Chen's theorem states that every large enough even integer is the sum of a prime and an almost prime, i.e. an integer that is either a prime or the product of two primes. As there are $s(n):=\pi(2n)-\pi(n)$primes between $n$ and $2n$, one gets that the number of almost primes $t(n)$ between $n$ and $2n$ is at most $s(n)+s(n)^2\sim s(n)^2$.

Moreover, denoting by $\pi_{k}(n)$ the number of integers $r\leq n$ such that $\Omega(n-r)\Omega(n+r)\leq k$ one has $\pi_{1}(n)\leq \pi_{2}(n)$. It has been proven that $\pi_{2}(n)\geq 0.867\frac{C(2n)2n}{\log^{2}2n}$ with $C(2n)=C_{2}\prod_{p>2, p\mid n}\frac{p-1}{p-2}$, $C_{2}=0.66016...$ being the twin prime constant. So is it possible to get a similar lower bound for $\pi_{1}(n)$?

  • $\begingroup$ Technically an n-almost prime has n prime factors. $\endgroup$ – Roddy MacPhee Mar 9 at 0:59

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