A368885 The number of unitary divisors of n that are squares of a squarefree number (A062503).

1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Jan 09 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A294932 at n = 32.

The largest of these divisors is A368884(n).

Links

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

Crossrefs

Cf. A034444, A062503, A294932, A337050, A368884.

Programs

Mathematica

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

PARI

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

Python

from sympy import factorint
def A368885(n): return 1<Chai Wah Wu, Jan 09 2024

Formula

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

a(n) >= 1, with equality if and only if n is in A337050.

a(n) <= A034444(n), with equality if and only if n is in A062503.

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

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + 1/p^2 - 1/p^3) = 1.30596827416754083231... .