Number of integers less than $x$ with $k$ prime divisors (not necessarily different)

Question

Let be $P_k = \{n | n = p_1 \cdot \ldots \cdot p_k, p_i = \text{prime numbers not necessary different} |\}$

$P_1 = \text{set of primes }$, $P_2 = \text{integers with exactly two primes factors}$, ....

We know that $|\{n \in P_1, n \leq x \}| = \pi(x) \sim x/\log x$-.

Is there any approximation for $$ | \{ n \leq x | n \in P_k \} | = |\{ n \leq x | n = p_1 \cdot \ldots \cdot p_k, p_i = \text{prime numbers not necessary different}\}|$$

Answer 1

$$\pi_k(x)\sim\frac{x(\log\log x)^{k-1}}{(k-1)!\log x}$$

Exact formula for $\pi_k(x)=P_k(x,0)$ here:

https://mathoverflow.net/questions/297785/prime-counting-meissel-lehmer-is-there-a-general-formula/300060#300060

Answer 2

Your function is the divisor function, $\sigma_0(n)$, given in OEIS A000005. I didn't find anything about the asymptotic growth rate easily. Note that it is $2$ whenever $n$ is prime, so you need to ask about the maximum up to $n$ or something like that. Mathworld gives $$\frac{\sum_{k=1}^n \sigma_0(k)}n\sim \log n+2\gamma-1$$