Spectra

The diagonal slice of a cube fractal is a hexagon with a tile grammar. Everything published about that grammar stops at base 3; this page asks what happens at every other odd base.

The base axis is the open thread. The dimension axis - the same log(fill) / log(base) - 1 comparison run along dim at fixed base 3 - is settled and lives on the shelf.
The spectra demo draws the Laplacian of a design, its degenerate families and the slope that reads the spectral dimension.
The modes demo lays a design's level-L mask over the torus, where every eigenvalue is a product of L rescaled copies of the tile's own transform and the field of eigenvalues is a picture of the tile.
Every number in THE CLAIM is regenerated by lab/py/odd-base-slice-grammar, a raster-free digit recursion cross-checked against a direct layer census, or by lab/py/spectra-from-cuts, which runs the carry automaton of cuts at every odd base. The grammar, the 2x2 map, its closed forms and the mod-4 side are theorems, proved under THE COMPOSITION; the tiles are cut out of cells as polygons by lab/py/spectra-from-cuts verb tiles.

WHAT IS KNOWN

Cited facts only.

The cut is Perez-Duarte's. "Slice of Menger", showing "a very interesting pattern of stars and hexagons" (Flickr), credited directly by Abel.
Hart popularised it in Mathematical Impressions, with a Scientific American mirror.
Cook published working code, applying the base-3 digit predicate to a plane through the cube's centre (Cook 2011). Its prose says the normal runs to (1, 1, 1) while the listing sets normal = (1, 1, 0.5), so the published plane is not the centroid diagonal.
The substitution rule is Abel's, verbatim: "replace each hexagon with 6 hexagons and 6 triangles, and replace each triangle with 1 hexagon and 3 triangles". As a matrix on (hexagons, triangles) that is [[6,1],[6,3]], trace 9, determinant 12.
The dimension is Abel's, and he states it as a computation rather than a proof: the slice dimension solves 3^x = (9+sqrt(33))/2, so it is log_3((9+sqrt(33))/2) = 1.8184, with the hedge "it takes a bit more work to turn the above computation into a full proof".
A299916 counts holes, not tiles. Its name is a(n) = A299914(2n+1), offset 0, terms 1, 6, 42, 306, 2250, 16578, 122202, signature (9,-12); the Menger reading is a comment on the entry, not the definition. That the two sequences are equal is Proved: A299914 is defined by its recurrence, the signature (9,-12) is a formula on the A299916 entry, and the hexagon census obeys the same recurrence by the carry automaton in cuts. That the entry counts the holes of this slice is the OEIS comment, not a proof here. The index shift is load-bearing: the mesh-triangle census is A299916(n+1).
The two-tile move is published at base 3. Hocking's Bridges paper, resolved in REFS, treats the base-3 slice as a closed fractal family on a hexagon and a triangle, which is a directed-graph iterated function system under another name. What is unpublished is every other base.
The nearest real theory looks elsewhere. Slice dimension is solved for almost every plane (Marstrand 1954, Mattila 1975); the centroid diagonal is a single maximally arithmetic plane, exactly the case those theorems exclude. So a plane landing off log(fill) / log(base) - 1 contradicts nothing and dodges no theorem.
The upstream is grey literature. A photograph, a video, three blog posts and an OEIS comment. Adjacent work does not close the gap: one generalisation runs along dimension, another changes the solid, and none touches base 5, 7 or 9.

THE CLAIM

Proved. For every odd base, the centroid diagonal slice of the parity solid bang dim 3, code 23 is a graph-directed set, a substitution tiling on two prototiles, a regular hexagon and an equilateral triangle, the triangle in two orientations exchanged by a half-turn, and its Hausdorff dimension is log_base(rho_base), rho_base the dominant root of the 2x2 matrix below. The proof is THE COMPOSITION, The tiles.
Proved. For every odd base, the (hexagons, triangles) census of that slice obeys a 2x2 integer matrix that is a fixed rational function of base within each class of base mod 4, and the growth exponent log_base(rho) of the census, the box-counting exponent that every rule below prints as its slice dimension, sits above log(fill) / log(base) - 1 when base = 3 mod 4 and below it when base = 1 mod 4, and by the line above it is the Hausdorff dimension. The proof is THE COMPOSITION below: the census is the carry automaton of cuts run at base base, its block is five coefficients of the digit polynomial, and the side is a polynomial inequality.
The method builds no raster. The plane x + y + z = 3*base^level/2 meets three diagonal layers and the coordinate sum splits as sum_k base^k * sigma_k over independently chosen digit triples, so the tile census is a memoised digit recursion with nothing allocated.
The four rules. Verified by lab/py/odd-base-slice-grammar; that they are tile grammars is the first line above. Two-term rules x9 -12 at base 3, x11 +62 at 5, x42 -288 at 7, x28 +693 at 9, with dimensions 1.8184 / 1.6869 / 1.8026 / 1.7204 against log(fill) / log(base) - 1 = 1.7268 / 1.7304 / 1.7430 / 1.7544.
Their arithmetic is only self-consistent: each printed dimension is log_base of the dominant root of its own printed rule, so a wrong rule and its own wrong dimension agree by construction.
The base-3 rung is the one external target, and it passes. Verified. A cold census returns Abel's [[6,1],[6,3]] and hexagons 1, 6, 42, 306, 2250, 16578, 122202; the grammar cannot be fitted to that.
The matrices. Verified by lab/py/odd-base-slice-grammar. [[7,3],[30,4]] at base 5, [[30,3],[24,12]] at 7, [[19,9],[96,9]] at 9, all non-negative, reproducing cell counts level after level; lab/py/spectra-from-cuts verb block returns the same four from the carry automaton. Each is the folded count of a substitution acting on tiles, read off polygons at base 5 by verb tiles.
The closed form. Proved for every odd base under THE COMPOSITION; Verified at base 3..21 by lab/py/odd-base-slice-grammar and at base 3..25 by lab/py/spectra-from-cuts verb block. On (hexagons, triangles): [[3(base+1)(3 base-1)/16, (base+1)(base+5)/32], [3(base+1)^2/8, 3(base+1)^2/16]] when base = 3 mod 4, and [[(3 base^2+6 base+7)/16, 3(base-1)(base+3)/32], [3(base-1)(3 base+5)/8, (base+3)^2/16]] when base = 1 mod 4. Both reproduce every census matrix at base 3..21.
What would falsify it. Nothing at any base: the grammar, the census recurrence, the closed forms and the mod-4 side are theorems below, and a base where the automaton's block disagrees with the closed form is exactly what lab/py/spectra-from-cuts verb block looks for and does not find.
What a result would be worth. No candidate row returns an OEIS hit, and no paper treats the diagonal cross-section at any base but 3. A confirmed row at base 5 is new; an unconfirmed one is a number in a file.

