Asymptotic approximation of integers with exactly $k$ divisors

Question

Is there any asymptotic approximation for the count of integers with $k$ divisors?

For example, for $k=2$, these are simply the prime numbers - only they have $2$ divisors precisely. And for a given $x$, the approximation is: $\frac {x}{\ln x}$

I think I've seen patterns for other values of $k$, but they are just observations and I can't really prove it, and I might be wrong too.

So given $x$ and $k$, is there asymptotic approximation for the count of integers smaller than $x$ with exactly $k$ divisors?

For $4$ factors you have the semiprimes. A count can be seen in this question so it is greater by a factor $\log \log x$ — Ross Millikan, Jan 21 '23 at 00:04
The last entry https://mathworld.wolfram.com/RiemannZetaFunction.html might be useful. — JMP, Jan 21 '23 at 00:13

score 1 · Answer 1 · answered Jan 21 '23 at 00:09

Write $k=2^\ell q_1q_2,\dots q_j$ where $\ell\ge1$ and the $q_i$ are (not necessarily distinct) odd primes. One of the ways that $n=p_1^{r_1} \cdots p_m^{r_m}$ can have $k$ divisors (where the $p_m$ are the distinct primes dividing $n$) is if $m=j+\ell$ and $r_i = q_i-1$ for $1\le i\le j$ and $r_{j+1}=\cdots=r_m = 1$. The number of such integers up to $x$ will be asymptotic to some constant (depending on $q_1,\dots,q_j$) times $\frac1{(\ell-1)!} x(\log\log x)^{\ell-1}/\log x$, which is the asymptotic number of squarefree integers with $\ell$ prime factors; and one can show that all other ways of achieving exactly $k$ divisors contribute a smaller order of magnitude to the counting function.

If $k$ is odd, then there will be a similar formula with $\ell$ being the multiplicity of the smallest prime $s$ dividing $k$, and with $x$ replaced by $x^{1/(s-1)}$. (For example, if $k=5^\ell$, then every such integer up to $x$ is the $4$th power of a squarefree integer up to $x^{1/4}$ with exactly $\ell$ prime factors.)

Asymptotic approximation of integers with exactly $k$ divisors

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