|
%I #34 Mar 15 2021 03:40:34
%S 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,
%T 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,
%U 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
%N Decimal expansion of Sum_{n>=1} 1/phi(n)^2.
%C 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
%C Sum_{n>=1} 1/phi(n)^k is convergent iff k > 1 (reference Monier). - _Bernard Schott_, Dec 13 2020
%D Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21, pp. 281 and 294.
%F Equals Product_p Sum_{k>=0} 1/phi(p^k)^2 = Product_p (1 + p^2/((p-1)^2*(p^2-1))).
%F Equals Sum{n>=1} 1/A127473(n). - _Amiram Eldar_, Mar 15 2021
%e 3.39064200557250391614259566300263079374053738121447169118...
%t $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 *)
%o (PARI) my(N=1000000000); prodeuler(p=2,N,1.+p^2/((p-1)^2*(p^2-1)))*(1+1/(N*log(N)))
%o (PARI) prodeulerrat(1 + p^2/((p-1)^2*(p^2-1))) \\ _Amiram Eldar_, Mar 15 2021
%Y Cf. A000010, A065484, A127473, A335818.
%K nonn,cons
%O 1,1
%A _Franklin T. Adams-Watters_, Aug 07 2005
%E Four more digits from _R. J. Mathar_, Feb 03 2009, 25 more Dec 18 2010
%E More digits from _Vaclav Kotesovec_, Jun 25 2020
|
