A365490 The number of divisors of the largest 4th power dividing n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Sep 05 2023

Keywords

• nonn
• easy
• mult

Comments

The number of divisors of the 4th root of the largest 4th power dividing n, A053164(n), is A063775(n).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000583, A046100, A008835, A053164, A063775.

Programs

Mathematica

f[p_, e_] := 4*Floor[e/4] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]

PARI

a(n) = vecprod(apply(x -> 4*(x\4) + 1, factor(n)[, 2]));

Python

from math import prod
from sympy import factorint
def A365490(n): return prod(e&-4|1 for e in factorint(n).values()) # Chai Wah Wu, Aug 08 2024

Formula

a(n) = A000005(A008835(n)).

Multiplicative with a(p^e) = 4*floor(e/4) + 1.

a(n) = 1 if and only if n is a biquadratefree number (A046100).

a(n) <= A000005(n) with equality if and only if n is a fourth power (A000583).

Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 3/p^(4*s)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 3/p^4) = 1.3414590511076... . In general, the asymptotic mean of the number of divisors of the largest k-th power dividing n is zeta(k) * Product_{p prime} (1 + (k-1)/p^k).