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.

Previous Showing 21-30 of 30 results.

A340710 Decimal expansion of Product_{primes p == 2 (mod 5)} (p^2+1)/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 16 2021

Keywords

Examples

			1.7551738411687377766074721228405237...
		

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, 2, 4]/Z[5, 2, 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) = this constant.
G = Product_{primes p == 3 (mod 5)} (p^2+1)/(p^2-1) = A340711.
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.
Formulas by Pascal Sebah, Jan 20 2021: (Start)
Let g = sqrt(Cl2(2*Pi/5)^2+Cl2(4*Pi/5)^2) = 1.0841621352693895..., where Cl2 is the Clausen function of order 2.
E = 15*sqrt(65)*g/(13*Pi^2).
H = 6*sqrt(13)*Pi^2/(195*g). (End)
Equals Sum_{q in A004616} 2^A001221(q)/q^2. - R. J. Mathar, Jan 27 2021

A326401 Expansion of Sum_{k>=1} k * x^k / (1 + x^k + x^(2*k)).

Original entry on oeis.org

1, 1, 3, 3, 4, 3, 8, 5, 9, 4, 10, 9, 14, 8, 12, 11, 16, 9, 20, 12, 24, 10, 22, 15, 21, 14, 27, 24, 28, 12, 32, 21, 30, 16, 32, 27, 38, 20, 42, 20, 40, 24, 44, 30, 36, 22, 46, 33, 57, 21, 48, 42, 52, 27, 40, 40, 60, 28, 58, 36, 62, 32, 72, 43, 56, 30, 68, 48, 66, 32
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 11 2019

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 70; CoefficientList[Series[Sum[k x^k/(1 + x^k + x^(2 k)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, # &, MemberQ[{1}, Mod[n/#, 3]] &] - DivisorSum[n, # &, MemberQ[{2}, Mod[n/#, 3]] &], {n, 1, 70}]
    f[p_, e_] := Which[p == 3, p^e, Mod[p, 3] == 1, (p^(e + 1) - 1)/(p - 1), Mod[p, 3] == 2, (p^(e + 1) + (-1)^e)/(p + 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Oct 25 2020 *)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 3, 3^f[i,2], if(f[i,1]%3 == 1, (f[i,1]^(f[i,2]+1) - 1)/(f[i,1] - 1), (f[i,1]^(f[i,2]+1) + (-1)^f[i,2])/(f[i,1] + 1))));} \\ Amiram Eldar, Nov 06 2022

Formula

a(n) = Sum_{d|n, n/d==1 (mod 3)} d - Sum_{d|n, n/d==2 (mod 3)} d.
a(n) = A326399(n) - A326400(n).
Multiplicative with a(3^e) = 3^e, a(p^e) = (p^(e+1) - 1)/(p - 1) if p == 1 (mod 3), and (p^(e+1) + (-1)^e)/(p + 1) if p == 2 (mod 3). - Amiram Eldar, Oct 25 2020
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{primes p == 1 (mod 3)} 1/(1 - 1/p^2) * Product_{primes p == 2 (mod 3)} 1/(1 + 1/p^2) = (1/2) * A175646 * (2*Pi^2/27)/A340577 = 0.3906512064... . - Amiram Eldar, Nov 06 2022

A334480 Decimal expansion of Product_{k>=1} (1 - 1/A007528(k)^3).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, May 02 2020

Keywords

Comments

In general, for s > 0, Product_{k>=1} (1 + 1/A007528(k)^(2*s+1)) / (1 - 1/A007528(k)^(2*s+1)) = (1 - 1/2^(2*s + 1)) * (3^(2*s + 1) - 1) * (2*s)! * zeta(2*s + 1) / (sqrt(3) * A002114(s) * Pi^(2*s + 1)).
For s > 1, Product_{k>=1} (1 + 1/A007528(k)^s) / (1 - 1/A007528(k)^s) = (2^s - 1) * (3^s - 1) * zeta(s) / (zeta(s, 1/6) - zeta(s, 5/6)).
For s > 1, Product_{k>=1} (1 - 1/A002476(k)^s) * (1 - 1/A007528(k)^s) = 6^s / ((2^s - 1)*(3^s - 1)*zeta(s)).

Examples

			0.990884145525213356563403173559432751643483121750... = 1/1.0091997177631243951237...
		

Crossrefs

Formula

A334479 / A334480 = 91*sqrt(3)*zeta(3)/(6*Pi^3).
A334478 * A334480 = 108/(91*zeta(3)).

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A334481 Decimal expansion of Product_{k>=1} (1 + 1/A002476(k)^2).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, May 02 2020

Keywords

Comments

Product_{k>=1} (1 - 1/A002476(k)^2) = 1/A175646 = 0.9671040753637981066150556834173635260473412207450...
Let Zeta_{6,1}(4) = 1/ Product_{k>=1}(1-1/A002476(k)^4) = 1.0004615089.. and Zeta_{6,1}(2)= A175646 as tabulated in arXiv:1008.2547. Then this constant equals Zeta_{6,1}(2)/Zeta_{6,1}(4). - R. J. Mathar, Jan 12 2021

Examples

			1.03353788846135284308284618497621833947517677481...
		

