1
$\begingroup$

Let $\mu(n)$ be the Mobius function. Let us define $\mu^+(n)$ to be $\mu(n)$ if $\mu(n)>0$ and $0$ otherwise. Is there a known asymptotic formula for $$ \sum_{n \leq N} \mu^+(n), $$ and similarly for $$ \sum_{ \substack{n \leq N \\ n \equiv a (\mod q)}} \mu^+(n). $$ I would greatly any appreciate references or comments. Thank you very much.

$\endgroup$
  • 7
    $\begingroup$ I think the results you seek probably follow easily from the identity $\mu^{+}(n) = \frac{\mu(n)^{2} + \mu(n)}{2}$ combined with known results about $\sum_{n \equiv a \pmod{q}, n \leq x} \mu(n)$, and the count of the number of squarefree integers in arithmetic progressions. $\endgroup$ – Jeremy Rouse Jun 1 '17 at 14:24
4
$\begingroup$

I'll have a stab at this. The relation $$ \sum_{n\leq X:n\equiv a~(mod~q)} \mu^2(n)=\frac{6}{\pi^2} \prod_{p|q} \left(1-\frac{1}{p^2} \right)\frac{X}{q}+E(X,q,a) $$ where the error term $E$ is $O_{\varepsilon}\left(\sqrt{X/q} +q^{\frac{1}{2}+\varepsilon}\right) $ provided $q\leq X^{\frac{2}{3}-\varepsilon},$ together with some recent results (assuming GRH) of the form

$$ \sum_{n\leq X:n\equiv a~(mod~q)} \mu(n) \ll_{\varepsilon} \sqrt{X} \exp\left( (\log X)^{3/5} (\log \log X)^{(16/5)+\varepsilon} \right) $$

(see the paper on arxiv here) and the observation $\mu^{+} (n)=\frac{1}{2}(\mu(n)+\mu^2(n))$ suffices to yield an answer.

Edit The paper here gives the bound $$ O\left( \frac{X}{q\log^C X}\right) $$

for any $C>0$, for $X$ large enough.

$\endgroup$
  • $\begingroup$ Is it possible to give me references where RH/GRH is not assumed? Thank you very much! $\endgroup$ – Johnny T. Jun 2 '17 at 15:12
  • $\begingroup$ Could you possibly provide me the reference for the first estimate also? Thank you very much! $\endgroup$ – Johnny T. Jun 2 '17 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.