# [seqfan] Re: A combinatorial problem

Fri Jan 14 19:26:41 CET 2011

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,

----- 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,
>
>
>
> ----- 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,
> >
> >
> > _______________________________________________
> >
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> >
>