# Why so many numbers with 12 divisors?

For some purposes I have been checking whether or not there were many numbers around 600000 with exactly 12 divisors. I have been struck by the fact that more than 10% of the numbers near 600000 (by less than 10000) have exactly 12 divisors. I know the divisor function $\tau(n)$ is really small, but I do not understand the distribution of the numbers such that $\tau(n)=k$ fixed. So:

What do we know about the set of numbers $n$ such that $\tau(n)=k$?

I would like to know if the situation above is exceptional or if it is that previsible by some analytic number theory. Maybe is it possible to address the following question:

Let $n \in \mathbb{N}$, $\delta < n$, and $k \in \mathbb{N}$. Is it possible to estimate the natural density $$\frac{\#\{n - \delta \leqslant m \leqslant n + \delta \ : \ \tau(m) = k \}}{2\delta} ?$$

• Interesting! But can you substantiate that frequency of $\tau(n)=12$ is uncharacteristically high in this region only? All the factorization patterns $n\in\{p_1^{11},p_1^5p_2, p_1^3p_2^2,p_1^2p_2p_3\}$, $p_i$ distinct primes, lead to $\tau(n)=12$. It may be possible to estimate (don't look at me) the frequency of any given pattern in a given range, and these patterns dominate for numbers in this range. It may, later on, happen that $24$ becomes more common. How far have you looked? Commented Mar 13, 2017 at 16:53
• Most numbers are highly composite in the sense that if $V(n,k)$ is the number of natural numbers $m\leq n$ such that $m$ has at most $k$ distinct prime divisors, then $V(n,k)/n\to 0$ as $n\to \infty.$ In particular $V(n,3)/n\to 0$ as $n\to \infty.$ Commented Jun 7, 2017 at 4:32

We know for a fact that, given $n=p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where $p_i$ are distinct primes and $e_i\geq 1$:

$$\tau(p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m})=(e_1+1)(e_2+1)\cdots(e_m+1)$$

So $\tau(n)=12$ just has so many solutions because $12$ is relatively small but has many divisors itself, so if we write $12=(e_1+1)(e_2+1)\cdots(e_m+1)$ then $n=p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ is a solution; and since each $e_i$ is small too, the primes $p_i$ (which are not relevant for the number of divisors) can get larger for $n$ to be under $600000$.

In short, small numbers $k$ with a lot of divisors have many solutions to $\tau(n)=k$.

The equation $\tau(n)=k$ for fixed positive integer $k$ has always infinitely many solutions, because $n=p^{k-1}$ is a solution to $\tau(n)=k$ for all primes $p$. But there are infinitely many primes. So in this respect $k=12$ is not special. Concerning density it depends on the definition (of density we want to consider).

• Thanks for the comment, I wonder specifically about density for it is more suitable to my question: we stuck to a fixed interval for the solutions, viz. $n$ around 600000 (or whatever else). So my question emphases on the possibility to be more precise, i.e. quantifying the answer of vrugtehagel. Commented Mar 13, 2017 at 12:26
• You are welcome. My "comment" shows that you can clarify your question " but I do not understand the distribution of the numbers such that $\tau(n)=k$ fixed". The is a natural density , but also several other definitions. Commented Mar 13, 2017 at 12:29
• The natural density is quite enough for having an idea, hence I edited my question accordingly. Commented Mar 13, 2017 at 16:18