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 11-19 of 19 results.

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

A340665 Decimal expansion of Product_{primes p == 3 (mod 5)} p^2/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 15 2021

Keywords

Examples

			1.135764878668921626868643009472082289511936413...
		

Crossrefs

Programs

  • Mathematica
    (* Using Vaclav Kotesovec's function Z from A301430. *)
    $MaxExtraPrecision = 100; digits = 50; (* Adjust as needed. *)
    digitize[c_] := RealDigits[Chop[N[c, digits+10]], 10, digits][[1]];
    digitize[Z[5, 3, 2]]

Formula

Equals Sum_{k>=1} 1/A004617(k)^2. - Amiram Eldar, Jan 24 2021

A340794 Decimal expansion of Product_{primes p == 2 (mod 5)} p^2/(p^2-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Jan 21 2021

Keywords

Examples

			1.36857205387664908586076389048310999017020782888589520500850402955633118881...
		

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, 2, 2]]

Formula

I = Product_{primes p == 0 (mod 5)} p^2/(p^2-1) = 25/24.
J = Product_{primes p == 1 (mod 5)} p^2/(p^2-1) = A340004.
K = Product_{primes p == 2 (mod 5)} p^2/(p^2-1) = this constant.
L = Product_{primes p == 3 (mod 5)} p^2/(p^2-1) = A340665.
M = Product_{primes p == 4 (mod 5)} p^2/(p^2-1) = A340127.
I*J*K*L*M = Pi^2/6 = zeta(2).
J*K*L*M = 4*Pi^2/25.
M = (Pi/2)*C(5,4)^(-2)*exp(-gamma/2)*sqrt(3/log(2+sqrt(5))), where gamma is the Euler-Mascheroni constant A001620 and C(5,4) is the Mertens constant = 1.29936454791497798816084...
Equals Sum_{k>=1} 1/A004616(k)^2. - Amiram Eldar, Jan 24 2021

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

A369566 Powerful numbers whose prime factors are all of the form 3*k + 2.

Original entry on oeis.org

1, 4, 8, 16, 25, 32, 64, 100, 121, 125, 128, 200, 256, 289, 400, 484, 500, 512, 529, 625, 800, 841, 968, 1000, 1024, 1156, 1331, 1600, 1681, 1936, 2000, 2048, 2116, 2209, 2312, 2500, 2809, 3025, 3125, 3200, 3364, 3481, 3872, 4000, 4096, 4232, 4624, 4913, 5000
Offset: 1

Views

Author

Amiram Eldar, Jan 26 2024

Keywords

Comments

Closed under multiplication.

Crossrefs

Intersection of A001694 and A004612.
Similar sequence: A352492, A369563, A369564, A369565.

Programs

  • Mathematica
    q[n_] := n == 1 || AllTrue[FactorInteger[n], Mod[First[#], 3] == 2 && Last[#] > 1 &]; Select[Range[5000], q]
  • PARI
    is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 1]%3 != 2 || f[i, 2] == 1, return(0))); 1;}

Formula

Sum_{n>=1} 1/a(n) = Product_{primes p == 2 (mod 3)} (1 + 1/(p*(p-1))) = (9/8) * A333240 * A334479 = 1.6053538210...

A333239 Decimal expansion of lim_{n->infinity} (Product_{k=1..n} p(k)^2 / Product_{k=1..n} (p(k)^2 - 1)) where p(k) = A167135(k) are the primes with p mod 12 != 1.

Original entry on oeis.org

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

Views

Author

Peter Luschny, May 13 2020

Keywords

Examples

			1.63250540290251307901036625893553760497201130044326292653142032442675746272540...
		

Crossrefs

Formula

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].

A234645 Sum of the divisors of n^3+1.

Original entry on oeis.org

1, 3, 13, 56, 84, 312, 256, 660, 800, 1332, 1344, 3458, 2240, 3792, 4836, 6572, 4356, 13440, 6160, 16800, 13312, 15192, 11136, 35685, 19840, 25284, 30976, 42560, 22740, 63648, 30464, 71820, 51792, 65664, 53952, 111440, 52136, 84480, 99008, 133560, 75264
Offset: 0

Views

Author

Vincenzo Librandi, Jan 01 2014

Keywords

Examples

			a(4) = 84 because 4^3+1 = 65 and the sum of the 4 divisors {1, 5, 13, 65} is 84.
		

Crossrefs

Programs

  • Magma
    [SumOfDivisors(n^3+1): n in [0..50]];
    
  • Mathematica
    Table[Total[Divisors[n^3 + 1]], {n, 0, 50}]
    DivisorSigma[1,Range[0,40]^3+1] (* Harvey P. Dale, Jul 27 2021 *)
  • PARI
    a(n) = sigma(n^3+1); \\ Michel Marcus, Jun 19 2015

Formula

a(n) = A000203(A001093(n)). - Michel Marcus, Jun 19 2015
Sum_{k=1..n} a(k) = c * n^4 + O((n*log(n))^3), where c = (83/288) * Product_{primes p == 1 (mod 3)} ((p^2+2)/(p^2-1)) * Product_{primes p == 2 (mod 3)} (p^2/(p^2-1)) = 0.449926279... . - Amiram Eldar, Dec 09 2024

A234860 Sum of the divisors of n^3 - 1.

Original entry on oeis.org

8, 42, 104, 224, 264, 780, 592, 1680, 1520, 2880, 1896, 5642, 2968, 5808, 8736, 9548, 7200, 15360, 8440, 19488, 19032, 23040, 14448, 49920, 23560, 31836, 32912, 53312, 34200, 77688, 38912, 70812, 62088, 74088, 67584, 152320, 56392, 107520, 99736
Offset: 2

Views

Author

Vincenzo Librandi, Jan 01 2014

Keywords

Crossrefs

Programs

  • Magma
    [SumOfDivisors(n^3-1): n in [2..50]];
    
  • Mathematica
    Table[Total[Divisors[n^3 - 1]], {n, 2, 50}]
  • PARI
    a(n) = sigma(n^3-1); \\ Amiram Eldar, Dec 09 2024

Formula

From Amiram Eldar, Dec 09 2024: (Start)
a(n) = A000203(A068601(n)).
Sum_{k=2..n} a(k) = c * n^4 + O((n*log(n))^3), where c = (83/288) * Product_{primes p == 1 (mod 3)} ((p^2+2)/(p^2-1)) * Product_{primes p == 2 (mod 3)} (p^2/(p^2-1)) = 0.4499262799... . (End)

Extensions

Offset corrected by Amiram Eldar, Dec 09 2024
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