# Double sum involving binomial coefficients

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!

• What do you mean by solution? What do you expext? This is most likely not simply a polynomial in your variables, or some nice function. If you split it into several cases, depending on which max/min is attained, you have better luck, and it might even be so that a CAS can solve it for you, using Zeliberger magic. – Per Alexandersson Mar 21 '14 at 18:28
• @PerAlexandersson Actually, I tried a random case (both sums from $0$ to the appropriate $m$) with Mathematica, and it cannot evaluate it, so I would view the prospects for a closed form solution as somewhat grim. – Igor Rivin Mar 21 '14 at 18:39
• I tried specializing to $n=m=n'=m', p=p'=0$. The result seems to be oeis.org/A186375. Can you suggest other natural specializations? – Douglas Zare Mar 21 '14 at 20:15
• If I am not mistaken, the range of summation indices is "natural"; that is, the ranges can be dropped under the convention $\binom{n}{k}=0$ unless $0 \le k \le n$. With this in mind, have you tried searching through hypergeometric identities? I will go against the crowd and predict/hope your double sum can be simplified to one sum (even in the general case). – Peter Dukes Mar 24 '14 at 9:11
• In fact, @DouglasZare found such a solution for a specific specialization of the expression above: $$F(n,n,0|n,n,0) = \sum_{k=0}^n \binom{n}{k}^2 \binom{2k}{k} 4^{n-k} = 4^n {}_3F_2\left(\frac{1}{2},-n,-n;1,1;1\right)$$ according to OEIS' sequence A186375. Another simple specializations are: $$F(n,n,-n|n,n,-n) = F(n,n,n|n,n,n) = \binom{2n}{n}^2$$ – FreeQuark Mar 24 '14 at 15:38

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 $u_{n,m,p,n',m',p'}$): $$\sum_{n,m,p,n',m',p' \geq 0} u_{n,m,p,n',m',p'} x_1^n y_1^m z_1^p x_2^{n'} y_2^{m'} z_2^{p'} = -\frac{(x_2-1)(x_1-1)}{(y_2z_2+x_2-1)(y_1x_2-y_2x_2-y_1-x_2+1)(y_1x_1-y_2x_1+y_2+x_1-1)(y_1z_1+x_1-1)}$$
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);

• Or, to make the symmetry clearer, $\prod\limits_{i=1}^2\dfrac{(x_i-1)}{\Bigr(y_iz_i+x_i-1\Bigr)\Bigl((x_i-1)(y_{3-i}-1)-x_iy_i\Bigr)}$. – Wolfgang Oct 12 '15 at 7:14