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%I #3 Mar 30 2012 18:37:08
%S 1,1,1,3,4,1,15,24,7,1,108,198,63,10,1,1036,2116,714,120,13,1,12569,
%T 28052,9884,1725,195,16,1,185704,446560,162729,29190,3393,288,19,1,
%U 3247546,8325700,3117660,571225,67756,5880,399,22,1,65762269,178284892
%N Triangle U, read by rows, where column k of U^(j+1) = column j of P^(3k+1) for j>=0, k>=0 and P=A136220.
%F This triangle U = P * [P^2 shift right one column] (see example), where P = A136220 and P^2 = A136225.
%e Triangle U begins:
%e 1;
%e 1, 1;
%e 3, 4, 1;
%e 15, 24, 7, 1;
%e 108, 198, 63, 10, 1;
%e 1036, 2116, 714, 120, 13, 1;
%e 12569, 28052, 9884, 1725, 195, 16, 1;
%e 185704, 446560, 162729, 29190, 3393, 288, 19, 1;
%e 3247546, 8325700, 3117660, 571225, 67756, 5880, 399, 22, 1; ...
%e where column k of U = column 0 of P^(3k+1) and
%e triangle P = A136220 begins:
%e 1;
%e 1, 1;
%e 3, 2, 1;
%e 15, 10, 3, 1;
%e 108, 75, 21, 4, 1;
%e 1036, 753, 208, 36, 5, 1;
%e 12569, 9534, 2637, 442, 55, 6, 1;
%e 185704, 146353, 40731, 6742, 805, 78, 7, 1; ...
%e where column k of P = column 0 of U^(k+1).
%e Also, this triangle U can be obtained by the matrix product:
%e U = P * [P^2 shift right one column]
%e where P^2 shift right one column begins:
%e 1;
%e 0, 1;
%e 0, 2, 1;
%e 0, 8, 4, 1;
%e 0, 49, 26, 6, 1;
%e 0, 414, 232, 54, 8, 1;
%e 0, 4529, 2657, 629, 92, 10, 1;
%e 0, 61369, 37405, 9003, 1320, 140, 12, 1; ...
%o (PARI) {T(n,k)=local(P=Mat(1),U=Mat(1),PShR);if(n>0,for(i=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));U=P*PShR^2; U=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,U[r,c], if(c==1,(P^3)[ #P,1],(P^(3*c-1))[r-c+1,1])))); P=matrix(#U, #U, r,c, if(r>=c, if(r<#R,P[r,c], (U^c)[r-c+1,1])))));U[n+1,k+1]}
%Y Cf. A136221 (column 0), A136229 (column 1); related tables: A136220 (P), A136225 (P^2), A136230 (V), A136231 (W=P^3), A136217, A136218.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Jan 28 2008
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