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

1,4

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)

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)

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A361236 A119350 A016528 * A350388 A056623 A038025

Adjacent sequences: A368881 A368882 A368883 * A368885 A368886 A368887

Keyword

nonn,easy,mult

Author

Amiram Eldar, Jan 09 2024

Status

approved