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A109695
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Decimal expansion of Sum_{n>=1} 1/phi(n)^2.
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5
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3, 3, 9, 0, 6, 4, 2, 0, 0, 5, 5, 7, 2, 5, 0, 3, 9, 1, 6, 1, 4, 2, 5, 9, 5, 6, 6, 3, 0, 0, 2, 6, 3, 0, 7, 9, 3, 7, 4, 0, 5, 3, 7, 3, 8, 1, 2, 1, 4, 4, 7, 1, 6, 9, 1, 1, 8, 0, 7, 3, 9, 8, 1, 5, 6, 8, 5, 7, 3, 8, 1, 3, 1, 1, 1, 7, 7, 6, 3, 3, 2, 1, 3, 6, 5, 0, 4, 1, 0, 2, 4, 4, 4, 9, 5, 2, 3, 7, 4, 2, 9, 8, 2, 5, 7
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OFFSET
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1,1
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COMMENTS
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The logarithm of the value can be expanded in a series Sum_{j>=2} c(j)*P(j) = P(2) + 2*P(3) + (7/2)*P(4) + ... where P(.) is the prime zeta function. The partial sums of the series are a slowly oscillating function of the upper limit of j, from which the bracketing interval [3.390642005572503655..., 3.390642005572504756...] for the constant can be computed. - R. J. Mathar, Feb 03 2009
Sum_{n>=1} 1/phi(n)^k is convergent iff k > 1 (reference Monier). - Bernard Schott, Dec 13 2020
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REFERENCES
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Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21, pp. 281 and 294.
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LINKS
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FORMULA
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Equals Product_p Sum_{k>=0} 1/phi(p^k)^2 = Product_p (1 + p^2/((p-1)^2*(p^2-1))).
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EXAMPLE
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3.39064200557250391614259566300263079374053738121447169118...
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MATHEMATICA
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$MaxExtraPrecision = 1000; f[p_] := (1 + p^2/((p - 1)^2*(p^2 - 1))); Do[cc = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; Print[f[2] * Exp[N[Sum[Indexed[cc, n]*(PrimeZetaP[n] - 1/2^n), {n, 2, m}], 120]]], {m, 100, 1000, 100}] (* Vaclav Kotesovec, Jun 25 2020 *)
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PROG
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(PARI) my(N=1000000000); prodeuler(p=2, N, 1.+p^2/((p-1)^2*(p^2-1)))*(1+1/(N*log(N)))
(PARI) prodeulerrat(1 + p^2/((p-1)^2*(p^2-1))) \\ Amiram Eldar, Mar 15 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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Four more digits from R. J. Mathar, Feb 03 2009, 25 more Dec 18 2010
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STATUS
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approved
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