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.

Showing 1-10 of 26 results. Next

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

Views

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

A050469 a(n) = Sum_{ d divides n, n/d=1 mod 4} d - Sum_{ d divides n, n/d=3 mod 4} d.

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 6, 8, 7, 12, 10, 8, 14, 12, 12, 16, 18, 14, 18, 24, 12, 20, 22, 16, 31, 28, 20, 24, 30, 24, 30, 32, 20, 36, 36, 28, 38, 36, 28, 48, 42, 24, 42, 40, 42, 44, 46, 32, 43, 62, 36, 56, 54, 40, 60, 48, 36, 60, 58, 48, 62, 60, 42, 64, 84, 40
Offset: 1

Views

Author

N. J. A. Sloane, Dec 23 1999

Keywords

Comments

Multiplicative with a(p^e)=p^e if p=2, (p^(e+1)-1)/(p-1) if p==1 (mod 4), else (p^(e+1)+(-1)^e)/(p+1). - Michael Somos, May 02 2005
Multiplicative because it is the Dirichlet convolution of A000027 = n and A101455 = [1 0 -1 0 1 0 -1 ...], which are both multiplicative. - Christian G. Bower, May 17 2005

Crossrefs

Programs

  • Mathematica
    max = 70; s = Sum[n*x^(n-1)/(1+x^(2*n)), {n, 1, max}] + O[x]^max; CoefficientList[s, x] (* Jean-François Alcover, Dec 02 2015 *)
    f[p_, e_] := Which[p == 2, p^e, Mod[p, 4] == 1, (p^(e + 1) - 1)/(p - 1), Mod[p, 4] == 3, (p^(e + 1) + (-1)^e)/(p + 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Nov 06 2022 *)
  • PARI
    a(n)=if(n<1,0,sumdiv(n,d,d*((n/d%4==1)-(n/d%4==3))))
    
  • PARI
    {a(n)=local(A,p,e); if(n<2, n==1, A=factor(n); prod(k=1,matsize(A)[1], if(p=A[k,1], e=A[k,2]; if(p==2, p^e, if(p%4==1, (p^(e+1)-1)/(p-1), (p^(e+1)+(-1)^e)/(p+1)))))) } /* Michael Somos, May 02 2005 */
    
  • PARI
    a(n)=if(n<1,0,polcoeff(sum(k=1,n,k*x^k/(1+x^(2*k)),x*O(x^n)),n))

Formula

G.f.: Sum_{n>=1} n*x^n/(1+x^(2*n)). - Vladeta Jovovic, Oct 16 2002
L.g.f.: Sum_{k>=1} arctan(x^k). - Ilya Gutkovskiy, Dec 16 2019
O.g.f.: Sum_{n >= 1} (-1)^(n+1) * x^(2*n-1)/(1 - x^(2*n-1))^2. - Peter Bala, Jan 04 2021
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{primes p == 1 (mod 4)} 1/(1-1/p^2) * Product_{primes p == 3 (mod 4)} 1/(1+1/p^2) = (1/2) * A175647 / A243381 = A006752/2 = 0.4579827970... . - Amiram Eldar, Nov 06 2022, Nov 05 2023
a(n) = Sum_{d|n} (n/d)*sin(d*Pi/2). - Ridouane Oudra, Sep 26 2024

A340004 Decimal expansion of Product_{primes p == 1 (mod 5)} p^2/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 15 2021

Keywords

Comments

This constant is called Euler product 2==1 modulo 5 (see Mathar's Definition 5 formula (38)) or equivalently zeta 2==1 modulo 5.

Examples

			1.01091516060101952260495658428951492...
		

Crossrefs

Programs

  • Mathematica
    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[5, 1, 2], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 15 2021, took 20 minutes *)

Formula

Equals Sum_{k>=1} 1/A004615(k)^2. - Amiram Eldar, Jan 24 2021
Equals exp(-gamma/2)*Pi/(A340839^2*sqrt(5*log((1 + sqrt (5))/2))). - Artur Jasinski, Jan 30 2021

A340127 Decimal expansion of Product_{primes p == 4 (mod 5)} p^2/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 15 2021

Keywords

Examples

			1.0049603239222975589937496248102521847955102941880228801995283785215071277...
		

