Let $\mathcal{F}_{Q;r,q}=\{\gamma=\frac{m}{n} | 0\leq m \leq n \leq Q, \gcd(m,n)=1, n \equiv r \mod q, \gcd(r,q)=1\}$. Usually, with no condition on arithmetic progression, then $\# \mathcal{F}_{Q}$ (cardinality) $=\varphi(1)+\varphi(2)+\varphi(3)+\cdots+\varphi(Q)$ $=\frac{3Q^2}{\pi^2}+O(Q\log Q)$ as $Q\rightarrow \infty$, but now how to get the cardinality for $\mathcal{F}_{Q;r,q}$ as $Q \rightarrow \infty$?
Farey fractions in arithmetic progression
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analytic-number-theory
additive-combinatorics
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0I suppose $\phi(\cdot)$ refers to the Euler phi (totient) function, sometimes denoted $\varphi(\cdot)$. – 2011-10-30
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0Have you tried looking at a proof of the estimate for the sum of the phi-values and adapting it to the sum over an arithmetic progression? – 2011-10-30
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0So I have tried $$\#\mathcal{F}_{Q;r,q}=\frac{1}{\varphi(q)}\sum_{n \leq Q} \varphi(Q) \sum_{\chi \mod q}\chi(n)\bar{\chi(r)}$$, which use $\varphi(n)=\sum_{d|n} \frac{\mu(d)}{d}$ becomes $$\frac{1}{\varphi(q)}\sum_{\chi \mod q} \bar{\chi(r)}\sum_{d \leq Q}\mu(d)\chi(d)\sum_{m \leq \frac{x}{d}}m \chi(m)$$, but then I don't know how to continue. – 2011-10-31
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0Are we given that $r\in(\frac{Z}{qZ})^{\times}$? If so we could use symmetry arguments. – 2013-01-23