[seqfan] Re: How many squares can you make from n points in the plane?

[Allan Wechsler]

Boy, doesn't it feel like we could prove that a(7) = 3?

A configuration for k squares on n points corresponds to a partition of 4k
into n parts, where each part gives the multiplicity of one of the n
points. For example, I can see two different configurations with 3 squares
on 7 points, one with the partition 3222111 and the other with the
partition 2222211. If a(7)=4, a configuration of 4 squares on 7 points
would have a partition of 16 into 7 parts. It feels like the first thing to
prove is that no point in a 7-point configuration could have a multiplicity
greater than 3.

If we can prove a few constraints on that partition, we might produce a
manageable number of cases for case analysis.

A couple of related sequences: the minimum number of points which are all
the vertices of a set of n squares; the maximum number of squares *sharing
a single vertex* that can be drawn on n points.

On Tue, Sep 28, 2021 at 6:13 PM Neil Sloane <njasloane at gmail.com> wrote:

> It seems that surprisingly little is known - see A051602. Even a(7) is an
> open question, although it is easy to see that a(7) >= 3.
>
> Best regards
> Neil
>
> Neil J. A. Sloane, Chairman, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
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