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> If we let $\omega\_Q(n)$ denote the number of distinct prime factors of $n$ less than a bound $Q$, then what asymptotic formulas exist for $\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]$ as $Q\to\infty$ if $k$ remains fixed (or perhaps very small with respect to n)?
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I am asking this question since my study led me to want to bound the quantity


$$\mathbf{E}\_{n\in\mathbb{N}}\left[\frac{2^{\omega\_Q(n)}}{\sqrt{\omega\_Q(n)}}\right]$$


as $Q\to\infty$. Since


$$\mathbf{E}\_{n\in\mathbb{N}}\left[\frac{2^{\omega\_Q(n)}}{\sqrt{\omega\_Q(n)}}\right]=\sum\_{n=1}^{\pi(Q)}\left(\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]\right)\left(\frac{2^{\omega\_Q(n)}}{\sqrt{\omega\_Q(n)}}\right)$$


and


$$\sum\_{n=1}^{\pi(Q)}\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]2^{\omega\_Q(n)}\sim\_{Q\to\infty} c\log(Q)$$


is well understood, good (upper) bounds on $\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]$ could help me in my effort.


For small values of $k$ computations can be done directly, like


$$\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=0]\sim\frac{c}{\log(Q)}$$


and


$$\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=1]\sim c\frac{\log(\log(Q))}{\log(Q)}$$


The main approach I have been using is noting that $\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]$ is exactly the coefficient of $x^k$ in the polynomial


$$\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)$$


Asymptotics of this full polynomial are easy to come by, for instance as $Q\to\infty$ we have that


$$\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\sim c \log^{x-1}(Q)$$


Heuristically this would suggest that


\begin{align\*}
\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]&=\frac{1}{k!}\left.\frac{d^k}{dx^k}\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|\_{x=0}\\
&\sim \frac{c}{k!}\left.\frac{d^k}{dx^k}\log^{x-1}(Q)\right|\_{x=0}\\
&=\frac{c}{k!}\frac{\log^k(\log(Q))}{\log(Q)}
\end{align\*}


This argument is however by no means rigorous so I would appreciate true asymptotics.



Asymptotic for the probability that a number has $k$ prime factors less than $Q$
As pointed out in the question, we have that


$$\prod\_{p<Q}\left(\frac{x-1}{p}+1\right)=\sum\_{k=0}^{\pi(Q)}\Pr\_{n\in\mathbb{N}}[\omega\_Q(n)=k]x^k$$


which can be derived by showing that on both the RHS and the LHS the coefficient of $x^k$ is equal to


$$\sum\_{\substack{S\subseteq \{p<Q\} \\ |S|=k}} \left(\prod\_{p\in S}\frac{1}{p}\right)\left(\prod\_{p\not\in S}\left(1-\frac{1}{p}\right) \right)$$


Treating the LHS with simple manipulation we get that


\begin{align\*}
\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left(\log\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right)\\
&=\exp\left(\sum\_{p<Q}\log\left(\frac{x-1}{p}+1\right)\right)\\
&=\exp\left(\sum\_{p<Q}\frac{x-1}{p}+\sum\_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)\right)\tag{1}
\end{align\*}


We now note that


$$\sum\_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=\sum\_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)-\sum\_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$


and thus we can set


$$f\_1(x)=\sum\_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$


and


$$g(x)=\sum\_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$


Morally, we can think of $g(x)$ as the "error" as $Q\to\infty$ which we must show is inconsequential. We thus get that


\begin{equation}
\sum\_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=f\_1(x)-g(x)\tag{2}
\end{equation}


By Merten's theorem, we have that


\begin{align\*}
\sum\_{p<Q}\frac{x-1}{p}&=(x-1)\left(\log(\log(Q))+M+\epsilon\_Q\right)\\
&=(x-1)\log(\log(Q))+Mx+\epsilon\_Q x-M-\epsilon\_Q\tag{3}
\end{align\*}


where $\epsilon\_Q\to 0$ and $M$ is the Meissel-Mertens constant. Substituting (2) and (3) into (1) we get that


\begin{align\*}
\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left((x-1)\log(\log(Q))+Mx+\epsilon\_Q x-M-\epsilon\_Q+f\_1(x)-g(x)\right)\\
&=e^{-M-\epsilon\_Q}\log^{x-1}(Q)e^{Mx}e^{f\_1(x)}e^{\epsilon\_Qx}e^{-g(x)}
\end{align\*}


For simplicity's sake, we now define


$$f\_2(x)=e^{Mx}e^{f\_1(x)}$$


and thus


\begin{equation}
\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)=e^{-M-\epsilon\_Q}\log^{x-1}(Q)f\_2(x)e^{\epsilon\_Qx}e^{-g(x)}\tag{4}
\end{equation}


Taking the derivative $k$ times yields


$$\frac{d^k}{dx^k}e^{-M-\epsilon\_Q}\log^{x-1}(Q)f\_2(x)e^{\epsilon\_Qx}e^{-g(x)}$$


As $Q\to\infty$, the only term that will matter in a product rule decomposition of this equation is the one that grows the fastest. It is easy to show that


$$g^{(n)}(x)=O\left(\frac{1}{x}\right)$$


for any order derivative $(n)$, and so the fastest growing term is the one where $\log^{(x-1)}(Q)$ is differentiated the full $k$ times. Since there are finitely many terms the others are inconsequential in terms of growth and so


\begin{align\*}
\frac{d^k}{dx^k}e^{-M-\epsilon\_Q}\log^{x-1}(Q)f\_2(x)e^{\epsilon\_Qx}e^{-g(x)}&\sim\_{Q\to\infty}e^{-M-\epsilon\_Q}f\_2(x)e^{\epsilon\_Qx}e^{-g(x)}\frac{d^k}{dx^k}\log^{x-1}(Q)\\
&=e^{-M-\epsilon\_Q}f\_2(x)e^{\epsilon\_Qx}e^{-g(x)}\log^k(\log(Q))\log^{x-1}(Q)
\end{align\*}


evaluating at $x=0$ and substituting into (4) yields that


\begin{equation}
\left.\frac{d^k}{dx^k}\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|\_{x=0}\sim e^{-M-\epsilon\_Q}f\_2(0)e^{-g(0)}\frac{\log^k(\log(Q))}{\log(Q)}
\end{equation}


As $Q\to\infty$ we have that $g(0)\to0$ and $f\_2(0)=e^{M-\gamma}$ and so


\begin{equation}
\left.\frac{d^k}{dx^k}\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|\_{x=0}\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q)}\tag{5}
\end{equation}


We also see that


\begin{align\*}
\left.\frac{d^k}{dx^k}\prod\_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|\_{x=0}&=\left.\frac{d^k}{dx^k}\sum\_{j=0}^{\pi(Q)} \mathrm{Pr}\_{n\in\mathbb{N}}[\omega\_Q(n)=j]x^j\right|\_{x=0}\\
&=k!\mathrm{Pr}\_{n\in\mathbb{N}}[\omega\_Q(n)=k]
\end{align\*}


and thus we conclude from (5) that


$$\mathrm{Pr}\_{n\in\mathbb{N}}[\omega\_Q(n)=k]\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q) k!}$$


which is the desired result


