%I #3 Mar 30 2012 18:36:57
%S 1,0,1,2,0,1,-2,2,0,1,4,-2,2,0,1,-6,4,-2,2,0,1,8,-6,4,-2,2,0,1,-10,8,
%T -6,4,-2,2,0,1,12,-10,8,-6,4,-2,2,0,1,-14,12,-10,8,-6,4,-2,2,0,1,16,
%U -14,12,-10,8,-6,4,-2,2,0,1,-18,16,-14,12,-10,8,-6,4,-2,2,0,1,20,-18,16,-14,12,-10,8,-6,4,-2,2,0,1
%N Triangle, read by rows, equal to the matrix square of triangle A118400; also equals the matrix inverse of triangle A118407.
%C This triangle has an integer matrix square-root (A118400) if the main diagonal of the square-root is allowed to be signed. Even though the columns of this triangle are all the same, the columns of the matrix square-root A118400 are all different.
%F G.f.: A(x,y) = (1 + 2*x + 2*x^2)*(1+x^2)/(1+x)^2/(1-x*y). Column g.f.: (1 + 2*x + 2*x^2)*(1+x^2)/(1+x)^2.
%e Triangle begins:
%e 1;
%e 0, 1;
%e 2, 0, 1;
%e -2, 2, 0, 1;
%e 4,-2, 2, 0, 1;
%e -6, 4,-2, 2, 0, 1;
%e 8,-6, 4,-2, 2, 0, 1;
%e -10, 8,-6, 4,-2, 2, 0, 1;
%e 12,-10, 8,-6, 4,-2, 2, 0, 1;
%e -14, 12,-10, 8,-6, 4,-2, 2, 0, 1;
%e 16,-14, 12,-10, 8,-6, 4,-2, 2, 0, 1; ...
%o (PARI) {T(n,k)=polcoeff(polcoeff((1+2*x+2*x^2)*(1+x^2)/(1+x)^2/(1-x*y+x*O(x^n)),n,x)+y*O(y^k),k,y)}
%Y Cf. A118400 (matrix square-root), A118402 (row sums), A118403 (unsigned row sums), A118407 (matrix inverse).
%K sign,tabl
%O 0,4
%A _Paul D. Hanna_, Apr 27 2006