A362412 The number of prime factors of the square root of the largest square dividing n, counted with multiplicity.

0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 3, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 0, 2, 2, 0, 0, 1, 0, 0, 0

Offset: 1

Author

Amiram Eldar, Sep 28 2023

Keywords

• nonn
• easy

Comments

First differs from A366073 at n = 64.

Links

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

Crossrefs

Cf. A000188, A000290, A001222, A004526, A005117, A008833, A154945, A162642, A349326, A366073.

Programs

Mathematica

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

PARI

a(n) = vecsum(factor(n)[, 2]\2);

Formula

a(n) = A001222(A000188(n)).

a(n) = A001222(A008833(n))/2.

Additive with a(p^e) = floor(e/2) = A004526(e).

a(n) >= 0, with equality if and only if n is squarefree (A005117).

a(n) <= A001222(n)/2, with equality if and only if n is square (A000290).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p^2-1) = 0.551693... (A154945).

a(n) = (A001222(n) - A162642(n))/2. - Ridouane Oudra, Apr 19 2025