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 1-8 of 8 results.

A082695 Decimal expansion of zeta(2)*zeta(3)/zeta(6).

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Apr 12 2003

Keywords

Comments

Equals the Dirichlet zeta-function Sum_{n>=1} A001615(n)/n^s at s=3. - R. J. Mathar, Apr 02 2011
Dressler shows that this is the average value of A014197, that is, the number of values m such that phi(m) <= n is asymptotically n times this constant. Erdős had shown earlier that this limit exists. - Charles R Greathouse IV, Nov 26 2013
From Stanislav Sykora, Nov 14 2014: (Start)
Equals lim_{n->infinity} (Sum_{k=1..n} k/phi(k))/n, i.e., the limit mean value of k/phi(k), where phi(k) is Euler's totient function.
Also equals lim_{n->infinity} (Sum_{k=1..n} 1/phi(k))/log(n).
Proofs are trivial using the formulas for Sum_{k=1..n} k/phi(k) and Sum_{k=1..n} 1/phi(k) listed in the Wikipedia link.
For the limit mean value of phi(k)/k, see A059956. (End)
The asymptotic mean of A005361. - Amiram Eldar, Apr 13 2020

Examples

			1.94359643682075920505707036257476343718785850176780571602663568890 ...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.7, p. 116.
  • Joe Roberts, Lure of the Integers, Mathematical Association of America, 1992. See p. 74.

Crossrefs

Programs

  • Mathematica
    First@RealDigits[ Zeta[2]*Zeta[3]/Zeta[6], 10, 100]
    RealDigits[ 315 Zeta[3]/(2 Pi^4), 10, 111][[1]] (* Robert G. Wilson v, Aug 11 2014 *)
  • PARI
    zeta(3)*315/2/Pi^4

Formula

Decimal expansion of Product_{p prime} (1+1/p/(p-1)) = zeta(2)*zeta(3)/zeta(6) = 1.94359643682075920505707...
The sum of the reciprocals of the powerful numbers, A001694. - T. D. Noe, May 03 2006
Equals A013661 * A002117 / A013664 = 1 / A068468 = zeta(3) * 315/(2*Pi^4) = zeta(3) * A157292.
Equals Sum_{k>=1} mu(k)^2/(k*phi(k)) (the sum of reciprocals of the squarefree numbers multiplied by their Euler totient function values, A000010). - Amiram Eldar, Aug 18 2020

Extensions

New definition from Eric W. Weisstein, May 04 2006

A365298 a(n) is the smallest number k such that k*n is a cubefull exponentially odd number (A335988).

Original entry on oeis.org

1, 4, 9, 2, 25, 36, 49, 1, 3, 100, 121, 18, 169, 196, 225, 2, 289, 12, 361, 50, 441, 484, 529, 9, 5, 676, 1, 98, 841, 900, 961, 1, 1089, 1156, 1225, 6, 1369, 1444, 1521, 25, 1681, 1764, 1849, 242, 75, 2116, 2209, 18, 7, 20, 2601, 338, 2809, 4, 3025, 49, 3249, 3364
Offset: 1

Views

Author

Amiram Eldar, Aug 31 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^If[OddQ[e], Max[e, 3] - e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^if(f[i, 2]%2, max(f[i, 2], 3) - f[i,2], 1))};

Formula

Multiplicative with a(p) = p^2, a(p^e) = p if e is even, and a(p^e) = 1 is e is odd and > 1.
a(n) = A356192(n)/n.
a(n) = 1 if and only if n is in A335988.
Dirichlet g.f.: zeta(2*s) * Product_{p prime} (1 + 1/p^(3*s) - 1/p^(3*s-2) - 1/p^(2*s) + 1/p^(2*s-1) + 1/p^(s-2)).
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = (2*Pi^4/315) * Product_{p prime} (1 - p^2 - p^3 + p^4 + p^8 + p^9)/(p^8*(p+1)) = 0.207915752545... .

