Consider the sum
$$ S(x)=\sum_{\substack{{d\mid P(\sqrt{x})}\\{d\leq x}}}|\mu(d)|, $$
where $P(x)$ is the product of all primes less than or equal to $x$, and $\mu(d)$ is the Möbius function.
Q: What is the asymptotic value of $S(x)$?
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4$\begingroup$ You're just asking for square-free numbers below $x$ with no prime factor larger than $\sqrt{x}$. The answer is $\sim (6/\pi^2)x (1-\log 2)$, where $(6/\pi^2)$ comes from the square-free-ness, and $1-\log 2$ from having no prime factor larger than $\sqrt{x}$. $\endgroup$– LuciaCommented Apr 12, 2016 at 1:19
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$\begingroup$ @Lucia: Thanks! Guess I should have seen this. My motivation for asking the question is trying to understand the behaviour of the sum $\sum_{\substack{ {d\mid P(\sqrt{x}) }\\ {d\leq x} \\ {|\mathcal{A}_d(a)| = \lfloor x/d \rfloor+1 } }} \mu(d)$, described in this MO question under the heading "Legendre sieve perspective". I'm a bit stuck here and would be grateful for any suggestions on how to attack the problem. $\endgroup$– user45947Commented Apr 12, 2016 at 11:25
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