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

1,2

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.

Links

Table of n, a(n) for n=1..55.

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

Example

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

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 *)

Crossrefs

Cf A120057, A000110, A056863.

Cf. A008275, A120095.

Cf. A134315.

Sequence in context: A209151 A125100 A128544 * A208532 A245334 A102756

Adjacent sequences: A120055 A120056 A120057 * A120059 A120060 A120061

Keyword

sign,tabl

Author

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

Status

approved