A385005 The sum of the cubefull divisors of n.

1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 28, 1, 1, 1, 1, 57, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 1, 28, 1, 9, 1, 1, 1, 1, 1, 1, 1, 121, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 109, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Jun 15 2025

Keywords

• nonn
• easy
• mult

Comments

The sum of the terms in A036966 that divide n.

The number of these divisors is A190867(n), and the largest of them is A360540(n).

Links

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

Crossrefs

Cf. A036966, A190867, A183097, A360540.

The sum of divisors d of n such that d is: A000593 (odd), A033634 (exponentially odd), A035316 (square), A038712 (power of 2), A048250 (squarefree), A072079 (3-smooth), A073185 (cubefree), A113061 (cube), A162296 (nonsquarefree), A183097 (powerful), A186099 (5-rough), A353900 (exponentially 2^n), this sequence (cubefull), A385006 (biquadratefree).

Programs

Mathematica

f[p_, e_] := (p^(e+1)-1)/(p-1) - p - If[e == 1, 0, p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; (p^(e+1)-1)/(p-1) - p - if(e == 1, 0, p^2));}

Formula

Multiplicative with a(p^e) = 1 if e <= 2, and a(p^e) = ((p^(e+1)-1) / (p-1)) - p - p^2 if e >= 3.

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