0,5

T(n,k) = A134531(n-k)*(2^k)^(n-k)*C(n,k), where A134531 is column 0 and satisfies: G.f.: Sum_{n>=0} A134531(n)*x^n/[n!*2^(n*(n-1)/2)] = LOG(Sum_{n>=0}x^n/[n!*2^(n*(n-1)/2)]).

Triangle begins:

0,

1, 0;

-1, 4, 0;

5, -12, 12, 0;

-79, 160, -96, 32, 0;

3377, -6320, 3200, -640, 80, 0;

-362431, 648384, -303360, 51200, -3840, 192, 0;

93473345, -162369088, 72619008, -11325440, 716800, -21504, 448, 0; ...

Matrix exponentiation yields triangle A111636, which begins:

1;

1, 1;

1, 4, 1;

1, 12, 12, 1;

1, 32, 96, 32, 1;

1, 80, 640, 640, 80, 1; ...

(PARI) {T(n, k)=local(M=matrix(n+1, n+1, r, c, if(r>=c, 2^((c-1)*(r-c))*binomial(r-1, c-1))), L); L=sum(i=1, #M, -(M^0-M)^i/i); L[n+1, k+1]}

Cf. A134531 (column 0); related triangles: A111636, A117401; A011266.

Adjacent sequences: A134527 A134528 A134529 this_sequence A134531 A134532 A134533

Sequence in context: A021965 A064520 A108174 this_sequence A126813 A056141 A125961

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Oct 30 2007