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

[Max Alekseyev]

No, I have not added this sequence - please do so as it was your
proposition.
Sage has been computing a(6) for over a day now.

Regards,
Max

On Mon, Dec 5, 2022 at 1:46 PM Pierre Abbat <phma at bezitopo.org> wrote:

> On Sunday, December 4, 2022 3:18:05 PM EST Max Alekseyev wrote:
> > 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.
>
> Sounds good. Have you started entering the sequence?
>
> I'd like to do a few more things to my program before I publish it. It
> computes a(5); I have to edit the program to get a(4) or a(6). And it
> gives no
> indication of how far along it is.
>
> Pierre
>
> --
> When a barnacle settles down, its brain disintegrates.
> Já não percebe nada, já não percebe nada.
>
>
>
>
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