Asymptotic approximation of integers with exactly $k$ divisors 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?
 A: 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.)
