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 $p_2(9000)/(\frac{9000\log 9000}{\log\log 9000})\approx 0.986.$