Crossrefs

Programs

  • 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

A340628 Decimal expansion of Product_{primes p == 4 (mod 5)} (p^2+1)/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 13 2021

Keywords

Examples

			1.009935934829401027349038488241778167715858547548801013...
		

Crossrefs

Programs

  • Maple
    evalf(Re(2*Pi^2/(5*sqrt(13*((I*Pi^2*(1/150)-I*polylog(2, (-1)^(2/5)))^2+((1/150)*(11*I)*Pi^2+I*polylog(2, (-1)^(4/5)))^2)))), 120) # Vaclav Kotesovec, Jan 20 2021, after formula by Pascal Sebah
  • Mathematica
    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; PrintTemporary["iteration = ", w, ", difference = ", N[difz, digits]]; w++]; Exp[sumz]);
    $MaxExtraPrecision = 1000; digits = 121; Chop[N[1/(Z[5,4,4]/Z[5,4,2]^2), digits]] (* Vaclav Kotesovec, Jan 15 2021, took over 20 minutes *)
    digits = 121; digitize[c_] := RealDigits[Chop[N[c, digits]], 10, digits][[1]];
    cl[x_] :=I(PolyLog[2,(-1)^x] - PolyLog[2,-(-1)^(1-x)]);
    A340628 :=(4 Pi^2)/(5 Sqrt[13])/ Sqrt[cl[2/5]^2 + cl[4/5]^2];
    digitize[A340628] (* Peter Luschny, Jan 23 2021 *)

Formula

Equals 6*sqrt(5)/(13*A340629).
Equals 6*sqrt(13)*Pi^2/(195*g) where g = sqrt(Cl2(2*Pi/5)^2 + Cl2(4*Pi/5)^2) = 1.0841621352693895..., and Cl2 is the Clausen function of order 2. Formula by Pascal Sebah (personal communication). - Artur Jasinski, Jan 20 2021
Equals A340127^2/A340809. - R. J. Mathar, Jan 22 2021
Equals Sum_{q in A004618} 2^A001221(q)/q^2. - R. J. Mathar, Jan 27 2021

Extensions

Corrected and more terms from Vaclav Kotesovec, Jan 15 2021

A088539 Decimal expansion of (4K/Pi)^2 where K is the Landau-Ramanujan constant.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Nov 16 2003

Keywords

Examples

			0.9468064071800793342160944131097562332500695...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, p. 100

Crossrefs

Programs

  • Mathematica
    digits = 104; LandauRamanujanK = 1/Sqrt[2]*NProduct[((1-2^(-2^n)) * Zeta[2^n] / DirichletBeta[2^n])^(1/2^(n+1)), {n, 1, 24}, WorkingPrecision -> digits+5]; (4*LandauRamanujanK/Pi)^2 // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Mar 04 2013, updated Mar 14 2018 *)

Formula

Equals prod(1-1/p^2) where p runs through the primes p==1 mod 4
A088539 * A243379 = 8 / Pi^2. - Vaclav Kotesovec, Apr 30 2020
Equals 1/A175647. - Vaclav Kotesovec, May 05 2020

A340576 Decimal expansion of Product_{primes p == 5 (mod 6)} 1/(1-1/p^2).

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jan 12 2021

Keywords

Comments

The four similar sequences for products of primes mod 6 are these:
A175646 for Product_{primes p == 1 (mod 6)} 1/(1-1/p^2),
A340576 for Product_{primes p == 5 (mod 6)} 1/(1-1/p^2),
A340577 for Product_{primes p == 1 (mod 6)} 1/(1+1/p^2),
A340578 for Product_{primes p == 5 (mod 6)} 1/(1+1/p^2).

Examples

			1.06054829316911072817412636430987203493077130204487163127994372...
		

