# Finite sum with three binomial coefficients

I need to find a closed form expression, if there is one, of the following sum:

$$\sum_{j=0}^m{n+1-k\choose j}{k-1\choose m-j}{A+2-k+m-j\choose m-j+2}$$

where all parameters are integers, $~1\leq k\leq n,\quad0\leq m< n,~$ and $~2n<A.$

I've tried a lot of binomial identities, from Wikipedia, Concrete Mathematics, A = B, and the Wolfram function site, without any real simplification.

I've also tried Zeilberger's algorithm, which does give me a second order recurrence with non-constant coefficients. Now if it was first order or with constant coefficients that would have given me the answer, but now I need to use Petrovšek's algorithm to solve the recurrence. This gets too hard by hand, and I can't get the only numerical implementation I've found working.

Any ideas?

• Mathematica say the sum is equal to "Binomial[-1 + k, m] Binomial[2 + A - k + m, 2 + m] HypergeometricPFQ[{-2 - m, -m, -1 + k - n}, {k - m, -2 - A + k - m}, 1]" – Giovanni Resta Feb 25 '16 at 14:26
• Yes I forgot to mention this; if you ask me that is more complicated that the original though (I guess it implies a discussion of what a "closed form" is) – jorgen Feb 25 '16 at 14:53
• Since you cited various sources but not software, I added it just to be sure, but I was confident you already know that. Btw, I hardly know what hypergeometric functions are... apart that Mma likes them a lot. – Giovanni Resta Feb 25 '16 at 14:59
• Thanks for that, I'm really looking for anything here! Yes it does, me not so much, hahaha – jorgen Feb 25 '16 at 15:21