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a(n) is the number of divisors of the powerful part of n.
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%I #12 Sep 10 2023 04:51:38

%S 1,1,1,3,1,1,1,4,3,1,1,3,1,1,1,5,1,3,1,3,1,1,1,4,3,1,4,3,1,1,1,6,1,1,

%T 1,9,1,1,1,4,1,1,1,3,3,1,1,5,3,3,1,3,1,4,1,4,1,1,1,3,1,1,3,7,1,1,1,3,

%U 1,1,1,12,1,1,3,3,1,1,1,5,5,1,1,3,1,1,1

%N a(n) is the number of divisors of the powerful part of n.

%C The corresponding sum of divisors of the powerful part of n is A295294.

%H Amiram Eldar, <a href="/A357669/b357669.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A000005(A057521(n)).

%F a(n) = A000005(n)/A056671(n).

%F a(n) = A000005(A064549(A003557(n))).

%F a(n) = 1 iff n is squarefree (A005117).

%F a(n) = A000005(n) iff n is powerful (A001694).

%F Multiplicative with a(p^e) = 1 if e = 1 and e+1 otherwise.

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} ((p^3 - p^2 + 2*p - 1)/(p^2*(p - 1))) = 2.71098009471568319328... .

%F Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^s + 2/p^(2*s) - 1/p^(3*s)). - _Amiram Eldar_, Sep 09 2023

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

%o (PARI) a(n) = {my(e = factor(n)[,2]); prod(i=1, #e, if(e[i] == 1, 1, e[i] + 1))};

%Y Cf. A000005, A001694, A003557, A005117, A056671, A057521, A064549, A295294.

%K nonn,easy,mult

%O 1,4

%A _Amiram Eldar_, Oct 08 2022