A367987 The number of square divisors of the largest unitary divisor of n that is a square.

1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 3, 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, 3, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Dec 07 2023

Keywords

• nonn
• easy
• mult

Comments

Also, the number of divisors of the square root of the largest unitary divisor of n that is a square.

Links

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

Crossrefs

Cf. A000005, A046951, A071974, A268335, A335762, A350388, A365401, A367988.

Programs

Mathematica

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

PARI

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

Formula

Multiplicative with a(p^e) = e/2 + 1 if e is even and 1 otherwise.

a(n) = A046951(A350388(n)).

a(n) = A000005(A071974(n)).

a(n) >= 1, with equality if and only if n is an exponentially odd number (A268335).

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

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + p/((p-1)*(p+1)^2)) = 1.450032... (A335762).