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 14 results. Next

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

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

A340711 Decimal expansion of Product_{primes p == 3 (mod 5)} (p^2+1)/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 16 2021

Keywords

Examples

			1.273986613206833925158...
		

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[1/(Z[5, 3, 4]/Z[5, 3, 2]^2)]

Formula

D = Product_{primes p == 0 (mod 5)} (p^2+1)/(p^2-1) = 13/12.
E = Product_{primes p == 1 (mod 5)} (p^2+1)/(p^2-1) = A340629.
F = Product_{primes p == 2 (mod 5)} (p^2+1)/(p^2-1) = A340710.
G = Product_{primes p == 3 (mod 5)} (p^2+1)/(p^2-1) = this constant.
H = Product_{primes p == 4 (mod 5)} (p^2+1)/(p^2-1) = A340628.
D*E*F*G*H = 5/2.
E*F*G*H = 30/13.
D*E*H = sqrt(5)/2.
D*F*G = 13*sqrt(5)/12.
F*G = sqrt(5).
E*H = 6*sqrt(5)/13.
Equals Sum_{q in A004617} 2^A001221(q)/q^2. - R. J. Mathar, Jan 27 2021

A189226 Curvatures in the nickel-dime-quarter Apollonian circle packing, ordered first by generation and then by size.

Original entry on oeis.org

-11, 21, 24, 28, 40, 52, 61, 157, 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397, 132, 156, 160, 189, 204, 205, 216, 237, 253, 285, 288, 309, 316, 336, 349, 405, 412, 421, 453, 460, 469, 472, 517, 544, 565, 616, 628, 685, 717, 741, 1084, 1093, 1104, 1125, 1128, 1140
Offset: 1

Views

Author

Jonathan Sondow, Apr 18 2011

Keywords

Comments

For a circle, curvature = 1/radius. The curvatures of a quarter, nickel, and dime are approximately proportional to 21, 24, and 28, respectively. Three mutually tangent circles with curvatures 21, 24, 28 can be inscribed in a circle of curvature 11.
Apollonius's and Descartes's Theorems say that, given three mutually tangent circles of curvatures a, b, c, there are exactly two circles tangent to all three, and their curvatures are a + b + c +- 2*sqrt(ab + ac + bc). (Here negative curvature of one of the two circles means that the three circles are inscribed in it.)
Fuchs (2009) says "An Apollonian circle packing ... is made by repeatedly inscribing circles into the triangular interstices in a Descartes configuration of four mutually tangent circles. Remarkably, if the original four circles have integer curvature, all of the circles in the packing will have integer curvature as well." That is because if a + b + c - 2s*qrt(ab + ac + bc) is an integer, then so is a + b + c + 2*sqrt(ab + ac + bc).
For n > 1, the n-th generation of the packing has 4*3^(n-2) circles.
Infinitely many of the curvatures are prime numbers A189227. In fact, in any integral Apollonian circle packing that is primitive (i.e., the curvatures have no common factor), the prime curvatures constitute a positive fraction of all primes (Bourgain 2012) and there are infinitely many pairs of tangent circles both of whose curvatures are prime (Sarnak 2007, 2011).
Fuchs and Sanden (2012) report on experiments with the nickel-dime-quarter Apollonian circle packing, which they call the coins packing P_C.

Examples

			The 1st-generation curvatures are -11, 21, 24, 28, the 2nd are 40, 52, 61, 157, and the 3rd are 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397. The 4th generation begins 132, 156, 160, 189, 204, 205, 216, ....
As 21 + 24 + 28 +- 2*sqrt(21*24 + 21*28 + 24*28) = 157 or -11, the sequence begins -11, 21, 24, 28, ... and 157 is in it.
The primes 157 and 397 are the curvatures of two circles that are tangent.
		

Crossrefs

Programs

  • Mathematica
    root = {-11, 21, 24, 28};
    triples = Subsets[root, {3}];
    a = {root};
    Do[
      ng = Table[Total@t + 2 Sqrt@Total[Times @@@ Subsets[t, {2}]], {t, triples}];
      AppendTo[a, Sort@ng];
      triples = Join @@ Table[{t, r} = tr; Table[Append[p, r], {p, Subsets[t, {2}]}], {tr, Transpose@{triples, ng}}]
      , {k, 3}];
    Flatten@a (* Andrey Zabolotskiy, May 29 2022 *)

Formula

a(n) == 0, 4, 12, 13, 16, or 21 (mod 24).

Extensions

Terms a(28) and beyond from Andrey Zabolotskiy, May 29 2022

A189227 Primes among the curvatures in the nickel-dime-quarter Apollonian circle packing A189226.

Original entry on oeis.org

-11, 61, 157, 181, 349, 373, 397, 421, 541, 661, 709, 733, 829, 853, 877, 997, 1021, 1069, 1093, 1213, 1237, 1381, 1429, 1597, 1621, 1669, 1693, 1741, 1861, 2029, 2221, 2293, 2341, 2389, 2557, 2677, 2749, 2917, 3037, 3061, 3109, 3181, 3229, 3253, 3301, 3373
Offset: 1

Views

Author

Jonathan Sondow, Apr 22 2011

Keywords

Comments

See A189226 for comments, references, links, examples, and crossrefs.

Crossrefs

Programs

  • Mathematica
    (* terms < 10^4 *) t = Range[9999]*0; w = {-11, 21, 24, 28}; s[1] = {{-1,2,2,2}, {0,1,0,0}, {0,0,1,0}, {0,0,0,1}}; s[2] = {{1,0,0,0}, {2,-1,2,2}, {0,0,1,0}, {0,0,0,1}}; s[3] = {{1,0,0,0}, {0,1,0,0}, {2,2,-1,2}, {0,0,0,1}}; s[4] = {{1,0,0,0}, {0,1,0,0}, {0,0,1,0}, {2,2,2,-1}}; r[m_, j_, p_] := Block[{v = (m.w)[[p]]}, If[v < 9999, t[[v]] = 1; Do[ If[i != j, r[m.s[i], i, p]], {i, 4}]]]; Do[ r[s[i], i, i], {i, 4}]; Prepend[ Select[ Flatten@ Position[t,1], PrimeQ], -11] (* Giovanni Resta, Jan 02 2014 *)

Formula

a(n) == 13 (mod 24) (because a(n) is prime, a(n) = A189226(k) for some k, and all terms of A189226 are == 0, 4, 12, 13, 16, or 21 (mod 24)).

Extensions

Corrected and extended by Steven Finch, Jan 02 2014
a(16)-a(46) from Giovanni Resta, Jan 02 2014
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