# The number of distinct prime factors of $n\in\mathbb N$

Let $$\omega(n)$$ be the number of distinct prime factors of a natural number $$n$$.

Note that $$\omega(n)=0\iff n=1$$, and that $$\omega(24)=\omega(2^3\cdot 3^1)=2\ (\not = 4)$$.

(For more details, you can see "Distinct Prime Factors" on wolfram math world)

Then, for $$m\in\mathbb N$$, let $$S_m=\{n|n\in\mathbb N, \omega(n)=m\}$$. Also, Let $$N_m(x)$$ be the counting function that gives the number of the elements of $$S_m$$ less than or equal to $$x$$, for any real number $$x$$. See the line graph below. This graph shows $$N_m(x)\ (m=1,2,3,4)$$ in $$1\lt x\le 2309=2\cdot3\cdot5\cdot7\cdot11-1.$$ Note that $$N_m(2309)=0$$ for $$m\ge 5$$, and that $$\omega(2310)=\omega(2\cdot 3\cdot 5\cdot 7\cdot 11)=5$$.

$$\ \ \ \ \ \ \ \ \ \$$ In addition to this, let $$P_m(x)=\frac{N_m(x)}{x}\times 100$$. See the line graph below. This graph shows $$P_m(x)\ (m=1,2,3,4)$$ in $$1\lt x\le 2309=2\cdot3\cdot5\cdot7\cdot11-1.$$

$$\ \ \ \ \ \ \ \ \ \$$ Then, here are my questions.

Question : Are the followings true?

$$(1)$$ For every $$m\in\mathbb N$$, $$N_m(x)=O(x)\ (x\to\infty)$$.

$$(2)$$ For every $$m\in\mathbb N$$, $$\lim_{x\to\infty} P_m(x)=0$$.

$$(3)$$ If $$x\ge 50$$, then $$P_2(x)\gt P_m(x)$$ for every $$m\ge 3$$.

The above graphs led me to the these conjectures, but I don't have any good ideas. I would like to know any relevant references. Can anyone help?

I don't quite understand question (1). Isn't $N_m(x) \leq x$ trivially? You seem to be interested in fixed $m$. Then an affirmative answer to (2) and a negative answer to (3) follows from a classical extension of the prime number theorem due to Landau. See, for example,
In fact, the answer to (2) is "yes, uniformly in $m$." Indeed, a theorem of Hardy and Ramanujan implies that $N_m(x) \ll x/\sqrt{\log\log{x}}$ uniformly in $m$.
• Thank you very much for this answer! I think what I was trying to ask as $(1)$ is actually $(2)$. – mathlove Nov 24 '13 at 5:01