|
%I #4 Mar 30 2012 18:36:51
%S 1,2,1,-3,0,1,4,4,4,1,-5,-10,-10,0,1,6,18,24,12,6,1,-7,-28,-49,-42,
%T -21,0,1,8,40,88,104,72,24,8,1,-9,-54,-144,-216,-198,-108,-36,0,1,10,
%U 70,220,400,460,340,160,40,10,1,-11,-88,-319,-682,-946,-880,-550,-220,-55,0,1
%N Triangle T, read by rows, where row n of T equals row n of matrix (n+1)-th power of triangle A112555.
%C Remarkably, the matrix logarithm (A113290) is an integer triangle. Matrix m-th power of A112555 = I + m*(A112555 - I) where I = identity matrix.
%F G.f.: A(x, y) = 1/(1-x*y) + x*(x+2)/((1-x*y)^2*(1+x+x*y)^2).
%e Triangle begins:
%e 1;
%e 2,1;
%e -3,0,1;
%e 4,4,4,1;
%e -5,-10,-10,0,1;
%e 6,18,24,12,6,1;
%e -7,-28,-49,-42,-21,0,1;
%e 8,40,88,104,72,24,8,1;
%e -9,-54,-144,-216,-198,-108,-36,0,1;
%e 10,70,220,400,460,340,160,40,10,1; ...
%o (PARI) {T(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k)); polcoeff(polcoeff(1/(1-x*y)+x*(x+2)/((1-x*y)^2*(1+x+x*y)^2),n,X),k,Y)}
%Y Cf. A112555, A113288 (inverse), A113290 (log), A113291, A072374.
%K sign,tabl
%O 0,2
%A _Paul D. Hanna_, Oct 23 2005
|
