linear combinations of binomial coefficients

[Roland Bacher]

writing (in a kind of tex notation) and using Wilsons identity
(p-1)!=-1 modulo p for p prime:

C(jp,p)=(jp-1)(jp-2)....(jp-(p-1))/(p-1)!=1-jp \sum_{h=1}^{p-1}
h\pm (jp)^2 \sum_{1\leq h_1<h_2<p} h_1h_2 \pm (jp)^3 \sum.... modulo p for prime p

(modulo errors,)

one can easily prove it for a given fixed value of n.
In order to get it for all n, one can perhaps prove it using
some generic divisibility properties of Stirling numbers but this 
needs surely some amount of work. If I can find some time, I will
think it over.   Roland


On Thu, Aug 03, 2006 at 04:37:47AM -0700, Max A. wrote:
> 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]