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 1-5 of 5 results.

A084218 a(n) = sigma_4(n^2)/sigma_2(n^2).

Original entry on oeis.org

1, 13, 73, 205, 601, 949, 2353, 3277, 5905, 7813, 14521, 14965, 28393, 30589, 43873, 52429, 83233, 76765, 129961, 123205, 171769, 188773, 279313, 239221, 375601, 369109, 478297, 482365, 706441, 570349, 922561, 838861, 1060033, 1082029
Offset: 1

Views

Author

Benoit Cloitre, Jun 21 2003

Keywords

Links

Crossrefs

Cf. A001157, A001159, A002117, A007434, A013663, A065827.
Cf. A068963, A372962, A372963, A373007.
Cf. A057660, A084220, A372966, A373105.

Programs

  • Maple
    with(numtheory): a:=n->sigma[4](n^2)/sigma[2](n^2): seq(a(n),n=1..40); # Muniru A Asiru, Oct 09 2018
  • Mathematica
    Table[DivisorSigma[4, n^2]/DivisorSigma[2, n^2], {n, 1, 50}] (* G. C. Greubel, Oct 08 2018 *)
f[p_, e_] := (p^(4*e + 2) + 1)/(p^2 + 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 35] (* Amiram Eldar, Sep 13 2020 *)
  • PARI
    a(n)=sumdiv(n^2,d,d^4)/sumdiv(n^2,d,d^2)
  • PARI
    a(n) = sigma(n^2, 4)/sigma(n^2, 2); \\ Michel Marcus, Oct 09 2018

Formula

Multiplicative with a(p^e) = (p^(4*e + 2) + 1)/(p^2 + 1). - Amiram Eldar, Sep 13 2020
Sum_{k>=1} 1/a(k) = 1.09957644430375183822287768590764825667080036406680891521221069625517483696... - Vaclav Kotesovec, Sep 24 2020
Sum_{k=1..n} a(k) ~ c * n^5, where c = zeta(5)/(5*zeta(3)) = 0.172525... . - Amiram Eldar, Oct 30 2022
From Peter Bala, Jan 18 2024: (Start)
a(n) = Sum_{d divides n} J_2(d^2) = Sum_{d divides n} d^2 * J_2(d), where the Jordan totient function J_2(n) = A007434(n).
a(n) = Sum_{1 <= j, k <= n} ( n/gcd(j, k, n) )^2.
Dirichlet g.f.: zeta(s) * zeta(s-4) / zeta(s-2) [Corrected by Michael Shamos, May 18 2025]. (End)
a(n) = Sum_{d|n} mu(n/d) * (n/d)^2 * sigma_4(d). - Seiichi Manyama, May 18 2024

A156733 Euler transform of n*A065958(n).

Original entry on oeis.org

1, 1, 11, 41, 176, 606, 2391, 8091, 28636, 95056, 316048, 1014240, 3237325, 10082015, 31109500, 94352346, 283209381, 838650191, 2458835711, 7127912979, 20471486368, 58224189612, 164181018330, 458982667630, 1273039111210, 3503609456548, 9572771822745, 25971150308985
Offset: 0

Views

Author

Paul D. Hanna and Vladeta Jovovic, Feb 14 2009

Keywords

Comments

Compare to the g.f. of planar partitions (A000219): exp( Sum_{n>=1} sigma(n,2)*x^n/n ) = Product_{n>=1} 1/(1-x^n)^n.

Links

Crossrefs

Cf. A001157, A156303, A065827, A301978, A301980.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*numtheory[sigma][2](j^2), j=1..n)/n)
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 24 2016
  • Mathematica
    a[0] = 1;
a[n_] := a[n] = (1/n) Sum[DivisorSigma[2, k^2] a[n-k], {k, 1, n}];
a /@ Range[0, 30] (* Jean-François Alcover, Nov 03 2020 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,sigma(m^2,2)*x^m/m)+x*O(x^n)),n)}
for(n=0,21,print1(a(n),", "))

Formula

a(n) = (1/n)*Sum_{k=1..n} sigma_2(k^2)*a(n-k) for n>0, with a(0) = 1.
G.f.: exp( Sum_{n>=1} A065827(n)*x^n/n ), where A065827(n) = sigma_2(n^2) is the sum of squares of the divisors of n^2. - Paul D. Hanna, Aug 09 2012

A060367 Average order of an element in a cyclic group of order n rounded down.

Original entry on oeis.org

1, 1, 2, 2, 4, 3, 6, 5, 6, 6, 10, 6, 12, 9, 9, 10, 16, 10, 18, 11, 14, 15, 22, 12, 20, 18, 20, 16, 28, 14, 30, 21, 23, 24, 25, 18, 36, 27, 28, 22, 40, 21, 42, 27, 28, 33, 46, 24, 42, 31, 37, 33, 52, 30, 42, 33, 42, 42, 58, 26, 60, 45, 41, 42, 50, 35, 66, 44, 51
Offset: 1

Views

Author

Avi Peretz (njk(AT)netvision.net.il), Apr 01 2001

Keywords

Links

Crossrefs

Cf. A000010, A000203, A001157, A018804, A057660, A065764, A065827, A253905.

