As $n$ grows sufficiently larger, $\pi(n)<\pi_{1}(n)$, where $\pi(n)$ and $\pi_{1}(n)$ is the number of prime and semiprime $\leq{n}$, respectively

Question

From $P_{12}=37$ the number of semiprime(s) appears to be higher than the number of prime(s). Though I couldn't check for a higher $n\geq{500}$ for several limitations, I could really use any proof or insight regarding this observation.

Thanks

Answer 1

This inequality has to hold eventually.

Counting only the semiprimes where one of the factors is $2$, $3$ or $5$, we have $$ \pi_1(n) \ge \pi(\tfrac12 n) + \pi(\tfrac13 n) + \pi(\tfrac15 n) - 3 $$ For any $\alpha\in(0,1)$ and $\varepsilon>0$ the prime number theorem implies that all sufficiently large $n$ will satisfy $\pi(\alpha n) > (1-\varepsilon)\alpha \pi(n)$. Applying this to $\alpha=\tfrac12,\tfrac13,\tfrac15$ in turn gives us $$ \pi_1(n) \ge (1-\varepsilon)\bigl(\tfrac12\pi(n)+ \tfrac13\pi(n) + \tfrac15\pi(n)\bigr) - 3 = (1-\varepsilon)\tfrac{31}{30}\pi(n) - 3 $$ and therefore eventually $\pi_1(n) > \pi(n) $.

In fact, by this argument we get $\pi_1(n) > k\pi(n)$ eventually for any constant $k$, since the sum of the reciprocals of primes diverges.