# Distribution of pre-images of the divisor function $\sigma$

If $$A\subseteq\mathbb{N}$$ is a subset of the positive integers, we let $$\mu^+(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$$ be the upper density of $$A$$.

For $$n\in\mathbb{N}$$ we let $$\sigma(n)$$ be the number of divisors of $$n$$, the numbers $$1$$ and $$n$$ included.

Do we have $$\mu^+\big(\sigma^{-1}(\{k\})\big) = 0$$ for all $$k\in\mathbb{N}$$? If not, what is the value of $$\sup\big\{\mu^+\big(\sigma^{-1}(\{k\})\big):k\in\mathbb{N}\big\}$$?

## 2 Answers

Notice that $$\sigma(p^{k-1}) = k$$ and so the image of $$\sigma$$ is all of $$\mathbb{N}$$.

By the way, $$\sigma$$ is usually used for the sum of divisors function, and it is more standard to use $$d$$ or $$\tau$$ for your function.

EDIT: I misread the question. I will use $$\tau$$ instead of $$\sigma$$.

I claim that $$\mu ^ {+}(\tau^{-1}(\{k\})) = 0$$. Take a number $$m$$ in this set, and let us look at $$m$$'s prime factorization: $$m = p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_r^{\alpha_r}$$. Notice that there are finitely many options for the $$\alpha_i$$ (up to a permutation), because $$(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_r + 1)=k$$, so it is enough to show that the upper density of numbers of the form $$p_1 ^ {\alpha_1} p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$$ where $$r, \alpha_i$$ are fixed is zero.

Let us look at numbers in this set that are at most $$x$$. Then if we fix $$p_1$$, we need to choose primes $$p_2, \cdots p_r$$ such that $$p_2 ^{\alpha_2} \cdots p_r^{\alpha_r} \leq \frac{x}{p_1 ^{\alpha_1}}$$.

By induction we can assume that the amount of numbers of the form $$p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$$ which are at most $$x$$ is $$o(x)$$, and if $$\alpha_1 \geq 2$$ then this shows that the amount of numbers of the form $$p_1 ^{\alpha_1} p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$$ is $$o(x)$$ by summing over the options of $$p_1$$ (and using the fact that $$\sum_{p} \frac{1}{p^2} \leq \sum_{n} \frac{1}{n^2}$$ converges. Therefore it is enough to solve in this case where all $$\alpha_i$$ are 1, that is to show that the amount of numbers of the form $$p_1 \cdots p_r$$ up to $$x$$ is $$o(x)$$ (for $$r$$ fixed).

Fixing $$p_1$$ we see that $$p_2$$ can be any prime that is at most $$\frac{x}{p_1}$$. and then $$p_3$$ can be anything that is at most $$\frac{x}{p_1 p_2}$$, ... and $$p_r$$ is at most $$\frac{x}{p_1 p_2 \cdots p_{r-1}}$$. So we see that the amount of numbers at most $$x$$ is

$$\sum_{p_1 \leq x} \sum_{p_2 \leq \frac{x}{p_1}} \cdots \sum_{p_{r-1} \leq \frac{x}{p_1 \cdots p_{r-2}}} \pi (\frac{x}{p_1 \cdots p_{r-1}})$$

From here we can use the simple bound $$\pi (x) \leq \frac{cx}{log x}$$ for some constant $$c$$ and see that this sum is small.

• This question is not about the image of $\sigma$, but about pre-images / fibers. Is the wording of the question bad, or is the question hard to understand? It is also possible that the question contains an error. I am open to amending the question – Dominic van der Zypen Oct 26 '20 at 10:44

$$\tau(n) \le k$$ implies that $$n=\prod_{i=1}^j p_i$$ with $$j\le k$$, thus $$\sum_{n=1,\tau(n)\le k}^\infty n^{-s} \le (1+\sum_{p \ prime} p^{-s})^k=\sum_n a_k(n) n^{-s}$$

(a coefficient-wise bound)

$$1+\pi(x)=O(x/\log x)=O(\sum_{n\le x} 1/\log n)$$ and $$x/\log x=O(\pi(x))$$ imply that $$f_k(x)=\sum_{n\le x} a_k(n) =\sum_{n\le x} \frac1{\log n}f_{k-1}(x/n)$$ $$=O(\sum_{n\le x} \frac1{\log n} \frac{x/n}{\log x/n}(\log \log x/n)^{k-1})=O(\frac{x (\log\log x)^{k-1}}{\log x})$$