Programs

  • Maple
    seq(floor(numtheory:-sigma[2](n^2)/numtheory:-sigma(n^2)/n), n=1..1000); # Robert Israel, Mar 24 2015
  • Mathematica
    f[n_] := Block[{i, j, k}, Reap@ For[j = 1, j <= n, j++, Sow[Floor[Sum[1/GCD[j, k], {k, 1, j}]]]]] // Flatten // Rest; f@ 49 (* Michael De Vlieger, Mar 24 2015 *)
a[n_] := Floor[DivisorSigma[2, n^2]/DivisorSigma[1, n^2]/n]; Array[a, 100] (* Amiram Eldar, Jul 25 2025 *)
  • PARI
    a(n) = {my(f = factor(n^2)); floor(sigma(f, 2)/(n * sigma(f)));} \\ Amiram Eldar, Jul 25 2025
  • Sage
    [floor(sum([1/gcd(n,k) for k in range(1,n+1)])) for n in range(1,50)] # Danny Rorabaugh, Mar 24 2015

Formula

Sequence A057660 gives the sum of the orders of the elements in a cyclic group with n elements so a(n) = floor(A057660(n) / n) = floor(Sum_{k=1..n} 1/GCD(n, k)) = floor(Sum of 1/d times phi(n/d)) for all divisors d of n, where phi is Euler's phi function. This sum may also be expressed as the product of (p^(2*e(p)+1)+1)/((p+1)*p^e(p)) over all prime divisors p of n where the canonical factorization of n is the product of p^e(p), the e(p) being the exponents of the power of p in the factorization.
From Amiram Eldar, Jul 25 2025: (Start)
a(n) = floor(sigma_2(n^2)/(n*sigma(n))) = floor(A001157(n^2)/(n*A000203(n^2))) = floor(A065827(n)/(n*A065764(n))).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(3)/zeta(2) (A253905). (End)

Extensions

Offset corrected and terms a(18)-a(50) added by Danny Rorabaugh, Mar 24 2015

A258331 Sum of the cubes of the divisors of n^3.

Original entry on oeis.org

1, 585, 20440, 299593, 1968876, 11957400, 40471600, 153391689, 402321277, 1151792460, 2359720584, 6123680920, 10609328380, 23675886000, 40243825440, 78536544841, 118612018980, 235357947045, 322734750520, 589861467468, 827239504000, 1380436541640
Offset: 1

Views

Author

Wesley Ivan Hurt, May 26 2015

Keywords

Examples

			For n=2, the divisors of 2^3 = 8 are 1, 2, 4 and 8. The sum of the cubes of these divisors is 1^3+2^3+4^3+8^3 = 585, therefore a(2) = 585.

Links

Crossrefs

Cf. A000578, A001158, A013668, A065827.

Programs

  • Magma
    [DivisorSigma(3, n^3): n in [1..50]]; // Vincenzo Librandi, May 27 2015
  • Maple
    with(numtheory): A258331:=n->sigma[3](n^3): seq(A258331(n), n=1..50);
  • Mathematica
    Table[DivisorSigma[3, n^3], {n, 50}]
  • PARI
    a(n)=sigma(n^3,3) \\ Charles R Greathouse IV, May 27 2015
  • Python
    from math import prod
from sympy import factorint
def A258331(n): return prod((p**((3*e+1)*3)-1)//(p**3-1) for p,e in factorint(n).items()) # Chai Wah Wu, Oct 25 2023
  • Sage
    [sigma(n^3, 3) for n in (1..50)] # Bruno Berselli, May 27 2015

Formula

a(n) = sigma_3(n^3) = A001158(A000578(n)).
From Amiram Eldar, Nov 05 2022: (Start)
Multiplicative with a(p^e) = (p^(9*e + 3) - 1)/(p^3 - 1).
Sum_{k=1..n} a(k) ~ c * n^10, where c = (zeta(10)/10) * Product_{p prime} (1 + 1/p^4 + 1/p^7) = 0.1087440273... . (End)

A307705 Expansion of Product_{k>=1} 1/(1 - x^k)^(k-phi(k)), where phi() is the Euler totient function (A000010).

Original entry on oeis.org

1, 0, 1, 1, 3, 2, 8, 5, 16, 15, 34, 30, 75, 66, 144, 150, 285, 292, 566, 585, 1062, 1170, 1988, 2205, 3729, 4159, 6755, 7785, 12214, 14147, 21957, 25560, 38709, 45839, 67884, 80747, 118332, 141244, 203614, 245330, 348396, 420971, 592439, 717659, 998248, 1215439, 1672544, 2040210, 2786687
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 22 2019

Keywords

Comments

Euler transform of A051953.

Crossrefs

Cf. A000010, A001157, A051953, A053650, A057660, A061255, A065764, A065827.

Programs

  • Mathematica
    nmax = 48; CoefficientList[Series[Product[1/(1 - x^k)^(k - EulerPhi[k]), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 48; CoefficientList[Series[Exp[Sum[(DivisorSigma[2, k] - DivisorSigma[2, k^2]/DivisorSigma[1, k^2]) x^k/k, {k, 1, nmax}]], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 - EulerPhi[d^2], {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 48}]

Formula

G.f.: exp(Sum_{k>=1} (sigma_2(k) - sigma_2(k^2)/sigma_1(k^2)) * x^k/k).
G.f.: exp(Sum_{k>=1} ( Sum_{d|k} cototient(d^2) ) * x^k/k).
a(n) ~ exp(3*((Pi^2 - 6)*Zeta(3))^(1/3) * n^(2/3) / (2*Pi)^(2/3) + 1/4) * ((Pi^2 - 6)*Zeta(3))^(1/4) / (A^3 * 2^(1/12) * 3^(1/2) * Pi^(5/6) * n^(3/4)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 06 2019
Showing 1-5 of 5 results.

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