A233808 Autosequence preceding A198631(n)/A006519(n+1). Numerators.

0, 0, 1, 3, 3, 5, 5, 7, 7, -3, -3, 121, 121, -1261, -1261, 20583, 20583, -888403, -888403, 24729925, 24729925, -862992399, -862992399, 36913939769, 36913939769, -1899853421885, -1899853421885

Offset: 0

Author

Paul Curtz, Dec 16 2013

Keywords

• sign
• frac

Comments

The fractions are g(n)=0, 0, 1, 3/2, 3/2, 5/4, 5/4, 7/4, 7/4, -3/8, -3/8, 121/8, 121/8, -1261/8, -1261/8, 20583/8, 20583/8, -888403/16, -888403/16,... . The denominators are 1, 1, followed by A053644(n+1).

g(n+2) - g(n+1) = A198631(n)/A006519(n+1).

The corresponding fractions to g(n) are f(n) in A165142(n).

g(n) differences table:

0, 0, 1, 3/2, 3/2, 5/4,

0, 1, 1/2, 0, -1/4, 0,

1, -1/2, -1/2, -1/4, 1/4, 1/2, Euler twin numbers (new),

-3/2, 0, 1/4, 1/2, 1/4, -1,

3/2, 1/4, 1/4, -1/4, -5/4, -5/8,

-5/4, 0, -1/2, -1, 5/8, 13/2, etc.

Like A198631(n)/A006519(n+1),g(n) is an autosequence of the second kind.

If we proceed, here for Euler polynomials, like in A233565 for Bernoulli polynomials, we obtain

1) A133138(n)/A007395(n) (unreduced form) or

2) A233508(n)/A232628(n) (reduced form),the first array in A133135.

The Bernoulli's corresponding fractions to 1) are A193815(n)/(A003056(n) with 1 instead of 0).

Crossrefs

Cf. A051716/A051717, Bernoulli twin numbers.

Programs

Mathematica

max = 27; p[0] = 1; p[n_] := (1 + x)*((1 + x)^(n - 1) + x^(n - 1))/2; t = Table[Coefficient[p[n], x, k], {n, 0, max + 2}, {k, 0, max + 2}]; a[n_] := (-1)^n*Inverse[t][[n, 2]] // Numerator; a[0] = 0; Table[a[n], {n, 0, max}] (* Jean-François Alcover, Jan 11 2016 *)

Formula

a(n) = 0, 0, followed by (-1)^n *A141424(n).