A134530 Matrix log of triangle A111636, where A111636(n,k) = (2^k)^(n-k)*C(n,k) for n>=k>=0.

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, -56272471039, 95716705280, -41566486528, 6196822016, -362414080, 9175040, -114688, 1024, 0

Offset: 0

Author

Paul D. Hanna, Oct 30 2007

Keywords

• sign
• tabl

Examples

			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; ...
		

Crossrefs

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

Programs

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]}

Formula

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)]).