A277471 Normalized values of the Fabius function: 2^binomial(n-1, 2) * (2*n)! * A005329(n) * F(1/2^n).

2, 1, 5, 105, 7007, 1298745, 619247475, 723733375365, 2006532782969715, 12889909959143502285, 188494585656727188486375, 6188497678605937441851529425, 451101946262511157576785806552415, 72341127537387548941434093006996374625, 25326487488712595887856341442148826764706875

Offset: 0

Author

Vladimir Reshetnikov, Oct 16 2016

Keywords

• nonn

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.

Comment from Vladimir Reshetnikov, Jan 25 2017: I just realized that I do not have a rigorous proof that all terms are integers. Could somebody suggest a proof? I would also be very interested to learn the asymptotics of this sequence.

Juan Arias de Reyna proved that all terms are indeed integers. - Vladimir Reshetnikov, Feb 28 2017

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. A005329, A272755, A272757.

Programs

Mathematica

c[0] = 1; c[n_] := c[n] = Sum[(-1)^k c[n - k]/(2 k + 1)!, {k, 1, n}] / (4^n - 1); Table[2^(1 - 2 n) (2 n)! QFactorial[n, 2] Sum[c[k] (-1)^k/(n - 2 k)!, {k, 0, n/2}], {n, 0, 15}]

Formula

a(n) = 2^binomial(n-1, 2) * (2*n)! * A005329(n) * A272755(n) / A272757(n).