OFFSET
1,4
COMMENTS
Cloutier et al. showed that the number of terms of A328014 below x is D0 * x^(3/4) + O(x^(2/3)*log(x)), where D0 is this constant.
LINKS
Maurice-Étienne Cloutier, Jean-Marie De Koninck, and Nicolas Doyon, On the powerful and squarefree parts of an integer, Journal of Integer Sequences, Vol. 17 (2014), Article 14.6.6.
FORMULA
Equals (4/3)*(zeta(3/2)/zeta(3)) * Product_{p prime} (1 - 1/p)*(1 + (1-1/p)/(p*(1 + 1/p^(3/2)))).
EXAMPLE
1.115436631111013698931934302941096327033286649113053...
MATHEMATICA
$MaxExtraPrecision = 500; m = 500; f[x_] := (1 - x)*(1 + (1 - x)*x/(1 + x^(3/2))); c = LinearRecurrence[{2, -3, 2, 1, -3, 3, -1}, {0, 0, 0, -8, -5, 6, 14}, m]; RealDigits[(4/3)*(Zeta[3/2]/Zeta[3])*f[1/2]*f[1/3]*Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/2] - 1/2^(n/2) - 1/3^(n/2))/n, {n, 3, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]]
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Amiram Eldar, Oct 01 2019
STATUS
approved