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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 $f(x) = x$?

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I would guess that the Piatetski-Shapiro sequence $[n^c]$ for some $c$ just a little larger than $1$ would have this property. –  Lucia Oct 28 '13 at 19:50
    
Note that if the density of n such that $d^2|f(n)$ is equal to $\prod_{p|d} g(p)$ for any squarefree d and some quantities $g(p)$ with $\sum_p g(p) < \infty$, then a simple application of the Bonferroni inequalities shows that the set $S$ here will have density $\prod_p (1-g(p))$. Here, we have $g(p) = O(1/p^2)$, so the hypothesis $\sum_p g(p) < \infty$ is certainly satisfied; the only issue is that of computing the product $\prod_p (1-g(p))$. –  Terry Tao Oct 28 '13 at 20:44
    
To Terry Tao: My understanding was that the question wants the density to be exactly $\pi^2/6$ (but the question is not completely clear on this point). I doubt that this holds for a polynomial (we can evaluate that only for polynomials of small degree; but assuming the appropriate conjecture I don't think the answer can be arranged to be $\pi^2/6$). For the PS example, the answer is exactly $\pi^2/6$. –  Lucia Oct 28 '13 at 20:49
    
So one does not expect to be able to find an explicit polynomial of degree at least 2 satisfying the problem? –  Stanley Yao Xiao Oct 28 '13 at 21:14
    
Ah, I see now. (And I realise also that my use of the Bonferroni inequalities isn't justified for polynomials of higher degree, although presumably this is still the correct formula for the density, as indicated by Granville's paper.) –  Terry Tao Oct 28 '13 at 22:03

1 Answer 1

up vote 7 down vote accepted

Indeed the Piatetski-Shapiro numbers are such an example: see Theorem 4 of this paper http://arxiv.org/pdf/1203.5884.pdf by Baker, Banks, Brudern, Shparlinski and Weingartner.

Note: I understood the question to ask for the density to be exactly $\pi^2/6$; which the PS example gives. One can also compute the density for polynomials of small degree (e.g. degrees $2$ and $3$) but the density won't be $\pi^2/6$. Finally a trivial example that also gives $\pi^2/6$ is to take $f(n) = n+ [\sqrt{n}]$ and use the distribution of squarefree numbers in short intervals. In my opinion, the PS example is perhaps the most interesting (at least the most interesting that I know).

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Oops -- should have been $6/\pi^2$ everywhere rather than $\pi^2/6$. –  Lucia Oct 29 '13 at 3:54

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