A335324 Square part of 4th-power-free part of n.
1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 1, 1, 1, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1
Offset: 1
Examples
Removing the 4th powers from 192 = 2^6 * 3^1 gives 2^(6 - 4) * 3^1 = 2^2 * 3 = 12. So the 4th-power-free part of 192 is 12. The square part of 12 (largest square dividing 12) is 4. So a(192) = 4.
Links
- Michel Marcus, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Biquadratefree.
- Eric Weisstein's World of Mathematics, Square part.
Programs
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Mathematica
f[p_, e_] := p^(2*Floor[e/2] - 4*Floor[e/4]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jun 01 2020 *)
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PARI
A053165(n)=my(f=factor(n)); f[, 2]=f[, 2]%4; factorback(f); a(n) = my(m=A053165(n)); m/core(m); \\ Michel Marcus, Jun 01 2020
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Python
from math import prod from sympy import factorint def A335324(n): return prod(p**(e&2) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 07 2024
Formula
a(n^2) = A007913(n)^2.
a(p^e) = p^(2*floor(e/2) - 4*floor(e/4)). - Amiram Eldar, Jun 01 2020
From Amiram Eldar, Sep 21 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * zeta(4*s)/(zeta(2*s) * zeta(4*s-4)).
Sum_{k=1..n} a(k) ~ (4*zeta(3/2)*zeta(4))/(21*zeta(3)) * n^(3/2). (End)
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