Concerning squarefree numbers with 2 primes and squarefrees with 3 primes. If a squarefree with two primes is a 2-prime and a squarefree with three primes is a 3-prime is there an integer N such that the number of 2-primes less than N  is equal to the number of 3-primes less than N. N>30. (the number of 2-primes < N being greater than 7)
 A: Yes. At $N=31$, the number of 2-primes up to $N$ is greater than the number of $3$-primes up to $N$. But the number of 3-primes up to $N$ is asymptotic to $N(\log\log N)^2/2\log N$, while the number of 2-primes up to $N$ is asymptotic to $N\log\log N/\log N$. Therefore eventually the former number must exceed the latter number, and since both quantities are integer-valued and can change by at most $1$ at a time, they must be equal somewhere. (But there will be only finitely many integers at which they are equal.)
As it turns out, the smallest $N>5$ for which the two counts are equal is $N=1279786$, up to which there are $265549$ 2-primes and $265549$ 3-primes. Moreover, $1279787$ is prime, $1279788$ is not squarefree, and $1279789$ is a 3-prime; so $N=1279789$ is the smallest $N$ for which the 3-prime count exceeds the 2-prime count. Calculations suggest that $N=1281378$ is the last $N$ for which the 2-prime count is tied with the 3-prime count, and that the most that the former ever exceeds the latter is $5661$, occurring at $11$ integers betwen $463200$ and $463300$.
