The set $E$ of perfect powers is defined as follows: $$E:=\{x^y:\ x,y\in\ \mathbb{N}_{\ge 2} \}.$$ (Thus, a positive integer $n\in E$ iff the exponents in the prime factorization of $n$ have GCD $>1$.)
I was curious what the counting function for $E$ would look like, i.e., $$\epsilon(n) := \mathrm{card}\{k\in E:\ k\le n \},$$ so I computed $\epsilon(n)$ for $0\le n\le 10^6$, obtaining the following plot:
This picture is a superposition of $\epsilon(n)$ (black) and a fitted function of form $\ a\,n^b\ $ (red), with $a\approx 1.4,\ b\approx 0.48.$ Empirically, as $n$ increases, such fitted functions suggest that $\ \epsilon(n)\sim \sqrt{n}$.
Question: Is it correct that $\ \epsilon(n)\sim \sqrt{n}\ ?\ $ How can the correct asymptotic form be derived?
EDIT: Upon learning that these are called "perfect powers", I located several online articles proving $\ \epsilon(n)\sim \sqrt{n},\ $ including the following:
M. A. Nyblom, A counting function for the sequence of perfect powers, Austral. Math. Soc. Gaz. 33 (2006), 338–343.
R. Jakimczuk, On the distribution of perfect powers, J. Integer Seq. 14 (2011), Article 11.8.5.
The point is that nearly all the members of $E$ are squares, and the number of squares $\le n$ is approximately $ \sqrt{n}$.
EDIT: More precisely: Every member of $E$ that is not a square is a $p$'th power for some $p \ge 3$. The number of $p$'th powers $k^p$ with $1 < k^p\le n$ is $\lfloor n^{1/p} - 1 \rfloor$; this is $0$ if $2^p > n$, i.e. $p > \log_2(n)$. So the number of members of $E$ that are not squares and $\le n$ is bounded by $$ \sum_{3 \le p \le \log_2(n)} (n^{1/p}-1) \le \log_2(n) n^{1/3} = o(n^{1/2})$$