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

[Vladimir Shevelev]

Thank you, Max. In addition, I was able to prove that this identity is equivalent to the following simpler one: sum{i=1,...,n-k}C(k+i,k)*s(n,k+i)=k*s(n-1,k).  But it is remains to be conjectural.
 
Regards,
Vladimir


----- Original Message -----
From: Max Alekseyev <maxale at gmail.com>
Date: Tuesday, August 28, 2012 19:20
Subject: [seqfan] Re: A conjectural relation for Stirling numbers of the 1-st kind
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> 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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> >
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 Shevelev Vladimir‎