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

[Pierre Abbat]

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.