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I am interested in the large n behaviour of this sum. Every sufficiently large integer can be written as the sum of two square free numbers so this is a well defined question.

I conjecture that this grows at most $O(n^{0.5+\epsilon})$

Any useful references on the Mobius function or a general discussion of this sum would be greatly appreciated !

user385459
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  • Shouldn't the representation function be given by $\sum_{r+s=n} \mu(t)^2 \mu(s)^2$ ? – Jack D'Aurizio Jun 23 '18 at 19:18
  • Square-free numbers have a positive density $(\frac{6}{\pi^2}=\frac{1}{\zeta(2)})$, hence the previous representation function is expected to grow like $C n+O(n)$. – Jack D'Aurizio Jun 23 '18 at 19:20
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    A useful identity for managing the sum without squares is $$ \sum_{p\text{ prime}}\frac{\log p}{p^s} = -P'(s) = -\sum_{k\geq 1} \mu(k)\frac{\zeta'}{\zeta}(ks),$$ holding for any $s$ such that $\text{Re}(s)>0$. I am slightly worried by the fact that a tight bound for your convolution might depend on the location of the non-trivial zeroes of the $\zeta$ function, like in Mertens' conjecture about $\sum_{n=1}^{N}\mu(n)$. – Jack D'Aurizio Jun 23 '18 at 19:23
  • I don't understand the remark about this being a well-defined question. Why wouldn't it be well-defined if not every sufficiently large integer could be written as a sum of two square-free numbers? – joriki Jul 20 '18 at 06:42
  • I think I just meant that the function of $n$ is well defined ! – user385459 Jul 21 '18 at 01:14

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