Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k \equiv 1 \pmod{2}$. 


What is the asymptotic density of $S$?


Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $\liminf\_{X \rightarrow \infty} \frac{\# (S \cap [1,X])}{X} \geq \frac{6}{\pi^2}$. 



Numbers whose prime factors all have odd exponent
The Dirichlet series $L (s)=\sum\_{n \in S} n^{-s} $ has the Euler product expression
$$L (s) = \prod\_p \frac{1+p^{-s}-p^{-2s}}{1-p^{-2s}}.$$
So $S $ has analytic density
$$\mu (S) = \operatorname{Res}\_{s=1} L(s) = \lim\_{s \to 1} \frac{L(s)}{\zeta(s)} = \prod\_p \frac{1+1/p-1/p^2}{1+1/p}$$
as can also be seen by an informal argument. We can also rewrite this as
$$ \mu(S) = \prod\_p \bigl(1-\frac{1}{p(p+1)}\bigr) = \prod\_p \frac{1-2/p^2+1/p^3}{1-1/p^2}$$
I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.


