[seqfan] Re: Number of orders of distances to vertices of n-dimensional cube

[Max Alekseyev]

Hi Pierre,

I'm glad to see you were able to confirm a(4) and a(5) by other means. Here
is a brief idea behind the hyperplane arrangements:
Equality of any two distances defines a hyperplane in R^n (although
different pairs of distances may define the same hyperplane). All these
hyperplanes partition the space into cells, and the interior of each
(n-dimensional) cell corresponds to a particular strong order of the
differences. Hence, the number of different orders equals the number of
cells in the partition of R^n by the hyperplanes.

As for computation, I do not know much about how Sage computes the number
of cells and my code uses its functionality as a blackbox. I do not know
even if it's able to compute a(6) in reasonable time, but I'll run it for a
few days to find that out.

Regards,
Max


On Sat, Dec 3, 2022 at 7:41 PM Pierre Abbat <phma at bezitopo.org> wrote:

> On Saturday, November 19, 2022 8:16:38 AM EST Max Alekseyev wrote:
> > It is easy to prove that the hyperplane arrangement I introduced does
> > produce your sequence.
> > There should not be any worries in this respect. Still, you are welcome
> to
> > double check the counts produced by Sage.
>
> I don't know hyperplane arrangements or Sage. Can you point me to an
> explanation of hyperplane arrangements?
>
> After more than a day of generating orders in six dimensions, the program
> spat
> out the number 124187. However, that factors as 7*113*157, and dumping the
> histogram of five dimensions (516) shows that the orders occur at widely
> varying frequencies, so I suspect that it missed something. The sequence I
> get
> is 1,2,8,96,5376,1981440,5722536960(?). What do you get with Sage?
>
> Pierre
> --
> Don't buy a French car in Holland. It may be a citroen.
>
>
>
>