Let $m$ an arbitrary integer. I would like to know if exists an estimation of $$\underset{\left(q,\, m\right)>1}{\underset{q>1}{\sum}}\frac{1}{\phi\left(q\right)\phi\left(q/\left(q,\, m\right)\right)}$$ or something similar, where $(q, m)$ is the g. c. d. of $q$ and $m$.

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    $\begingroup$ You can fix $d=(q,m)$ and sum over $q\equiv 0\pmod d$. It gives an estimate $\ll\log\log m$. $\endgroup$ – Alexey Ustinov Dec 28 '13 at 12:15
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    $\begingroup$ "or something similar"? should we ask the question as well as answering it? (half-joking) $\endgroup$ – Greg Martin Dec 28 '13 at 16:22

\begin{align*} \sum_{\substack{q>1 \\ (q,m)=1}} \frac{1}{\phi(q)\phi\big(q/(q,m)\big)} &= \sum_{\substack{d\mid m \\ d>1}} \sum_{\substack{q>1 \\ (q,m)=d}} \frac{1}{\phi(q)\phi(q/d)} \\ &\le \sum_{\substack{d\mid m \\ d>1}} \sum_{r=1}^\infty \frac{1}{\phi(rd)\phi(r)} \\ &\le \sum_{\substack{d\mid m \\ d>1}} \frac1{\phi(d)} \sum_{r=1}^\infty \frac{1}{\phi(r)^2} < C \sum_{d\mid m} \frac1{\phi(d)}, \end{align*} where $C = \sum_{r=1}^\infty \frac{1}{\phi(r)^2}$. (The first inequality is because we've dropped the condition $(r,m/d)=1$; the second inequality comes from the elementary $\phi(ab) \ge \phi(a)\phi(b)$.)

As Alexey commented, this last sum is $\ll \log\log m$, although seeing so takes a few steps. Simplest perhaps is to note that $f(m) = \sum_{d\mid m} \frac1{\phi(d)}$ is a multiplicative function of $m$ satisfying $f(p^k) \le \frac p{p-1} \frac{p^2+2}{p^2}$ for every prime power $p^k$, whence $$ f(m) = \prod_{p^k\|m} f(p^k) \le \prod_{p^k\|m} \frac p{p-1} \prod_{p^k\|m} \frac{p^2+2}{p^2} < \frac m{\phi(m)} \prod_p \frac{p^2+2}{p^2}; $$ it's a standard estimate that $\frac m{\phi(m)} \ll \log\log m$, while the last product converges to some constant.

It would even be possible to get an explicit evaluation of the original sum, via: \begin{align*} \sum_{\substack{q>1 \\ (q,m)=1}} \frac{1}{\phi(q)\phi\big(q/(q,m)\big)} &= \sum_{\substack{d\mid m \\ d>1}} \sum_{\substack{r\ge1 \\ (r,m/d)=1}} \frac{1}{\phi(rd)\phi(r)} \\ &= \sum_{\substack{d\mid m \\ d>1}} \frac1{\phi(d)} \sum_{r=1}^\infty \begin{cases} \frac{\phi(d)}{\phi(rd)\phi(r)}, &\text{if } (r,m/d)=1, \\ 0, &\text{if } (r,m/d)>1. \end{cases} \end{align*} For a given $m,d$, the inner summand is a multiplicative function of $r$, and so the inner sum can be converted into its Euler product. That would leave a sum of the form $\sum_{d\mid m} g(d)$ for some multiplicative function $g$ (or something similar), which can be evaluated as an explicit multiplicatve function of $m$.

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  • $\begingroup$ Thank you for your answer, it's very useful. Or something similar ;) $\endgroup$ – Marco Cantarini Dec 28 '13 at 23:50
  • $\begingroup$ 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 $\underset{q>P}{\sum}\frac{1}{\phi\left(q\right)^{2}}\ll P^{-1}$)? $\endgroup$ – Marco Cantarini Dec 29 '13 at 9:31
  • $\begingroup$ 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.) $\endgroup$ – Greg Martin Dec 29 '13 at 18:03
  • $\begingroup$ Can you explain what do you mean by "something a little worse than the number of divisor of $m$"? $\endgroup$ – Marco Cantarini Jan 3 '14 at 22:47
  • $\begingroup$ 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. $\endgroup$ – Greg Martin Jan 4 '14 at 0:30

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