A328015 Decimal expansion of the growth constant for the number of terms of A328014 (numbers whose powerful part is larger than their powerfree part).

1, 1, 1, 5, 4, 3, 6, 6, 3, 1, 1, 1, 1, 0, 1, 3, 6, 9, 8, 9, 3, 1, 9, 3, 4, 3, 0, 2, 9, 4, 1, 0, 9, 6, 3, 2, 7, 0, 3, 3, 2, 8, 6, 6, 4, 9, 1, 1, 3, 0, 5, 3, 1, 6, 1, 6, 7, 1, 1, 4, 7, 6, 3, 9, 5, 7, 6, 8, 0, 3, 0, 7, 0, 3, 2, 1, 1, 7, 2, 4, 6, 8, 3, 7, 7, 2, 3

Offset: 1

Author

Amiram Eldar, Oct 01 2019

Keywords

• nonn
• cons

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.

Examples

			1.115436631111013698931934302941096327033286649113053...
		

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.

Crossrefs

Cf. A057521, A055231, A328014.

Programs

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]]

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)))).