A362984 Decimal expansion of the asymptotic mean of the abundancy index of the powerful numbers (A001694).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, May 12 2023

Keywords

Comments

The abundancy index of a positive integer k is A000203(k)/k = A017665(k)/A017666(k).
The asymptotic mean of the abundancy index over all the positive integers is lim_{m->oo} (1/m) * Sum_{k=1..m} A000203(k)/k = Pi^2/6 = zeta(2) = 1.644934... (A013661).

Examples

			2.14968690301526765128219042105109416145987653275100999873...
		

Crossrefs

Similar constants (the asymptotic mean of the abundancy index of other sequences): A013661 (all positive integers), A082020 (cubefree), A111003 (odd), A157292 (5-free), A157294 (7-free), A157296 (9-free), A240976 (squares), A245058 (even), A306633 (squarefree), A362985 (cubefull).

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{2, -3, 4, -6, 7, -7, 7, -6, 5, -3, 2, -1}, {0, 0, 0, 4, 5, 6, 0, -12, -9, -5, 0, 22}, m]; RealDigits[(2^4 + 2^2 + 2^(3/2) - 1)/(2^4 - 2)*(3^4 + 3^2 + 3^(3/2) - 1)/(3^4 - 3) * Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/2] - 1/2^(n/2) - 1/3^(n/2))/n, {n, 4, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    prodeulerrat((p^8 + p^4 + p^3 - 1)/(p^8 - p^2), 1/2)

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A180114(k)/A001694(k).
Equals Product_{p prime} (p^4 + p^2 + p^(3/2) - 1)/(p^4 - p) = Product_{p prime} (1 + (p^2 + p^(3/2) + p - 1)/(p^4 - p)) (Jakimczuk and Lalín, 2022).

A362985 Decimal expansion of the asymptotic mean of the abundancy index of the cubefull numbers (A036966).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, May 12 2023

Keywords

Examples

			2.48217919642235952546167643674687698536368940971930468354...
		

Crossrefs

Similar constants (the asymptotic mean of the abundancy index of other sequences): A013661 (all positive integers), A082020 (cubefree), A111003 (odd), A157292 (5-free), A157294 (7-free), A157296 (9-free), A245058 (even), A240976 (squares), A306633 (squarefree), A362984 (powerful).

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{2, -1, -2, 3, -2, -1, 3, -2, -2, 3, -1, -2, 3, -1, -1, 1}, {0, 0, 0, -4, 0, 6, 7, 4, 9, 0, -11, -22, -26, -21, -15, 20}, m]; RealDigits[((2^5 + 2^(10/3) + 2^3 + 2^(8/3) - 1)/(2^(10/3)*(2^(5/3) + 2^(1/3) + 1)))*((3^5 + 3^(10/3) + 3^3 + 3^(8/3) - 1)/(3^(10/3)*(3^(5/3) + 3^(1/3) + 1))) * Zeta[4/3] * Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/3] - 1/2^(n/3) - 1/3^(n/3))/n, {n, 4, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    zeta(4/3) * prodeulerrat((p^15 + p^10 + p^9 + p^8 - 1)/(p^10 * (p^5 + p + 1)), 1/3)

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A362986(k)/A036966(k).
Equals zeta(4/3) * Product_{p prime} ((p^5 + p^(10/3) + p^3 + p^(8/3) - 1)/(p^(10/3) * (p^(5/3) + p^(1/3) + 1))).

A372937 a(n) = Sum_{1 <= x_1, x_2, x_3, x_4 <= n} gcd(x_1, x_2, x_3, x_4, n)^5.

Original entry on oeis.org

1, 47, 323, 1744, 3749, 15181, 19207, 59648, 84969, 176203, 175691, 563312, 399853, 902729, 1210927, 1970176, 1503377, 3993543, 2606419, 6538256, 6203861, 8257477, 6716183, 19266304, 12105625, 18793091, 21172347, 33497008, 21218429, 56913569, 29552671, 64028672
Offset: 1

Views

Author

Seiichi Manyama, May 17 2024

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(4*e-4)*(p^e*(p^5-1) - (p^4-1))/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 21 2024 *)
  • PARI
    a(n) = sumdiv(n, d, moebius(n/d)*d^4*sigma(d));

