Haw can I estimate $$ \min_{q\le\sqrt{x}}\frac{x}{q}\sum_{\substack{1\le a<q,\\(a,q)=1}}\sum_{\substack{1\le b<a,\\ (b,a)=1}} \frac{\varphi(a+bq)}{a(a+bq)}, $$ where $\varphi(n)$ is Euler Totient function, as $x$ goes to infinity?
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$\begingroup$ What have you tried? In general, for almost all n, $\phi(a+bq)$ will be close to $a+bq$ on average. So one should expect that a very good approximation will come from just replacing everything in the final term with $\frac{1}{a}$. My guess is they will have the same asymptotics up to a constant factor. $\endgroup$– JoshuaZCommented Dec 6, 2020 at 15:58
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1$\begingroup$ The idea here is that for most numbers $\phi(n)$ is close to a constant times $n$. Classically we have $\sum_{k=1}^n \phi(k) \approx \frac{3}{\pi^2}n^2$.. Many other sums involving $\phi$ end up looking up to a constant the same as if one replaces $\phi(n)$ with $n$. So this should give you a good guess for the growth rate of the function. $\endgroup$– JoshuaZCommented Dec 6, 2020 at 18:09
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1$\begingroup$ Because I’m providing the complete details, I’d leave this here as (an extended) comment. The double sum is equivalent to $$\frac{x}{q}\sum_{\substack{1\le a\le q\\(a,q)=1}}\frac{1}{a}\sum_{\substack{n\equiv a\pmod q\\(a,n)=1\,,n<(a+1)q}}\frac{\varphi(n)}{n}= \frac{x}{q^2}\sum_{\chi\mod q}\sum_{a=1}^q\frac{\overline{\chi(a)}}{a}\sum_{(a,n)=1\,,n<(a+1)q}\frac{\chi(n)\varphi(n)}{n}\,,$$ where $\chi$ is Dirichlet character......:. $\endgroup$– Jack L.Commented Dec 6, 2020 at 18:23
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2$\begingroup$ .....Then by considering the $L$-function $$ L_a(s,\chi):= \sum_{\substack{n\ge 1\\,(a,n)=1}}\frac{\chi(n)\varphi(n)}{n}=\prod_{\substack{p~prime\\p\nmid a}}\left(1+(1-p^{-1}\frac{\chi(p)p^{-s}}{1-\chi(p)p^{-s}}\right)\,$$ it appears that the main term will come from the estimates springing from the L-series modulo the principal character, that is from $L_a(s,\chi_0)=\frac{\zeta(s-1)}{\zeta(s)}$. $\endgroup$– Jack L.Commented Dec 6, 2020 at 18:24
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2$\begingroup$ In the first comment above, I meant to say, “because I’m not providing complete details...” and in the Euler product, a parenthesis is ommited; should be $$1+(1-p^{-1})\frac{\chi(p)p^{-s}}{1-\chi(p)p^{-s}}\,.$$ $\endgroup$– Jack L.Commented Dec 6, 2020 at 18:50
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