Why so many numbers with 12 divisors?

Question

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 $$\frac{\#\{n - \delta \leqslant m \leqslant n + \delta \ : \ \tau(m) = k \}}{2\delta} ?$$

Answer 1

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).

Answer 2

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 $\tau(n)=k$.