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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.

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    This can be done, and you can look it up in Hardy and Wright.(that is, asymptotics for numerator and denominator.)2012-06-01
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    @mike, I should have thought about that. I will try again. Thanks.2012-06-01
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    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.2012-06-01
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    @GerryMyerson, I meant what you get when you add the word "exactly", without concerns on multiplicity or uniqueness.2012-06-02
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    So, do you know the asymptotics on the number of numbers up to $n$ with exactly $m$ prime factors?2012-06-02
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    @GerryMyerson, thank you. My problem is with the sum of the almost primes up to $n$. According to the following paper, http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077306714 there is an asymptotic formula which answers your question.2012-06-02
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    I understand that your problem concerns the sum. But I can't see any reason for the average prime to differ much from the average almost-prime, so if you know how many, you know the sum (relative to the sum for primes).2012-06-02
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    @GerryMyerson, would you write out your answer or explain your heuristics more precisely?2012-06-03
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    What I'm thinking is this. Suppose among the first $N$ numbers there are $P$ primes. If the primes were scattered randomly, I'd expect their sum to be about $PN/2$. Well, they aren't scattered quite randomly, they are biased toward the smaller numbers, so let's make it $PNf(N)/2$ for some $f(N)$ strictly less than 1. Now let $A$ be the number of almost-primes up to $N$. I'm inclined to believe that for large enough $N$ the almost-primes have the same kind of bias toward the smaller numbers the primes have, giving sum $ANf(N)/2$. Then the number you're asking for is $A/P$.2012-06-03
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    @GerryMyerson, thanks a lot. In addition to trying to have a paper and pen proof attempt, I had better try writing some programs and make better guesses as well.2012-06-04

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

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    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.2012-06-05