What is the relative density of the prime numbers among the set of prime powers? In particular, let $\Pi(x)$ be the number of prime powers less than $x$ and let $\pi(x)$ be the number of primes less than $x$. What can be said of the limit
$$\lim_{x \to \infty} \frac{\pi(x)}{\Pi(x)}?$$
In other words, what is the density of prime powers which are square-free? My intuition says the answer should be $1$, but I'm not sure. -- in $\mathbb{Z}$, the square-free integers make up only $\frac{6}{\pi^2}$. On the other hand, in $\mathbb{F}_q[x]$, the square-free polynomials of degree $n$ are $1 + o_n(1)$.
$\endgroup$2 Answers
$\begingroup$The argument below is not elegant, but it works.
The number of perfect powers below $x$ is easily bounded by $ \sum_{n = 2}^{\log_2 x} \lfloor x^{1/n} \rfloor \le x^{1/2} \log x$.
On the other hand the number of primes below $x$ is about $x/\log x$.
Bounding the number of prime powers (including primes) by the number of primes plus the number of perfect powers, the claim follows.
$\endgroup$ 6 $\begingroup$We can observe that if $p^{n}\leq x$ then $p\leq x^{1/n}$ so we can write $$\Pi\left(x\right)=\pi\left(x\right)+\pi\left(x^{1/2}\right)+\dots+\pi\left(x^{1/n}\right) $$ hence $$\Pi\left(x\right)=\pi\left(x\right)+O\left(\frac{\sqrt{x}}{\log\left(x\right)}\right).$$
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