[seqfan] Re: Improved lower bound for A250000

[Rob Pratt]

Please share your n =24 solution.  Under the central symmetry constraint, I get a maximum of 80, not 84.

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Benoît Jubin
Sent: Tuesday, February 24, 2015 11:50 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Improved lower bound for A250000

Dear seqfans,

Let a(n)=A250000(n) be the maximum size of coexisting armies of queens on an (n,n) chessboard.

By modifying the Pratt--Selcoe configuration, I improved the best known lower bound from
a(n) > (9/4)*(n/4)^2
to
a(n) > (7/3)*(n/4)^2.
I have been sloppy with side effects, but to be on the safe side, let's say
a(n) > (7/3)*(floor(n/4))^2 - (3+8*sqrt(2)/3)*ceiling(n/4), where the coefficient 3+8*sqrt(2)/3 is a perimeter that you can compute from the following description.

The configuration in the limit n = infinity is as follows: denoting by x,y in [0,1] the coordinates on the chessboard, the queens of one color are in the two regions x<1/4, y<1/2, x<y<x+1/3 and 1/2<x<3/4, y<x-1/3, y<1-x and the queens of the other color are obtained by central symmetry. As you can guess, I obtained these coefficients by equalizing the lengths of the "opposite" boundaries of the armies (this already improves (by 1) on the "Board 4" example of the webpage).

Using an easy upper bound, one has asymptotically
(2+1/3)*(n/4)^2 < a(n) < 4*(n/4)^2.
Anyone to help fill the gap?

Best,
Benoit

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