A365487 The number of divisors of the largest cube dividing n.

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 4, 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 cube root of the largest cube dividing n, A053150(n), is A061704(n).

Links

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

Crossrefs

Cf. A000005, A000578, A004709, A008834, A053150, A061704.

Programs

Mathematica

f[p_, e_] := 3*Floor[e/3] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]

PARI

a(n) = vecprod(apply(x -> 3*(x\3) + 1, factor(n)[, 2]));

Formula

a(n) = A000005(A008834(n)).

Multiplicative with a(p^e) = 3*floor(e/3) + 1.

a(n) = 1 if and only if n is cubefree (A004709).

a(n) <= A000005(n) with equality if and only if n is a cube (A000578).

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

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(3) * Product_{p prime} (1 + 2/p^3) = 1.6552343865608... .