Volumes of polytopes in hypercubes
Paul D. Hanna
pauldhanna at juno.com
Sat Aug 19 15:32:57 CEST 2006
Roland,
Consider A011818 :
Normalized volume of center slice of n-dimensional cube:
2^n * n! * Vol({ x\in [ 0,1 ]: sum_{i=1}^n n/2 <= x_i <= (n+1)/2 }).
FORMULA
a(n) = Sum_{k=1..n} C(n,k-1)*A008292(n,k) for n>=1.
Looks related to your interesting study, and
the formula involves Eulerian numbers.
Paul
On Sat, 19 Aug 2006 12:41:28 +0200 Roland Bacher
<Roland.Bacher at ujf-grenoble.fr> writes:
> On Fri, Aug 18, 2006 at 09:13:00AM -0400, N. J. A. Sloane wrote:
> > Roland said:
> >
> > It would be interesting to know them:
> > The first value $n! Vol(P(0))=n! Vol(P(n-1))$ is always one.
> > The triangle of these Volumes starts thus
> >
> > 1: 1
> > 2: 1 1
> > 3: 1 4 1
> > 4: 1 11 11 1
> > 5: 1 x y x 1
> >
> > where 2+2x+y=120 and y/120 is the volume of the convex hull
> > P(2)=P(2)_5 of the 20 integral points with coordinates in {0,1}^5
> and
> > coordinate sum either 2 or 3. (Remark that this polytope is
> symmetric
> > with respect to central symmetry in its barycenter
> 1/5(1,1,1,1,1),
> > this holds of course for all "middle" polytopes P(n)_{2n+1}
> > in odd dimension 2n+1).
> >
> > Me: Looks like the Eulerian numbers, A008292 !
> >
> > NJAS
>
> A random computations for n=5,6 (choosing 100000 points uniformly
> at random in [0,1] and computing the proportion of these points
> in a given such polytope) seem to confirm the Eulerian numbers.
>
> Roland
>
>
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