cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A175646 Decimal expansion of the Product_{primes p == 1 (mod 3)} 1/(1 - 1/p^2).

Original entry on oeis.org

1, 0, 3, 4, 0, 1, 4, 8, 7, 5, 4, 1, 4, 3, 4, 1, 8, 8, 0, 5, 3, 9, 0, 3, 0, 6, 4, 4, 4, 1, 3, 0, 4, 7, 6, 2, 8, 5, 7, 8, 9, 6, 5, 4, 2, 8, 4, 8, 9, 0, 9, 9, 8, 8, 6, 4, 1, 6, 8, 2, 5, 0, 3, 8, 4, 2, 1, 2, 2, 2, 2, 4, 5, 8, 7, 1, 0, 9, 6, 3, 5, 8, 0, 4, 9, 6, 2, 1, 7, 0, 7, 9, 8, 2, 6, 2, 0, 5, 9, 6, 2, 8, 9, 9, 7
Offset: 1

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Author

R. J. Mathar, Aug 01 2010

Keywords

Comments

The Euler product of the Riemann zeta function at 2 restricted to primes in A002476, which is the inverse of the infinite product (1-1/7^2)*(1-1/13^2)*(1-1/19^2)*...
There is a complementary Product_{primes p == 2 (mod 3)} 1/(1-1/p^2) = A333240 = 1.4140643908921476375655018190798... such that (this constant here)*1.4140643.../(1-1/3^2) = zeta(2) = A013661.
Because 1/(1-p^(-2)) = 1+1/(p^2-1), the complementary 1.414064... also equals Product_{primes p == 2 (mod 3)} (1+1/(p^2-1)), which appears in Eq. (1.8) of [Dence and Pomerance]. - R. J. Mathar, Jan 31 2013

Examples

			1.03401487541434188053903064441304762857896...
		

Crossrefs

Programs

  • Maple
    z := n -> Zeta(n)/Im(polylog(n, (-1)^(2/3))):
    x := n -> (z(2^n)*(3^(2^n)-1)*sqrt(3)/2)^(1/2^n) / 3:
    evalf(4*Pi^2 / (27*mul(x(n), n=1..8)), 106); # Peter Luschny, Jan 17 2021
  • Mathematica
    digits = 105;
    precision = digits + 5;
    prodeuler[p_, a_, b_, expr_] := Product[If[a <= p <= b, expr, 1], {p, Prime[Range[PrimePi[a], PrimePi[b]]]}];
    Lv3[s_] := prodeuler[p, 1, 2^(precision/s), 1/(1 - KroneckerSymbol[-3, p]*p^-s)] // N[#, precision]&;
    Lv4[s_] := 2*Im[PolyLog[s, Exp[2*I*Pi/3]]]/Sqrt[3];
    Lv[s_] := If[s >= 10000, Lv3[s], Lv4[s]];
    gv[s_] := (1 - 3^(-s))*Zeta[s]/Lv[s];
    pB = (3/4)*Product[gv[2^n*2]^(2^-(n+1)), {n, 0, 11}] // N[#, precision]&;
    pA = Pi^2/9/pB ;
    RealDigits[pA, 10, digits][[1]]
    (* Jean-François Alcover, Jan 11 2021, after PARI code due to Artur Jasinski *)
    S[m_, n_, s_] := (t = 1; sums = 0; difs = 1; While[Abs[difs] > 10^(-digits - 5) || difs == 0, difs = (MoebiusMu[t]/t) * Log[If[s*t == 1, DirichletL[m, n, s*t], Sum[Zeta[s*t, j/m]*DirichletCharacter[m, n, j]^t, {j, 1, m}]/m^(s*t)]]; sums = sums + difs; t++]; sums);
    P[m_, n_, s_] := 1/EulerPhi[m] * Sum[Conjugate[DirichletCharacter[m, r, n]] * S[m, r, s], {r, 1, EulerPhi[m]}] + Sum[If[GCD[p, m] > 1 && Mod[p, m] == n, 1/p^s, 0], {p, 1, m}];
    Z[m_, n_, s_] := (w = 1; sumz = 0; difz = 1; While[Abs[difz] > 10^(-digits - 5), difz = P[m, n, s*w]/w; sumz = sumz + difz; w++]; Exp[sumz]);
    $MaxExtraPrecision = 1000; digits = 121; RealDigits[Chop[N[Z[3,1,2], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 15 2021 *)
    z[n_] := Zeta[n] / Im[PolyLog[n, (-1)^(2/3)]];
    x[n_] := (z[2^n] (3^(2^n) - 1) Sqrt[3]/2)^(1/2^n) / 3;
    N[4 Pi^2 / (27 Product[x[n], {n, 8}]), 106] (* Peter Luschny, Jan 17 2021 *)

Formula

Equals 2*Pi^2 / (3^(7/2) * A301429^2). - Vaclav Kotesovec, May 12 2020
Equals Sum_{k>=1} 1/A004611(k)^2. - Amiram Eldar, Sep 27 2020

Extensions

More digits from Vaclav Kotesovec, May 12 2020 and Jun 27 2020