%I #13 Apr 08 2016 08:44:20
%S 1,2,-1,3,-4,2,4,-9,10,-4,5,-16,28,-24,8,6,-25,60,-80,56,-16,7,-36,
%T 110,-200,216,-128,32,8,-49,182,-420,616,-560,288,-64,9,-64,280,-784,
%U 1456,-1792,1408,-640,128,10,-81,408,-1344,3024,-4704,4992,-3456,1408,-256
%N Coefficients for obtaining A120057 from Bell numbers.
%C 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.
%C Alternating column sums appear to be 3^n.
%F A120057(n,k) = sum_{i=1,k} T(n,i)*B(n-i+1).
%F T(n,k) = Sum_j A120095(n,j) * S1(j,n-k+1), where S1 is the Stirling numbers of the first kind (A008275).
%F Unsigned version, as an infinite lower triangular matrix, equals A007318 * A134315. - _Gary W. Adamson_, Oct 19 2007
%F 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
%e Table starts:
%e 1
%e 2,-1
%e 3,-4,2
%e 4,-9,10,-4
%e 5,-16,28,-24,8
%e 6,-25,60,-80,56,-16
%t 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_ *)
%Y Cf A120057, A000110, A056863.
%Y Cf. A008275, A120095.
%Y Cf. A134315.
%K sign,tabl
%O 1,2
%A _Franklin T. Adams-Watters_, Jun 06 2006, Jun 07 2006