login
Matrix square of triangle A136220, read by rows.
8

%I #2 Mar 30 2012 18:37:08

%S 1,2,1,8,4,1,49,26,6,1,414,232,54,8,1,4529,2657,629,92,10,1,61369,

%T 37405,9003,1320,140,12,1,996815,627435,153276,22606,2385,198,14,1,

%U 18931547,12248365,3031553,450066,47500,3904,266,16,1,412345688,273211787

%N Matrix square of triangle A136220, read by rows.

%C Column 0 of this triangle = column 1 of square array A136217.

%F Let P=A136220, V=A136230, then column k of P^2 (this triangle) = column 0 of V^(k+1) while column j of V = column 0 of P^(3j+2).

%e Let P = A136220, then this triangle is P^2 and begins:

%e 1;

%e 2, 1;

%e 8, 4, 1;

%e 49, 26, 6, 1;

%e 414, 232, 54, 8, 1;

%e 4529, 2657, 629, 92, 10, 1;

%e 61369, 37405, 9003, 1320, 140, 12, 1;

%e 996815, 627435, 153276, 22606, 2385, 198, 14, 1;

%e 18931547, 12248365, 3031553, 450066, 47500, 3904, 266, 16, 1; ...

%e where column k of P^2 = column 0 of V^(k+1) and

%e triangle V = A136230 begins:

%e 1;

%e 2, 1;

%e 8, 5, 1;

%e 49, 35, 8, 1;

%e 414, 325, 80, 11, 1;

%e 4529, 3820, 988, 143, 14, 1;

%e 61369, 54800, 14696, 2200, 224, 17, 1; ...

%e where column k of V = column 0 of P^(3k+2).

%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) and U = A136228.

%o (PARI) {T(n,k)=local(P=Mat(1),U,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])))));(P^2)[n+1,k+1]}

%Y Cf. columns: A136226, A136227; related tables: A136228 (U), A136230 (V), A136231 (W=P^3), A136217, A136218.

%K nonn,tabl

%O 0,2

%A _Paul D. Hanna_, Jan 28 2008