How to prove that the following double sum is always an integer？ I have veriﬁed the following double sum is always an integer for $s$ up to $1000$ via Maple.
But I can not prove it. Proofs, hints, or references are all welcome.
Thanks!
$$\sum_{m=s}^{2s}\sum_{k=0}^{s} {2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} \frac{1}{(s+1)(2k-1)(2m-2k-1)}$$
What I have known is that: 
(1) Every term is not always an integer, but I can prove that ${2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} \frac{1}{(2k-1)(2m-2k-1)}$ is always an integer.
(2) $\sum_{k=0}^{s} {2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} ={2s\choose s}^2{s\choose m-s}$. This combinatorial identity may be helpful to solve this problem.
Note:- The problem has also been posted  here.
 A: Here is an attempt at an answer. We assume that the recurrence from my comment above holds [with a small correction] (a proof was obtained by Kevin using Zeilberger's algorithm, see the comment below):
$$ (7s+8)(s+4)(s+3)^2 a_{s+3} - 4(56s^2+127s+57)(s+3)(s+2) a_{s+2} $$
$$ - 16(7s^4-6s^3-121s^2-210s-90) a_{s+1} + 128(7s+15)(2s+3)(2s+1)(s-1) a_s = 0 $$
Write $a_s = b_s/(s+1)$. According to Kevin's claim (1), $b_s \in \mathbb Z$. Now
the sequence $(b_s)$ satisfies the recurrence
$$ (7s+8)(s+3)^2(s+2)(s+1) b_{s+3} - 4(56s^2+127s+57)(s+2)^2(s+1) b_{s+2} $$
$$     - 16(7s^4-6s^3-121s^2-210s-90)(s+1) b_{s+1}
     + 128(7s+15)(2s+3)(2s+1)(s+2)(s-1) b_s = 0 .$$
We want to show that $s+1$ divides $b_s$. By the recurrence, we have that
$$ (s+1) \mid 128(7s+15)(2s+3)(2s+1)(s+2)(s-1) b_s . $$
Since the gcd of $s+1$ with the factor in front of $b_s$ is a power of 2, we can conclude that the odd part of $s+1$ divides $b_s$. On the other hand, it is easy to see that each term in the sum for $a_s$ is 2-adically integral (it is a 
Catalan number times some binomial coefficients times a fraction with odd
denominator), so there is no need to consider the 2-power part of $s+1$.
