0,8

Triangle begins:

1;

0, 1;

0, 1, 1;

0, 2, 2, 1;

0, 6, 7, 3, 1;

0, 25, 34, 15, 4, 1;

0, 138, 215, 99, 26, 5, 1;

0, 970, 1698, 814, 216, 40, 6, 1;

0, 8390, 16220, 8057, 2171, 400, 57, 7, 1;

0, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1; ...

This triangle equals matrix product P^-1*R,

which equals triangle P shifted right one column,

where P = A135880 begins:

1;

1, 1;

2, 2, 1;

6, 7, 3, 1;

25, 34, 15, 4, 1;

138, 215, 99, 26, 5, 1;

970, 1698, 814, 216, 40, 6, 1; ...

and Q = P^2 = A135885 begins:

1;

2, 1;

6, 4, 1;

25, 20, 6, 1;

138, 126, 42, 8, 1;

970, 980, 351, 72, 10, 1;

8390, 9186, 3470, 748, 110, 12, 1; ...

and R = A135894 begins:

1;

1, 1;

2, 3, 1;

6, 12, 5, 1;

25, 63, 30, 7, 1;

138, 421, 220, 56, 9, 1;

970, 3472, 1945, 525, 90, 11, 1; ...

where column k of R equals column 0 of P^(2k+1),

and column k of Q=P^2 equals column 0 of P^(2k+2), for k>=0.

(PARI) {T(n, k)=local(P=Mat(1), R=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])))); 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^-1*R)[n+1, k+1]}

Cf. A135880 (P), A135885 (Q=P^2), A135894 (R); A135899 (P*R^-1*P), A135900 (R^-1*Q).

Sequence in context: A152882 A130167 A084938 this_sequence A131182 A093729 A113080

Adjacent sequences: A135895 A135896 A135897 this_sequence A135899 A135900 A135901

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Dec 15 2007