Every 33-length subsequence of $1,2,\dotsc,122$ contains a three term arithmetic progression Is it possible to prove that every 33-length subsequence of the sequence $1,2,3,\dotsc,122$ contains a three term arithmetic progression?
Maybe I should post it on mathoverflow
 A: This is hardly an answer, but I'd like to direct you to OEIS A065825.
This sequence, beginning
$$S_3=\{1, 2, 4, 5, 9, 11, 13, 14, 20, 24, 26, 30, 32, 36, 40, 41, 51, 54, 58, 63, 71, 74, 82, 84, 92, 95, 100, 104, 111, 114, 121, 122, 137, 145, 150, 157, 163, 165, 169, 174, 194\}$$
gives for its $n$th term the minimum $k$ such that $[1,k]$ that has an $n$-term subset that avoids $3$-term arithmetic progressions (typically called a $3$-free set).  Since the $32$ term of this sequence is $122$ and the $33$rd is $137$, it follows that no $33$ term sequence in $[1,122]$ is $3$-free.  Not much is known about the growth of this sequence, and I would not be surprised if the sequence above had been found by brute force calculations.
It was at one point conjectured that the sequence $G_3=\{1,2,4,5,10,\ldots\}$ (i.e. the sequence obtained by always appending the smallest element that retains $3$-freeness) would yield competitive bounds to $S_3$ infinitely often. This was been disproved by work of F. Behrend in 1946, who crafted examples of $3$-free sets of length $n$ that fit in the interval $n^{1+\epsilon}$ (for fixed $\epsilon >0$ and sufficiently large $n$).  In contrast, we can prove that the "greedy" version of this packing requires space like
$$n^{\log_2 3},$$
by recognizing it as the set of integers whose base $3$ representation omits the digit $2$, increased by $1$.
