[seqfan] Re: A combinatorial problem

[Vladimir Shevelev]

Dear SeqFans,

I found a simple necessary condition for a suitable n. It is
Sum{k=0,...,n-3}(-1)^k*C(2n-k-4, k)*(n-k-2)!*(n-k-2)=A000179/(n-2)

I ask anyone to verify this much more simple condition for n>6. It seems that, possibly, no other solutions except of n=3,4,6.

Regards,
Vladimir

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Saturday, January 15, 2011 22:19
Subject: [seqfan] Re: A combinatorial problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> I would like to add to a solution of the problem the following: 
> if to take into account all trivial restrictions on binomial 
> coefficients, then  we have:
> max(r+k-n-1, 2r+k-2n-2, 0)<=i<=r-2. ( It seems that I 
> wrote r=1,...,n, but it should be r=3,...,n).
>  
> With these restrictions one can calculate n-2 sums for 
> r=3,4,...,n. Moreover,  using a symmetry, one can calculate 
> floor((n+3) / 2) sums for r=1,...,floor((n+3)/2) only. Every 
> (full) sum should be equal to  A000179(n) / (n-2).
>  
> Regards,
> Vladimir
> 
> 
> ----- Original Message -----
> From: Vladimir Shevelev <shevelev at bgu.ac.il>
> Date: Friday, January 14, 2011 18:00
> Subject: [seqfan] Re: A combinatorial problem
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> 
> > Correction. I wrote that "the interior sums B does not depend 
> on 
> > r."  I ask to ignore this phraze: of course, instead of 
> B, 
> > I ment the full sum: Sum{k=0,...,n-1}(...) does not depend on 
> r, 
> > since equals to
> > A000179(n)/(n-2).
> >  
> > Regards,
> > Vladimir
> > 
> > 
> > 
> > ----- Original Message -----
> > From: Vladimir Shevelev <shevelev at bgu.ac.il>
> > Date: Friday, January 14, 2011 14:19
> > Subject: [seqfan] A combinatorial problem
> > To: seqfan at list.seqfan.eu
> > 
> > > Dear SeqFans,
> > >  
> > > I ask anyone to extend a sequence which is connected with 
> the 
> > > following modification of the menage problem. A well known 
> > > mathematician N found himself with his wife among the 
> guests, 
> > > which were
> > > n(>=3) married couples. After seating the ladies on every 
> > other 
> > > chair at a circular table, N was the first offered to choose 
> > an 
> > > arbitrary chair but not side by side with his wife. For 
> which 
> > > values of n the number of ways of seating of other men ( 
> under 
> > > the condition that no husband is beside his wife) does not 
> > > depend on how far N takes his seat from his wife?
> > >  
> > > The first terms of this sequence are 3,4,6.  I proved 
> > that 
> > > the problem reduces to description the values of n>=3 for 
> > which, 
> > > for every r=1,...,n, we have
> > > Sum{k=0,...,n-1}((-1)^k)*(n-k-1)!*B=A000179(n)/(n-2),
> > > where B=Sum{i=0,...,k}C(2r-i-4, i)*C(2n-2r-k+i+2, k-i), 
> i.e., 
> > > for such an n, B does not depend on r  (here C-binomial 
> > > coefficients).In addition, I proved that A000179(n)/(n-2) is 
> > > integer, if n has the form 2^t+2 ( and I conjecture that 
> here 
> > > one can write "iff").
> > >  
> > > E.g., if n=3, then, for every r,  if k=0, then B=1; if 
> > k=1, 
> > > then B=2; if k=2, then B=1. Thus
> > > 1*2!-2*1!+1*0!=A000179(3)/1=1.
> > >  
> > > Regards,
> > > Vladimir
> > > 
> > >  Shevelev Vladimir‎
> > > 
> > > _______________________________________________
> > > 
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > > 
> > 
> >  Shevelev Vladimir‎
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