%I #8 Jun 13 2017 22:13:21
%S 1,1,1,0,2,1,0,2,4,1,0,2,12,8,1,0,2,44,56,16,1,0,2,236,504,240,32,1,0,
%T 2,2028,6776,4720,992,64,1,0,2,29164,146552,139120,40672,4032,128,1,0,
%U 2,719340,5314680,6583152,2500832,337344,16256,256,1,0,2,30943724
%N Triangle T, read by rows, such that the matrix square shifts T left one column and up one row, with T(0,0)=T(1,0)=1 and T(n,0)=0 for n>1 and T(n,n)=1 for n>=0.
%C Column 2 forms A098543. Row sums form A098544. The absolute value of the matrix inverse equals A098539.
%F T(n+1, 1) = 2 for n>0; T(n+1, n) = 2^n, T(n+2, n) = 4^n - 2^n for n>=0. Matrix square: [T^2](n, k) = T(n+1, k+1). Matrix inverse: [T^-1](n, k) = (-1)^(n-k)*A098539(n, k). Matrix square inverse: [T^-2](n, k) = (-1)^(n-k)*A098539(n+1, k+1).
%e Rows of T begin:
%e [1],
%e [1,1],
%e [0,2,1],
%e [0,2,4,1],
%e [0,2,12,8,1],
%e [0,2,44,56,16,1],
%e [0,2,236,504,240,32,1],
%e [0,2,2028,6776,4720,992,64,1],
%e [0,2,29164,146552,139120,40672,4032,128,1],
%e [0,2,719340,5314680,6583152,2500832,337344,16256,256,1],...
%e Rows of T^2 begin:
%e [1],
%e [2,1],
%e [2,4,1],
%e [2,12,8,1],
%e [2,44,56,16,1],
%e [2,236,504,240,32,1],...
%e showing that T shifts left and up under matrix square.
%e The matrix inverse of T begins:
%e [1],
%e [ -1,1],
%e [2,-2,1],
%e [ -6,6,-4,1],
%e [26,-26,20,-8,1],
%e [ -166,166,-140,72,-16,1],...
%e the absolute value of which equals triangle A098539.
%o (PARI) T(n,k)=local(A,B);A=matrix(1,1);A[1,1]=1;for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,j]=(A^0)[i-1,1],B[i,j]=(A^2)[i-1,j-1]));));A=B);A[n+1,k+1]
%Y Cf. A098543, A098544, A098539 (absolute inverse).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Sep 16 2004