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 18 results. Next

A049532 Numbers k such that k^2 + 1 is not squarefree.

Original entry on oeis.org

7, 18, 32, 38, 41, 43, 57, 68, 70, 82, 93, 99, 107, 117, 118, 132, 143, 157, 168, 182, 193, 207, 218, 232, 239, 243, 251, 257, 268, 282, 293, 307, 318, 327, 332, 343, 357, 368, 378, 382, 393, 407, 408, 418, 432, 437, 443, 457, 468, 482, 493, 500, 507, 515
Offset: 1

Views

Author

Keywords

Comments

The sequence is infinite. For instance, it contains all numbers of the form 7 + 25m. - Emmanuel Vantieghem, Oct 25 2016
More generally, the sequence contains all numbers of the form a(n) + (a(n)^2 + 1) * m for even a(n) and a(n) + (a(n)^2 + 1) * m / 2 for odd a(n). - David A. Corneth, Oct 25 2016
The asymptotic density of this sequence is 1 - A335963 = 0.1051587754... - Amiram Eldar, Jul 08 2020

Examples

			a(1) = 7 because 7^2 + 1 = 49 + 1 = 50 is divisible by 25, a square.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..6*10^2]| not IsSquarefree(n^2+1)]; // Bruno Berselli, Oct 15 2012
  • Mathematica
    n=1;Reap[Do[While[SquareFreeQ[n^2+1],n++];Sow[n];n++,{c,10000}]][[2,1]] (* Zak Seidov, Feb 24 2011 *)
  • PARI
    for(n=1,1e4,if(!issquarefree(n^2+1),print1(n", "))) \\ Charles R Greathouse IV, Feb 24 2011
    

Formula

A059592(a(n)) > 1; A124809(n) = a(n)^2 + 1. - Reinhard Zumkeller, Nov 08 2006

Extensions

Definition rewritten by Bruno Berselli, Oct 15 2012
Mathematica updated by Jean-François Alcover, Jun 19 2013

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

A049533 Numbers k such that k^2+1 is squarefree.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 40, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80
Offset: 1

Views

Author

Keywords

Comments

Estermann proved that a(n) ~ kn with k = 1.117...; more precisely, there are cx + O(x^(2/3) log x) terms up to x, where c = 1/k = Product (1 - 2/p^2) where the product is over primes p which are 1 mod 4. Heath-Brown improves the error term to O(x^(7/12) log x). - Charles R Greathouse IV, Oct 16 2017, corrected by Amiram Eldar, Jul 08 2020
There are 89489 terms up to 10^5, 894856 terms up to 10^6, 8948417 up to 10^7, 89484102 up to 10^8, and 894841314 up to 10^9. - Charles R Greathouse IV, Nov 26 2017, corrected and extended by Amiram Eldar, Jul 08 2020

Examples

			10 is a member because 10^2 + 1 = 100 + 1 = 101 is squarefree.
Reasons why certain numbers are excluded: 7^2+1 = 2*5^2, 18^2+1 = 13*5^2, 32^2+1 = 41*5^2, 38^2+1 = 5*17^2, 41^2+1 = 2*29^2, 43^2+1 = 74*5^2, 57^2+1 = 130*5^2, 82^2+1 = 269*5^2. - Neven Juric, Oct 06 2008
		

Crossrefs

Complement of A049532.

Programs

  • Magma
    [ n: n in [1..100] | IsSquarefree(n^2+1) ]; // Vincenzo Librandi, Dec 25 2010
    
  • Mathematica
    Select[Range@ 80, SquareFreeQ[#^2 + 1] &] (* Michael De Vlieger, Aug 09 2017 *)
  • PARI
    isok(n) = issquarefree(n^2+1); \\ Michel Marcus, Feb 09 2016

Formula

Numbers k such that A059592(k) = 1. - Reinhard Zumkeller, Nov 08 2006
Showing 1-10 of 18 results. Next