# Error term for the summatory function of $k$-free numbers indicator and RH

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?

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 $$c>1/(2k)$$.
• 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 $1/(2k)$. – asahay Oct 11 at 11:26