A366125 The number of prime factors of the cube root of the largest unitary divisor of n that is a cube (A366126), counted with multiplicity.

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Amiram Eldar, Sep 30 2023

Keywords

• nonn
• easy

Comments

First differs from A295662 at n = 32, and from A295883, A318673, A359472 and A366124 at n = 64.

One third of the sum of exponents that are divisible by 3 in the prime factorization of n.

Links

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

Crossrefs

Cf. A001222, A175676, A350386, A366124, A366126.

Cf. A295662, A295883, A318673, A359472.

Programs

Mathematica

f[p_, e_] := If[Divisible[e, 3], e/3, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = vecsum(apply(x -> if(!(x%3), x/3, 0), factor(n)[, 2]));

Formula

a(n) = A001222(A366126(n))/3.

Additive with a(p^e) = A175676(e) = e if e is divisible by 3, and 0 otherwise.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} p^2*(p-1)/(p^3-1)^2 = 0.35687351842962928035... .