[seqfan] Re: A conjectural relation for Stirling numbers of the 1-st kind

Max Alekseyev maxale at gmail.com
Wed Aug 29 09:19:51 CEST 2012


Vladimir,

Just a quick remark - your identity can be written in the following
symmetric form:
sum{i=0,...,n-k} C(k+i,k) s(n,k+i) (n-1)^i = (-1)^(n+k) s(n,k).

I'll check later if it is known or follows from known results.

Regards,
Max

On Aug 28, 2012 11:49 PM, "Vladimir Shevelev" <shevelev at bgu.ac.il> wrote:
>
>
> Dear SeqFans,
>
> For Stirling numbers of the 1-st kind, it is known the following formula
(Abramowitz and Stegun, the third control relation):
> sum{k=m,m+1,...,n}n^(k-m)s(n+1,k+1)=s(n,m),
> or, after simple transformations,
> sum{i=1,...,n-k}s(n,k+i)(n-1)^i=(n-1)s(n-1,k).
> Recently I observed that a close sum
> sum{i=1,...,n-k}C(k+i,k)s(n,k+i)(n-1)^i=0, if n+k is even, and  -2s(n,k),
if n+k is odd.  If anyone saw it anywhere or can prove it?
>
> Regards,
> Vladimir
>
>  Shevelev Vladimir‎
>
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