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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