A120058 Coefficients for obtaining A120057 from Bell numbers.

1, 2, -1, 3, -4, 2, 4, -9, 10, -4, 5, -16, 28, -24, 8, 6, -25, 60, -80, 56, -16, 7, -36, 110, -200, 216, -128, 32, 8, -49, 182, -420, 616, -560, 288, -64, 9, -64, 280, -784, 1456, -1792, 1408, -640, 128, 10, -81, 408, -1344, 3024, -4704, 4992, -3456, 1408, -256

Offset: 1

Author

Franklin T. Adams-Watters, Jun 06 2006, Jun 07 2006

Keywords

• sign
• tabl

Comments

Appears to be essentially the same as A056863, but (as of Jun 06 2006) that sequence definition is unclear and there are discrepencies in the signs.

Alternating column sums appear to be 3^n.

Examples

			Table starts:
1
2,-1
3,-4,2
4,-9,10,-4
5,-16,28,-24,8
6,-25,60,-80,56,-16
		

Crossrefs

Cf A120057, A000110, A056863.

Cf. A008275, A120095.

Cf. A134315.

Programs

Mathematica

T[n_, 1] := n; T[n_, n_] := (-1)^(n+1)*2^(n-2); T[n_, k_] /; 2 <= k <= n-1 := T[n, k] = 2*T[n-1, k] - 2*T[n-1, k-1] + 2*T[n-2, k-1] - T[n-2, k]; T[, ] = 0; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 08 2016, after Philippe Deléham *)

Formula

A120057(n,k) = sum_{i=1,k} T(n,i)*B(n-i+1).

T(n,k) = Sum_j A120095(n,j) * S1(j,n-k+1), where S1 is the Stirling numbers of the first kind (A008275).

Unsigned version, as an infinite lower triangular matrix, equals A007318 * A134315. - Gary W. Adamson, Oct 19 2007

T(n,k) = 2*T(n-1,k) - 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k). - Philippe Deléham, Feb 27 2012