A number $n \in \mathbb{N}$ is squarefree if for every divisor $d | n, d > 1$, we have $d^2 \nmid n$. It is known that the squarefree numbers have a density of $6/{\pi^2}$ over $\mathbb{N}$. It is a question of interest to know whether polynomials take on infinitely many squarefree values.

I am asking for an explicit example of a function (perhaps a polynomial) that is known to take on infinitely many squarefree values with the correct density. This is trivial if $f(x) = x$, and it is known that all polynomials $f(x) \in \mathbb{Z}[x]$ satisfying basic non-degeneracy conditions and of degree 2 or 3 take on infinitely many squarefree values. It is conjectured to hold for all $f(x) \in \mathbb{Z}[x]$ satisfying non-degeneracy conditions (that is, the content of $f$ should be square-free, and that for all primes $p$, there exist $n \in \mathbb{Z}$ such that $f(n)$ is not divisible by $p^2$), and this conjecture was proved to be true by Granville if one assumes the abc conjecture.

However, the question of whether $f$ assumes squarefree values with the 'correct' density is not investigated as often, at least to my knowledge. That is, the density of the set $$\displaystyle S = \{n \in \mathbb{Z}: f(n) \text{ is squarefree. }\}$$ over $\mathbb{Z}$. The case of interest would be when the density of $S$ is exactly $6/\pi^2$, which would show that $f$ has no bias towards squarefree values.

Are there any arithmetic function where this statement is known to be true, other than obvious ones like $f(x) = x$?