THE COMPOSITION

The two-tile census is the carry automaton of cuts with the base as the variable and the dimension fixed at 3. Every symbol below is lab/py/spectra-from-cuts's and every number is regenerated there.

The objects

b is an odd base, n = (b+1)/2 its count of even digits, g = 3(b-1)/2 the target digit, and fill = n^2 (4n-3) the count of digit triples with at most one odd digit.
P_b(t) = E^2 (E + 3 O) with E = 1 + t^2 + .. + t^(b-1) and O = t + t^3 + .. + t^(b-2) is the digit polynomial, the sum of t^(digit sum) over those triples; P_b[s] is its coefficient, supported on 0 <= s <= 2g.
M[c, c'] = P_b[c + g - b c'] on abs(c) <= 1 is the carry automaton of cuts, c' = (c + g - s)/b, with base 3 replaced by b and the target digit dim = 3 replaced by g, which is 3 again at b = 3.
h_n = 3(b^n - 1)/2 = g (1 + b + .. + b^(n-1)) is the central height of cuts at level n; the plane x + y + z = 3 b^n / 2 = h_n + 3/2 of THE CLAIM cuts the cells of corner sum h_n - 1, h_n, h_n + 1 at offsets 5/2, 3/2, 1/2, a triangle, a hexagon, a triangle.
The correspondence is exact: hexagons_n = (M^n)[0, 0] and triangles_n = (M^n)[1, 0] + (M^n)[-1, 0], the level-n cells at heights h_n and h_n +- 1, with no index shift and no factor, hexagons_0 = 1.

