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In this post $\sigma(n):=\sum_{d|x}{d}$ which is called the divisor function or sometimes the sum-of-the-divisors function.

Is there an asymptotic formula for $$f(a,b,x):=\sum_{an+b\le x}\sigma(an+b)$$

For the case $a=1, b=0$ we have that the $$\sum_{n\le x}\sigma(n)=\frac{\pi^2}{12}x^2+O(x\log(x))$$

And I am looking for if there is a quick and easy way to modify the constant $\frac{\pi^2}{12}$ so that we can get bounds on a slightly broader class. If I had to eyeball it: It looks to me like $$\sum_{2n+1\le x}\sigma(2n+1)\approx \frac{\pi^2}{32}x^2$$

Which would mean that $\sum_{2n\le x}\sigma(2n)\approx \frac{5\pi^2}{96}x^2$ but I am really just speculating based on graphs here.

It should be the case however that

$\sum_{b=1}^a f(a,b,x)=f(1,0,x)=\frac{\pi^2}{12}x^2$. That may prove helpful in this puzzle I am not so sure.

Thanks

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    $\begingroup$ For a Dirichlet character $\sum_{n\le x}\chi(n)\sigma(n)=Res(\frac{L(s,\chi)L(s-1,\chi)}{s},2)+O(x\log(x))$ whence for $(a,q)=1$, $\sum_{n \le x} \sigma(n) 1_{n \equiv a \bmod q} = \frac{1}{\varphi(q)}\sum_{\chi \bmod q} \overline{\chi(a)} \sum_{n\le x}\chi(n)\sigma(n)=\frac{\frac{\pi^2}{12} \prod_{p | q}(1-p^{-1})(1-p^{-2})}{\varphi(q)}x^2+O(x \log(x))$ $\endgroup$ – reuns Oct 27 '18 at 1:49
  • $\begingroup$ The idea is similar with $(a,q) = d$ because $\frac{\sigma(dn)}{\sigma(d)}$ is multiplicative and its Dirichlet series is close to $\zeta(s)\zeta(s-1)$. $\sum_{\chi \bmod q}$ is a sum over the $\varphi(q)$ Dirichlet characters modulo $q$. $\endgroup$ – reuns Oct 27 '18 at 2:06
  • $\begingroup$ Yes but it would help if you show us your proof that $\sum_{n\le x}\sigma(n)=Ax^2+O(x\log(x))$ $\endgroup$ – reuns Oct 27 '18 at 23:13
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    $\begingroup$ If you understand math.stackexchange.com/questions/2954097/… then can you adapt it for $\sum_{n \le x} \chi(n)\sigma(n)$ and $\sum_{n \le x} \chi(n)\frac{\sigma_k(dn)}{\sigma_k(d)}$ $\endgroup$ – reuns Oct 27 '18 at 23:35
  • $\begingroup$ In your linked post there is no complex analysis and residues. The tools needed to answer "yes" as I did without taking care of the details appear naturally in the proof of Dirichlet's theorem on arithmetic progressions and of the PNT. $\endgroup$ – reuns Oct 28 '18 at 0:11

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