2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – vrugtehagel Jan 27 '16 at 16:15
  • $\begingroup$ 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 $P_1$. $\endgroup$ – Erick Wong Jan 27 '16 at 16:58
4
$\begingroup$

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 $k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.