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I started to read this preprint: https://arxiv.org/abs/2010.03696

In it, the author states that $\sum_{n\leq x}\mu_{k}(n)=\zeta(k)^{-1}x+O(x^{1/k})$ and that under RH, the exponent in the error term becomes $\frac{1}{k+1}$ (where $\mu_{k}$ is the indicator of $k$-free numbers).

What would an exponent of the form $\frac{1}{\sqrt{k(k+1)}}$ imply towards RH? Conversely, assuming the supremum of the real parts of the non trivial zeros of the Riemann zeta function is $1-\varepsilon$ for some $\varepsilon >0$, what would it imply for the value of the considered exponent?

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The Dirichlet series of the indicator function of $k$-free numbers is $\zeta(s)/\zeta(ks)$. Hence any exponent less than $1/k$ in the error term implies a quasi-Riemann Hypothesis. More precisely, if the number of $k$-free numbers is $x/\zeta(k)+O(x^c)$, then $s=1$ is the only pole of $\zeta(s)/\zeta(ks)$ in the half-plane $\Re(s)>c$, whence all the zeros of $\zeta(s)$ satisfy $\Re(s)\leq ck$.

This also shows that $c\geq 1/(2k)$, so the best bound one can hope for is $x/\zeta(k)+O(x^{1/(2k)})$. It is likely that even this bound is provably false (just like an error term $O(x^{1/2})$ in the Prime Number Theorem is provably false), so a more realistic hope is $x/\zeta(k)+O(x^c)$ for any $c>1/(2k)$.

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    $\begingroup$ Thank you very much! $\endgroup$ – Sylvain JULIEN Oct 10 at 15:49
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    $\begingroup$ mat.uniroma3.it/users/pappa/papers/allahabad2003.pdf This is a ~15-year old survey on this topic. In particular, as GH from MO mentions, Vaidya proved that the exponent in the error term necessarily requires an epsilon more than $1/(2k)$. $\endgroup$ – asahay Oct 11 at 11:26

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