A365549 The number of exponentially odd divisors of the square root of the largest square dividing n.

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

Offset: 1

Author

Amiram Eldar, Sep 08 2023

Keywords

• nonn
• easy
• mult

Comments

First differs from A278908, A307848, A323308 and A358260 at n = 64.

The number of exponentially odd divisors of the largest square dividing n is the same as the number of squares dividing n, A046951(n).

Links

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

Crossrefs

Cf. A000188, A005117, A013662, A046951, A322483.

Cf. A278908, A307848, A323308, A358260.

Programs

Mathematica

f[p_, e_] := 2 + Floor[(e-2)/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = vecprod(apply(x -> 2 + (x-2)\4, factor(n)[, 2]));

Formula

a(n) = A322483(A000188(n)).

a(n) >= 1 with equality if and only if n is squarefree (A005117).

Multiplicative with a(p^e) = 2 + floor((e-2)/4).

Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 1/p^(2*s) - 1/p^(4*s)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 1/p^2 - 1/p^4) = 1.54211628314015874165... .