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

1,2

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

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... .

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]

Prog

(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

Crossrefs

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

Sequence in context: A385043 A366743 A191750 * A394255 A346613 A378433

Adjacent sequences: A384551 A384552 A384553 * A384555 A384556 A384557

Keyword

nonn,easy,mult

Author

Amiram Eldar, Jun 03 2025

Status

approved