%I #2 Mar 30 2012 18:37:08
%S 1,4,1,26,10,1,232,110,16,1,2657,1435,248,22,1,37405,22135,4240,440,
%T 28,1,627435,397820,81708,9295,686,34,1,12248365,8203057,1773156,
%U 214478,17248,986,40,1,273211787,191405232,43039532,5442349,463267,28747,1340
%N Matrix square of triangle V = A136230, read by rows.
%F Column k of V^2 (this triangle) = column 1 of P^(3k+2), where P = triangle A136220.
%e This triangle, V^2, begins:
%e 1;
%e 4, 1;
%e 26, 10, 1;
%e 232, 110, 16, 1;
%e 2657, 1435, 248, 22, 1;
%e 37405, 22135, 4240, 440, 28, 1;
%e 627435, 397820, 81708, 9295, 686, 34, 1;
%e 12248365, 8203057, 1773156, 214478, 17248, 986, 40, 1;
%e 273211787, 191405232, 43039532, 5442349, 463267, 28747, 1340, 46, 1; ...
%e where column 0 of V^2 = column 1 of P^2 = triangle A136225.
%o (PARI) {T(n,k)=local(P=Mat(1),U=Mat(1),V=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;V=P^2*PShR; 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])))); V=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,V[r,c], if(c==1,(P^3)[ #P,1],(P^(3*c-2))[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]))))); (V^2)[n+1,k+1]}
%Y Cf. A136227 (column 0); related tables: A136220 (P), A136228 (U), A136230 (V), A136231 (W=P^3), A136237 (V^3).
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Feb 07 2008