linear combinations of binomial coefficients

[Max A.]

On 8/3/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:

> up to a factor, the following formula (in maple notation) seems to
> work:
>
> formula:=(n,p)->sum('(-1)^(j+1)*(binomial(2*n-3,n-j)-binomial(2*n-3,n-j-2))*binomial(j*p,p)/j','j'=1..n+1);

Nice catch, but unfortunately formula(n,p) is divisible only by
p^(2n-1) while it consists of n+1 summands. It would be a perfect
answer if it were divisible by p^(2(n+1)-1).

Please note that we are seeking for a linear combination of the form
k1*C(p,p) + k2*C(2*p,p) + ... + kn*C(n*p,p)     (consisting of n terms)
that is divisible by p^(2n-1) for all large enough prime p.

Max