%I #6 Jun 14 2017 00:39:22
%S 1,1,1,1,3,1,1,12,6,1,1,75,51,10,1,1,744,636,145,15,1,1,11181,11406,
%T 2980,330,21,1,1,242835,284541,83890,10095,651,28,1,1,7330930,9580386,
%U 3133720,411510,27741,1162,36,1,1,297801232,424183235,151359220
%N Triangle, read by rows, where column k of T = column 0 of matrix power T^{(k+1)(k+2)/2} for k>=0, with T(n,0)=1 for n>=0.
%e Triangle T begins:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 12, 6, 1;
%e 1, 75, 51, 10, 1;
%e 1, 744, 636, 145, 15, 1;
%e 1, 11181, 11406, 2980, 330, 21, 1;
%e 1, 242835, 284541, 83890, 10095, 651, 28, 1;
%e 1, 7330930, 9580386, 3133720, 411510, 27741, 1162, 36, 1; ...
%e Matrix cube, T^3, begins:
%e 1;
%e 3, 1;
%e 12, 9, 1;
%e 75, 90, 18, 1;
%e 744, 1224, 333, 30, 1; ...
%e where column 0 of T^3 = column 1 of T.
%e Matrix 6th power, T^6, begins:
%e 1;
%e 6, 1;
%e 51, 18, 1;
%e 636, 342, 36, 1;
%e 11406, 8145, 1206, 60, 1; ...
%e where column 0 of T^6 = column 2 of T.
%e 1;
%e 10, 1;
%e 145, 30, 1;
%e 2980, 930, 60, 1;
%e 83890, 34635, 3210, 100, 1; ...
%e where column 0 of T^10 = column 3 of T.
%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, 1]=1, B[i, j]=(A^(j*(j+1)/2))[i-j+1, 1])); )); A=B); A[n+1, k+1]
%Y Cf. columns: A134524, A134525, A134526.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 15 2007