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Am I correct in assuming that the same result: $$ N_k(x):=\ \mid\{n\leq x : \Omega(n)=k\}\mid \ \sim \frac{x}{\log x}\frac{(\log_2 x)^{k-1}}{(k-1)!}\ (x \rightarrow \infty) $$ also holds for: $$ \pi_k(x):=\ \mid\{n\leq x : \omega(n)=k\}\mid \ $$ even though they are wildly different for high $k$? See here.

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$\pi(n)$ is a nondecreasing function, so $\pi_k(x)$ stabilizes, as a function of $x$, while your other function goes to infinity. Did you by any chance mean $\omega(n)$? – Gerry Myerson Dec 25 '13 at 0:21
Yes - I have updated as suggested :) – martin Jan 4 '14 at 7:41
There may be something useful in the references and links at and at – Gerry Myerson Jan 4 '14 at 15:30

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