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