A369427 The number of unitary divisors of n that are squares of primes.

0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0

Offset: 1

Author

Amiram Eldar, Jan 23 2024

Keywords

• nonn
• easy

Comments

The number of exponents in the prime factorization of n that are equal to 2.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001248, A038109, A056169, A056170, A061742, A085541, A085548.

Similar sequences: A125070, A293439, A366074, A367512, A369426, A369428.

Programs

Mathematica

f[p_, e_] := If[e == 2, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = vecsum(apply(x -> if(x == 2, 1, 0), factor(n)[, 2]));

Formula

Additive with a(p^e) = 1 if e = 2, and 0 otherwise.

a(n) > 0 if and only if n is in A038109.

a(A061742(n)) = n, and a(k) < n for all k < A061742(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (1/p^2 - 1/p^3) = A085548 - A085541 = 0.27748478074162196208... .