Crossrefs

Programs

  • Maple
    a := n -> 3^(2^(-n-2))*((1-3^(-2^(n+1)))/2)^(2^(-n-1)):
    b := n -> Zeta(n)/Im(polylog(n, (-1)^(2/3))):
    c := n -> a(n)*b(2^(n+1))^(1/2^(n+1)):
    Digits := 107: evalf((3/4)*mul(c(n), n=0..9)); # Peter Luschny, Jan 14 2021
  • Mathematica
    digits = 105;
    precision = digits + 10;
    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]&;
    RealDigits[pB, 10, digits][[1]] (* Most of this code is 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[6,5,2], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 15 2021 *)

Formula

g = A143298 = (9 - PolyGamma(1, 2/3) + PolyGamma(1, 4/3))/(4 sqrt(3));
h = A301429;
Equals (3*sqrt(3)*h^2)/2.
Equals (3/4)*A333240.
A340577 = Pi^4/(243*g*h^2);
A340578 = (45*g*h^2)/(2*Pi^2).
Equals Pi^2/(9*A175646). - Artur Jasinski, Jan 11 2021
Equals Sum_{k>=1} 1/A259548(k)^2. - Amiram Eldar, Jan 24 2021

A340577 Decimal expansion of Product_{primes p == 1 (mod 6)} 1/(1+1/p^2).

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jan 12 2021

Keywords

Examples

			0.96755040251956188660947077043906773001524912960304386356302398...
		

Crossrefs

Programs

  • 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]&;
    pC = (2*Pi^4)/(243*pB*Lv[2]);
    RealDigits[pC, 10, digits][[1]](* Most of this code is 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[6, 1, 4]/Z[6, 1, 2], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 15 2021 *)

Formula

Equals 1.0004615.../A175646, both constants taken from p 27 of arXiv:1008.2537v2. - R. J. Mathar, Aug 21 2022

Extensions

a(104) corrected by Vaclav Kotesovec, Jan 15 2021

A340578 Decimal expansion of Product_{primes p == 5 (mod 6)} 1/(1+1/p^2).

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jan 12 2021

Keywords

Examples

			0.94450093450470098673429109419144434254611078086906676955735771...
		

Crossrefs

Programs

  • 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]&;
    pD = (45*pB*Lv[2])/(4*Pi^2);
    RealDigits[pD, 10, digits][[1]] (* Most of this code is 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[6, 5, 4]/Z[6, 5, 2], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 15 2021 *)

A340629 Decimal expansion of Product_{primes p == 1 (mod 5)} (p^2+1)/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 13 2021

Keywords

Examples

			1.0218780604187566757444489146002708261704607377325...
		

Crossrefs

Programs

  • Maple
    evalf(Re(15*sqrt((1/13)*(5*((I*Pi^2*(1/150)-I*polylog(2, (-1)^(2/5)))^2+((1/150)*(11*I)*Pi^2+I*polylog(2, (-1)^(4/5)))^2)))/Pi^2), 120) # Vaclav Kotesovec, Jan 20 2021, after formula by Pascal Sebah.
  • Mathematica
    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; PrintTemporary["iteration = ", w, ", difference = ", N[difz, digits]]; w++]; Exp[sumz]);
    $MaxExtraPrecision = 1000; digits = 121; Chop[N[1/(Z[5,1,4]/Z[5,1,2]^2), digits]] (* Vaclav Kotesovec, Jan 15 2021, took over 20 minutes *)
    digits = 121; digitize[c_] := RealDigits[Chop[N[c, digits]], 10, digits][[1]];
    cl[x_] := I (PolyLog[2, (-1)^x] - PolyLog[2, -(-1)^(1 - x)]);
    A340629 := (15 Sqrt[65]/(26 Pi^2)) Sqrt[cl[2/5]^2 + cl[4/5]^2];
    digitize[A340629] (* Peter Luschny, Jan 23 2021 *)

Formula

Equals 6*sqrt(5)/(13*A340628).
Equals A340004^2/A340808. - R. J. Mathar, Jan 15 2021
Equals 15*sqrt(65)*g/(13*Pi^2) where g = sqrt(Cl2(2*Pi/5)^2 + Cl2(4*Pi/5)^2) = 1.0841621352693895..., and Cl2 is the Clausen function of order 2. Formula by Pascal Sebah (personal communication). - Artur Jasinski, Jan 20 2021
Equals Sum_{q in A004615} 2^A001221(q)/q^2. - R. J. Mathar, Jan 27 2021

Extensions

Corrected and more terms from Vaclav Kotesovec, Jan 15 2021
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