# Is this expression for $p_2(n)$, the nth composite of two primes, correct?

The PNT gives an expression for the n$$^{th}$$ prime: $$n\log n.$$

My question is whether

$$p_2(n) \sim \frac{n\log n}{\log\log n}$$

is the correct analogous form for 2-primes $$p_2(n)$$ comprised of two prime factors (possibly the same).

Examples of 2-primes: 4,6,9 10,...

The notation $$\pi_2(x)$$ denotes the number of 2-primes not exceeding x. This is my derivation of $$p_2(n)$$ above. Any corrections/simplifications appreciated.

From the GPNT we have

$$\pi_2(x)\sim \frac{x \log\log x}{\log x}.$$

If $$y = \pi_2(x)$$ then $$\frac{y~ \log x}{x \log\log x}\sim 1,$$

then

$$\log y + \log\log x - \log x -\log\log\log x \to 0$$ and

$$\frac{\log y}{\log x}\sim 1.$$

Multiplying by $$y/y$$ and by $$x\log\log x,$$

$$x\log\log x \sim \frac{ y \log y \cdot x \log\log x}{y \log x} \sim \frac{\pi_2(x)~y \log y}{y}$$

$$\frac{\pi_2(x)~ y \log y}{\pi_2(x)} = y \log y \sim x \log\log x$$

Letting $$x = p_2(n),$$ noting that $$y = \pi_2(p_2(n)) = n,$$

$$p_2(n)\log\log p_2(n) \sim n\log n$$

We can easily show that $$\log p_2(n)\sim \log n$$ so that

$$p_2(n)\log\log p_2(n)\sim p_2(n)\log\log n\sim n\log n$$ so finally

$$p_2(n) \sim \frac{n\log n}{\log\log n}$$

As a sanity check, $$p_2(9000)=36591$$ and $$p_2(9000)/(\frac{9000\log 9000}{\log\log 9000})\approx 0.986.$$

• What is the question exactly? Oct 26, 2014 at 7:43
• @Travis: Is the proof okay, can it be simplified...? Oct 26, 2014 at 7:43
• @Travis: Thanks I didn't know that tag existed. Oct 26, 2014 at 7:46
• Cheers, glad to help. Oct 26, 2014 at 7:48
• The derivation looks fine. Oct 29, 2014 at 18:12

In the question I tried to adapt Ingham's method for $k=1$ but this is much easier and gives the analogous form for all $k.$
$$\pi_k(p_k(n)) = n \sim \frac{p_k(n)(\log\log p_k(n))^{k-1} }{\log p_k(n)\cdot(k-1)! }$$
and using $\log p_k(n)\sim \log n$ we get for fixed $k$ and suff. large n
$$p_k(n) \sim \frac{n\log n (k-1)!}{(\log\log n)^{k-1}}$$