Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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 $(\log\log x)^{m-1}/(m-1)!$

share|improve this answer
    
AT Gerry Myerson, thank you for the clear reference and the discussion of your ideas. I will also try to check your conjecture in the last paragraph. –  Mathing being Jun 5 '12 at 11:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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