{x-1}\right)
&=\frac{1}{2^m}\sum_{j=0}^m\binom{m}{j}^2\left(\frac{x+1}{x-1}-1\right)^{m-j}\left(\frac{x+1}{x-1}+1\right)^j\\
&=\frac{1}{2^m}\sum_{j=0}^m\binom{m}{j}^2\frac{2^{m-j}}{(x-1)^{m-j}}\cdot
\frac{(2x)^j}{(x-1)^j}\\
&=\frac{1}{(x-1)^m}\sum_{j=0}^m\binom{m}{j}^2x^j\tag{1}
\end{align*}

Multiplication of (1) with $(x-1)^m$ gives
\begin{align*}
S_m(x)=(x-1)^mP_m\left(\frac{x+1}{x-1}\right)\qquad\qquad m\geq 0
\end{align*}
and we finally conclude
\begin{align*}
S_{n,m}&=[x^m]\left(S_m(x)\right)^n\\
&=[x^m](x-1)^{mn}P_m\left(\frac{x+1}{x-1}\right)^n\qquad\qquad n\geq 1,m\geq 0
\end{align*}

Example: $S_{3,1}$ to $S_{3,3}$

At first wie calculate $S_{3,1}$ to $S_{3,3}$ according to the OPs definition of $S_{n,m}$. The we look at the coefficients of the corresponding transformed Legendre polynomials.

\begin{align*}
S_{3,1}&=\sum_{{x_1+x_2+x_3=1}\atop{x_j\geq 0}}\prod_{i=1}^3\binom{1}{x_i}^2
=\frac{3!}{2!1!}\binom{1}{0}^2\binom{1}{0}^2\binom{1}{1}^2\\
&=\color{blue}{3}\\

S_{3,2}&=\sum_{{x_1+x_2+x_3=2}\atop{x_j\geq 0}}\prod_{i=1}^3\binom{2}{x_i}^2\\
&=\frac{3!}{2!1!}\binom{2}{0}^2\binom{2}{0}^2\binom{2}{2}^2
+\frac{3!}{1!2!}\binom{2}{0}^2\binom{2}{1}^2\binom{2}{1}^2\\
&=3+48\\
&=\color{blue}{51}\\
S_{3,3}&=\sum_{{x_1+x_2+x_3=3}\atop{x_j\geq 0}}\prod_{i=1}^3\binom{3}{x_i}^2\\
&=\frac{3!}{2!1!}\binom{3}{0}^2\binom{3}{0}^2\binom{3}{3}^2
+\frac{3!}{1!1!1!}\binom{3}{0}^2\binom{3}{1}^2\binom{3}{2}^2\\
&\qquad+\frac{3!}{3!}\binom{3}{1}^2\binom{3}{1}^2\binom{3}{1}^2\\
&=3+486+729\\

&=\color{blue}{1218}
\end{align*}

On the other hand we obtain with some help of Wolfram Alpha
\begin{align*}
S_1(x)&=(x-1)P_1\left(\frac{x+1}{x-1}\right)\\
&=1+x\\
S_1^2(x)&=1+2x+x^2\\
S_1^3(x)&=1+\color{blue}{3}x+3x^2+x^3\\
\\

S_2(x)&=(x-1)^2P_2\left(\frac{x+1}{x-1}\right)\\
&=1+4x+x^2\\
S_2^2(x)&=1+8x+18x^2+8x^3+x^4\\
S_2^3(x)&=1+12x+\color{blue}{51}x^2+88x^3+51x^4+12x^5+x^6\\
\\
S_3(x)&=(x-1)^3P_3\left(\frac{x+1}{x-1}\right)\\
&=1+9x+9x^2+x^3\\
S_3^2(x)&=1+18x+99x^2+164x^3+99x^4+18x^5+x^6\\
S_3^3(x)&=1+27x+270x^2+\color{blue}{1218}x^3+2484x^4+2484x^5+1218x^6+270x^7+27x^8+x^9
\end{align*}

with corresponding coefficients marked in $\color{blue}{\text{blue}}$.