[seqfan] Improved lower bound for A250000

[Benoît Jubin]

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