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A366076
The number of primes factors of the largest divisor of n that is a cubefull number (A036966), counted with multiplicity.
4
0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 3, 0, 3, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 4, 0, 0, 0, 0, 0, 0
OFFSET
1,8
COMMENTS
The sum of exponents larger than 2 in the prime factorization of n.
The number of distinct primes factors of the largest divisor of n that is a cubefull number is A295659(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) = A001222(A360540(n)).
a(n) = A001222(n) - A366077(n).
Additive with a(p^e) = 0 if e <= 2, and a(p^e) = e for e >= 3.
a(n) >= 0, with equality if and only if n is cubefree (A004709).
a(n) <= A001222(n), with equality if and only if n is cubefull (A036966).
a(n) >= 3*A295659(n), with equality if and only if n is a biquadratefree number (A046100).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (2/p^3 + 1/(p^2*(p-1))) = 2 * A085541 + A152441 = 0.67043452760761670220... .
MATHEMATICA
f[p_, e_] := If[e < 3, 0, e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) a(n) = vecsum(apply(x -> if(x < 3, 0, x), factor(n)[, 2]));
CROSSREFS
Similar sequence: A275812 (number of primes factors of the powerful part).
Sequence in context: A151795 A375199 A008437 * A324326 A377131 A341743
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Sep 28 2023
STATUS
approved