TITLE: What is known about the minimal number $f(n)$ of geometric progressions needed to cover $\{1,2,\ldots,n\}$, as a function of $n$?
QUESTION [6 upvotes]: What is known about the minimal number $f(n)$ of geometric progressions needed to cover $\{1,2,\ldots,n\}$, as a function of $n$?

So a geometric progression can contain at most two primes. This automatically gives a lower bound on the minimal number $f(n)$ of geometric progressions needed to cover the integers $\{1,2,\ldots,n\}$, at least asymptotically, because of the prime number theorem. But this seems like a very crude bounding technique. I don't expect a closed-form formula for $f(n)$ is known. But what about its asymptotics? Can we say that $f(n) = \Theta (g(n))$ for some easy to understand function $g(n)$? Or give some fairly tight asymptotic upper and lower bounds for $f(n)$, even if there is an asymptotic growth rate gap between the two bounds?
UPDATE: From the link given in the comment, we have $f(n) \geq cn$ always for some constant $0 < c < 1$. And clearly $f(n) \leq n$. So then the interesting question seems to become, what is the best constant $d \leq 1$ we can find so that we can say $f(n) \leq dn$ holds for large enough $n$?

REPLY [3 votes]: I wrote a program that computes $f(n)$ for larger $n$. It generates feasible geometric progressions (sets) and then uses simulated annealing to solve (near optimally) the resulting set covering problem. From other answers (like http://math.stackexchange.com/questions/1386381/cover-1-2-100-with-minimum-number-of-geometric-progressions) we already know that $f(100)=36$. For larger $n$, I was able to compute some good upper bounds on $f(n)$. I found that $f(1000) \leq 362$ and $f(10000) \leq 3620$. After seeing these numbers I suddenly realised that $f(n) \approx n/e$, which was rather surprising! I also submitted a sequence to OEIS: https://oeis.org/A309095
Here is my Java code to compute $f(n)$: https://pastebin.com/b23ugGZN