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It is known that the average number of divisors, calculated over all positive integers between $1$ and $N$, can be expressed using the classical Dirichlet formula as

$$\frac{1}{N} \sum_{n=1}^N d(n)= \log(N)+2 \gamma -1+O(N^{-\frac{1}{2}})$$

where $\gamma$ is the Euler's constant and $d(n)$ is the divisor function. I would like to know whether there is a similar asymptotic formula if we restrict the calculation, for any $n$, to a narrower range for the divisors. In particular, given $n$, we can consider only the divisors $<c \sqrt{n}\,$, where $c$ is a positive real number.

Let us call this restricted divisor function $d(n,c)\,$. For $c=1\,\,$, the resulting summatory function trivially becomes

$$\frac{1}{N} \sum_{n=1}^{N} d(n,1)= \frac{1}{2} \log(N)+ \gamma -\frac{1}{2}+O(N^{-\frac{1}{2}})$$

However, for $c \neq 1 \,$ I was not able to prove a general asymptotic formula. After some calculations, I guess that the constant term varies by $\log(c)$, but I would be happy to have a formal proof.

Anatoly
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  • Sorry, I don't quite follow. How is it any way a "restriction" to sum up to a different value of $N$? I don't see anywhere that the heights of the divisors are restricted, even though I'm guessing that's what you meant to write. – Erick Wong Sep 19 '17 at 21:42
  • how is $\frac{1}{N} \sum \limits_{n=1}^{\sqrt{N}} d(n) = \frac{1}{2} \ln N +\gamma -\frac{1}{2}+O(N^{-0.5}) $ is obvious ??! , numerical calculation suggest that it goes to $0$. – Ahmad Sep 19 '17 at 21:52
  • Thank you for your comments. I edited the question to better explain the problem. – Anatoly Sep 19 '17 at 22:16
  • It seems that the first term will not change $0.5 \ln N$ and what will change is the constant, being smaller for $c>1$ and bigger for $c<1$. – Ahmad Sep 19 '17 at 22:27
  • Yes, you are right. It seems that the constant term varies according to $\log(c),$. I am looking for a formal demonstration of this. – Anatoly Sep 19 '17 at 22:36
  • @Anatoly I think the main term should follow fairly easily by changing the order of summation. So is it the error term that you are having difficulty with? – Erick Wong Sep 20 '17 at 16:46
  • Yes, this is the case. – Anatoly Sep 21 '17 at 19:01
  • @Anatoly Unless I made a mistake, the most straightforward approach already gives the required error term (which would give a nice alternative proof of Dirichlet's hyperbola bound), except that the implied constant is not uniform in $c$. – Erick Wong Sep 22 '17 at 08:49

2 Answers2

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A simple change of summations gives a usable, albeit crude result:

$$\sum_{n=1}^N d(n,c) = \sum_{n=1}^N \sum_{\substack{d\mid n\\d<c\sqrt{n}}} 1 = \sum_{d=1}^{c\sqrt{N}} \sum_{\substack{d^2/c^2 < n \le N\\ n \equiv 0 \pmod d}} 1 = \sum_{d=1}^{\lfloor c\sqrt{N} \rfloor} \big\lfloor \frac{N}{d} \big\rfloor - \big\lfloor \frac{\lceil d^2/c^2 \rceil}{d} \big\rfloor \\ = \sum_{d=1}^{\lfloor c\sqrt{N} \rfloor} (\frac{N}{d} - \frac{d}{c^2}) + O(c\sqrt{N}) \\ = N(\log(c\sqrt{N}) + \gamma + O((c\sqrt{N})^{-1}) - (\tfrac12 N + O(c^{-1}\sqrt{N}) ) + O(c\sqrt{N}).$$

Thus the average value of $d(n,c)$ is $\frac12 \log N + \log c + \gamma - \frac12 + O_c(N^{-1/2})$, where the subscript in $O_c$ connotes that the implied constant may depend on $c$ (in this case it looks to be bounded by $O(c + c^{-1})$). Perhaps the error term can be made more uniform in $c$ by refining this calculation via Dirichlet's hyperbola method.

Erick Wong
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For the special case when $c = 1$ then from Olivier Bordellès "On restricted divisor sums associated to the Chowla-Walum conjecture", International Journal of Number Theory, V18 #01, pp 19-25, 2022 we have a more accurate summation where

\begin{equation*} \sum_{n \le x} \sum_{d \mid n, d \le \sqrt{n}} 1 \sim \frac{1}{2}\, x\, \log \left({x}\right) + \frac{1}{2} \left({2\, \gamma - 1}\right) x + \frac{1}{2} \sqrt{x} + O \left({{x}^{{\theta}_{0} + \epsilon}}\right). \end{equation*}

where for $x$ sufficiently large and all $\epsilon \in \left({0, 1/2}\right]$

\begin{equation*} {\theta}_{\alpha} = \frac{1}{2}\, \alpha + \begin{cases} \frac{1}{4}, & \text{ if the Chowla-Walum conjecture is true}, \\ \frac{517}{1{,}648}, & \text{ otherwise}. \end{cases} \end{equation*}

The average sum becomes

\begin{equation*} \frac{1}{x} \sum_{n \le x} \sum_{d \mid n, d \le \sqrt{n}} 1 \sim \frac{1}{2}\, \log \left({x}\right) + \frac{1}{2} \left({2\, \gamma - 1}\right) + \frac{1}{2} \frac{1}{\sqrt{x}} + O \left({{x}^{{\theta}_{0} - 1 + \epsilon}}\right). \end{equation*}

Lorenz H Menke
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