Let $f(n)$ denotes number of subsets of set $\{1,2,...,n\}$ that contain at least one arithmetic progression of length $3$.
I made some calculation in c++ and there are some first values: $$f(3)=1,f(4)=3,f(5)=9, f(6)= 24, f(7)=63, f(8)=150, f(9)=343, f(10)=746$$ ... The thing i saw is that sequence $a_{n}=\frac{f(n+1)}{f(n)}$ is bounded by $2$ and $3$, and it is at some point strictly decreasing.
Is this true that $\lim_{n\to\infty} a_{n}=2$?
If so, then how to calculate $f(n)$?
How to find an upper bound of $f(n)$?