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 1-2 of 2 results.

A365498 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 1, 2, 4, 2, 4, 2, 4, 4, 1, 2, 4, 2, 4, 4, 4, 2, 2, 2, 4, 1, 4, 2, 8, 2, 1, 4, 4, 4, 4, 2, 4, 4, 2, 2, 8, 2, 4, 4, 4, 2, 2, 2, 4, 4, 4, 2, 2, 4, 2, 4, 4, 2, 8, 2, 4, 4, 1, 4, 8, 2, 4, 4, 8, 2, 2, 2, 4, 4, 4, 4, 8, 2, 2, 1, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Vaclav Kotesovec, Sep 06 2023

Keywords

Comments

The number of unitary divisors of n that are cubefree numbers (A004709). - Amiram Eldar, Sep 06 2023

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e <= 2, 2, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2023 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * (1 + X - X^3))[n], ", "))

Formula

Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,
f'(1) = f(1) * Sum_{p prime} (-4 + 3*p + 2*p^2) * log(p) / (1 - p - p^2 + p^4) = f(1) * 1.4525924794451595590371439593828547341482465114411929136723476679...
and gamma is the Euler-Mascheroni constant A001620.
Multiplicative with a(p^e) = 2 if e <= 2, and 1 otherwise. - Amiram Eldar, Sep 06 2023
From Vaclav Kotesovec, Jan 27 2025: (Start)
Following formulas have been conjectured for this sequence by Sequence Machine, with each one giving the first 1000000 terms correctly:
a(n) = A056671(n) * A368885(n).
a(n) = A034444(n) / A368248(n).
a(n) = A158522(n) / A307428(n).
a(n) = A369310(n) / A190867(n).
a(n) = A286324(n) / A368172(n). (End)

A368884 The largest unitary divisor of n that is the square of a squarefree number (A062503).

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 1, 25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 49, 25, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 9, 1, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 25, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jan 09 2024

Keywords

Comments

The number of these divisors is A368885(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 2, p^2, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 2, f[i,1]^f[i,2], 1));}
    
  • Python
    from math import prod
    from sympy import factorint
    def A368884(n): return prod(p**e for p, e in factorint(n).items() if e==2) # Chai Wah Wu, Jan 09 2024

Formula

Multiplicative with a(p^e) = p^2 if e = 2, and 1 otherwise.
a(n) = n / A368886(n).
a(n) >= 1, with equality if and only if n is in A337050.
a(n) <= n, with equality if and only if n is in A062503.
Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(2*s-2) - 1/p^(2*s) - 1/p^(3*s-2) + 1/p^(3*s)).
From Vaclav Kotesovec, Jan 09 2024: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (p^(2*s) - p^2) * (1 + (p^s - 1) * (p^2 + p^s + p^(2*s))) / p^(5*s).
Sum_{k=1..n} a(k) ~ c * zeta(3/2) * n^(3/2)/3, where c = Product_{p prime} (1 - 1/p^(11/2) + 1/p^(9/2) + 1/p^4 + 1/p^(7/2) - 1/p^3 - 1/p^(5/2) - 1/p^2) = 0.45021226373776124069206513259105992151602618717147857709105849... (End)
Showing 1-2 of 2 results.