The theorem

Lemma, the carry contracts. abs(c + g - s) <= g + 1 < 2b for abs(c) <= 1 and 0 <= s <= 2g, so no carry leaves abs(c) <= 1 and M closes on three states; verb block checks that the two outer columns vanish at odd b = 3..25.
Lemma, the reflection. d -> b - 1 - d keeps the parity of a digit at odd b, so P_b[s] = P_b[2g - s] and M[-c, -c'] = M[c, c']; hence (M^n)[1, 0] = (M^n)[-1, 0] and M^n e_0 = hexagons_n e_0 + (triangles_n / 2)(e_1 + e_(-1)).
Proved, the block. (hexagons, triangles)_(n+1) = S_b (hexagons, triangles)_n with S_b = [[P_b[g], P_b[g+b]], [2 P_b[g+1], P_b[g+b-1] + P_b[g+b+1]]], which is the even block M_even of cuts written on the basis {e_0, (e_1 + e_(-1))/2}, whose coordinates are (hexagons, triangles); on {e_0, e_1 + e_(-1)} itself it reads [[P_b[g], 2 P_b[g+b]], [P_b[g+1], P_b[g+b-1] + P_b[g+b+1]]], same trace and determinant. Apply M to the reflection lemma's vector and read the coefficients of e_0 and e_1 + e_(-1), using P_b[g - b] = P_b[g + b] and P_b[g +- 1 - b] = P_b[g + b -+ 1]. At b = 3 the digit polynomial reads 1, 3, 3, 6, 3, 3, 1, S_3 = [[6,1],[6,3]] is Abel's matrix, and its recurrence x^2 - 9x + 12 is the one cuts proves; nothing is re-proved here.
Proved, the five coefficients. E^3 carries the even exponents of P_b and 3 O E^2 the odd ones, so P_b[s] is [x^(s/2)] ((1 - x^n)/(1 - x))^3 for even s and 3 [x^((s-1)/2)] (1 - x^(n-1))(1 - x^n)^2/(1 - x)^3 for odd s, each a signed sum of at most six binomials C(j, 2) by inclusion-exclusion. When b = 3 mod 4: P_b[g] = 3(b+1)(3b-1)/16, P_b[g+1] = 3(b+1)^2/16, P_b[g+b] = (b+1)(b+5)/32, P_b[g+b-1] = 3(b+1)(b+5)/32, P_b[g+b+1] = 3(b-3)(b+1)/32. When b = 1 mod 4: P_b[g] = (3b^2+6b+7)/16, P_b[g+1] = 3(b-1)(3b+5)/16, P_b[g+b] = 3(b-1)(b+3)/32, P_b[g+b-1] = (b+3)(b+7)/32, P_b[g+b+1] = (b-1)(b+3)/32. Verb forms derives each from the binomials that are active for all b past a threshold, b >= 11 at the latest, and matches every form against the digit polynomial at odd b = 3..101, which covers every base below its threshold.
Corollary, the closed forms of THE CLAIM, by substitution into S_b: trace 3b(b+1)/4 and determinant 3(b-1)(b+1)^3/32 when b = 3 mod 4, trace (b^2+3b+4)/4 and determinant -(b+3)(3b^3-3b^2-11b+3)/32 when b = 1 mod 4.
Corollary, the mod-4 split of the matrix. g = 3(b-1)/2 is odd exactly when b = 3 mod 4, the parity of the middle layer Proved on CLAIMS, so P_b[g], P_b[g+b-1], P_b[g+b+1] are read from 3 O E^2 and P_b[g+1], P_b[g+b] from E^3 in that class, since g + 1 and g + b have the parity opposite to g, and all five swap parts in the other. That is the whole mechanism.
Proved, the side. Let q = fill/b = n^2(4n-3)/(2n-1) and chi(x) = x^2 - trace x + det, whose discriminant is positive in both classes. In both classes (2n-1)(trace - 2q) < 0, reading -n(4n^2-3)/2 and -(n-1)(12n^2-1)/2, so q sits right of the vertex and rho_b > q if and only if chi(q) < 0. Then (2n-1)^2 chi(q) reads -n^3(4n^3 - 12n^2 + 15n - 6)/2 < 0 for even n >= 2 and (n-1)^2(4n^4 - 4n^3 + 7n^2 - 1)/2 > 0 for odd n >= 3, every real root of every factor lying below the first n of its class. So rho_b > fill/b when b = 3 mod 4 and rho_b < fill/b when b = 1 mod 4, which is the split of THE CLAIM once log_b is taken; verb split isolates the roots exactly.
Proved, the residue classes. P_b(w) = -2/(1+w)^3 at every b-th root of unity w != 1, since E(w) = 1/(1+w) and O(w) = -1/(1+w) there, and the residue-class sums of P_b modulo b are the row sums of M: the class of g sums to (b+1)(5b+1)/8 or (3b^2+6b-1)/8 and each class of g +- 1 to 3(b+1)^2/8 or (5b^2+6b-3)/8, by class 3, 1 mod 4. The block itself is five single coefficients, not class sums; verb classes checks both.
What stays open. Nothing of the census or of the grammar, which The tiles below reads off the same automaton.

The tiles

The census counts cut cells, and a cut cell is a tile: the substitution acting on tiles is the cube substitution read on the plane. Every number below is regenerated by lab/py/spectra-from-cuts verb tiles, which cuts each cell by the plane as a polygon and counts cut cells by their offset.

