Let $$S_a(N)=\sum_{n\le N}\frac{\varphi(an)}{n^2}.$$ The usual machinery gives an asymptotic formula $$S_a(N)=\frac1{\zeta(2)}\cdot\frac{a^2}{\varphi_+(a)}\log N+C(a)+O(N^{-1+\varepsilon}a^{1+\varepsilon}),$$ where $C(a)$ some complicated function and $$\varphi_+(a)=a\prod_{p\mid a}\left(1+\frac1p\right).$$ Is it possible to give a reference on this asymptotic formula? (My proof is rather long.)
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asked Jan 3, 2014 at 5:03
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1$\begingroup$ I know you asked for a reference, not a proof, and I don't know of one. But I will note that $$S_a(N) = \phi(a) \sum_{n\le N} \frac{\phi(an)}{n^2\phi(a)},$$ where the summand is a multiplicative function of $n$. And sums of this type can be evaluated by off-the-shelf theorems. See for example equation (A.27) of my paper math.ubc.ca/~gerg/papers/downloads/AFNSVP.pdf , although the literature cited at the beginning of Appendix A might serve you better if you went there directly (particularly if you want an explicit secondary term $C(a)$). $\endgroup$– Greg MartinJan 3, 2014 at 6:14
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$\begingroup$ I can prove it using Perron's formula as well. But I want to save a space in the article. $\endgroup$– Alexey UstinovJan 3, 2014 at 7:34
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I have found a proof of more general formula in the book Postnikov, A. G. Introduction to analytic number theory American Mathematical Society, 1988, (section 4.2). This proof is simple but it has a small mistake inside. (For arithmetic progression starting from $0$ this mistake vanishes.)
Please give more references if you know ones.
answered Jan 3, 2014 at 13:53