New twist(?) on old point-counting problem

[N. J. A. Sloane]

I am adding this entry to the OEIS based
on David's message:

%I A123937
%S A123937 1,1,2,1,2,4,1,2,4,8,1,4,4,8,16,1,4,12,8,16,32,1,4,12,32,16,32,64,1,
%T A123937 4,12,32,80,32,64,128,1,4,16,32,80,192,64,128,256,1,6,16,56,80,192,
%U A123937 448,128,256,512,1,6,24,56,176,192,448,1024,256,512,1024
%N A123937 Triangle read by rows: T(x, y) = 0 if y > x, = 1 if y = 0, or = 2*Sum_{k >= 1 , x-k^2 >= y} T(x-k^2, y-1) otherwise. The zeros are omitted from the sequence.
%e A123937 Triangle begins:
%e A123937 1
%e A123937 1 2
%e A123937 1 2 4
%e A123937 1 2 4 8
%e A123937 1 4 4 8 16
%e A123937 1 4 12 8 16 32
%e A123937 1 4 12 32 16 32 64
%e A123937 1 4 12 32 80 32 64 128
%O A123937 0,3
%K A123937 nonn
%A A123937 David Wilson (davidwwilson(AT)comcast.net), Oct 30 2006


But I've not included any mention of the point counting
connection - perhaps David or one of the people
who posted follow-up messages would add that?

Neil