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Numerator of the integral of the n-th power of the Cantor function.
7

%I #38 Jan 14 2025 06:02:09

%S 1,1,3,1,33,5,75,611,97653,83057,22018179,9625216,20894487717,

%T 93120706729,411117020063871,297434062421057,6650181371241300777,

%U 6082551300359191981,2198073713661546055399083,53388901948383223161199,31122898898234908646386438959

%N Numerator of the integral of the n-th power of the Cantor function.

%H Amiram Eldar, <a href="/A095844/b095844.txt">Table of n, a(n) for n = 0..117</a>

%H Harold G. Diamond and Bruce Reznick, <a href="http://www.jstor.org/stable/2975295">Problem 10621</a>, Problems and Solutions, The American Mathematical Monthly, Vol. 104, No. 9 (1997), p. 870; <a href="https://www.jstor.org/stable/2589071">Cantor's Singular Moments</a>, Solutions to Problem 10621 by Kenneth F. Andersen and Omran Kouba, ibid., Vol. 106, No. 2 (1999), pp. 175-176.

%H E. A. Gorin and B. N. Kukushkin, <a href="https://doi.org/10.1090/S1061-0022-04-00817-9">Integrals related to the Cantor function</a>, St. Petersburg Math. J., 15, 449-468, 2004.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CantorFunction.html">Cantor Function</a>.

%F 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

%F 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

%F 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

%e 1, 1/2, 3/10, 1/5, 33/230, 5/46, 75/874, 611/8740, 97653/1673710, ...

%p 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

%t 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_ *)

%t 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 *)

%Y Cf. A095845 (denominators).

%K nonn,frac

%O 0,3

%A _Eric W. Weisstein_, Jun 08 2004