A368884 The largest unitary divisor of n that is the square of a squarefree number (A062503).

1, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 1, 25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 49, 25, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 9, 1, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 25, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1

Offset: 1

Author

Amiram Eldar, Jan 09 2024

Keywords

• nonn
• easy
• mult

Comments

The number of these divisors is A368885(n).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)

Crossrefs

Cf. A005117, A062503, A337050, A368885, A368886.

Programs

Mathematica

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

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 2, f[i,1]^f[i,2], 1));}

Python

from math import prod
from sympy import factorint
def A368884(n): return prod(p**e for p, e in factorint(n).items() if e==2) # Chai Wah Wu, Jan 09 2024

Formula

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

a(n) = n / A368886(n).

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

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

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

From Vaclav Kotesovec, Jan 09 2024: (Start)

Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (p^(2*s) - p^2) * (1 + (p^s - 1) * (p^2 + p^s + p^(2*s))) / p^(5*s).

Sum_{k=1..n} a(k) ~ c * zeta(3/2) * n^(3/2)/3, where c = Product_{p prime} (1 - 1/p^(11/2) + 1/p^(9/2) + 1/p^4 + 1/p^(7/2) - 1/p^3 - 1/p^(5/2) - 1/p^2) = 0.45021226373776124069206513259105992151602618717147857709105849... (End)