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A109695
Decimal expansion of Sum_{n>=1} 1/phi(n)^2.
5
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
OFFSET
1,1
COMMENTS
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
REFERENCES
Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21, pp. 281 and 294.
FORMULA
Equals Product_p Sum_{k>=0} 1/phi(p^k)^2 = Product_p (1 + p^2/((p-1)^2*(p^2-1))).
Equals Sum{n>=1} 1/A127473(n). - Amiram Eldar, Mar 15 2021
EXAMPLE
3.39064200557250391614259566300263079374053738121447169118...
MATHEMATICA
$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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
EXTENSIONS
Four more digits from R. J. Mathar, Feb 03 2009, 25 more Dec 18 2010
More digits from Vaclav Kotesovec, Jun 25 2020
STATUS
approved