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%I #17 Oct 05 2013 04:45:48
%S 0,0,0,1,2,2,1,1,4,4,-13,-13,142,142,-1931,-1931,36296,36296,-893273,
%T -893273,27927346,27927346,-1081725559,-1081725559,50861556172,
%U 50861556172,-2854289486309,-2854289486309
%N Autosequence preceding -A226158(n).
%C Extension of the difference table of Genocchi numbers A226158(n). The signs are changed.
%C Consider the difference table of -A226158:
%C 0, 1, 1, 0 -1, 0,
%C 1, 0, -1, -1, 1, 3,
%C -1, -1, 0, 2, 2, -6,
%C 0, 1, 2, 0, -8, -8,
%C 1, 1, -2, -8, 0, 56,
%C 0, -3, -6, 8, 56, 0, etc.
%C Upon the table, we prolonged the main diagonal by 0 followed by 0 on the same row. Hence
%C 0, 0, 0, 1, 2, 2, 1, 1,
%C 0, 0, 1, 1, 0, -1, 0, 3, = 0 followed by -A226158.
%C 0, 1, 0, -1, -1, 1, 3, -3,
%C 1, -1, -1, 0, 2, 2, -6, -14,
%C -2, 0, 1, 2, 0, -8, -8, 48,
%C 2, 1, 1, -2, -8, 0, 56, 56,
%C -1, 0, -3, -6, 8, 56, 0, -608,
%C 1, -3, -3, 14, 48, -56, -608, 0, etc.
%C The first row, a(n), is equal to its inverse binomial transform signed, the main diagonal of the difference table is composed of 0's, so it is an autosequence of the first kind.
%F a(n) = 0, 0, 0 followed by (-1)^n * A133135(n).
%t max = 24; p[0, _] = 1; p[n_, x_] := (1+x)*((1+x)^(n-1) + x^(n-1))/2; t = Table[Coefficient[p[n, x], x, k], {n, 0, max+2}, {k, 0, max+2}]; a[n_] := Inverse[t][[All, 3]][[n+3]]; A133135 = Table[a[n], {n, 0, max}]; Join[{0, 0, 0}, Table[(-1)^n*A133135[[n+1]], {n, 0, max}]]
%t (* or *)
%t g[n_ /; n < 3] = 0; g[3] = -1; g[n_] := (n-2)*EulerE[n-3, 0]; Table[-g[n], {n, 0, 27}] // Accumulate (* _Jean-François Alcover_, Sep 12 2013 *)
%K sign
%O 0,5
%A _Paul Curtz_, Sep 12 2013
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