# [seqfan] Re: A combinatorial problem

Fri Jan 14 16:56:52 CET 2011

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,

----- 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,