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A366078
The number of distinct prime factors of the cubefree part of n (A360539).
4
0, 1, 1, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 2, 0, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 0, 2, 1, 3, 1, 0, 2, 2, 2, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 1, 2, 2, 0, 2, 3, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 0, 2, 1, 3, 2, 2, 2
OFFSET
1,6
COMMENTS
The number of exponents smaller than 3 in the prime factorization of n.
The number of prime factors of the cubefree part of n (A360539), counted with multiplicity is A366077(n).
LINKS
Rafael Jakimczuk and Matilde LalĂ­n, Sums of omega(n) and Omega(n) over the k-free parts and k-full parts of some particular sequences, Integers, Vol. 22 (2022), Article #A113.
FORMULA
a(n) = A001221(A360539(n)).
a(n) = A001221(n) - A295659(n).
Additive with a(p^e) = 1 if e <= 2, and a(p^e) = 0 for e >= 3.
a(n) >= 0, with equality if and only if n is cubefull (A036966).
a(n) <= A001221(n), with equality if and only if n is cubefree (A004709).
a(n) <= A366077(n), with equality if and only if n is squarefree (A005117).
Sum_{k=1..m} a(k) = n * (log(log(n)) + B - C) + O(n/log(n)), where B is Mertens's constant (A077761) and C = Sum_{p prime} 1/p^3 = 0.174762... (A085541).
MATHEMATICA
f[p_, e_] := If[e < 3, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) a(n) = vecsum(apply(x -> if(x < 3, 1, 0), factor(n)[, 2]));
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Sep 28 2023
STATUS
approved