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A362985
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Decimal expansion of the asymptotic mean of the abundancy index of the cubefull numbers (A036966).
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2
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2, 4, 8, 2, 1, 7, 9, 1, 9, 6, 4, 2, 2, 3, 5, 9, 5, 2, 5, 4, 6, 1, 6, 7, 6, 4, 3, 6, 7, 4, 6, 8, 7, 6, 9, 8, 5, 3, 6, 3, 6, 8, 9, 4, 0, 9, 7, 1, 9, 3, 0, 4, 6, 8, 3, 5, 4, 3, 6, 3, 9, 3, 2, 8, 1, 4, 4, 4, 2, 3, 3, 8, 8, 5, 7, 6, 7, 5, 0, 4, 6, 3, 4, 1, 1, 5, 0, 7, 3, 1, 0, 3, 9, 8, 0, 4, 4, 7, 4, 0, 3, 7, 3, 1, 0
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OFFSET
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1,1
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LINKS
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FORMULA
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Equals zeta(4/3) * Product_{p prime} ((p^5 + p^(10/3) + p^3 + p^(8/3) - 1)/(p^(10/3) * (p^(5/3) + p^(1/3) + 1))).
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EXAMPLE
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2.48217919642235952546167643674687698536368940971930468354...
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MATHEMATICA
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$MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{2, -1, -2, 3, -2, -1, 3, -2, -2, 3, -1, -2, 3, -1, -1, 1}, {0, 0, 0, -4, 0, 6, 7, 4, 9, 0, -11, -22, -26, -21, -15, 20}, m]; RealDigits[((2^5 + 2^(10/3) + 2^3 + 2^(8/3) - 1)/(2^(10/3)*(2^(5/3) + 2^(1/3) + 1)))*((3^5 + 3^(10/3) + 3^3 + 3^(8/3) - 1)/(3^(10/3)*(3^(5/3) + 3^(1/3) + 1))) * Zeta[4/3] * Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/3] - 1/2^(n/3) - 1/3^(n/3))/n, {n, 4, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
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PROG
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(PARI) zeta(4/3) * prodeulerrat((p^15 + p^10 + p^9 + p^8 - 1)/(p^10 * (p^5 + p + 1)), 1/3)
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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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