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Jan 4, 2014 at 0:30 comment added Greg Martin A multiplicative function that is bounded by something like the number of divisors of $m$ times $m/\phi(m)$, or a constant times $(m/\phi(m))^2$, or something like that. I think if you go through the proof, you'll see how it arises.
Jan 3, 2014 at 22:47 comment added Marco Cantarini Can you explain what do you mean by "something a little worse than the number of divisor of $m$"?
Dec 29, 2013 at 18:03 comment added Greg Martin It's not quite that simple, unfortunately: the bound $q>P$ turns only into $r>P/d$ when we change variables. So you can get the $P^{-1}$ but at a cost of changing the dependence on $m$ from $\log\log m$ to something a little worse than the number of divisors of $m$. (I don't know whether that's inevitable or just an artifact of the proof.)
Dec 29, 2013 at 9:31 comment added Marco Cantarini in the first sums you obtain $C\underset{d\mid m}{\sum}\frac{1}{\phi\left(d\right)}$. If we fix $P>0$ and if we consider $\underset{\left(q,\, m\right)>1}{\underset{q>P}{\sum}}\frac{1}{\phi\left(q\right)\phi\left(q/\left(q,\, m\right)\right)}$ can we obtain a bound like $P^{-1}\log\log m$ (using the elemntary fact that $\underset{q>P}{\sum}\frac{1}{\phi\left(q\right)^{2}}\ll P^{-1}$)?
Dec 28, 2013 at 23:50 vote accept Marco Cantarini
Dec 28, 2013 at 23:50 comment added Marco Cantarini Thank you for your answer, it's very useful. Or something similar ;)
Dec 28, 2013 at 16:45 history answered Greg Martin CC BY-SA 3.0