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%I #16 Feb 06 2023 14:22:26
%S 0,0,1,3,3,5,5,7,7,-3,-3,121,121,-1261,-1261,20583,20583,-888403,
%T -888403,24729925,24729925,-862992399,-862992399,36913939769,
%U 36913939769,-1899853421885,-1899853421885
%N Autosequence preceding A198631(n)/A006519(n+1). Numerators.
%C 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).
%C g(n+2) - g(n+1) = A198631(n)/A006519(n+1).
%C The corresponding fractions to g(n) are f(n) in A165142(n).
%C g(n) differences table:
%C 0, 0, 1, 3/2, 3/2, 5/4,
%C 0, 1, 1/2, 0, -1/4, 0,
%C 1, -1/2, -1/2, -1/4, 1/4, 1/2, Euler twin numbers (new),
%C -3/2, 0, 1/4, 1/2, 1/4, -1,
%C 3/2, 1/4, 1/4, -1/4, -5/4, -5/8,
%C -5/4, 0, -1/2, -1, 5/8, 13/2, etc.
%C Like A198631(n)/A006519(n+1),g(n) is an autosequence of the second kind.
%C If we proceed, here for Euler polynomials, like in A233565 for Bernoulli polynomials, we obtain
%C 1) A133138(n)/A007395(n) (unreduced form) or
%C 2) A233508(n)/A232628(n) (reduced form),the first array in A133135.
%C The Bernoulli's corresponding fractions to 1) are A193815(n)/(A003056(n) with 1 instead of 0).
%F a(n) = 0, 0, followed by (-1)^n *A141424(n).
%t 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 *)
%Y Cf. A051716/A051717, Bernoulli twin numbers.
%K sign,frac
%O 0,4
%A _Paul Curtz_, Dec 16 2013