TITLE: Closed form or/and asymptotics of a hypergeometric sum
QUESTION [6 upvotes]: Dear mathematicians,
I am a computer scientist wandering in the deep sea of combinatorics and asymptotics to pursue a recent interest in average case analysis of algorithms. In doing so, I designed the following double summation:
$$
\sum_{h = 1}^{n}\sum_{j=0}^{n-h}h{2j+h-1\choose j}{2n - 2j - h\choose n-j},
$$
which, divided by the $n$th Catalan number, yields the statistics I want. Since the variable $j$ occurs in many places, I thought that I could exchange the summations without worrying about the bounds and then work on $h$ instead:
$$
\sum_{j\geq 0}\sum_{h>0}h{(2j-1)+h\choose j}{2(n-j)-h\choose n -j}.
$$
Does anyone know if there is a closed form for
$$
\sum_{h>0}{h}{p+h\choose q}{2r - h \choose r}?
$$
I read the relevant chapter in Concrete Mathematics, to no avail. Perhaps generating functions would help? Anyhow, I would be interested in the main term of the asymptotic expansion of the original double summation.
Thanks for reading me.

REPLY [4 votes]: For your sequence
s:=[0, 1, 7, 39, 198, 955, 4458, 20342, 91276, 404307, 1772610, 7707106, 33278292, 142853854, 610170148, 2594956620, 10994256152, 46425048451, 195456931506, 820725032042, 3438011713540]

I use with(gfun) in Maple and then guessgf(s, x) to find out that the generating function for your sequence is
$$
\frac{x(1+\sqrt{1-4x})}{2(1-4x)^2}.
$$
This isn't now hard to establish rigorously by verifying that your double sum satisfies a polynomial recurrence (a multivariate version of the Gosper-Zeilberger creative telescoping).