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%I #42 Jun 11 2017 05:22:14
%S 1,2,72,288,2073600,33177600,561842749440,179789679820800,
%T 704200217922109440000,180275255788060016640000,
%U 1246394851358539387238350848000,6381541638955721662660356341760000,292214732887898713986916575925267070976000000
%N Denominators of the Fabius function F(1/2^n).
%C The Fabius function F(x) is the smooth monotone increasing function on [0, 1] satisfying F(0) = 0, F(1) = 1, F'(x) = 2*F(2*x) for 0 < x < 1/2, F'(x) = 2*F(2*(1-x)) for 1/2 < x < 1. It is infinitely differentiable at every point in the interval, but is nowhere analytic. It assumes rational values at dyadic rationals.
%C From _Juan Arias-de-Reyna_, Jun 08 2017: (Start)
%C It is true that n! divides a(n) for all n? This is true for the first 200 terms.
%C If this is true A272755, the sequence of numerators of F(2^(-n)) is also the sequence of numerators of the half moments of Rvachëv function. (Cf. A288161). (End)
%D Rvachev V. L., Rvachev V. A., Non-classical methods of the approximation theory in boundary value problems, Naukova Dumka, Kiev (1979) (in Russian), pages 117-125.
%H Vincenzo Librandi, <a href="/A272757/b272757.txt">Table of n, a(n) for n = 0..64</a>
%H Juan 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 Juan Arias de Reyna, <a href="https://arxiv.org/abs/1702.06487">On the arithmetic of Fabius function</a>, arXiv:1702.06487 [math.NT], 2017.
%H Yuri Dimitrov, G. A. Edgar, <a href="http://people.math.osu.edu/edgar.2/preprints/dimitrov/paper.pdf">Solutions of Self-differential Functional Equations</a>
%H G. A. Edgar, <a href="http://people.math.osu.edu/edgar.2/selfdiff/">Examples of self differential functions</a>
%H J. Fabius, <a href="http://dx.doi.org/10.1007/BF00536652">A probabilistic example of a nowhere analytic C^infty-function</a>, Z. Wahrscheinlichkeitstheorie verw. Gebiete (1966) 5: 173.
%H Jan Kristian Haugland, <a href="http://arxiv.org/abs/1609.07999">Evaluating the Fabius function</a>, arXiv:1609.07999 [math.GM], 23 Sep 2016.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Fabius_function">Fabius function</a>
%F Recurrence: d(0) = 1, d(n) = (1/(n+1)! + Sum_{k=1..n-1} (2^(k*(k-1)/2)/(n-k+1)!)*d(k))/((2^n-1)*2^(n*(n-1)/2)), where d(n) = A272755(n)/A272757(n). - _Vladimir Reshetnikov_, Feb 27 2017
%e A272755/A272757 = 1/1, 1/2, 5/72, 1/288, 143/2073600, 19/33177600, 1153/561842749440, 583/179789679820800, ...
%t c[0] = 1; c[n_] := c[n] = Sum[(-1)^k c[n - k]/(2 k + 1)!, {k, 1, n}] / (4^n - 1); Denominator@Table[Sum[c[k] (-1)^k / (n - 2 k)!, {k, 0, n/2}] / 2^((n + 1) n/2), {n, 0, 15}] (* _Vladimir Reshetnikov_, Oct 16 2016 *)
%Y Cf. A272755 (numerators), A272343.
%K nonn,frac
%O 0,2
%A _Vladimir Reshetnikov_, May 05 2016
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