I am looking for the analytic continuation of
\begin{align\*}
& f\_m(v,w) := \sum\limits\_{k,l=0}^\infty v^k w^l {k+l+m \choose k} {k+l+m \choose l} \ ,
\end{align\*}
where $m \in \{1,2,...\}$ is fixed. The sums converge for small enough $|v|$ and $|w|$ and I have already made a few observations:


We can write
\begin{align\*}
& f\_m(v,w) = \sum\limits\_{k=0}^\infty v^k \frac{k!}{(k+m)!} \partial\_w^m [\phantom{|}\_2F\_1(k+1,k+1;1;w)]
\end{align\*}
and the hypergeometric function $\phantom{|}\_2F\_1(k+1,k+1;1;w)$ has the analytic continuation $(1-w)^{-k-1} P\_k(\frac{1+w}{1-w})$, where $P\_k$ is the $k$-th Legendre Polynomial. I think the property $\sum\limits\_{k=0}^\infty t^k P\_k(x) = \frac{1}{\sqrt{1-2xt+t^2}}$ or maybe the analogon for Jacobi-Polynomials (see [here](https://en.wikipedia.org/wiki/Jacobi_polynomials#Generating_function)) might be helpful.


Another observation is that the Chu-Vandermonde identity allows us to split up the binomial coefficients ${k+l+m \choose k}$ and ${k+l+m \choose l}$ into sums, which leads to the (not quite recursive) porperty
\begin{align\*}
& f\_m(v,w) = \sum\limits\_{i,j=0}^m w^i v^j {m \choose i} {m \choose j} f\_{i+j}(v,w) \ .
\end{align\*}


Finally the special case $v=w$ by Chu-Vandermonde identity yields
\begin{align\*}
& f\_m(v,v) = \sum\limits\_{a=0}^\infty v^a \sum\limits\_{k=0}^a {a+m \choose k} {a+m \choose k+m} = \sum\limits\_{a=0}^\infty v^a {2(a+m) \choose a} = \phantom{|}\_2{F}\_{1}\Big( m+\frac{1}{2} , m+1 ; 2m+1 ; 4v \Big) \ .
\end{align\*}


Any help in finding the analytic continuation is greatly appreciated!



Help finding an analytic continuation
Your $f\_m(v,w)$ is a special case of [Appell's $F\_4$ hypergeometric function](https://mathworld.wolfram.com/AppellHypergeometricFunction.html),
$$f\_m(v,w) = F\_4(m+1;m+1;m+1,m+1;v,w).$$
Some information about analytic continuation of $F\_4$ can be found in <https://arxiv.org/abs/2005.07170> and the references cited there.


