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It is well know that the density of positive square-free integers up to $x$ is asymptotically $x/\zeta(2)$. The error term
$$ E(x)=\sum_{1\leq n \leq x } \mu(n)^2 -\frac{x}{\zeta(2)} $$ can easily prove $$ E(x)=O(x^{1/2}).$$ Using arguments like those in the proof of the prime number theorem, one can show that $$ E(x)=o(x^{1/2})$$ with an explicit saving.
There are several papers that prove $$ E(x)=O(x^{1/2-c})$$ for some small positive constant $c$ under the assumption of the Riemann Hypothesis.

My question is whether there is any reference or heuristics for the smallest $a\geq 0 $ such that, for all $\epsilon>0$, we have $$ E(x)=O_\epsilon(x^{a+\epsilon}) ?$$ It is tempting to guess $a=1/4$ motivated by the Gauss circle problem conjecture by Hardy... but the heuristic behind Hardy's conjecture is irrelevant to square-frees.

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    $\begingroup$ the generating function is $\zeta(s)/\zeta(2s)$ so by moving the usual Perron integral to the left of $1$ and picking up the $x/\zeta(2)$ residue it is natural to go as much as we can which is $1/4+\epsilon$ under RH etc - of course one still needs to prove that the relevant integrals work out $\endgroup$
    – Conrad
    Commented Nov 14 at 0:15

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Your guess is correct! It is indeed conjectured that $a=1/4$. A good recent reference is [1]. In particular, it is known that

$$E(x)=\Omega(x^{1/4})$$

and computations have shown

$$|E(x)|<1.12543x^{1/4}$$

for all $0<x\leq 10^{18}$.

[1] M. Mossinghoff, T. Oliviera e Silva, T. Trudgian, The distribution of $k$-free numbers, Math. Comp., 2021.

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