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A252505 Number of biquadratefree (4th power free) divisors of n.

%I A252505 #51 May 19 2025 15:56:44
%S A252505 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 A252505 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 A252505 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 A252505 Number of biquadratefree (4th power free) divisors of n.
%C A252505 Equivalently, a(n) is the number of divisors of n that are in A046100.
%C A252505 a(n) is also the number of divisors d such that the greatest common square divisor of d and n/d is 1.
%C A252505 The number of divisors d of n such that gcd(d, n/d) is squarefree. - _Amiram Eldar_, Aug 25 2023
%D A252505 Paul J. McCarthy, Introduction to Arithmetical Functions, Springer Verlag, 1986, page 37, Exercise 1.27.
%H A252505 Antti Karttunen, <a href="/A252505/b252505.txt">Table of n, a(n) for n = 1..10000</a>
%H A252505 Jon Maiga, <a href="http://sequencedb.net/s/A252505">Computer-generated formulas for A252505</a>, Sequence Machine.
%H A252505 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Biquadratefree.html">Biquadratefree</a>.
%F A252505 Dirichlet g.f.: zeta(s)^2/zeta(4*s).
%F A252505 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 A252505 a(n) = Sum_{d|n} mu(gcd(d, n/d))^2. - _Ilya Gutkovskiy_, Feb 21 2020
%F A252505 Multiplicative with a(p^e) = min(e, 3) + 1. - _Amiram Eldar_, Sep 19 2020
%F A252505 From _Antti Karttunen_, May 14 2025: (Start)
%F A252505 Following formulas have been generated for this sequence by Sequence Machine:
%F A252505 a(n) = A000005(A058035(n)).
%F A252505 a(n) = Sum_{d|n} A307430(d).
%F A252505 a(n) = Sum_{d|n} A034444(d)*A227291(n/d).
%F A252505 a(n) = Sum_{d|n} A007427(d)*A286779(n/d).
%F A252505 a(n) = Sum_{d|n} A008966(d)*A323308(n/d).
%F A252505 a(n) = Sum_{d|n} A048691(d)*A363552(n/d).
%F A252505 a(n) = Sum_{d|n} A271102(d)*A322327(n/d).
%F A252505 a(n) = Sum_{d|n} A307445(d)*A370296(n/d).
%F A252505 a(n) = Sum_{d|n} A018892(d)*A378214(n/d). [Conjectured]
%F A252505 (End)
%e A252505 a(16) = 4 because there are 4 divisors of 16 that are 4th power free: 1,2,4,8.
%e A252505 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 A252505 Prepend[Table[Apply[Times, (FactorInteger[n][[All, 2]] /. x_ /; x > 3 -> 3) + 1], {n, 2, 100}], 1]
%o A252505 (PARI) isA046100(n) = (n==1) || vecmax(factor(n)[, 2])<4;
%o A252505 a(n) = {d = divisors(n); sum(i=1, #d, isA046100(d[i]));} \\ _Michel Marcus_, Mar 22 2015
%o A252505 (PARI) a(n) = vecprod(apply(x->min(x, 3) + 1, factor(n)[, 2])); \\ _Amiram Eldar_, Aug 25 2023
%Y A252505 Cf. A046100 (biquadratefree numbers).
%Y A252505 Cf. A034444 (squarefree divisors), A073184 (cubefree divisors).
%Y A252505 Cf. A001620.
%Y A252505 Cf. A000005, A058035, A307430.
%Y A252505 Also obtained as a Dirichlet convolution of the following pairs: A034444 and A227291, A007427 and A286779, A008966 and A323308, A048691 and A363552, A271102 and A322327, A307445 and A370296, and A018892 and A378214 (conjectured).
%K A252505 nonn,easy,mult
%O A252505 1,2
%A A252505 _Geoffrey Critzer_, Mar 21 2015