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 1-3 of 3 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

Links

Crossrefs

Cf. A001620, A004709, A056671, A365488, A365499.
Cf. A034444, A158522, A190867, A286324, A307428, A368172, A368248, A368885, A369310.

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)

A362852 The number of divisors of n that are both bi-unitary and exponential.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, May 05 2023

Keywords

Comments

First differs from A061704 at n = 128, and from A304327 and abs(A307428) at n = 64.
If e > 0 is the exponent of the highest power of p dividing n (where p is a prime), then for each divisor d of n that is both a bi-unitary and an exponential divisor, the exponent of the highest power of p dividing d is a number k such that k | e but k != e/2.
The least term that is higher than 2 is a(64) = 3.
This sequence is unbounded. E.g., a(2^(2^prime(n))) = prime(n).

Examples

			a(8) = 2 since 8 has 2 divisors that are both bi-unitary and exponential: 2 and 8.

Links

Crossrefs

Cf. A000005, A004709, A049419, A188999, A222266, A286324, A322791, A359411, A362853 (indices of records), A362854.
Cf. A061704, A304327, A307428.

Programs

  • Mathematica
    f[p_, e_] := DivisorSigma[0, e] - If[OddQ[e], 0, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, numdiv(f[i, 2]) - !(f[i, 2] % 2));}

Formula

Multiplicative with a(p^e) = d(e) if e is odd, and d(e)-1 if e is even, where d(k) is the number of divisors of k (A000005).
a(n) = 1 if and only if n is cubefree (A004709).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + Sum_{k>=1} (d(k)+(k mod 2)-1)/p^k) = 1.1951330849... .

A368248 The number of unitary divisors of the cubefull part of n (A360540).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Dec 19 2023

Keywords

Comments

First differs from A061704 and A362852 at n = 64, and from A304327 at n = 72.
Also, the number of squarefree divisors of the cubefull part of n.
Also, the number of cubes of squarefree numbers (A062838) that divide n.
The number of unitary divisors of n that are cubefull numbers (A036966). - Amiram Eldar, Jun 19 2025

Links

Crossrefs

Cf. A004709, A034444, A036966, A062838, A157289, A307428, A323308, A360540, A366145.
Cf. A061704, A304327, A362852.

Programs

  • Mathematica
    f[p_, e_] := If[e > 2, 2, 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> if(x < 3, 1, 2), factor(n)[, 2]));

Formula

a(n) = A034444(A360540(n)).
a(n) = abs(A307428(n)).
Multiplicative with a(p) = 1 for e <= 2, and 2 for e >= 3.
a(n) >= 1, with equality if and only if n is cubefree (A004709).
a(n) <= A034444(n), with equality if and only if n is cubefull (A036966).
Dirichlet g.f.: zeta(s)*zeta(3*s)/zeta(6*s).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(3)/zeta(6) = 1.181564... (A157289).
In general, the asymptotic mean of the number of unitary divisors of the k-full part of n is zeta(k)/zeta(2*k).
Showing 1-3 of 3 results.

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