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A229054
Autosequence preceding -A226158(n).
1
0, 0, 0, 1, 2, 2, 1, 1, 4, 4, -13, -13, 142, 142, -1931, -1931, 36296, 36296, -893273, -893273, 27927346, 27927346, -1081725559, -1081725559, 50861556172, 50861556172, -2854289486309, -2854289486309
OFFSET
0,5
COMMENTS
Extension of the difference table of Genocchi numbers A226158(n). The signs are changed.
Consider the difference table of -A226158:
0, 1, 1, 0 -1, 0,
1, 0, -1, -1, 1, 3,
-1, -1, 0, 2, 2, -6,
0, 1, 2, 0, -8, -8,
1, 1, -2, -8, 0, 56,
0, -3, -6, 8, 56, 0, etc.
Upon the table, we prolonged the main diagonal by 0 followed by 0 on the same row. Hence
0, 0, 0, 1, 2, 2, 1, 1,
0, 0, 1, 1, 0, -1, 0, 3, = 0 followed by -A226158.
0, 1, 0, -1, -1, 1, 3, -3,
1, -1, -1, 0, 2, 2, -6, -14,
-2, 0, 1, 2, 0, -8, -8, 48,
2, 1, 1, -2, -8, 0, 56, 56,
-1, 0, -3, -6, 8, 56, 0, -608,
1, -3, -3, 14, 48, -56, -608, 0, etc.
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.
FORMULA
a(n) = 0, 0, 0 followed by (-1)^n * A133135(n).
MATHEMATICA
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}]]
(* or *)
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 *)
CROSSREFS
Sequence in context: A293208 A247364 A301895 * A133135 A292189 A284992
KEYWORD
sign
AUTHOR
Paul Curtz, Sep 12 2013
STATUS
approved