For any real number $x$, let's define $Om_{k}(x)$ as the number of positive integers $m$ below $x$ such that $\Omega(m)\omega(m)=k$, where $\omega(n)$ is the number of distinct primes dividing $n$, and $\Omega(n)$ the total number of prime factors of $n$ counted with multiplicity. Obviously $Om_{0}(x)$ is just the number of squarefree integers below $x$.
Do we know asymptotics and (maybe conditional) error terms for $Om_{k}(x)$? What would be the consequences of RH on such error terms?
Thanks in advance.

1$\begingroup$ There is a nice discussion of this question in Kac, Statistical Independence in Probability, Analysis, and Number Theory in the MAA Carus Monograph series, pages 6471. $\endgroup$– Gerry MyersonCommented Jul 22, 2014 at 0:21
2 Answers
This problem was studied by Renyi, in
On the density of certain sequences of integers
Publ. Inst. Math. Belgrade 8 (1955) 157162.
http://elib.mi.sanu.ac.rs/files/journals/publ/14/13.pdf
Let $d_k = \lim_{x\to\infty} Om_k(x)/x$. Then the $d_k$ are the coefficients in the following beautiful identity, valid for $z < 2$: $$ \sum_{k=0}^{\infty} d_k z^k = \prod_{p} \left(1\frac{1}{p}\right)\left(1+\frac{1}{pz}\right). $$
There have been several papers concerned with error estimates; a recent specimen is the following article by J. Wu:
Sur un problème de Rényi
Monatsh. Math. 117 (1994), no. 34, 303–322.
For the case $k=1$, I get $$\dfrac{Om_1(x)}{x} \to \dfrac{6}{\pi^2} \sum_p \dfrac{1}{p(p+1)} \approx 0.20076$$ the sum being over all primes. For $k = 2$, I get $$ \dfrac{Om_2(x)}{x} \to \dfrac{6}{\pi^2} \left( \sum_p \dfrac{1}{p^2(p+1)} + \sum_{p,q} \dfrac{1}{p(p+1)q(q+1)}\right) $$ where the second sum is over all unordered pairs $(p,q)$ of distinct primes.
Similarly, for each positive integer $k$, $Om_k(x)/x$ should approach a constant as $x \to \infty$, and I expect that constant to be expressible in terms of sums of rational functions over $j$tuples of primes for $j \le k$, the expressions getting complicated as $k$ increases.