A369889 The sum of squarefree divisors of the cubefree numbers.

1, 3, 4, 3, 6, 12, 8, 4, 18, 12, 12, 14, 24, 24, 18, 12, 20, 18, 32, 36, 24, 6, 42, 24, 30, 72, 32, 48, 54, 48, 12, 38, 60, 56, 42, 96, 44, 36, 24, 72, 48, 8, 18, 72, 42, 54, 72, 80, 90, 60, 72, 62, 96, 32, 84, 144, 68, 54, 96, 144, 72, 74, 114, 24, 60, 96, 168

Offset: 1

Author

Amiram Eldar, Feb 15 2024

Keywords

• nonn
• easy

Comments

The number of squarefree divisors of the n-th cubefree number is A366536(n).

Links

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

Crossrefs

Cf. A004709, A048250, A062822, A366440, A366536, A366537.

Cf. A002117, A013663.

Programs

Mathematica

f[p_, e_] := p + 1; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; cubefreeQ[n_] := Max[FactorInteger[n][[;; , 2]]] < 3; s /@ Select[Range[100], cubefreeQ]
(* or *)
f[p_, e_] := If[e > 2, 0, p + 1]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Array[s, 100], # > 0 &]

PARI

lista(kmax) = {my(f, s, p, e); for(k = 1, kmax, f = factor(k); s = prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(e < 3, p + 1, 0)); if(s > 0, print1(s, ", ")));}

Python

from math import prod
from sympy import mobius, integer_nthroot, primefactors
def A369889(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return prod(p+1 for p in primefactors(m)) # Chai Wah Wu, Aug 12 2024

Formula

a(n) = A048250(A004709(n)).

Sum_{j=1..n} a(j) ~ c * n^2, where c = zeta(3)^2/(2*zeta(5)) = 0.6967413068... .

In general, the formula holds for the sum of squarefree divisors of the k-free numbers with c = zeta(k)^2/(2*zeta(2*k-1))..., for k >= 2.