[Math] Double sum involving binomial coefficients

binomial-coefficientsco.combinatoricscombinatorial-identities

I came across a sum of binomial coefficients while trying to solve a problem involving $SU(2)$ group integrals. I am not able to solve it, nor I found a similar identity in the literature. I would like to ask if anyone knows of any techniques/references that I might use in order to find a solution. The expression is the following

$$
\sum_{i=\max(0,p)}^{\min(n+p,m)}
\sum_{i'=\max(0,p')}^{\min(n'+p',m')}
\binom{n+p}{i}
\binom{n'+p'}{i'}
\binom{i-p+m'-i'}{m'-i'}
\binom{i'-p'+m-i}{m-i}
=F(m,n,p|m',n',p')
$$

It is very symmetric, and that gives me hope to find a solution, even though it depends on 6 independent non-negative integers, $m,n,p$ and $m',n',p'$. (Edit: in fact it should be $p\geq-n$ and $p'\geq-n'$).

I noticed that in most (if not all) binomial identities, there are more instances of the summation index (e.g. $i$) on the lower part of the binomial coefficients than on the top, in any given identity. Here the situation is the opposite, and it is the reason why I cannot solve this expression.

Thanks!

Best Answer

Using my Maple package BinomSums it is easy to compute the generating function of the left-hand side of your identity (let's call it $u_{n,m,p,n',m',p'}$): $$\sum_{n,m,p,n',m',p' \geq 0} u_{n,m,p,n',m',p'} x_1^n y_1^m z_1^p x_2^{n'} y_2^{m'} z_2^{p'} = -\frac{(x_2-1)(x_1-1)}{(y_2z_2+x_2-1)(y_1x_2-y_2x_2-y_1-x_2+1)(y_1x_1-y_2x_1+y_2+x_1-1)(y_1z_1+x_1-1)}$$

Thus a rational generating function usually gives all we need about a sequence.

Here is the code:

S := Sum(Sum(Binomial(n1+p1,i1)*Binomial(n2+p2,i2)*Multinomial([i1-p1,m2-i2])*Multinomial([i2-p2,m1-i1]), i1=0..infinity), i2=0..infinity);
vars := [n1,m1,p1,n2,m2,p2]:
gfunS := foldl(Sum, S*mul((t||v)^v, v in vars), seq(v=0..infinity, v in vars));
BinomSums[sumtores](gfunS, u);
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