Call a square-free a 2-prime if it has exactly two prime divisors. Call a square-free a 3-prime if it has exactly three prime divisors,etc. Does there exist an integer sufficiently large N such that the number of 3-primes less than N is two times (or more) as large as the number of 2-primes less than N ?

  • $\begingroup$ You seem to have dropped most of your question... in case you were wondering why you didn't get any answers... $\endgroup$ – David Apr 11 '14 at 3:54
  • $\begingroup$ The number of products of 3 primes is asymptotically greater than the number of products of 2 primes, so there is some number $n$ beyond which you never get more 2-primes than 3-primes. Whether $n=230$ works, I couldn't say. $\endgroup$ – Gerry Myerson Apr 13 '14 at 11:48
  • $\begingroup$ Indeed, the quotient (number of 3-primes up to $N$) divided by (number of 2-primes up to $N$) tends to infinity as $N$ tends to infinity. $\endgroup$ – Greg Martin Apr 16 '14 at 3:07

There are about $\frac{x\log\log x}{\log x}$ 2-primes up to $x$ and about $\frac{x(\log\log x)^2}{2\log x}$ 3-primes up to $x$, so the crossover should happen roughly when $\log\log x=4$ which is $e^{e^4}\approx2^{78.8}.$


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