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A362984
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Decimal expansion of the asymptotic mean of the abundancy index of the powerful numbers (A001694).
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2
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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
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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2.14968690301526765128219042105109416145987653275100999873...
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MATHEMATICA
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$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]]
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PROG
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(PARI) prodeulerrat((p^8 + p^4 + p^3 - 1)/(p^8 - p^2), 1/2)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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