%I
%S 1,1,1,3,2,1,11,15,3,1,25,44,18,4,1,41,115,110,50,5,1,43,246,
%T 375,220,45,6,1,29,315,861,805,385,105,7,1,335,232,1204,2296,
%U 1750,616,84,8,1,1199,3033,1044,3780,5166,2898,924,180,9,1
%N Triangle T, read by rows, equal to the matrix product T = H*[C^1]*H, where H is the selfinverse triangle A118433 and C is Pascal's triangle.
%C The matrix inverse of H*[C^1]*H is H*C*H = A118438, where H^2 = I (identity). The matrix log, log(T) = A118441, is a matrix square root of a triangular matrix with a single diagonal (two rows down from the main diagonal).
%F Since T + T^1 = C + C^1, then [T^1](n,k) = (1+(1)^(nk))*C(n,k)  T(n,k) is a formula for the matrix inverse T^1 = A118438.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 3, 2, 1;
%e 11, 15, 3, 1;
%e 25,44,18, 4, 1;
%e 41,115,110, 50, 5, 1;
%e 43, 246, 375,220,45, 6, 1;
%e 29, 315, 861,805,385, 105, 7, 1;
%e 335, 232,1204, 2296, 1750,616,84, 8, 1;
%e 1199, 3033, 1044, 3780, 5166,2898,924, 180, 9, 1; ...
%e The matrix log, log(T) = A118441, starts:
%e 0;
%e 1, 0;
%e 4, 2, 0;
%e 12, 12, 3, 0;
%e 32,48,24, 4, 0;
%e 80,160,120, 40, 5, 0; ...
%e where matrix square, log(T)^2, is a single diagonal:
%e 0;
%e 0,0;
%e 2,0,0;
%e 0,6,0,0;
%e 0,0,12,0,0;
%e 0,0,0,20,0,0; ...
%o (PARI) {T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r1,c1)*(1)^(r\2 (c1)\2+rc))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r1,c1))));(M*C^1*M)[n+1,k+1]}
%Y Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (selfinverse H).
%K sign,tabl
%O 0,4
%A _Paul D. Hanna_, Apr 28 2006
