A362984 Decimal expansion of the asymptotic mean of the abundancy index of the powerful numbers (A001694).

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

Offset: 1

Author

Amiram Eldar, May 12 2023

Keywords

• nonn
• cons

Comments

The abundancy index of a positive integer k is A000203(k)/k = A017665(k)/A017666(k).

The asymptotic mean of the abundancy index over all the positive integers is lim_{m->oo} (1/m) * Sum_{k=1..m} A000203(k)/k = Pi^2/6 = zeta(2) = 1.644934... (A013661).

Examples

			2.14968690301526765128219042105109416145987653275100999873...
		

Links

• Rafael Jakimczuk and Matilde Lalín, Asymptotics of sums of divisor functions over sequences with restricted factorization structure, Notes on Number Theory and Discrete Mathematics, Vol. 28, No. 4 (2022), pp. 617-634, eq. (6).

Crossrefs

Cf. A000203, A001694, A017665, A017666, A180114.

Similar constants (the asymptotic mean of the abundancy index of other sequences): A013661 (all positive integers), A082020 (cubefree), A111003 (odd), A157292 (5-free), A157294 (7-free), A157296 (9-free), A240976 (squares), A245058 (even), A306633 (squarefree), A362985 (cubefull).

Programs

Mathematica

$MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{2, -3, 4, -6, 7, -7, 7, -6, 5, -3, 2, -1}, {0, 0, 0, 4, 5, 6, 0, -12, -9, -5, 0, 22}, m]; RealDigits[(2^4 + 2^2 + 2^(3/2) - 1)/(2^4 - 2)*(3^4 + 3^2 + 3^(3/2) - 1)/(3^4 - 3) * Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/2] - 1/2^(n/2) - 1/3^(n/2))/n, {n, 4, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]

PARI

prodeulerrat((p^8 + p^4 + p^3 - 1)/(p^8 - p^2), 1/2)

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A180114(k)/A001694(k).

Equals Product_{p prime} (p^4 + p^2 + p^(3/2) - 1)/(p^4 - p) = Product_{p prime} (1 + (p^2 + p^(3/2) + p - 1)/(p^4 - p)) (Jakimczuk and Lalín, 2022).