0,3

Sum_{k=0,2n} (-1)^k*C(2n,k)*T(n,k) = (-6)^n.

Triangle begins:

1;

1, 4, 1;

6, 24, 36, 24, 6;

96, 384, 636, 744, 636, 384, 96;

2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976;

151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416; ...

If we write the triangle like this:

.......................... ....1;

................... ....1, ....4, ....1;

............ ....6, ...24, ...36, ...24, ....6;

..... ...96, ..384, ..636, ..744, ..636, ..384, ...96;

.2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, .2976;

then the first term in each row is the sum of the previous row:

2976 = 96 + 384 + 636 + 744 + 636 + 384 + 96

the next term is 4 times the first:

11904 = 4*2976,

and the remaining terms in each row are obtained by the rule

illustrated by:

20256 = 2*11904 - 2976 - 6*96;

26304 = 2*20256 - 11904 - 6*384;

28536 = 2*26304 - 20256 - 6*636;

26304 = 2*28536 - 26304 - 6*744;

20256 = 2*26304 - 28536 - 6*636;

11904 = 2*20256 - 26304 - 6*384;

2976 = 2*11904 - 20256 - 6*96.

An alternate recurrence is illustrated by:

11904 = 2976 + 3*(96 + 384 + 636 + 744 + 636 + 384 + 96);

20256 = 11904 + 3*(384 + 636 + 744 + 636 + 384);

26304 = 20256 + 3*(636 + 744 + 636);

28536 = 26304 + 3*(744);

and then for k>n, T(n,k) = T(n,2n-k).

(PARI) {T(n, k)=local(p=3); if(2*n<k|k<0, 0, if(n==0&k==0, 1, if(k==0, sum(j=0, 2*n-2, T(n-1, j)), if(k==1, (p+1)*T(n, 0), if(k<=n, 2*T(n, k-1)-T(n, k-2)-2*p*T(n-1, k-2), T(n, 2*n-k))))))} (PARI) /* Alternate Recurrence: */ {T(n, k)=local(p=3); if(2*n<k|k<0, 0, if(n==0&k==0, 1, if(k==0, sum(j=0, 2*n-2, T(n-1, j)), if(k<=n, T(n, k-1)+p*sum(j=k-1, 2*n-1-k, T(n-1, j)), T(n, 2*n-k)))))}

Cf. A126151 (column 0); diagonals: A126152, A126153; A126154; variants: A008301, A125053, A126155.

Sequence in context: A094264 A056140 A140895 this_sequence A096966 A140703 A157259

Adjacent sequences: A126147 A126148 A126149 this_sequence A126151 A126152 A126153

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Dec 19 2006