The objects. Scale the solid into the unit cube and cut it with x + y + z = 3/2. A level-n cell of corner sum sigma, in units of b^-n, meets the plane at doubled offset o = 3 b^n - 2 sigma, an odd integer by parity, so the plane never passes through a cell corner. A closed cell is cut in a polygon of positive area exactly when o is 1, 3 or 5: an equilateral triangle near the low corner T+, a regular hexagon H, an equilateral triangle near the high corner T-, all of side sqrt(2)/2 cell units. In the carry states of THE COMPOSITION, T+, H, T- are c = 1, 0, -1.
Proved, the substitution. A cut level-(n+1) cell sits inside its parent cell, which is therefore cut. Its local digit triple u is a filled digit triple, and its local doubled offset is b o - 2 |u|, with o the parent's offset and |u| the digit sum. So the children of a tile form one list of (u, type) pairs fixed by the parent's type alone, placed by the homothety of ratio 1/b that carries the unit cell onto the parent. Every tile is replaced by the same pattern at every level, with no rotation, and the count matrix on (T-, H, T+) is M itself, rows the child's type and columns the parent's, so the incidence matrix of the graph in the parent-to-child convention is M^T.
Proved, the half-turn. The complement d -> b - 1 - d keeps digit parity at odd b, so it fixes the design. It is the point reflection of the cube through its centre, which lies on the plane, so on the plane it is a half-turn. It fixes H, so the pattern of H is half-turn symmetric, swaps T+ with T- and carries the pattern of T+ onto the pattern of T-. Up to that half-turn the substitution runs on two prototiles, the hexagon and the triangle, and folding T+ with T- gives S_b.
Proved, the slice is graph-directed. Take the solid as the nested intersection of its closed level-n unions, the attractor of its digit maps; its slice is the nested intersection of the level-n tile unions. That is the attractor of a graph-directed system of ratio-1/b homotheties on the vertices T+, H, T- with incidence M. The open set condition holds with the relative interiors of the tiles: a child's open tile lies inside its parent's, distinct cells have disjoint interiors, and the relative interior of a cut lies inside its open cell.
Proved, the dimension. The Hausdorff and box-counting dimensions of the slice are both log_b(rho_b), where rho_b is the dominant root of S_b. M is irreducible, since M[1, 0] = P_b[g + 1] > 0 and M[0, 1] = P_b[g - b] > 0, and the reflection gives the same for T-. So the Perron vector of M is unique, hence fixed by the reflection and even, which gives rho(M) = rho(S_b). Upward: the level-n tiles number hexagons_n + triangles_n, of order rho_b^n, and each has diameter at most sqrt(2) b^-n. Downward: give a level-n tile of type c the mass w_c rho_b^-n / w_0, with w > 0 the left Perron vector, w M = rho_b w, which is consistent from parent to children. A disc of radius b^-n lies in a cube of side 2 b^-n, whose shadow on each axis is an interval of length 2 b^-n meeting at most 4 closed grid intervals of length b^-n, so it meets at most 64 level-n cells and its mass is at most C rho_b^-n. A disc of radius r with b^-(n+1) < r <= b^-n sits in the disc of radius b^-n, so its mass is at most C rho_b^-n = C (b^-n)^s <= C b^s r^s with s = log_b(rho_b), and the mass distribution principle closes it.
Verified at base 5, the finite check (tiles, levels 1..3): H -> 7 H + 15 T+ + 15 T-, T+ -> 3 H + 3 T+ + 1 T-, T- -> 3 H + 1 T+ + 3 T-, folded to [[7,3],[30,4]]. The T- pattern is the half-turn of the T+ pattern. Every cut cell has doubled offset in {1, 3, 5}, every tile of a type is a translate of one hexagon or triangle, and every child polygon lies inside its grown parent polygon. The children of every parent match their type's pattern, and the census of cut cells, (7, 30), (139, 330), (1963, 5490), equals S_5^n (1, 0). At base 3, levels 1..4, the same verb returns Abel's H -> 6 H + 3 T+ + 3 T-, T+ -> 1 H + 3 T+, and the census (6, 6), (42, 54), (306, 414), (2250, 3078).
What this settles. THE CLAIM's first line is a theorem at every odd base, and the slice dimension it prints is the Hausdorff dimension. What base 5 adds to base 3 is a triangle of the other orientation inside the triangle's pattern, T+ -> 1 T-, since M[-1, 1] = P_b[(b - 5)/2] is 0 at b = 3 and positive from b = 5.

OPEN QUESTIONS

Which sponge is the sponge at a given base? Cook's predicate is "at most one coordinate in the middle third"; this page inherits "at most one odd coordinate" from bang dim 3, code 23. The two agree at base 3 and nowhere else, and there is no canonical base-5 Menger sponge, so the whole generalisation rests on an unstated choice.
The choice is not cosmetic, and the two rules land on opposite sides. Verified. At base 5, dim 3 the middle-digit rule fills 112 of 125 and its central diagonal slice sits ABOVE log(fill) / log(base) - 1, excess +2.888e-02, at slice dimension 1.960651 against 1.931768; the odd-coordinate rule fills 81 of 125 and sits BELOW, 1.6869 against 1.7304. No such claim at base 5 can be quoted without naming the rule.
The mod-4 split has a mechanism. Proved. It holds at every odd base up to 401 by exact rational comparison in lab/py/odd-base-slice-grammar, and at every odd base by THE COMPOSITION: the middle diagonal layer sits at coordinate sum 3*(base-1)/2, odd exactly when base = 3 mod 4, so P_b[g], P_b[g+b-1], P_b[g+b+1] are read from the odd part 3 O E^2 of the digit polynomial and P_b[g+1], P_b[g+b] from the even part E^3 in that class, all five swapping parts in the other, and the side of rho against fill/base follows from those forms as a polynomial inequality. The parity of cell forced into the middle layer is that flip.
The split is a property of the rule as well as the base. The mod-4 statement is about the odd-coordinate solid, and the middle-digit solid contradicts it at base 5. Among the 256 parity designs the rule enters only through sgn(o - e), odd-weight patterns against even-weight ones, at every odd base: THE SLICE AT LARGE SIDE, Every parity design. Outside the parity designs a statement of it must still pin base, dimension and digit rule.
Is the two-tile grammar geometric, or only arithmetic? Geometric, Proved. A cut cell is a tile, the children of a tile depend on its type alone, and the count matrix is the carry automaton; THE COMPOSITION, The tiles.
The base axis factors. Proved. cuts walks the dimension axis at fixed base 3 and has a theorem there: the order law, from a factorisation of the digit polynomial. The same three-step argument - digit polynomial E^2 (E + 3 O), carry contraction to abs(c) <= 1, palindromic symmetry P_b[s] = P_b[2g - s] - is THE COMPOSITION above, and it decides the mod-4 split. The geometry comes with it: a cut cell is a tile, THE COMPOSITION, The tiles.

