# [seqfan] Re: A combinatorial problem

Mon Jan 17 18:14:37 CET 2011

```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,

----- 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,
>
>
> ----- 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,
> > >
> > >
> > > _______________________________________________
> > >
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>