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A $k$-almost prime is a positive integer having exactly $k$ prime factors, not necessarily distinct. Let $\mathbb{P}_k$ be the set of the $k$-almost primes and let $$ \rho_k(n):=\sum\limits_{\substack{q\in \mathbb{P_k}\\q\le n}}\frac1q. $$ So with $k=1$ and $n\to\infty$ we have the prime harmonic series, which diverges like $\log\log n+ B_1$, $B_1$ being the Meissel-Mertens the constant.. $\rho_2(n)$ diverges at a rate of $\frac{\log^2\log n}{2}+\frac15$. In general, is the asymptotical behaviour of $\rho_k(n)$ known?

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  • $\begingroup$ The counting function of the $k$-almost primes is known, so a straightforward partial summation should give you an asymptotic for the leading term of the sum of reciprocals. Are you asking for an expansion that is accurate up to $o(1)$? $\endgroup$
    – Erick Wong
    Nov 6, 2015 at 4:53
  • $\begingroup$ @VincenzoOliva do you know a proof for $\rho_2(n) \sim \frac{(\log \log n)^2}{2}+\frac15$. $\endgroup$
    – kodlu
    Sep 14, 2016 at 22:26

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The first-order asymptotics are easy. The counting function of $\mathbb P_k$ satisfies the Poisson-like asymptotic:

$$P_k(x) := \#\{n \le x : n \in \mathbb P_k\} \sim \frac{x}{\log x}\cdot \frac{ (\log \log x)^{k-1}}{(k-1)!}.$$

By partial summation, the reciprocal sum is $$ \sum_{\substack{n \in \mathbb P_k \\ n \le x}} \frac{1}{n} = \int_{t=2}^x \frac{d P_k(t)}{t} = \frac{P_k(t)}{t} \bigg|_{t=2}^x + \int_{t=2}^x \frac{P_k(t)}{t^2} dt.$$

The first term on the right is $O(1)$ since $P_k(x)/x \to 0$, so all the divergence in the sum is coming from the integral term, which is asymptotic to:

$$ \frac{1}{(k-1)!}\int_{t=2}^x \frac{(\log \log x)^{k-1}}{t \log t} dt = \frac{1}{(k-1)!} \int_{u=\log \log 2}^{\log \log x} u^{k-1} du = \frac{1}{k!}(\log \log x)^k + O(1).$$

Edit: The above is asymptotic to the reciprocal sum, which is not to say that $O(1)$ is the correct error term. In fact, the asymptotic for $P_k(x)$ has a relative error of $O(1/\log \log x)$ (the error term must depend in some way on the specific definition of $k$-almost prime, like whether duplicates are allowed), so the same is true of the estimate above. I would expect an error term of size $O(\log \log x)^{k-1}$.

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  • $\begingroup$ Thank you, Erick. Is something known about those constants? For example, $B_1>1/5$. Does this go on indefinitely? Do they get smaller and smaller? $\endgroup$ Nov 6, 2015 at 12:01
  • $\begingroup$ If you'd be so kind, would you please explain a bit more about how you got the asymptotic? Intuitively, it reminds me of $\pi(x)\simeq\dfrac{\ln x}x,$ and of $\displaystyle\sum_{n=1}^x\frac1{p_n}\simeq\sum_{n=2}^x\frac1{n\ln n}\simeq\int_2^x\frac{dt}{t\ln t}\simeq\ln\ln x,$ but I'm afraid I can't really grasp the specifics. $\endgroup$
    – Lucian
    Nov 6, 2015 at 15:00
  • $\begingroup$ @Lucian Which asymptotics are you referring to specifically? I starting to suspect there's an elementary approach to this, by just expanding out products of $\sum 1/p$. $\endgroup$
    – Erick Wong
    Nov 6, 2015 at 16:21
  • $\begingroup$ @Erick: Would you please answer the question in my comment? $\endgroup$ Nov 6, 2015 at 16:23
  • $\begingroup$ @VincenzoOliva I don't know the answer to the question in your comment. Actually, come to think of it, the format of my answer is a bit misleading. Let me fix. $\endgroup$
    – Erick Wong
    Nov 6, 2015 at 16:44

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