THE TENT IDENTITY

Law E's window length is a distance measured inside a slot, and that is an identity among Law E's own closed forms rather than a fact about the module. Every symbol below is lab/py/smith-window's and every number is regenerated there.

The objects

D = 2R + 1 is odd with D >= 5, so R >= 2.
J is Jacobsthal: J(n) = 0 for n < 0 and J(n) = (2^n - (-1)^n)/3 otherwise, so J(0) = 0, J(1) = J(2) = 1, J(3) = 3, J(4) = 5, with J(n) = J(n-1) + 2 J(n-2) for n >= 2.
b is the least integer with 2^b >= 3R - 1, so b >= 3 and 2^(b-1) < 3R - 1 <= 2^b.
g = |2R - 2^(b-1) - 1| is odd and at least 1, and s = (g + 1)/2 >= 1 is its half.
e = min{e >= 1 : J(e) >= s} is the slot index and k = b - 1 - e its complement.
The window box is i0 = max(2, 4R - 2^b) and hi = hi0 - (hi0 mod 2) with hi0 = floor((6R + 2 - 2^b)/3), and K = (hi - i0)/2 is the top index of the family X_0, ..., X_K, never its length.
t = (J(k) - 1)/2, and c_t are the F_2 Fibonacci polynomials c_0 = 1, c_1 = 1 + y, c_t = y c_(t-1) + c_(t-2).
N = J(e) - J(e-1) is the slot length, u = s - J(e-1) - 1 the offset inside it, and p = u above the octave centre R = 2^(b-2) while p = N - 1 - u at or below it.
C_D = K - 2 J(e-1) when k is even and C_D = K when k is odd is Law E's ceiling, W = C_D - t 2^e its offset, chi = 2 J(e-2) - 1 for e >= 3 and chi = 1 otherwise, m = max(0, 2W - chi), and g_D = z^m c_t(z^(2^e)) its generator.

The statement

Proved. For every odd D = 2R + 1 >= 5, min(p, N - 1 - p) = C_D - deg g_D.
The proof is exact arithmetic in b, e, k, R and uses no property of V_2, only the formulas above. C_D and g_D are taken here as those closed forms and not as the measured ceiling and generator, so the theorem is an identity among Law E's formulas and says something about V_2 only where Law E itself holds.
Law E is a swept law, Verified at 1199/1199 rows of odd D = 5..2401 by lab/py/smith-window, and the identity inherits that standing hypothesis. Against the formulas themselves it is a theorem, re-checked as a transcription at 999999/999999 rows of odd D = 5..2000001 (lab/py/smith-window).

The proof

Throughout 3 J(n) = 2^n - (-1)^n, J(n) is odd for n >= 1, and J is nondecreasing on n >= 0.

Lemma 1, no e = 2. e = 2 would need J(1) < s <= J(2), that is 1 < s <= 1, which is empty, so e = 1 or e >= 3.
Lemma 2, the bracket. J(e-1) < s <= J(e): the right inequality defines e, the left is its minimality for e >= 2 and reads 0 < s at e = 1.
Lemma 3, the box length. K = J(b-2) - s in both octave halves: above centre s = R - 2^(b-2) gives i0 = 4s and hi = 2s + J(b-1) + 1 - 2[b even], below or at centre s = 2^(b-2) + 1 - R gives i0 = 2 and hi = J(b-1) + 3 - 2s - 2[b even], both landing on K = (J(b-1) - (-1)^b)/2 - s, and J(b-1) - (-1)^b = 2 J(b-2).
Lemma 4, k >= 1. The defining bound 3R - 1 <= 2^b gives 3s <= 2^(b-2) + 1 above centre and the minimality 2^(b-1) < 3R - 1 gives 3s < 2^(b-2) + 2 below, so s <= J(b-2) either way, hence K >= 0, e <= b - 2, k >= 1, J(k) odd and t >= 0 an integer; Law E's standing hypothesis k >= 1 is therefore a theorem.
Lemma 5, the collapse. W = J(e) - s whether k is even or odd: with b - 2 = k + e - 1 and t 2^e = (J(k) - 1) 2^(e-1), 3(J(b-2) - t 2^e) = (-1)^(k+e) + ((-1)^k + 3) 2^(e-1), which is 3 J(e) for k odd, where C_D = K, and 3 J(e+1) for k even, where the ceiling's -2 J(e-1) turns J(e+1) - 2 J(e-1) back into J(e); the ceiling deficit and the parity of k cancel exactly.
Lemma 6, the slot. For e >= 2, 3(J(e) - J(e-1)) = 2^(e-1) - 2(-1)^e = 6 J(e-2), so N = 2 J(e-2) and chi = N - 1 on the live range e >= 3; at e = 1 the slot is N = 1 while chi = 1, and that bridge fails.
Lemma 7, the reflection. u + W = (s - J(e-1) - 1) + (J(e) - s) = N - 1 with both terms nonnegative by Lemma 2, so {u, W} = {p, N - 1 - p} in both halves and min(p, N - 1 - p) = min(W, N - 1 - W).
Lemma 8, parity. p == R mod 2 whenever e >= 3, since then b >= 5 makes 2^(b-2) even and J(e-1), J(e) are odd; it is sharp, failing exactly on the e = 1 rows above centre, R = 2^(b-2) + 1, that is exactly on D = 2^j + 3 for j >= 2.
The theorem. deg c_t = t, because y c_(t-1) has degree t against t - 2 for c_(t-2), so deg g_D = m + t 2^e and C_D - deg g_D = W - max(0, 2W - chi) = min(W, chi - W); by Lemma 1 three cases exhaust, at e >= 3 Lemma 6 reads chi as N - 1 and Lemma 7 closes it, at e = 2 there is nothing to prove, and at e = 1 Lemma 2 forces s = 1, so W = u = p = 0 and N = 1 make both sides 0.

