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Joseph O'Rourke
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The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.


**Answered** by Gjergji Zaimi and Douglas Zare and Noam Elkies: the density indeed grows as $1/\zeta(k)$, and this has been established.

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.


**Answered** by Gjergji Zaimi and Douglas Zare: the density indeed grows as $1/\zeta(k)$, and this has been established.

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.


**Answered** by Gjergji Zaimi and Douglas Zare and Noam Elkies: the density indeed grows as $1/\zeta(k)$, and this has been established.
Answered.
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Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.


**Answered** by Gjergji Zaimi and Douglas Zare: the density indeed grows as $1/\zeta(k)$, and this has been established.

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.


**Answered** by Gjergji Zaimi and Douglas Zare: the density indeed grows as $1/\zeta(k)$, and this has been established.
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Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958

Square-free grows as $6n/\pi^2$: $k$-th free?

The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$.

OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here).

Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?

I ask this in (obvious) number-theoretic naiveté.