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

[Max Alekseyev]

Hi Pierre,

The next term is 5376 as computed by this quick'n'dirty Sage code based on
hyperplane arrangements:
https://sagecell.sagemath.org/?q=gbndab

It may also be possible to compute a(5) but not online.

Regards,
Max


On Mon, Nov 14, 2022 at 9:08 AM Pierre Abbat <phma at bezitopo.org> wrote:

> Let C be an n-dimensional cube and p be a point in R^n such that the
> distances
> from p to the 2^n vertices of C are all different. Write the vertices in
> order
> of their distance from p. How many different orders of vertices are there?
>
> I'm writing an algorithm to search an octree for the two closest points to
> a
> given point p. Usually one of them will be p. I search the eight subcubes
> in
> order of their distance from p, so that I can ignore farther subcubes
> entirely, knowing that the distance to the closest points is less than the
> distance to the subcube. There are 96 orders, but I'm using only 48 to
> simplify the code, using e.g. {0,1,2,3,4,5,6,7} but not {0,1,2,4,3,5,6,7}.
>
> The sequence begins 1,2,8,96. There are four sequences beginning 1,2,8,96,
> of
> which one has offset 1,2 so I discard it. I suspect it's the map-folding
> sequence A001417, since the next term 4608 is divisible by 2^4×4!, but my
> four-dimensional intuition is not good enough to verify it without writing
> some code, and I need to write only the code for n=3 because someone needs
> it
> for a device.
>
> Take a set of n real numbers {x[0]..x[n-1]}. Take all 2^n subsets of the
> set,
> sum each subset, and make sure they're all different. The nth term of the
> sequence (starting at n=0) is the number of different orders the sums can
> be
> in.
>
> Pierre
> --
> Don't buy a French car in Holland. It may be a citroen.
>
>
>
>
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