alternating sum of binomial coefficients

Question

I would like to know a closed formula for $\sum_{j=0}^{p-n } (-1)^j\binom{n^2}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1}$, especially in the case $p$ is near $n^2/2$. Similarly, I would like a closed formula for: setting $q=2\cdot\lceil\frac{n(n+1)}{4}\rceil -1$, and setting $p=\lceil\frac{q}{2}\rceil-1$, what is the sum $ \sum_{j=0}^{p-n } (-1)^j\binom{q}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1} $?

In either case I would be happy for an estimate of the growth of the sum (divided by $\binom {n^2-1}p$ in the first case, and divided by $\binom{q-1}p$ in the second).

Answer 1

I played around with your sum in Maple and got

$$ \frac{2n}{n+1}{2n-1 \choose n-1}{n^2 \choose p-n} 3F_{2}([n,n-p,2n+1],[n+2,n^2+n+1-p],1) $$

I make no guarantees that this is correct (especially as the original answer contained a $\binom{n^2}{-1}$ in it).

Answer 2

$\sum_{j=0}^{p-n} (-1)^j a_{n,p}(j) = \sum_{j=0}^{\frac{p-n}{2}} a_{n,p}(2j)-a_{n,p}(2j+1)$

so examine this difference $a_{n,p}(2j)-a_{n,p}(2j+1)$, we can factor out $c_{n,p}(j):=\frac{(n^2)!(2n+2j)!(n+2j)!}{(n-1)!(n-1)!(2j+1)!(n+2j+2)!(p-n-2j)!(n^2-p+n+2j)!}$

giving $c_{n,p}(j) \left[ \frac{(n+2j+2)(2j+1)}{(n+2j)} - \frac{(2n+2j+1)(p-n-2j)}{(n^2-p+n+2j+1)}\right]$

$=c_{n,p}(j) \left[ (2j+1)\left(1+\frac{2}{n+2j}- \frac{p-n-2j}{n^2-p+n+2j+1}\right)-\frac{2n(p-n-2j)}{n^2-p+n+2j+1}\right]\approx -2nc_{n,p}(j)$

this is assuming $p\approx \frac{n^2}{2} $ is large. not sure if this helps