A118407 - OEIS

%I #3 Mar 30 2012 18:36:57

%S 1,0,1,-2,0,1,2,-2,0,1,0,2,-2,0,1,-2,0,2,-2,0,1,4,-2,0,2,-2,0,1,-6,4,

%T -2,0,2,-2,0,1,4,-6,4,-2,0,2,-2,0,1,6,4,-6,4,-2,0,2,-2,0,1,-20,6,4,-6,

%U 4,-2,0,2,-2,0,1,26,-20,6,4,-6,4,-2,0,2,-2,0,1,-12,26,-20,6,4,-6,4,-2,0,2,-2,0,1

%N Triangle, read by rows, equal to the matrix square of triangle A118404; also equals the matrix inverse of triangle A118401.

%C This triangle has an integer matrix square-root (A118404) 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 A118404 are all different.

%F G.f.: A(x,y) = (1+x)^2/(1+x^2)/(1+2*x+2*x^2)/(1-x*y). Column g.f.: (1+x)^2/(1+x^2)/(1+2*x+2*x^2).

%e Triangle begins:

%e 1;

%e 0, 1;

%e -2, 0, 1;

%e 2,-2, 0, 1;

%e 0, 2,-2, 0, 1;

%e -2, 0, 2,-2, 0, 1;

%e 4,-2, 0, 2,-2, 0, 1;

%e -6, 4,-2, 0, 2,-2, 0, 1;

%e 4,-6, 4,-2, 0, 2,-2, 0, 1;

%e 6, 4,-6, 4,-2, 0, 2,-2, 0, 1;

%e -20, 6, 4,-6, 4,-2, 0, 2,-2, 0, 1;

%e 26,-20, 6, 4,-6, 4,-2, 0, 2,-2, 0, 1; ...

%o (PARI) {T(n,k)=polcoeff(polcoeff((1+x)^2/(1+x^2)/(1+2*x+2*x^2)/(1-x*y+x*O(x^n)),n,x)+y*O(y^k),k,y)}

%Y Cf. A118404 (matrix square-root), A118401 (matrix inverse), A118408 (row sums), A118409 (unsigned row sums).

%K sign,tabl

%O 0,4

%A _Paul D. Hanna_, Apr 27 2006