What is the asymptotic number of square-free numbers less than $x$ with exactly $k$ prime divisors?



Asymptotics of special square-free numbers
All this is taken from Section 7.4 of Montgomery-Vaughan's *Multiplicative Number Theory I. Classical Theory*.


**Theorem 7.19**: The number of integers up to $x$ with exactly $k$ prime divisors counted with multiplicity is
$$
\frac{F((k-1)/\log\log x)}{\Gamma(1+(k-1)/\log\log x)} \frac{x(\log\log x)^{k-1}}{(k-1)!\log x} \bigg( 1 + O\bigg( \frac k{(\log\log x)^2} \bigg) \bigg). \tag{$\*$}
$$
Here $\Gamma$ is the Euler Gamma function, and
$$
F(z) = \prod\_p \bigg( 1 - \frac zp \bigg)^{-1} \bigg( 1 - \frac1z \bigg)^p.
$$
(This is true uniformly for $k\le 1.99\log\log x$, say.) Note that $F(0)=1$, which gives the result Will quoted for fixed $k$. Note also that $F(1)=1$; this is relevant because most integers near $x$ have about $k=\log\log x$ prime factors.


**Problem 3**: The number of integers with exactly $k$ *distinct* prime factors is also given by ($\*$), except one must change $F$ to
$$
F(z) = \prod\_p \bigg( 1 + \frac z{p-1} \bigg) \bigg( 1 - \frac1z \bigg)^p.
$$
(This is true in a wider range - uniformly for $k\le 1000\log\log x$ or whatever constant you want.) Note that again $F(0)=1$ and $F(1)=1$.


**Problem 4**: The number of *squarefree* integers with exactly $k$ *distinct* prime factors is also given by ($\*$), except one must change $F$ to
$$
F(z) = \prod\_p \bigg( 1 + \frac zp \bigg) \bigg( 1 - \frac1z \bigg)^p.
$$
(This is also uniform for $k\le 1000\log\log x$.) Note that again $F(0)=1$, but now $F(1) = 6/\pi^2$.


