Number of integers for which $\Omega=k$

Question

for every $x>0$, I'm trying to find out the number of positive integers $n\leq x$ such that $\Omega(n)=k$ for all $k\geq 1$ in $N$, where $\Omega(n)=\Omega(p_1^{\mu_1}\cdot\ldots\cdot p_j^{\mu_j})=\mu_1+\ldots+\mu_j$.

Or, at least, I need informations about the order of growth of the function $f(k,x)=$ number of positive integers $n\leq x$ such that $\Omega(n)=k$.

Maybe someone can give me a hint to know in what direction I must go.

Answer 1

One of the easy consequences of the prime number theorem is that $$\lim_{x\to+\infty} \frac{\#\{1\leq n \leq x: \Omega(n)=k\}(\log x )}{x (\log \log x )^{k-1} } =\frac{1}{(k-1)!}.$$ You can read about it in Montgomery & Vaughan 'Multiplicative Number Theory' volume 1, chapter 7, exercise 3.