%I #5 Mar 30 2012 18:36:44
%S 1,1,1,1,2,1,1,1,3,1,1,4,-2,4,1,1,-13,26,-10,5,1,1,142,-229,116,-25,6,
%T 1,1,-1931,3181,-1567,371,-49,7,1,1,36296,-59700,29464,-6922,952,-84,
%U 8,1,1,-893273,1469380,-725108,170398,-23358,2100,-132,9,1,1,27927346,-45938639,22669816,-5327198,730252,-65526,4152
%N Triangular matrix, read by rows, where T(n,k) = T(n-1,k) - [T^-1](n-1,k-1); also equals the matrix inverse of A060083 (Euler polynomials).
%C Column 1 forms A102055. Column 2 forms A102056.
%F T(n, k) = T(n-1, k) - A060083(n-1, k-1), for n>0, with T(0, 0)=1.
%e T(5,3) = -10 = T(4,3) - A060083(4,2) = 4 - 14.
%e T(6,2) = -229 = T(5,2) - A060083(5,1) = 26 - 255.
%e Rows begin:
%e [1],
%e [1,1],
%e [1,2,1],
%e [1,1,3,1],
%e [1,4,-2,4,1],
%e [1,-13,26,-10,5,1],
%e [1,142,-229,116,-25,6,1],
%e [1,-1931,3181,-1567,371,-49,7,1],
%e [1,36296,-59700,29464,-6922,952,-84,8,1],...
%e The matrix inverse is equal to A060083:
%e [1],
%e [ -1,1],
%e [1,-2,1],
%e [ -3,5,-3,1],
%e [17,-28,14,-4,1],
%e [ -155,255,-126,30,-5,1],...
%o (PARI) {T(n,k)=local(M=matrix(n+1,n+1));M[1,1]=1;if(n>0,M[2,1]=1;M[2,2]=1); for(r=3,n+1, for(c=1,r,M[r,c]=if(c==1,M[r-1,1], if(c==r,1,M[r,c]=M[r-1,c]-((matrix(r-1,r-1,i,j,M[i,j]))^-1)[r-1,c-1])))); return(M[n+1,k+1])}
%Y Cf. A060083, A102055, A102056.
%K sign,tabl
%O 0,5
%A _Paul D. Hanna_, Dec 28 2004