editing
approved
a(n) = Sum_{k = 1..A000005(n)} A212793(A027750(n,k)). - _Reinhard Zumkeller, _, May 27 2012
proposed
The divisors of 56 are {1, 2, 4, 7, 8, 14, 28, 56}, 8=2^3 and 56=7*2^3 are not cubefree, therefore a(56) = 6.
Cf. A000005, A000012, A001620, A004709, A027750, A034444, A073180, A073183, A073185, A212793.
Cf. A000005, A073185, A001620, A004709, A073183, A073180, A027750, A034444, A073180, A073183, A073185, A212793.
a(n) <= A073182(n).
a(n) = sum number of divisors of the cubefree kernel of n: a(n) = A073184A000005(A007948(n)); [corrected by _Amiram Eldar_, Oct 08 2022]
a(n) = sumSum_{k = 1..A000005(n)} A212793(A027750(n,k)): k = 1..A000005(n)). - _Reinhard Zumkeller_, , May 27 2012
Sum_{k=1..n} a(k) ~ n / Zetazeta(3) * (log(n) - 1 + 2*gamma - 3*Zetazeta'(3)/Zetazeta(3)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jan 31 2019
(PARI) a(n) = {my(e = factor(n)[, 2]); prod(i = 1, #e, if(e[i] == 1, 2, 3))}; \\ Amiram Eldar, Oct 08 2022