# 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.

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

• Yes, your guess of $1/\zeta(k)$ is correct :) en.wikipedia.org/wiki/… – Gjergji Zaimi Feb 22 '15 at 1:02
• The analogue holds for the probability that $k$ randomly chosen positive integers (up to $n$, then let $n\to \infty$) are relatively prime. J.E Nymann. "On the probability that k positive integers are relatively prime." Journal of Number Theory Volume 4, Issue 5, October 1972, Pages 469–473. A similar result is known for the probability that $k$ positive integers are pairwise coprime. – Douglas Zare Feb 22 '15 at 2:42

Yes, it works in much the same way for any $k$. Here's an elementary proof.
Let $Q_k(n)$ be the number of $k$-th power free integers $\leq n$. Then $$Q_k(n) = \sum_{d^k \leq n} \mu(d) \lfloor n/d^k \rfloor = \sum_{d^k \leq n} \mu(d) \, (n/d^k + \theta_d)$$ for some $\theta_d \in [0,1)$. Hence $$\Bigl| \, Q_k(n) - \sum_{d^k \leq n} \frac{\mu(d)}{d^k} n \, \Bigr| < n^{1/k}.$$ But $\sum_{d^k \leq n} \mu(d)/d^k$ is a partial sum of a series that converges to $1/\zeta(k)$, with error bounded by $\sum_{d^k > n} 1/d^k \ll n^{1/k}/n$. Therefore $Q_k(n) = n/\zeta(k) + O(n^{1/k})$, QED.
• It's even true for $k=1$ (and with a much better error term) ;-) – Noam D. Elkies Feb 22 '15 at 2:10