Crossrefs

Formula

A334481 * A334482 = 54/(5*Pi^2).

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A334482 Decimal expansion of Product_{k>=1} (1 + 1/A007528(k)^2).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, May 02 2020

Keywords

Comments

Product_{k>=1} (1 - 1/A007528(k)^2) = 9*A175646/Pi^2 = 0.9429084997268899069451546585312672145658112624159...

Examples

			1.058760201782545491315895454572153336734712663249...
		

Crossrefs

Formula

A334481 * A334482 = 54/(5*Pi^2).

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A369565 Powerful numbers whose prime factors are all of the form 3*k + 1.

Original entry on oeis.org

1, 49, 169, 343, 361, 961, 1369, 1849, 2197, 2401, 3721, 4489, 5329, 6241, 6859, 8281, 9409, 10609, 11881, 16129, 16807, 17689, 19321, 22801, 24649, 26569, 28561, 29791, 32761, 37249, 39601, 44521, 47089, 49729, 50653, 52441, 57967, 58081, 61009, 67081, 73441
Offset: 1

Views

Author

Amiram Eldar, Jan 26 2024

Keywords

Comments

Closed under multiplication.

Crossrefs

Intersection of A001694 and A004611.
Similar sequence: A352492, A369563, A369564, A369566.

Programs

  • Mathematica
    q[n_] := n == 1 || AllTrue[FactorInteger[n], Mod[First[#], 3] == 1 && Last[#] > 1 &]; Select[Range[75000], q]
  • PARI
    is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 1]%3 != 1 || f[i, 2] == 1, return(0))); 1;}

Formula

Sum_{n>=1} 1/a(n) = Product_{primes p == 1 (mod 3)} (1 + 1/(p*(p-1))) = A175646 * A334477 = 1.0377399555...

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].

A188596 Decimal expansion of Product_{primes p} (1-1/p)^(-2)*(1-(2+A102283(p))/p).

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Apr 05 2011

Keywords

Comments

This is the principal scale factor in an estimate of the number of primes p not exceeding N such that p^2+p+1 is also prime [Bateman-Horn].
A102283 in the definition plays the role of the Dirichlet character modulo 3.
After splitting the product into the three modulo-3 classes of primes, this constant turns out to be the product of four factors.
One factor as mentioned by Bateman and Horn is the inverse of A073010.
The second factor is 3/4 arising from the prime 3 which is the sole prime in the class == 0 (mod 3).
The third factor is product_{p == 1 (mod 3)} (1-(3p-1)/(p-1)^3) = 0.8675121817.. which is the constant C(m=3,n=1,s=3) of the arXiv preprint, basically the C(3) variant of A065418 reduced to the modulo class.
The final factor is product_{p == 2 (mod 3)} (1+1/(p^2-1)) = 1/product_{p == 2 (mod 3)} (1-1/p^2) = 1.41406439089214763.. which is the constant zeta(m=3,n=2,s=2) of the preprint and mentioned in A175646.

Examples

			Equals 1.5217315350757058188419... = 0.92003856361849186... / A073010 .
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.1, p. 86.

Crossrefs

Cf. A053182.

Programs

  • Maple
    a073010 := evalf(Pi/3/sqrt(3)) ;
    Cm3n0s2 := 1-1/(3-1)^2 ;
    Cm3n1s3 := 0.867512181712394919089076584762888869720269526863 ;
    Zm3n2s2 := 1.4140643908921476375655018190798293799076950693931 ;
    Cm3n0s2*Cm3n1s3*Zm3n2s2/a073010 ;
  • 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]);
    Zs[m_, n_, s_] := (w = 2; sumz = 0; difz = 1; While[Abs[difz] > 10^(-digits - 5), difz = (s^w - s) * P[m, n, w]/w; sumz = sumz + difz; w++]; Exp[-sumz]);
    $MaxExtraPrecision = 1000; digits = 121; RealDigits[Chop[N[3^(5/2)*Zs[3, 1, 3]*Z[3, 2, 2]/(4*Pi), digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 16 2021 *)

Extensions

More terms from Vaclav Kotesovec, Jan 16 2021

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

A340565 Decimal expansion of the Product_{lesser twin primes p == 5 (mod 6)} 1/(1 - 1/p^2).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 11 2021

Keywords

Comments

Lesser twin primes A001359 (with the exception of the first prime, 3) are congruent to 5 mod 6: this constant is smaller than A340576.
By extrapolating method most probably the next two decimal digits are 1.056932291(46).
The known high-precision algorithms for Euler products are based on the Dirichlet L function and the Moebius inversion formula (see Mathematica procedure of Jean-François Alcover in A175646).
The constant is between 1.056932291453... and 1.056932291494. - R. J. Mathar, Feb 14 2025

Examples

			1.0569322914...
		

Crossrefs

Extensions

One more digit confirmed by a bracketing of partial products - R. J. Mathar, Feb 14 2025
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