# Density of numbers with exactly $n$ distinct prime factors in $\mathbb{N}$

It is quite well known that the density of the primes in $\mathbb{N}$ is $0$, that is, $$\lim_{n\to\infty}\frac{|\{p\mid p\leq n, p \text{ prime}\}|}{|\mathbb{N}_{\leq n}|}=0$$ It is less well-known, but true, that $$\lim_{n\to\infty}\frac{|\{p^k\mid p^k\leq n, p \text{ prime}, k\in\mathbb{N}\}|}{|\mathbb{N}_{\leq n}|}=0$$ I'm looking at the sets with numbers with the same amount of prime factors. We define for $n\in\mathbb{N}$ the following: $$P_n:=\{k\mid k\text{ has exactly }n\text{ distinct prime factors}\}$$ So \begin{align} P_1&=\{2,3,4,5,7,8,9,11,13,16,17,19,23,25,\cdots\}\\ P_2&=\{6,10,12,14,15,18,20,21,22,24,26,28,\cdots\}\\ P_3&=\{30,42,60,66,70,78,84,90,102,105,110,\cdots\} \end{align} It feels like the density of any $P_n$ in $\mathbb{N}$ should be $0$ give that the density of $P_1$ in $\mathbb{N}$ is $0$, but has this been proven for any $n$? For all $n$? Thanks for your effort in advance.

• @hardmath, excuse me, I forgot the vertical bars to denote the number of elements in the sets. So for example $|\mathbb{N}_{\leq n}|=n$. I edited my answer, hope this clears up things. – vrugtehagel Jan 27 '16 at 16:15
• Not only has this been proven, but the first-order asymptotics for every $P_n$ are known, in analogy with the Prime Number Theorem for $P_1$. – Erick Wong Jan 27 '16 at 16:58

## 1 Answer

The Hardy–Ramanujan theorem says that the number of distinct prime factors of $n$ has normal order $\log \log n$... i.e., most numbers have about this many distinct prime factors. In particular, since this estimate diverges as $n\rightarrow\infty$, the theorem tells you that the density of numbers with exactly $k$ distinct prime factors is zero for any $k$.