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A175645
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Decimal expansion of the sum 1/p^3 over primes == 1 (mod 3).
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12
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0, 0, 3, 6, 0, 0, 4, 2, 3, 3, 4, 6, 9, 4, 2, 9, 5, 8, 9, 5, 7, 4, 7, 6, 9, 4, 7, 6, 2, 9, 2, 3, 8, 4, 6, 4, 9, 4, 2, 4, 9, 5, 1, 6, 5, 1, 3, 6, 9, 4, 3, 9, 1, 5, 4, 8, 1, 0, 3, 5, 8, 7, 3, 5, 1, 0, 7, 4, 1, 2, 0, 2, 5, 3, 5, 0, 4, 4, 6, 1, 2, 9, 2, 7, 0, 6, 8, 5, 0, 9, 7, 5, 9, 5, 3, 2, 0, 7, 9, 1, 7, 2, 9, 6, 7, 7
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OFFSET
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0,3
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COMMENTS
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The Prime Zeta modulo function at 3 for primes of the form 3k+1, which is Sum_{prime p in A002476} 1/p^3 = 1/7^3 + 1/13^3 + 1/19^3 + 1/31^3 + ...
The complementary sum, Sum_{prime p in A003627} 1/p^3 is given by P_{3,2}(3) = A085541 - 1/3^3 - (this value here) = 0.13412517891546354042859932999943119899...
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LINKS
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EXAMPLE
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P_{3,1}(3) = 0.00360042334694295895747694762923846494249516...
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MATHEMATICA
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(* A naive solution yielding 12 correct digits: *) s1 = s2 = 0.; Do[Switch[Mod[n, 3], 1, If[PrimeQ[n], s1 += 1/n^3], 2, If[PrimeQ[n], s2 += 1/n^3]], {n, 10^7}]; Join[{0, 0}, RealDigits[(PrimeZetaP[3] + s1 - s2 - 1/27)/2, 10, 12][[1]]] (* Jean-François Alcover, Mar 15 2018 *)
With[{s=3}, Do[Print[N[1/2 * Sum[(MoebiusMu[2*n + 1]/(2*n + 1)) * Log[(Zeta[s + 2*n*s]*(Zeta[s + 2*n*s, 1/6] - Zeta[s + 2*n*s, 5/6])) / ((1 + 2^(s + 2*n*s))*(1 + 3^(s + 2*n*s)) * Zeta[2*(1 + 2*n)*s])], {n, 0, m}], 120]], {m, 100, 500, 100}]] (* Vaclav Kotesovec, Jan 13 2021 *)
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}];
$MaxExtraPrecision = 1000; digits = 121; Join[{0, 0}, RealDigits[Chop[N[P[3, 1, 3], digits]], 10, digits-1][[1]]] (* Vaclav Kotesovec, Jan 22 2021 *)
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PROG
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(PARI) s=0; forprimestep(p=1, 1e8, 3, s+=1./p^3); s \\ for illustration only: primes up to 10^N give about 2N+2 correct digits. - M. F. Hasler, Apr 22 2021
A175645_upto(N=100)=localprec(N+5); digits((PrimeZeta31(3)+1)\.1^N)[^1] \\ Cf. A175644 for PrimeZeta31. - M. F. Hasler, Apr 23 2021
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CROSSREFS
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Cf. A086033 (P_{4,1}(3): same for p==1 (mod 4)), A175644 (P_{3,1}(2): same for 1/p^2), A343613 (P_{3,2}(3): same for p==2 (mod 3)), A085541 (PrimeZeta(3)).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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