What it buys

Proved. The upper half of the layer-2 window law, that z^(C_D - deg g_D + 1) g_D does not lift, is free wherever C_D = K: every element of V_2 has coefficient degree at most K, while that candidate has degree C_D + 1 whatever deg g_D is, so at C_D = K it leaves the family outright. The tent identity is not used; the cut costs the ceiling law alone.
Proved. By the ceiling law C_D = K exactly when k is odd or e = 1, and C_D < K exactly when k is even and e >= 3, where the deficit is K - C_D = 2 J(e-1) > 0. So the upper half is unconditional at every row with k odd or e = 1, and what stays open is the rows with k even and e >= 3: 448 of the 1199 rows of odd D = 5..2401, and 29116 of 99999 over odd D = 5..200001 (lab/py/smith-window).
Conjecture. At those open rows the family element of coefficient degree C_D + 1 has mod-4 obstruction outside the image of the mod-2 symbol on the same coefficient box, for a deficit of exactly 2 J(e-1) steps. That is a rank statement about the corrector image, not arithmetic in b, e, k, R, and it is all that remains of the upper half.
Proved. Lemma 8's parity fails exactly on D = 2^j + 3 for j >= 2, rows carrying k = j - 1, so the reach law's escaping family D = 4^m + 3 is the k odd half of that set and nothing more: D = 11 fails the parity and is not of that form. Why the reach law excepts that half and not the k even rows D = 11, 35, 131, ... is open.

THE SLICE AT LARGE SIDE

The central hexagon of bang dim 3, code 23 as the side grows, at one base and on words of growing sides. Every number below is regenerated by lab/py/slice-at-large-side, verb by verb.

The objects

A word (b_1, .., b_L) of odd sides, coarsest first, cuts the cube of side N = b_1 .. b_L into b_1^3 blocks, each block into b_2^3, and so on: a coordinate's digit at level k is in base b_k, and the last letter carries the units digit. A cell is kept when its digit triple at every level has at most one odd digit.
The plane x + y + z = 3(N-1)/2 meets [0, N)^3 in (3N^2 + 1)/4 lattice points, the hexagon. count is the number of kept points on it and ink = count / ((3N^2 + 1)/4) their share.
S_b is the block of THE COMPOSITION and X_b = S_b / b^2; A_3 and A_1 are its limits on the classes b = 3, 1 mod 4, lambda a Perron root, gamma a drift exponent and C the constant in front of lambda^L L^gamma.
A parity design is a set Omega of digit-parity patterns in {0,1}^3; e and o count its patterns of even and odd weight, P_Omega is its digit polynomial and fill_Omega its number of digit triples. Code 23 has e = 1, o = 3.
The census exponent of a design at base b is limsup_L log_b(count) / L on the word (b, .., b). For the 30 codes whose block has a zero diagonal on a class the count vanishes at every odd level there, and the limsup is read on the even levels.

The word product

Proved. count = (S_(b_L) .. S_(b_1))[0, 0]. The target splits as 3(N-1)/2 = sum_k g_k b_(k+1) .. b_L with g_k = 3(b_k - 1)/2, so the carry automaton of THE COMPOSITION reads the units digit first, letter b_L, from carry 0 back to carry 0; its carry and reflection lemmas hold letter by letter, which puts the whole product on the even block.
Verified against brute force by verb words: (3,5) 60, (5,3) 72, (3,5,7) 2412, (7,5,3) 2688, (5,7) 300, seven more words, and four words for each of seven other codes. The letters do not commute, so the order of a word moves its count.

