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

[Vladimir Shevelev]

Very sorry for a misprint in my previous message: it should be
sum{i=1,...,n-k}C(k+i,k)*s(n,k+i)=n*s(n-1,k).  Now I proved this identity.

Regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Sunday, September 2, 2012 16:59
Subject: [seqfan] Re: A conjectural relation for Stirling numbers of the 1-st kind
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

>  
> 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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> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