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%I #3 Mar 30 2012 18:59:08
%S 1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,
%T 1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,0,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1
%N A mod 2 triangle based on floor((n+2)/2).
%C Row sums are A108355. Diagonal sums are A108356.
%F Number triangle T(n, k)=(sum{j=0..n-k, C(k, j)C(n-j, k)*floor((j+2)/2)} mod 2); Column k has g.f. x^k(1+x(mod(a(k+2), 4)))/(1-x^a(k)) where a(n)=-3cos(pi*n/2)/2-3sin(pi*n/2)/2+5/2=(1, 1, 4, 4, 1, 1, 4, 4, ...). T(n, k)=A108359(n, k) mod 2.
%e Rows begin
%e 1;
%e 1,1;
%e 1,1,1;
%e 1,1,1,1;
%e 1,1,0,1,1;
%e 1,1,0,0,1,1;
%e 1,1,1,0,1,1,1;
%e 1,1,1,1,1,1,1,1;
%K easy,nonn,tabl
%O 0,1
%A _Paul Barry_, May 31 2005
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