3
$\begingroup$

I am interested in the function $f_n(m)$ which can be defined by the Dirichlet generating function $$\zeta(s)\zeta(s - 1) \cdots\zeta(s - n + 1) = \sum\limits_{m = 1}^\infty \frac{f_n(m)}{m^s} $$

This is called the number-of-sublattices function. It is straightforward to get the standard estimate of $\sum\limits_{m \leqslant x} f_n(m)$:

$$\sum\limits_{m \leqslant x} f_n(m) = \frac{x^n}{n} \prod\limits_{\ell = 2}^{n - 1} \zeta(\ell) + O\left( x^{n - 1}\log x \right) \tag 1$$

I am looking for an improvement in the error term. The sharpest known estimate for $\sum\limits_{m \leqslant x} \sigma_1(m) $ is known due to Walfisz (Chapter III in Weylsche Exponentialsummen in der neueren Zahlentheorie):

$$\sum\limits_{m \leqslant x} \sigma_1 (m) = \frac{\pi^2}{12} x^2 + O(x\log^{2/3}x) \tag 2$$

Since $\zeta(s)\zeta(s - 1) = \sum\limits_{m = 1}^\infty \frac{\sigma_1(m)}{m^s}$, I have been wondering if it is possible to get a sharper bound for the error term of $(1)$ using $(2).$

So far, I have tried estimating $f_3(m) = \sum\limits_{d\mid m} \sigma_1(d)\left( \tfrac{m}{d} \right)^2 $ but without success. I have tried summation interchange identities as well as the Dirichlet hyperbola method. I also tried to do some work involving the Fourier series representation of the fractional part, again unsuccessfully. Does anybody know of any theorems or techniques which could be use $(2)$ to obtain a better error term in $(1)$?

Improve this question
$\endgroup$
4
  • $\begingroup$ Have you tried reading Walfisz's article? $\endgroup$
    – Daniel Weber
    Commented Apr 5 at 10:40
  • $\begingroup$ You can't expect a much better error: if the error in (1) is $o(x^{n-1})$ then it follows that $f_n(m)=c_n(m^n-(m-1)^n) +o(m^{n-1}) \sim c_n n m^{n-1}$ holds for $c_n=\prod_{\ell=2}^{n}\zeta(\ell)/n$. And this asymptotic formula is false. $\endgroup$
    – Ofir Gorodetsky
    Commented Apr 5 at 19:50
  • 1
    $\begingroup$ To try and exploit (2), try to use partial summation first to estimate the partial sums of the function $g_n$ whose Dirichlet series is $\zeta(s-n+1)\zeta(s-n+2)$, which is nothing but $\sigma_1(m)$ times $m^{n-2}$. Then, express $f_n$ as $g_n$ convolved with a simple function. $\endgroup$
    – Ofir Gorodetsky
    Commented Apr 5 at 19:53
  • $\begingroup$ We have $f_3(p)=p^2+p+1$ when $p$ is prime. Maybe we can use the "powerful number trick" (see gbroxey.github.io/blog/2023/04/30/mult-sum-1.html). To do that you need a function $g$ which we can sum relatively quickly and such that $g(n)=n^2+n+1$. Maybe you've already tried... $\endgroup$
    –  Babar
    Commented Apr 5 at 20:39

0

Reset to default

You must log in to answer this question.