linear combinations of binomial coefficients

[Max A.]

On 8/3/06, Max A. <maxale at gmail.com> 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).

As Roland mentioned in private email, the (n+1)-th term in his formula
is always zero. In other words, it can be re-defined as:

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);

Moreover, its numerical results match those from my original message
on the topic. Therefore, the j-th coefficient in n-th row of the table
seems to be explicitly expressed as

k(n,j) = (-1)^(j+1) * (C(2*n-3,n-j)-C(2*n-3,n-j-2)) / j *
lcm(1,2,...,2*(n-1)) / (C(2*n-3,n-1)-C(2*n-3,n-3))

PARI/GP verification code along with results is listed below.

Now, thanks Roland, we have the explicit formula. But the divisibility
results are still empirical. Can anybody *prove* that formula(n,p) is
divisible by p^(2n-1) ?

Thanks,
Max

PARI/GP verification code:

{ k(n,j) = (-1)^(j+1)*(binomial(2*n-3,n-j)-binomial(2*n-3,n-j-2)) / j
* lcm(vector(2*(n-1),i,i)) / (binomial(2*n-3,n-1)-binomial(2*n-3,n-3))
}

? matrix(9,10,n,j,k(n+1,j))
%1 =
[2 -1 0 0 0 0 0 0 0 0]
[12 -9 2 0 0 0 0 0 0 0]
[60 -54 20 -3 0 0 0 0 0 0]
[840 -840 400 -105 12 0 0 0 0 0]
[2520 -2700 1500 -525 108 -10 0 0 0 0]
[27720 -31185 19250 -8085 2268 -385 30 0 0 0]
[360360 -420420 280280 -133770 45864 -10780 1560 -105 0 0]
[720720 -864864 611520 -321048 127008 -36960 7488 -945 56 0]
[12252240 -15036840 11138400 -6297480 2776032 -942480 238680 -42525 4760 -252]