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%I #41 Aug 25 2023 08:40:44
%S 1,2,2,3,2,4,2,4,3,4,2,6,2,4,4,4,2,6,2,6,4,4,2,8,3,4,4,6,2,8,2,4,4,4,
%T 4,9,2,4,4,8,2,8,2,6,6,4,2,8,3,6,4,6,2,8,4,8,4,4,2,12,2,4,6,4,4,8,2,6,
%U 4,8,2,12,2,4,6,6,4,8,2,8,4,4,2,12,4,4,4,8,2,12,4,6,4,4,4,8,2,6,6,9
%N Number of biquadratefree (4th power free) divisors of n.
%C Equivalently, a(n) is the number of divisors of n that are in A046100.
%C a(n) is also the number of divisors d such that the greatest common square divisor of d and n/d is 1.
%C The number of divisors d of n such that gcd(d, n/d) is squarefree. - _Amiram Eldar_, Aug 25 2023
%D Paul J. McCarthy, Introduction to Arithmetical Functions, Springer Verlag, 1986, page 37, Exercise 1.27.
%H Antti Karttunen, <a href="/A252505/b252505.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Biquadratefree.html">Biquadratefree</a>.
%F Dirichlet g.f.: zeta(s)^2/zeta(4*s).
%F Sum_{k=1..n} a(k) ~ 90*n/Pi^4 * (log(n) - 1 + 2*gamma - 360*zeta'(4)/Pi^4), where gamma is the Euler-Mascheroni constant A001620. - _Vaclav Kotesovec_, Feb 02 2019
%F a(n) = Sum_{d|n} mu(gcd(d, n/d))^2. - _Ilya Gutkovskiy_, Feb 21 2020
%F Multiplicative with a(p^e) = min(e, 3) + 1. - _Amiram Eldar_, Sep 19 2020
%e a(16) = 4 because there are 4 divisors of 16 that are 4th power free: 1,2,4,8.
%e a(16) = 4 because there are 4 divisors d of 16 such that the greatest common square divisor of d and 16/d is 1: 1,2,8,16.
%t Prepend[Table[Apply[Times, (FactorInteger[n][[All, 2]] /. x_ /; x > 3 -> 3) + 1], {n, 2, 100}], 1]
%o (PARI) isA046100(n) = (n==1) || vecmax(factor(n)[, 2])<4;
%o a(n) = {d = divisors(n); sum(i=1, #d, isA046100(d[i]));} \\ _Michel Marcus_, Mar 22 2015
%o (PARI) a(n) = vecprod(apply(x->min(x, 3) + 1, factor(n)[, 2])); \\ _Amiram Eldar_, Aug 25 2023
%Y Cf. A046100 (biquadratefree numbers).
%Y Cf. A034444 (squarefree divisors), A073184 (cubefree divisors).
%Y Cf. A001620.
%K nonn,easy,mult
%O 1,2
%A _Geoffrey Critzer_, Mar 21 2015
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