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A382969
The excess of the n-th noncubefree number.
1
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
1,1
LINKS
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... .
EXAMPLE
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.
MATHEMATICA
f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[Max[e] < 3, Nothing, Total[e] - Length[e]]]; Array[f, 100]
PROG
(PARI) list(lim) = {my(e); for(k = 2, lim, e = factor(k)[, 2]; if(vecmax(e) > 2, print1(vecsum(e) - #e, ", "))); }
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Apr 10 2025
STATUS
approved