

A252505


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


2



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, 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, 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
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OFFSET

1,2


COMMENTS

Equivalently, a(n) is the number of divisors of n that are in A046100.
a(n) is also the number of divisors d such that the greatest common square divisor of d and n/d is 1.


REFERENCES

Paul J. McCarthy, Introduction to Arithmetical Functions, Springer Verlag, 1986, page 37, Exercise 1.27


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Biquadratefree.


FORMULA

Dirichlet g.f.: zeta(s)^2/zeta(4*s).
Sum_{k=1..n} a(k) ~ 90*n/Pi^4 * (log(n)  1 + 2*gamma  360*Zeta'(4)/Pi^4), where gamma is the EulerMascheroni constant A001620.  Vaclav Kotesovec, Feb 02 2019
a(n) = Sum_{dn} mu(gcd(d, n/d))^2.  Ilya Gutkovskiy, Feb 21 2020
Multiplicative with a(p^e) = min(e, 3) + 1.  Amiram Eldar, Sep 19 2020


EXAMPLE

a(16) = 4 because there are 4 divisors of 16 that are 4th power free: 1,2,4,8.
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.


MATHEMATICA

Prepend[Table[Apply[Times, (FactorInteger[n][[All, 2]] /. x_ /; x > 3 > 3) + 1], {n, 2, 100}], 1]


PROG

(PARI) isA046100(n) = (n==1)  vecmax(factor(n)[, 2])<4;
a(n) = {d = divisors(n); sum(i=1, #d, isA046100(d[i])); } \\ Michel Marcus, Mar 22 2015


CROSSREFS

Cf. A046100 (biquadratefree numbers).
Cf. A034444 (squarefree divisors), A073184 (cubefree divisors).
Sequence in context: A286605 A035149 A074848 * A325560 A318412 A322986
Adjacent sequences: A252502 A252503 A252504 * A252506 A252507 A252508


KEYWORD

nonn,mult


AUTHOR

Geoffrey Critzer, Mar 21 2015


STATUS

approved



