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 $$\frac{y~ \log x}{x \log\log x}\sim 1, $$


$$\log y + \log\log x - \log x -\log\log\log x \to 0 $$ 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 $p_2(9000)/(\frac{9000\log 9000}{\log\log 9000})\approx 0.986.$

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

1 Answer 1


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

$$ p_k(n) \sim \frac{n\log n (k-1)!}{(\log\log n)^{k-1}}$$


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