Formula

a(n) = Sum_{1 <= x_1, x_2, x_3, x_4, x_5 <= n} gcd(x_1, x_2, x_3, x_4, x_5, n)^4.
a(n) = Sum_{d|n} mu(n/d) * d^4 * sigma(d), where mu is the Moebius function A008683.
From Amiram Eldar, May 21 2024: (Start)
Multiplicative with a(p^e) = p^(4*e-4)*(p^e*(p^5-1) - (p^4-1))/(p-1).
Dirichlet g.f.: zeta(s-4)*zeta(s-5)/zeta(s).
Sum_{k=1..n} a(k) ~ c * n^6 / 6, c = zeta(2)/zeta(6) = 315/(2*Pi^4) = 1.616892... (A157292). (End)
Mobius transformation of A280022. - R. J. Mathar, Jul 14 2025

A372965 a(n) = Sum_{k = 1..n} ( n/gcd(k, n) )^4.

Original entry on oeis.org

1, 17, 163, 529, 2501, 2771, 14407, 16913, 39529, 42517, 146411, 86227, 342733, 244919, 407663, 541201, 1336337, 671993, 2345779, 1323029, 2348341, 2488987, 6156503, 2756819, 7815001, 5826461, 9605467, 7621303, 19803869, 6930271, 27705631, 17318417, 23864993
Offset: 1

Views

Author

Seiichi Manyama, May 18 2024

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(5*e+5) - p^(5*e+4) + p^4 - 1)/(p^5-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 21 2024 *)
  • PARI
    a(n) = sumdiv(n, d, moebius(n/d)*(n/d)^4*sigma(d, 5));
    
  • PARI
    a(n) = sumdiv(n, d, eulerphi(d^5));

Formula

a(n) = Sum_{d|n} mu(n/d) * (n/d)^4 * sigma_5(d).
a(n) = Sum_{d|n} d^(5-m) * phi(d^m) for m > 0.
G.f.: Sum_{k>=1} k^(5-m) * phi(k^m) * x^k/(1 - x^k) for m > 0.
From Amiram Eldar, May 21 2024: (Start)
Multiplicative with a(p^e) = (p^(5*e+5) - p^(5*e+4) + p^4 - 1)/(p^5-1).
Dirichlet g.f.: zeta(s)*zeta(s-5)/zeta(s-4).
Sum_{k=1..n} a(k) ~ c * n^6 / 6, where c = zeta(6)/zeta(2) = 2*Pi^4/315 = 0.6184704192... (1/A157292). (End)

A365170 The sum of divisors d of n such that gcd(d, n/d) is squarefree.

Original entry on oeis.org

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 27, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 51, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 108, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 99, 84, 144
Offset: 1

Views

Author

Amiram Eldar, Aug 25 2023

Keywords

Comments

The number of these divisors is A252505(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Switch[e, 1, 1 + p, 2, 1 + p + p^2, , (1 + p)*(1 + p^(e - 1))]; a[1] = 1; a[n] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), p , e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(e == 1, 1 + p, if(e == 2, 1 + p + p^2, (1 + p)*(1 + p^(e - 1)))));}

Formula

Multiplicative with a(p) = 1 + p, a(p^2) = 1 + p + p^2, and a(p^e) = (1 + p)*(1 + p^(e - 1)) if e >= 3.
a(n) >= A034448(n), with equality if and only if n is squarefree number (A005117).
a(n) <= A000203(n), with equality if and only if n is biquadratefree number (A046100).
Sum_{k=1..n} a(k) ~ c * n^2, where c = 315/(4*Pi^4) = A157292 / 2 = 0.808446... .

A373702 Decimal expansion of (2 - zeta(2))*zeta(2)*zeta(3)/zeta(6).

Original entry on oeis.org

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

Views

Author

Stefano Spezia, Jun 13 2024

Keywords

Examples

			0.69010488251022497818773002567827532644066623131...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[(2-Zeta[2])Zeta[2]Zeta[3]/Zeta[6],10,100][[1]]
Showing 1-8 of 8 results.