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%I #2 Mar 30 2012 18:37:07
%S 1,3,1,12,6,1,63,39,9,1,421,300,81,12,1,3472,2741,816,138,15,1,34380,
%T 29380,9366,1716,210,18,1,399463,363922,122148,23647,3105,297,21,1,
%U 5344770,5135894,1795481,362116,49880,5088,399,24,1,81097517,81557270
%N Triangle, read by rows, equal to the matrix cube of triangle P = A135880.
%C Matrix square equals triangle A135893.
%e Triangle P^3 begins:
%e 1;
%e 3, 1;
%e 12, 6, 1;
%e 63, 39, 9, 1;
%e 421, 300, 81, 12, 1;
%e 3472, 2741, 816, 138, 15, 1;
%e 34380, 29380, 9366, 1716, 210, 18, 1;
%e 399463, 363922, 122148, 23647, 3105, 297, 21, 1;
%e 5344770, 5135894, 1795481, 362116, 49880, 5088, 399, 24, 1;
%e 81097517, 81557270, 29478724, 6138746, 875935, 93306, 7770, 516, 27, 1;
%e where P = A135880 begins:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 6, 7, 3, 1;
%e 25, 34, 15, 4, 1;
%e 138, 215, 99, 26, 5, 1;
%e 970, 1698, 814, 216, 40, 6, 1; ...
%e where column k of P^2 equals column 0 of P^(2k+2)
%e such that column 0 of P^2 equals column 0 of P shift left.
%o (PARI) {T(n,k)=local(P=Mat(1),R,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]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1])))));(P^3)[n+1,k+1]}
%Y Cf. columns: A135889, A135890; related tables: A135880 (P), A135894 (R), A135893 (P^6).
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Dec 15 2007
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