Generating function for product of binomial coefficients In general, if $a(n)$ is an integer sequence with generating function $A(t)$ and $b(n)$ is an integer sequence with generating function $B(t)$, it is not easy to find the generating function $C(t)$ for $c(n)=a(n)b(n)$ in terms of $A$ and $B$, i.e., to find the Hadamard product of $A$ and $B$. However, it is not impossible to do so in some special cases.
I am interested in the case where $a(n) = \binom{N}{n}$ and $b(n) = \binom{M}{n}$, i.e., where the sequences are binomial coefficients. In this case $A(t) = (1+t)^N$ and $B(t) = (1+t)^M$. But what is $C$?
Thanks in advance for any advice or references!
 A: Here is a "solution" that might be useful; but it is not wholly satisfactory.
(as usual read carefully before using (sigh!))
We use the “Snake Oil Method” from Wilf's book “generatingfunctionology” (second edition) Amazon
or free: Free version
A good book in either form. Chapter 4, page 125 and following the example on 126.  We get:
$\left(\begin{array}{c}
N\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)= \left[t^{n}\right]\left[s^{N}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$
Let
$F(N,M,n;t)=\displaystyle \sum_{n=0}\left(\begin{array}{c}
N\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}$
We “dodge the bullet” and sum on an independent parameter N and variable s. In this case it's easy to undo afterwards.
$f(M,N;s,t)={\displaystyle \sum_{n=0}\left(\begin{array}{c}
i\\
n
\end{array}\right)\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}s^{i}={\displaystyle {\displaystyle \sum_{n=0}}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)t^{n}\cdot{\displaystyle \sum_{i=0}}\left(\begin{array}{c}
i\\
n
\end{array}\right)s^{i}$
$={\displaystyle {\displaystyle \sum_{n=0}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot t^{n}\cdot\frac{s^{n}}{\left(1-s\right)^{n+1}}=\frac{1}{1-s}\cdot{\displaystyle {\displaystyle \sum_{n=0}}}\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot\frac{\left(t\cdot s\right)^{n}}{\left(1-s\right)^{n}}=\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$
Now it's obvious how to pick out $t^{n}$
$\left[t^{n}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}=\frac{1}{1-s}\cdot\left(\begin{array}{c}
M\\
n
\end{array}\right)\cdot\left(\frac{\left(s\right)}{\left(1-s\right)}\right)^{n}$
And  $s^{N}$
$\left[s^{N}\right]\frac{1}{1-s}\left(\frac{\left(s\right)}{\left(1-s\right)}\right)^{n}=\left[s^{N-n}\right]\frac{1}{\left(1-s\right)^{n+1}}=\left(\begin{array}{c}
n+\left(N-n\right)\\
N-n
\end{array}\right)=\left(\begin{array}{c}
N\\
N-n
\end{array}\right)=\left(\begin{array}{c}
N\\
n
\end{array}\right)$
But not what we would prefer.
$\left[s^{N}\right]\frac{1}{1-s}\cdot\left(1+\frac{\left(t\cdot s\right)}{\left(1-s\right)}\right)^{M}$ to leave $\left[t^{n}\right]$
Which I can't see how to do.
