N-th k-almost prime satisfying a certain condition

Question

Let $G(n,k)$ be the n-th k-almost prime. Prove that for every for every $n \in N$ there exists infinitely many $k \in N$ satisfying $2*G(n,k) = G(n,k+1)$.

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Answer 1

$G(n,1)=p_n$, the $n^{\text{th}}$ prime.
$G(n,k) \le 2^{k-1}p_n$ because we can display $n-1$ numbers that must be smaller; $2^{k-1}$ times all the smaller primes.
Given $n$, we can find $m$ such that $3^m \gt 2^{m-1}p_n \ge G(n,m)$
Then for all $k \ge m, 2G(n,k)=G(n,k+1)$