A384554 The sum of the infinitary divisors of n that are cubefree.

1, 3, 4, 5, 6, 12, 8, 7, 10, 18, 12, 20, 14, 24, 24, 1, 18, 30, 20, 30, 32, 36, 24, 28, 26, 42, 13, 40, 30, 72, 32, 3, 48, 54, 48, 50, 38, 60, 56, 42, 42, 96, 44, 60, 60, 72, 48, 4, 50, 78, 72, 70, 54, 39, 72, 56, 80, 90, 60, 120, 62, 96, 80, 5, 84, 144, 68, 90

Offset: 1

Author

Amiram Eldar, Jun 03 2025

Keywords

• nonn
• easy
• mult

Comments

The number of these divisors is A368883(n), and the largest of them is A384555(n).

The sum of the infinitary divisors of n that are squarefree (A005117) is A367991(n).

Links

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

Crossrefs

Cf. A000203, A005117, A004709, A013666, A049417, A073185, A077609, A367991, A368883, A384555.

Programs

Mathematica

f[p_, e_] := Switch[Mod[e, 4], 0, 1, 1, p+1, 2, p^2+1, 3, p^2+p+1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; [1, p+1, p^2+1, p^2+p+1][e%4+1]);}

Python

from math import prod
from sympy import factorint
def A384554(n): return prod((1,p+1,p**2+1,p*(p+1)+1)[e&3] for p,e in factorint(n).items()) # Chai Wah Wu, Jun 03 2025

Formula

Multiplicative with a(p^e) = 1 if e == 0 (mod 4), p + 1 if e == 1 (mod 4), p^2 + 1 if e == 2 (mod 4), and p^2 + p + 1 if e == 3 (mod 4).

a(n) <= A000203(n), with equality if and only if n is squarefree (A005117).

a(n) <= A049417(n), with equality if and only if n is cubefree (A004709).

Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^s + 1/p^(2*s-2) + 1/p^(2*s) + 1/p^(3*s-1) + 1/p^(3*s-2) + 1/p^(3*s)).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(8) * Product_{p prime} (1 + 1/p^2 - 2/p^3 + 2/p^4 - 1/p^5 - 1/p^7) = 1.2351002232125595782019... .