The letter at infinite side

Proved. X_b = A + B/b + B_2/b^2 exactly on each class, read off the five coefficients of THE COMPOSITION: A_3 = [[9/16, 1/32], [3/8, 3/16]], A_1 = [[3/16, 3/32], [9/8, 1/16]], and B = (3/8)[[1, 1/2], [2, 1]] on both classes (verb letter).
The limit is a two-state matrix, not a number. On the three carries M_b[c, c'] / b^2 -> f(c') q(c + c' + g): f = 1/8, 3/4, 1/8 at c' = -1, 0, 1 is the Irwin-Hall density of three uniform digits at 5/2, 3/2, 1/2, and q is 1/4 on an even digit sum and 3/4 on an odd one, the share of the four parity patterns of that sum with at most one odd digit. On the block A = [[3u/4, v/8], [3v/2, u/4]] with u = q(g) and v = q(g + 1), and u = 3/4 exactly when g is odd, which is b = 3 mod 4.
diag(1, 1/sqrt(12)) conjugates A to the symmetric F^(1/2) Q F^(1/2), where F = diag(3/4, 1/4) is the Irwin-Hall mass of the middle and side carries and Q = [[u, v], [v, u]], so lambda = (2u + sqrt(u^2 + 3v^2))/4: (3 + sqrt(3))/8 = 0.591506 at 3 mod 4 and (1 + sqrt(7))/8 = 0.455719 at 1 mod 4. The limits do not commute, [A_3, A_1] = [[0, 1/32], [-3/8, 0]], and A_1 A_3 has Perron root 3(1 + sqrt(2))/32, 0.475744 per letter.
Proved, the exact roots. rho_b = (b+1)(3b + sqrt(3b^2 + 6))/8 at 3 mod 4 and rho_b = (b^2 + 3b + 4 + (b-1) sqrt(7b^2 + 32b + 34))/8 at 1 mod 4, from the trace and determinant of THE COMPOSITION, so rho_b / b^2 = lambda (1 + mu/b + O(b^-2)) with mu = 1 and mu = 1 + 2/sqrt(7).
Proved, the side at infinite side. With fill/b^3 = (1/2)(1 + 3/(2b) + O(b^-2)), (log b)(log_b rho_b - log_b fill + 1) = log(2 lambda) + (mu - 3/2)/b + O(b^-2): 0.168064 - 0.5/b at 3 mod 4 and -0.092732 + 0.255929/b at 1 mod 4. The mod-4 side of THE COMPOSITION survives at infinite side with gap log(2 lambda) / log b.

The drift

Lemma, the slow product. Proved. Let Y_j be positive 2x2 matrices with Y_j = Y + R/j + O(j^-2), the limit Y positive with Perron root Lambda and Perron vectors r, l, and nu = l R r / (Lambda l r). Then x^T Y_m .. Y_1 y = Lambda^m m^nu (kappa + o(1)) with kappa > 0 for every positive y and every x >= 0 with x^T r > 0.
Proof. From some j_0 on, the Perron root lambda_j of Y_j is simple and analytic in the entries, so lambda_j = Lambda (1 + nu/j + O(j^-2)) and prod_(j<=m) lambda_j = Lambda^m m^nu (kappa' + o(1)) with kappa' > 0. Write Y_j = V_j diag(lambda_j, lambda'_j) V_j^-1 with the Perron vector r_j first and l_j the first row of V_j^-1; then V_(j+1)^-1 V_j = I + Delta_j with sum ||Delta_j|| < inf, since Y_(j+1) - Y_j = O(j^-2). The vector z_m = V_m^-1 Y_m .. Y_1 y / prod_(j<=m) lambda_j obeys z_m = diag(1, theta_m)(I + Delta_(m-1)) z_(m-1) with abs(theta_m) <= theta < 1, because the positive Y has its second eigenvalue strictly inside Lambda in modulus; so z_m stays bounded, its second coordinate tends to 0, and its first, p_m = l_m Y_m .. Y_1 y / prod_(j<=m) lambda_j, moves by O(m^-2) and converges to some p. Positivity gives p > 0: l_m - l_(m-1) = O(m^-2) against the positive vector Y_(m-1) .. Y_1 y and l_(m-1) bounded below give p_m >= (1 - O(m^-2)) p_(m-1). Hence x^T Y_m .. Y_1 y = prod_(j<=m) lambda_j (p x^T r + o(1)) with p x^T r > 0.
Proved, the drift. ink = lambda^L L^gamma (C + o(1)) with C > 0 on three words: (3, 7, .., 4L-1) with lambda = (3 + sqrt(3))/8 and gamma = 1/4; (5, 9, .., 4L+1) with lambda = (1 + sqrt(7))/8 and gamma = 1/4 + 1/(2 sqrt(7)) = 0.438982; and (3, 5, .., 2L+1) in either order with lambda = sqrt(3(1 + sqrt(2))/32), gamma = (2 + sqrt(2))/4 = 0.853553 and C depending on the parity of L.
Proof. ink = (4/3) e_0^T X_(b_L) .. X_(b_1) e_0 (1 + O(N^-2)) and every X_b is positive, so the lemma runs with x = e_0, e_0^T r > 0, and y the first factor applied to e_0, a positive vector. On one class the letter at b_k = 4k -+ 1 is A + B/(4k) + O(k^-2), so R = B/4 and the lemma's nu is gamma = mu/4, where mu = l B r / (lambda l r) is 1 and 1 + 2/sqrt(7). On (3, 5, ..) coarsest first the pair X_(4j+1) X_(4j-1) = A_1 A_3 + (B A_3 + A_1 B)/(4j) + O(j^-2) has R = (B A_3 + A_1 B)/4 and the lemma's nu = (2 + sqrt(2))/4, so L = 2m letters give m^nu = (L/2)^gamma, and an odd L adds one letter tending to A_3. Finest first is the transposed product on the pairs X_(4j-1) X_(4j+1), with the same nu because a product and its reverse have one Perron root as a function of 1/j.
Verified. The local exponent (log ink(2L) - log ink(L) - L log lambda) / log 2 at L = 4000, 8000, 16000 reads 0.249992, 0.249996, 0.249998 at 3 mod 4, 0.438890, 0.438936, 0.438959 at 1 mod 4 and 0.853306, 0.853430, 0.853492 on (3, 5, ..), its error halving per doubling; one Richardson step returns 0.250000, 0.438982, 0.853553 at both of the last two lengths (verb drift).

