A351394 Number of divisors of n that are either squarefree, prime powers, or both.

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 5, 2, 4, 4, 5, 2, 5, 2, 5, 4, 4, 2, 6, 3, 4, 4, 5, 2, 8, 2, 6, 4, 4, 4, 6, 2, 4, 4, 6, 2, 8, 2, 5, 5, 4, 2, 7, 3, 5, 4, 5, 2, 6, 4, 6, 4, 4, 2, 9, 2, 4, 5, 7, 4, 8, 2, 5, 4, 8, 2, 7, 2, 4, 5, 5, 4, 8, 2, 7, 5, 4, 2, 9, 4, 4, 4, 6, 2, 9, 4, 5, 4, 4, 4

Offset: 1

Author

Wesley Ivan Hurt, Feb 09 2022

Keywords

• nonn
• easy

Examples

			a(36) = 6; 36 has 4 squarefree divisors 1,2,3,6 (where the primes 2 and 3 are both squarefree and 1st powers of primes) and 2 (additional) divisors that are powers of primes, 2^2 and 3^2.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A001221, A008683, A046660, A048105, A246655.

Cf. Similar to A327527.

Programs

Mathematica

a[n_] := Module[{e = FactorInteger[n][[;;, 2]], nu, omega}, nu = Length[e]; omega = Total[e]; 2^nu + omega - nu]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Oct 06 2023 *)

PARI

a(n) = {my(f = factor(n), nu = omega(f), om = bigomega(f)); 2^nu + om - nu;} \\ Amiram Eldar, Oct 06 2023

Formula

a(n) = Sum_{d|n} sign(mu(d)^2 + [omega(d) = 1]).

a(n) = Sum_{d|n} (mu(d)^2 + [omega(d) = 1]*(1 - mu(d)^2)).

a(n) = A048105(n) + A046660(n). - Amiram Eldar, Oct 06 2023