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A272757
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Denominators of the Fabius function F(1/2^n).
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6
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1, 2, 72, 288, 2073600, 33177600, 561842749440, 179789679820800, 704200217922109440000, 180275255788060016640000, 1246394851358539387238350848000, 6381541638955721662660356341760000, 292214732887898713986916575925267070976000000
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OFFSET
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0,2
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COMMENTS
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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.
It is true that n! divides a(n) for all n? This is true for the first 200 terms.
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)
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REFERENCES
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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A272755/A272757 = 1/1, 1/2, 5/72, 1/288, 143/2073600, 19/33177600, 1153/561842749440, 583/179789679820800, ...
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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