A340127 Decimal expansion of Product_{primes p == 4 (mod 5)} p^2/(p^2-1).
1, 0, 0, 4, 9, 6, 0, 3, 2, 3, 9, 2, 2, 2, 9, 7, 5, 5, 8, 9, 9, 3, 7, 4, 9, 6, 2, 4, 8, 1, 0, 2, 5, 2, 1, 8, 4, 7, 9, 5, 5, 1, 0, 2, 9, 4, 1, 8, 8, 0, 2, 2, 8, 8, 0, 1, 9, 9, 5, 2, 8, 3, 7, 8, 5, 2, 1, 5, 0, 7, 1, 2, 7, 7, 0, 0, 7, 0, 0, 7, 6, 9, 8, 8, 5, 4, 3, 2, 4, 9, 1, 3, 6, 1, 1, 8, 0, 0, 6, 1, 9
Offset: 1
Examples
1.0049603239222975589937496248102521847955102941880228801995283785215071277...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..501
- S. Ettahri, O. Ramare, L. Surel, Fast multi-precision computation of some Euler products, arxiv:1908.06808 (2019), Section 9.
- Steven Finch and Pascal Sebah, Residue of a Mod 5 Euler Product, arXiv:0912.3677 [math.NT], 2009 (C(5,n) = mu(n,5) formulas p.2).
- Alessandro Languasco and Alessandro Zaccagnini, Computation of the Mertens constants mod q; 3 <= q <= 100, (2007) (GP-PARI procedure 100 digits accuracy).
- Alessandro Languasco and Alessandro Zaccagnini, On the constant in the Mertens product for arithmetic progressions. I. Identities, Funct. Approx. Comment. Math. Volume 42, Number 1 (2010), 17-27.
- For other links see A340711.
Crossrefs
Programs
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Mathematica
(* Using Vaclav Kotesovec's function Z from A301430. *) $MaxExtraPrecision = 1000; digits = 121; digitize[c_] := RealDigits[Chop[N[c, digits]], 10, digits - 1][[1]]; digitize[Z[5, 4, 2]]
Formula
Equals (1/C(5,4))*Pi*sqrt(3*C(5,1)*C(5,2)*C(5,3)/(5*C(5,4)*log(2+sqrt(5)))).
for definitions of Mertens constants C(5,n) see A. Languasco and A. Zaccagnini 2010.
for high precision numerical values C(5,n) see A. Languasco and A. Zaccagnini 2007.
C(5,1)=1.225238438539084580057609774749220527540595509391649938767...
C(5,2)=0.546975845411263480238301287430814037751996324100819295153...
C(5,3)=0.8059510404482678640573768602784309320812881149390108979348...
C(5,4)=1.29936454791497798816084001496426590950257497040832966201678...
Equals (1/C(5,4)^2)*Pi*sqrt(3*exp(-gamma)/(4*log(2 + sqrt(5)))), where gamma is the Euler-Mascheroni constant A001620.
Equals Sum_{k>=1} 1/A004618(k)^2. - Amiram Eldar, Jan 24 2021