[seqfan] Re: A combinatorial problem

[Vladimir Shevelev]

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