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%I #17 Jun 26 2017 01:15:02
%S 2,18,6,1350,270,23814,17010,65063250,7229250,9762090030,4437313650,
%T 8267713725521250,635977978886250,81188783595533250,
%U 297692206516955250,22510683177794610356250,1564913803803903393750,40011216302189267004656036250,10529267447944543948593693750
%N Denominator of half moments of Rvachëv function.
%C a(n) is equal to the denominator of the integral over (0,1) of n*t^(n-1)*up(t).
%C These numbers are the half moments of the Rvachëv function. The Rvachëv function is related to the Fabius function, up(x)=F(x+1) for |x|<1 and up(x)=0 for |x|>=1.
%C The sequence of numerators is not in the OEIS because it appears t coincide with A272755: Numerators of Fabius function F(1/2^n). In fact d(n) = n! 2^binomial(n,2)F(1/2^n). The coincidence depends on the fact that n! 2^binomial(n,2) divides the denominator of F(1/2^n). It is true that 2^binomial(n,2) divides this denominator, but I do not see any reason for n! to divide this denominator.
%H J. Arias de Reyna, <a href="https://arxiv.org/abs/1702.05442">An infinitely differentiable function with compact support: Definition and properties</a>, arXiv:1702.05442 [math.CA], 2017.
%H J. Arias de Reyna, <a href="https://arxiv.org/abs/1702.06487">Arithmetic of the Fabius function</a>, arXiv:1702.06487 [math.NT], 2017.
%F Recurrence d(0)=1; d(n)=Sum_{k=0..n-1}(binomial(n+1,k)d(k))/((n+1)*(2^n-1)) with a(n) are the denominators of d(n).
%F It may also be defined to be the only sequence d(n) with d(0)=1 and such that the function f(x)=Sum_{n>=0} d(n) x^n/n! satisfies x*f(2x)=(e^x-1)*f(x).
%e The rationals d(n) are 1/2, 5/18, 1/6, 143/1350, 19/270, ...
%t d[0] = 1;
%t d[n_] := d[n] =
%t Sum[Binomial[n + 1, k] d[k], {k, 0, n - 1}]/((n + 1)*(2^n - 1));
%t Table[Denominator[d[n]], {n, 1, 20}]
%Y Cf. A287936, A287937, A287938.
%K nonn,frac
%O 1,1
%A _Juan Arias-de-Reyna_, Jun 06 2017
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