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What is the best known bound for the Mertens function along arithmetic progressions? More specifically, what is the best bound known for

$$\sum_{n<x}\mu(kn)$$

as $k,x\to\infty$. This paper of Lynelle Ye gives a very complete solution, but only under the RH so it is not much use to me. This paper about the Mertens function along arithmetic progressions $an+b$ is also promising, but it works under the assumption $\gcd(a,b)=1$ the whole time, meaning that $b=0$ can not be applied.

Working out simple bounds using Perron's formula does not look very hard, but seeing as this seems like an incredibly natural result to already have been proved it feels like it would not be the best idea to put all this time into proving a weaker version of a result that already exists.

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  • $\begingroup$ This is the classical problem of summing a multiplicative function -- you sum $\mu$ up to $y:=x/k$ along integers coprime to $k$ (a multiplicative condition). $\endgroup$ – Ofir Gorodetsky Jan 18 at 21:28
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We use the bound given by my previous answer: $$\sum_{n<x}\mu(kn)\ll_A \frac{kx}{(\log (kx))^A}.$$ Now we improve this bound by using that the sum $$\sum_{d \in \mathbb N\atop d\mid n, d\mid k}\mu(d)$$ is $1$ or $0$ according to if $\gcd(n,k)=1$ or not. We get $$\sum_{n<x}\mu(kn)=\mu(k)\sum_{n<x\atop\gcd(k,n)=1}\mu(n)=\mu(k)\sum_{d\mid k}\mu(d)\sum_{n<x\atop d\mid n}\mu(n)=\mu(k)\sum_{d\mid k}\mu(d)\sum_{t<x/d}\mu(td),$$where we wrote $n=td$. Now observe that the last sum over $t$ is one that we can bound by using what we proved in the previous answer. This gives the bound $$\mu(k)\sum_{d\mid k}\mu(d)O_A(\frac{d \cdot x/d}{(\log x)^A})\ll_A \tau(k)\frac{x}{(\log x)^A},$$ where $\tau(k)=\sum_{d\mid k}1$ is the sum of the divisors function and the implied constant is independent of $x$ and $k$. The bound $\tau(k)\frac{x}{(\log x)^A}$ is exactly what the Perron approach would give, furthermore, this bound is ok for very large values of $k$, for example, if $k<x^{100}$ the average value of $\tau(k)$ is $\log (x^{100})\ll \log x$ so you could take simply $A>1$.

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    $\begingroup$ I'm not quite sure if I'd agree that $\tau(k)$ is the constant that Perron's formula would give, since my idea would be to use Lemma 3 in the paper Mean Values of Multiplicative Functions by Montgomery and Vaughan which would give a k independent bound (deepblue.lib.umich.edu/bitstream/handle/2027.42/43188/…). This results does look very good though, and I think it will work for my purposes $\endgroup$ – Milo Moses Jan 19 at 23:22
  • $\begingroup$ Interesting, thanks, I didn't know about Lemma 3. $\endgroup$ – Dr. Pi Jan 19 at 23:29
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As Ofir says this shouldn't be a problem to prove directly using Perron. But let me describe another way to get the same result: we shall use the well-known result of Davenport that for any $A>0$ one has $$ \sup_{\alpha \in \mathbb R} \left|\sum_{n <y} \mathrm{e}^{2\pi i \alpha n}\mu(n)\right|\ll_A\frac{y}{(\log y)^A}.$$ We now write $$ \sum_{n<x} \mu(kn)=\sum_{m< kx\atop m\equiv 0\mod k}\mu(m).$$ Then we use the fact that the geometric series $$\frac{1}{k}\sum_{t=0}^{k-1} \mathrm{e}^{2\pi imt/k}$$ is $1$ or $0$ respectively when $k$ divides $m$ or not.Therefore, $$\sum_{m< kx\atop m\equiv 0\mod k}\mu(m)=\frac{1}{k}\sum_{t=0}^{k-1} \sum_{m<kx }\mu(m)\mathrm{e}^{2\pi i t m/k }.$$ Now the sum over $m$ is $O_A(kx/(\log (kx))^A)$ independently of $t$ and $k$. Hence, $$\frac{1}{k}\sum_{t=0}^{k-1} O_A(kx/(\log (kx))^A)=O_A(kx/(\log (kx))^A).$$ * Note that if $k$ is as small as $(\log x)^C$ for some fixed $C>0$ then you can replace $A$ by $A+C$ so that the bound becomes $O_A(x/(\log x)^A)$ for any fixed $A$.

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  • $\begingroup$ I have heard this result of Davenport stated many times, but I do not know its original source. Do you have a reference? $\endgroup$ – Milo Moses Jan 19 at 22:14
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    $\begingroup$ Do you mean $m<kx$ in your second line of math? $\endgroup$ – Milo Moses Jan 19 at 22:39
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    $\begingroup$ H. Davenport. On some infinite series involving arithmetical functions, II. Quart. J. Math. 8 (1937), pages 313–320. $\endgroup$ – Dr. Pi Jan 19 at 22:55
  • $\begingroup$ And yes I fixed the $m<kx$ problem. $\endgroup$ – Dr. Pi Jan 19 at 23:01
  • $\begingroup$ The applications I am thinking of require us to be able to take $k$ large w.r.t $x$, and so this bound isn't entirely useful to me anymore which is why I unaccepted it. I still really do appreciate the technique. $\endgroup$ – Milo Moses Jan 19 at 23:04

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