A351414 Number of divisors of n that are either prime or have at least 1 square divisor > 1 and at least two distinct prime factors.

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 3, 1, 1, 2, 2, 2, 5, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 6, 1, 2, 3, 1, 2, 3, 1, 3, 2, 3, 1, 7, 1, 2, 3, 3, 2, 3, 1, 5, 1, 2, 1, 6, 2, 2, 2, 4, 1, 6, 2, 3, 2, 2, 2, 6

Offset: 1

Author

Wesley Ivan Hurt, Feb 10 2022

Keywords

• nonn
• easy

Examples

			a(24) = 4; 24 has divisors 2,3 (primes) and 12,24 (which both have at least 1 square divisor > 1 and at least two distinct prime factors).
a(36) = 5; 36 has divisors 2,3 (primes) and 12,18,36 (which all have at least 1 square divisor > 1 and at least two distinct prime factors).
		

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 (omega), A008683 (mu), A048105, A361205.

Programs

Mathematica

a[n_] := Module[{e = FactorInteger[n][[;;, 2]], d, nu, omega}, d = Times @@ (e+1); nu = Length[e]; omega = Total[e]; d - 2^nu - omega + 2*nu]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Oct 06 2023 *)

PARI

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

Formula

a(n) = Sum_{d|n} [[omega(d) = 1] = mu(d)^2], where [ ] is the Iverson bracket.

a(n) = A048105(n) + A361205(n). - Amiram Eldar, Oct 06 2023