A095844 Numerator of the integral of the n-th power of the Cantor function.
1, 1, 3, 1, 33, 5, 75, 611, 97653, 83057, 22018179, 9625216, 20894487717, 93120706729, 411117020063871, 297434062421057, 6650181371241300777, 6082551300359191981, 2198073713661546055399083, 53388901948383223161199, 31122898898234908646386438959
Offset: 0
Examples
1, 1/2, 3/10, 1/5, 33/230, 5/46, 75/874, 611/8740, 97653/1673710, ...
Links
- Amiram Eldar, Table of n, a(n) for n = 0..117
- Harold G. Diamond and Bruce Reznick, Problem 10621, Problems and Solutions, The American Mathematical Monthly, Vol. 104, No. 9 (1997), p. 870; Cantor's Singular Moments, Solutions to Problem 10621 by Kenneth F. Andersen and Omran Kouba, ibid., Vol. 106, No. 2 (1999), pp. 175-176.
- E. A. Gorin and B. N. Kukushkin, Integrals related to the Cantor function, St. Petersburg Math. J., 15, 449-468, 2004.
- Eric Weisstein's World of Mathematics, Cantor Function.
Crossrefs
Cf. A095845 (denominators).
Programs
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Maple
seq(numer(1/(n+1)-sum(binomial(n,2*k)*(2^(2*k-1)-1)*bernoulli(2*k)/(3*2^(2*k-1)-1)/(n-2*k+1),k = 1 .. floor(1/2*n))),n=1..18); # Emeric Deutsch, Feb 22 2005
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Mathematica
a[n_] := Numerator[ 1/(n+1) - Sum[Binomial[n, 2 k]*Floor[2^(2k - 1) - 1]*BernoulliB[2k]/Floor[(3*2^(2k - 1) - 1)*(n - 2k + 1)], {k, 1, Floor[n/2]}]]; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 23 2012, after Emeric Deutsch *) f[0] = 1; f[1] = 1/2; f[n_] := f[n] = (1/(3*2^n - 2))*(2 + Sum[Binomial[n, k]*f[k], {k, 1, n - 1}]); Numerator[Array[f, 20, 0]] (* Amiram Eldar, Jan 26 2024 *)
Formula
The integral, a rational number, is given by J(n) = 1/(n+1) - Sum_{k = 1..floor(n/2)} binomial(n,2*k)*(2^(2*k-1)-1)*Bernoulli(2*k)/((3*2^(2*k-1)-1)*(n-2*k+1)). - Emeric Deutsch, Feb 22 2005
Note that the Cantor function C(x) satisfies C(x) = C(3*x)/2 for x in [0,1/3], 1/2 for x in [1/3,2/3] and (1+C(3*x-2))/2 for x in [2/3,1]. Integrating both sides yields J(n) = (1 + Sum_{k=0..n-1} binomial(n,k)*J(k))/(3*2^n - 2) with J(0) = 1, where J(n) := Integral_{x=0..1} (C(x))^n dx. - Jianing Song, Nov 19 2023
J(n) = (2 + Sum_{k=1..n-1} binomial(n,k) * J(k))/(3*2^n-2) (Diamond and Reznick, 1997). - Amiram Eldar, Jan 26 2024