A272755 Numerators of the Fabius function F(1/2^n).

1, 1, 5, 1, 143, 19, 1153, 583, 1616353, 132809, 134926369, 46840699, 67545496213157, 4068990560161, 411124285571171, 1204567303451311, 73419800947733963069, 4146897304424408411, 86773346866163284480799923, 18814360006695807527868793, 539741515875650532056045666422369

Offset: 0

Author

Vladimir Reshetnikov, May 05 2016

Keywords

• nonn
• frac

Comments

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.

Examples

			A272755/A272757 = 1/1, 1/2, 5/72, 1/288, 143/2073600, 19/33177600, 1153/561842749440, 583/179789679820800, ...
		

References

• 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.

Links

• Juan Arias de Reyna, An infinitely differentiable function with compact support: Definition and properties, arXiv:1702.05442 [math.CA], 2017.
• Juan Arias de Reyna, On the arithmetic of Fabius function, arXiv:1702.06487 [math.NT], 2017.
• Yuri Dimitrov, G. A. Edgar, Solutions of Self-differential Functional Equations
• G. A. Edgar, Examples of self differential functions
• J. Fabius, A probabilistic example of a nowhere analytic C^infty-function, Probability Theory and Related Fields, June 1966, Volume 5, Issue 2, pp 173-174.
• Jan Kristian Haugland, Evaluating the Fabius function, arXiv:1609.07999 [math.GM], 23 Sep 2016.
• Wikipedia, Fabius function

Crossrefs

Cf. A272757 (denominators), A272343.

Programs

Mathematica

c[0] = 1; c[n_] := c[n] = Sum[(-1)^k c[n - k]/(2 k + 1)!, {k, 1, n}] / (4^n - 1); Numerator@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 *)

Formula

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