If we would try to prove the simplest case, that is: if we color the integers with 2 colors, then the coloring must contain an arithmetic progression of length 3.
Let $R_n$ be a randomly generated coloring of $n$ elements. Define $f(n) = \mathbb{P}( R_n \text{ is AP-3 free})$, the chance that a randomly colored sequence of length $n$ does not have any arithmnetic progressions of length 3.
We can argue that $f(2n)$ equals the chance that we don't have AP-3 over the $n$ even elements, multiplied by the chance that the odds are AP-3 free as well, multiplied by the chance that there aren't AP-3 on a combination of even and odd elements, given that the odd elements are AP-3 free and the even elements are AP-3 free.
Now if we could find a sufficiently tight bound for the last probability, for example $\mathbb{P}(\text{ AP-3 free | odds and evens AP-3 free}) < f(n)^{1/2}$ Then we would have: $$f(2n) < f(n)^2 f(n)^{1/2} = f(n)^{5/2}$$ $$f(n) < C^{n^{1.21}}$$ But then, for sufficiently large $n$ : $f(n) < \frac{1}{2}^n$ But this would be problematic since there are only $2^n$ possible sequences. Therefore, we could conclude that there are $0$ feasible sequences of length $n$
Does this make any sense at all?