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Is there an explicit estimate in the literature bounding from above the number of square-free numbers in a short interval $x<n\leq x y$? I can easily do this by means of the Selberg sieve, but I do not want to reinvent the wheel.

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  • $\begingroup$ $\sum\limits_{n=1}^x | \mu(n)|=\sum\limits_{n=1}^x \mu(n)^2 \approx\frac{6 x}{\pi ^2}+1$ but I'm not sure about the error term. Note that $\sum\limits_{n=1}^\infty \frac{| \mu(n)|}{n^s}=\sum\limits_{n=1}^\infty \frac{\mu(n)^2}{n^s}=\frac{\zeta(s)}{\zeta(2 s)}$ where $\Re(s)>1$, so perhaps there's an error term predicted by the Riemann hypothesis. Note that $\sum\limits_{n=1}^x \mu(n)=O(x^{\frac{1}{2}+\epsilon})$ for every positive $\epsilon$ is equivalent to the RH. $\endgroup$ Aug 29, 2022 at 15:08
  • $\begingroup$ Notice I am talking about short intervals (i.e. $y = 1+\epsilon$, where $\epsilon$ may be very small). $\endgroup$ Aug 29, 2022 at 15:17
  • $\begingroup$ Yes, but the number of square-free numbers in the interval $x<n\le x\, y$ is $\sum\limits_{k=1}^{x\, y} |\mu(k)|-\sum\limits_{k=1}^x |\mu(k)|=\sum\limits_{k=x}^{x\, y} |\mu(k)|$, so I was suggesting you search for results associated with the Möbius function $\mu(n)$ (e.g. see some of the results under the Formula section at oeis.org/A013928). $\endgroup$ Aug 29, 2022 at 15:31
  • $\begingroup$ But that leads to adding error terms - one won't get an error term better than $O(\sqrt{x})$ that way. $\endgroup$ Aug 29, 2022 at 15:35
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    $\begingroup$ @HAHelfgott The result appears in exercise 6 of section 3.2 of Montgomery & Vaughan's Multiplicative Number Theory I: Classical Theory. $\endgroup$
    – TravorLZH
    Sep 3, 2022 at 16:05

2 Answers 2

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Let me just show how to derive a simple bound that has been mentioned in the comments. We are trying to bound the estimate the number $Q(x,x+u)$ of squarefree integers in $(x,x+u]$.

We can now apply Selberg's sieve (as in Iwaniec-Kowalski (say), Thm. 6.4, with $g(d) = 1/d^2$, $X = u$, $r_d = 2$ and $|\{x<m\leq x+u: \max \{d: d^2|m\} = n\}|$, $D$ to be set later). We obtain $$Q(x,x+u)\leq \frac{u}{H} + 2 \sum_{d\leq D} |\lambda_d|, $$ where $\lambda_d = \sum_{[d_1,d_2]=d} \rho_{d_1} \rho_{d_2}$ and $|\rho_d|\leq 1$ for all $d$, $\rho_d=0$ for $d>\sqrt{D}$ or $d$ not square-free, and $$\begin{aligned} H &= \sum_{d\leq \sqrt{D}} \frac{\mu^2(d)}{\prod_{p|d} p^2 (1-1/p^2)} = \sum_{d\leq \sqrt{D}} \mu^2(d) \sum_{m: \textrm{rad}(m)=d} \frac{1}{m^2}\\&\geq \sum_{d\leq \sqrt{D}} \frac{1}{d^2} \geq \zeta(2) - \frac{1}{\sqrt{D}} - \frac{1}{D}\end{aligned}$$ with $\textrm{rad}(m) = \prod_{p|m} p$. It does not take much work (just some casework for $D<4$) to show that in fact $H\geq \zeta(2)-1/D$ for $D\geq 1$. Clearly $\sum_{d\leq D} |\lambda_d| \leq \left(\sum_{d\leq \sqrt{D}} \rho_d\right)^2 \leq D.$ Since $1/H\leq (\zeta(2) (1-1/(\zeta(2) \sqrt{D})))^{-1}$, it follows that $$\begin{aligned}Q(x,x+u)&\leq \frac{u}{\zeta(2)} + 2 D + \frac{u}{\zeta(2)^2 \left(1- \frac{1}{\zeta(2) \sqrt{D}}\right)\sqrt{D} }\\ &< \frac{u}{\zeta(2)} + 2 D + \frac{u}{\sqrt{D}}.\end{aligned}$$ We set $D = (u/4)^{2/3}$, and obtain $$Q(x,x+u) < \frac{u}{\zeta(2)} + \frac{3}{2^{1/3}} u^{2/3}.$$

(Please feel free to point how this proof can be made simpler or shorter.)

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  • $\begingroup$ Euler product expansion gives $\sum_{d\ge1}{\mu^2(d)\over\prod_{p|d}(p^2-1)}=\prod_p\left(1+{1\over p^2-1}\right)=\zeta(2)$, so there is no need to perform the ordinary lower-bounding trick here. $\endgroup$
    – TravorLZH
    Sep 3, 2022 at 16:02
  • $\begingroup$ Why? We do need a lower bound on $H$. We don't have the right to complete the product just like that. $\endgroup$ Sep 3, 2022 at 18:38
  • $\begingroup$ $H=\zeta(2)-\sum_{d>\sqrt D}{\mu^2(d)\over\prod_{p|d}(p^2-1)}$, and the latter sum might allow you to replace $3\cdot2^{-1/3}$ with a better constant. $\endgroup$
    – TravorLZH
    Sep 4, 2022 at 7:33
  • $\begingroup$ I don't see a simple way (and it's not different from what I was saying). You are right, though, in that that sum should be about 2/(sqrt(D) log D), not 1/sqrt(D). $\endgroup$ Sep 4, 2022 at 11:29
  • $\begingroup$ One way would be to bound the sum from above by $\sum_{p>\sqrt{D}} 1/p^2$. Estimating that well requires PNT, but I don't see how to obtain the right leading term without using PNT anyhow. $\endgroup$ Sep 4, 2022 at 12:13
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Also: see Cohen-Dress, "Estimations numériques du reste de la fonction sommatoire relative aux entiers sans facteurs carrés", 1988 (https://www.imo.universite-paris-saclay.fr/~biblio/numerisation/docs/Colloque-theorie-analytique-nombres_J.Coquet_1985/pdf/Colloque-theorie-analytique-nombres_J.Coquet_1985.pdf), which gives several estimates, none of them strictly better or worse than the one in my answer (each one seems to be best in some range).

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