A382969 The excess of the n-th noncubefree number.

2, 3, 2, 2, 4, 2, 3, 2, 2, 5, 3, 3, 3, 2, 4, 2, 3, 3, 2, 2, 6, 2, 2, 4, 2, 4, 3, 2, 3, 2, 2, 5, 3, 3, 4, 4, 2, 3, 4, 2, 2, 7, 2, 2, 3, 2, 5, 2, 2, 3, 2, 5, 4, 2, 3, 2, 2, 2, 4, 3, 3, 2, 2, 2, 6, 3, 4, 3, 2, 4, 2, 5, 2, 5, 2, 2, 3, 2, 4, 4, 2, 3, 3, 3, 8, 2, 2

Offset: 1

Author

Amiram Eldar, Apr 10 2025

Keywords

• nonn
• easy

Examples

			a(1) = 2 since the 1st noncubefree number is A046099(1) = 8 = 2^3. It has 3 prime factors when counted with multiplicity, and 1 distinct prime factor, so a(1) = 3 - 1 = 2.
		

Links

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

Crossrefs

Cf. A002117, A046099, A046660, A136141, A275699, A376366.

Programs

Mathematica

f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[Max[e] < 3, Nothing, Total[e] - Length[e]]]; Array[f, 100]

PARI

list(lim) = {my(e); for(k = 2, lim, e = factor(k)[,2]; if(vecmax(e) > 2, print1(vecsum(e) - #e, ", ")));}

Formula

a(n) = A046660(A046099(n)).

a(n) >= 2.

Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = ((Sum_{p prime} 1/(p*(p-1))) - (1/zeta(3)) * (Sum_{p prime} (p-1)/(p^3-1))) / (1-1/zeta(3)) = 3.12223294188308957729... .