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A semiprime is a number that is the product of two prime numbers.

What is an upper bound for the number of numbers of the form $pq$ less than $n$?

$p,q$ are prime numbers smaller than $n$.

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1 Answers1

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Recall that by the PNT we have $\pi(n) \sim \frac{n}{\log n}$ where $\pi(n)$ is the number of primes less than or equal to $n$. The number of semiprimes $\pi_2(n)$ is approximately

$$\pi_2(n) \sim \sum_{p \le \sqrt{n}} \pi \left( \left\lfloor \frac{n}{p} \right\rfloor \right)$$

where the sum runs over primes. The summand is approximately $\frac{n}{p \log n}$, so overall this sum is approximately

$$\pi_2(n) \sim \frac{n}{\log n} \sum_{p \le \sqrt{n}} \frac{1}{p}.$$

The sum $\sum_{p \le \sqrt{n}} \frac{1}{p}$ is known to be asymptotically $\log \log n$, so overall we get

$$\pi_2(n) \sim \frac{n \log \log n}{\log n}$$

at least heuristically. This is in fact the correct asymptotic by a result of Landau.

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Qiaochu Yuan
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    $\pi_2(n) = \frac{1}{2} \left(-\pi(\sqrt{n}) + \sum_{p \le n} \pi(n/p)\right)$ where $\pi_2(n)$ is the number of integers of the form $pq, p \ne q$ which are $\le n$ – reuns Jan 20 '16 at 18:25
  • Yes, but that expression is more annoying to extract an asymptotic from; the problem is that the approximation $\pi \left( \lfloor \frac{n}{p} \rfloor \right) \sim \frac{n}{p \log n}$ gets worse as $p$ gets large. If you try extracting an asymptotic from that using the same reasoning as above and without accounting for this, you'll be off by a factor of $2$. – Qiaochu Yuan Jan 20 '16 at 19:19
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    atfer that I write $\sum_{p \le n} \pi(n/p) = \pi \ast \delta_p(n) = \int \delta_p \ast \delta_p (n)$ while $\sum \frac{\mu_n}{ n} \ln \zeta(sn) = \sum p^{-s} = \sum \delta_p(n) n^{-s}$, $\ \ s \sum \frac{\mu_n}{ n} \ln \zeta(sn) = s \int_1^\infty \pi(x) x^{-s-1}dx$, $\int_1^\infty \pi_2(x) x^{-s-1} dx = \ldots$ – reuns Jan 20 '16 at 19:21