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.

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A340857 Decimal expansion of constant K5 = 29*log(2+sqrt(5))*(Product_{primes p == 1 (mod 5)} (1-4*(2*p-1)/(p*(p+1)^2)))/(15*Pi^2).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 24 2021

Keywords

Comments

Finch and Sebah, 2009, p. 7 (see link) call this constant K_5. K_5 is related to the Mertens constant C(5,1) (see A340839). For more references see the links in A340711. Finch and Sebah give the following definition:
Consider the asymptotic enumeration of m-th order primitive Dirichlet characters mod n. Let b_m(n) denote the count of such characters. There exists a constant 0 < K_m < oo such that Sum_{n <= N} b_m(n) ∼ K_m*N*log(N)^(d(m) - 2) as N -> oo, where d(m) is the number of divisors of m.

Examples

			0.262652188720536766675962011472088346530204393064744739106825510587...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; digits = 121; f[p_] := (1 - 4*(2*p-1)/(p*(p+1)^2));
    coefs = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, 1000}], x]];
    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}];
    m = 2; sump = 0; difp = 1; While[Abs[difp] > 10^(-digits - 5) || difp == 0, difp = coefs[[m]]*P[5, 1, m]; sump = sump + difp; PrintTemporary[m]; m++];
    RealDigits[Chop[N[29*Log[2+Sqrt[5]]/(15*Pi^2) * Exp[sump], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 25 2021, took over 50 minutes *)

Formula

Equals (29/25)*(Product_{primes p} (1-1/p)^2*(1+gcd(p-1,5)/(p-1))) [Finch and Sebah, 2009, p. 10].

A326575 Expansion of Sum_{k>=1} k * x^k * (1 + x^(2*k)) / (1 + x^(2*k) + x^(4*k)).

Original entry on oeis.org

1, 2, 3, 4, 4, 6, 8, 8, 9, 8, 10, 12, 14, 16, 12, 16, 16, 18, 20, 16, 24, 20, 22, 24, 21, 28, 27, 32, 28, 24, 32, 32, 30, 32, 32, 36, 38, 40, 42, 32, 40, 48, 44, 40, 36, 44, 46, 48, 57, 42, 48, 56, 52, 54, 40, 64, 60, 56, 58, 48, 62, 64, 72, 64, 56, 60
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 12 2019

Keywords

Examples

			G.f. = x + 2*x^2 + 3*x^3 + 4*x^4 + 4*x^5 + 6*x^6 + 8*x^7 + 8*x^8 + ... - _Michael Somos_, Oct 23 2019
		

Crossrefs

Cf. A003586 (fixed points), A035178, A050469, A122373, A326401.

Programs

  • Mathematica
    nmax = 66; CoefficientList[Series[Sum[k x^k (1 + x^(2 k))/(1 + x^(2 k) + x^(4 k)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, # &, MemberQ[{1}, Mod[n/#, 6]] &] - DivisorSum[n, # &, MemberQ[{5}, Mod[n/#, 6]] &], {n, 1, 66}]
    f[p_, e_] := Which[p < 5, p^e, Mod[p, 6] == 5, (p^(e + 1) - (-1)^(e + 1))/(p + 1), Mod[p, 6] == 1, (p^(e + 1) - 1)/(p - 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 02 2020 *)
  • PARI
    a(n) = { sumdiv(n, d, d*((n/d%6==1)-(n/d%6==5))) } \\ Andrew Howroyd, Sep 12 2019
    
  • PARI
    {a(n) = if( n<0, 0, sumdiv( n, d, n/d * kronecker( -12, d)))}; /* Michael Somos, Oct 23 2019 */

Formula

a(n) = Sum_{d|n, n/d==1 (mod 6)} d - Sum_{d|n, n/d==5 (mod 6)} d.
G.f.: Sum_{k>=0} x^(6*k+1) / (1 - x^(6*k+1))^2 - x^(6*k+5) / (1 - x^(6*k+5))^2. - Michael Somos, Oct 23 2019
Multiplicative with a(p^e) = p^e if p < 5, (p^(e+1)-(-1)^(e+1))/(p+1) if p == 5 (mod 6), and (p^(e+1)-1)/(p-1) if p == 1 (mod 6). - Amiram Eldar, Dec 02 2020
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{primes p == 5 (mod 6)} 1/(1+1/p^2) * Product_{primes p == 1 (mod 3)} 1/(1 - 1/p^2) = A340578 * A175646 / 2 = 0.48831400806... . - Amiram Eldar, Nov 06 2022
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