A368247 - OEIS

%I #10 Dec 19 2023 09:19:36

%S 1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,3,1,1,1,1,3,1,1,

%T 1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,1,1,1,3,1,3,1,1,1,1,1,1,1,3,1,1,1,1,

%U 1,1,1,3,1,1,1,1,1,1,1,3,3,1,1,1,1,1,1

%N The number of cubefree divisors of the cubefull part of n (A360540).

%H Amiram Eldar, <a href="/A368247/b368247.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A073184(A360540(n)).

%F Multiplicative with a(p^e) = 1 if e <= 2, and 3 otherwise.

%F a(n) >= 1, with equality if and only if n is cubefree (A004709).

%F a(n) <= A073184(n), with equality if and only if n is cubefull (A036966).

%F Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 2/p^(3*s)).

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + 2/p^3) = 1.37700168952903630206... .

%F In general, the asymptotic mean of the number of k-free divisors of the k-full part of n is Product_{p prime} (1 + (k-1)/p^k).

%t f[p_, e_] := If[e > 2, 3, 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) a(n) = vecprod(apply(x -> if(x < 3, 1, 3), factor(n)[, 2]));

%Y Cf. A004709, A036966, A073184, A323308, A360540.

%K nonn,easy,mult

%O 1,8

%A _Amiram Eldar_, Dec 19 2023