# Limit of ratios of numbers with $m$ factors and primes

This is my first question.

Let $a_1, a_2,\ldots, a_k$ be natural numbers $\leq n$ with $m$ prime factors.

Let $p_1, p_2, \ldots, p_r$ be the prime numbers $\leq n$.

Let $$C_{m,n} = \frac{\sum_{i=1}^{k}a_i}{\sum_{i=1}^{r}p_i}$$ Does the following limit exist when $m\geq 2$? Obviously, when $m=1$, $C_{1,n}=1$.

$$\lim_{n\to \infty} C_{m,n}$$

I have some guesses but not defendable enough.

My best,

The FM.

• This can be done, and you can look it up in Hardy and Wright.(that is, asymptotics for numerator and denominator.) – mike Jun 1 '12 at 21:29
• @mike, I should have thought about that. I will try again. Thanks. – Mathing being Jun 1 '12 at 22:07
• I take it you want $a_1,a_2,\dots,a_k$ to be all the numbers up to $n$ with $m$ prime factors, but I'm not sure whether you mean exactly $m$ prime factors, or at least $m$ prime factors, nor whether you are talking about distinct prime factors, or prime factors counted with multiplicity. – Gerry Myerson Jun 1 '12 at 23:40
• @GerryMyerson, I meant what you get when you add the word "exactly", without concerns on multiplicity or uniqueness. – Mathing being Jun 2 '12 at 9:21
• So, do you know the asymptotics on the number of numbers up to $n$ with exactly $m$ prime factors? – Gerry Myerson Jun 2 '12 at 12:53

For $m\gt1$, there are more numbers with $m$ prime factors than there are primes, in the following sense:
Hardy and Wright, Theorem 437, prove that the number of integers up to $x$ with exactly $m$ prime factors (and it doesn't matter whether the primes are distinct or not) is asymptotic to $${x(\log\log x)^{m-1}\over(m-1)!\log x}$$ They attribute the result to Landau, 1900. Now the number of primes is asymptotic to $x/(\log x)$, so the ratio is $(\log\log x)^{m-1}/(m-1)!$, which of course goes to infinity with $x$.
Meanwhile, the sum of the primes up to $x$ is asymptotically $x^2/(2\log x)$ although I haven't found a wholly trustworthy cite for this. I'm sure (although I'm not up to proving it) that the sum of the almost-primes will be bigger by that same factor of $(\log\log x)^{m-1}/(m-1)!$