The constants

Proved, a closed form at sides 3 mod 4. On (3, 7, .., 4L-1), S_(4k-1) = (k/2)(k K_1 + K_0) with K_1 = [[18, 1], [12, 6]] and K_0 = [[-6, 1], [0, 0]], and count = (L!)^2 2^-L [z^L] (1 - 6z)(1 - 24z + 96z^2)^(-3/4), which reads 6, 198, 15552, 2218104. Proof: count = (L!)^2 2^-L e_0^T w_L with k w_k = (k K_1 + K_0) w_(k-1) and w_0 = e_0, which is (1 - z K_1) W' = (K_1 + K_0) W for W = sum_k w_k z^k with W(0) = e_0, a linear system regular at 0; W = (1 - 24z + 96z^2)^(-3/4) (1 - 6z, 12z) solves it, a polynomial identity checked by verb closed, which also matches the series against the integer product at L = 1..60.
Proved, the constant. C_3 = Gamma(3/4)(1 + sqrt(3)) / (3 (sqrt(3) - 1)^(3/4)) = 1.410085329792638597969, rounded. Here 1 - 24z + 96z^2 = (1 - alpha z)(1 - alpha' z) with alpha = 4(3 + sqrt(3)), and h(z) = (1 - 6z)(1 - alpha' z)^(-3/4) is analytic on abs(z) < 1/alpha'; splitting the convolution of h with (1 - alpha z)^(-3/4) at half its length gives [z^L] = alpha^L L^(-1/4) (h(1/alpha) + o(1)) / Gamma(3/4) with h(1/alpha) = (1 + sqrt(3)) / (4 (sqrt(3) - 1)^(3/4)). With N = 4^L Gamma(L + 3/4) / Gamma(3/4) and alpha/32 = lambda that is the constant, and the extrapolation of verb constants agrees to 19 digits.
Proved, the blink. On (3, 5, ..) in either order C_odd / C_even = sqrt((5 sqrt(2) - 1)/3) = 1.4225643291682. By the lemma an odd length adds one letter tending to A_3 against the Perron vector r of A_1 A_3, so the ratio is e_0^T A_3 r / (lambda e_0^T r); finest first reads the same, because diag(1, 1/12) conjugates every limit letter [[3u/4, v/8], [3v/2, u/4]] to its transpose and fixes e_0.
Conjecture, the other constants. Neville extrapolation in 1/L from L = 256 .. 16384, two depths agreeing to 19 digits (verb constants): C_1 = 0.72001825738796 on (5, 9, ..); on (3, 5, ..) coarsest first 0.53693769481512 at even L and 0.76382841162981 at odd L, finest first 0.66052225470496 and 0.93963539816504, finest over coarsest 1.2301655500875 at both parities. No closed form is known: at 1 mod 4 the letter is quadratic in k and the drift exponent is irrational, which no algebraic series of the kind above can carry.

Every parity design

Proved, the letter of every design. For a parity design Omega the same reading gives A_Omega = [[3u/4, v/8], [3v/2, u/4]] with u = q(g) and v = q(g + 1), where q is e/4 on even digit sums and o/4 on odd ones: u = o/4 and v = e/4 at b = 3 mod 4, u = e/4 and v = o/4 at b = 1 mod 4. So lambda_Omega = (2u + sqrt(u^2 + 3v^2))/4 against fill_Omega / b^3 -> (u + v)/2, and log b times the census exponent minus log_b fill_Omega - 1 tends to log(8 lambda_Omega / (e + o)), positive exactly when u > v, since sqrt(u^2 + 3v^2) > 2v reads u > v. Each coefficient P_Omega[c + g - b c'] counts lattice points of a box by inclusion-exclusion and is a quadratic in b on each class from b = 11, its leading term the Irwin-Hall density times q; verb designs derives all 63 weight classes and checks them against the digit polynomial at odd b up to 61.
Proved, the ties. When e = o, P_Omega(zeta) = (e - o)/(1 + zeta)^3 = 0 at every b-th root of unity zeta != 1, by E(zeta) = 1/(1 + zeta) and O(zeta) = -1/(1 + zeta) of THE COMPOSITION. So every residue class of P_Omega mod b sums to fill_Omega / b, every row of the automaton is one class by the carry lemma, and the automaton's Perron root is fill_Omega / b exactly; 69 of the 255 nonempty codes, row sums checked at odd b = 3..41 by verb designs.
Proved, computer-assisted, the sign law of every design. For each of the 255 nonempty codes and every odd base, the census exponent minus log_b fill_Omega - 1 has the sign of o - e when b = 3 mod 4 and of e - o when b = 1 mod 4, an empty slice counting as below: 93 codes sit above at 3 mod 4 and below at 1 mod 4, 93 the reverse, and 69 on the line. Proof: verb designs writes the block and fill_Omega as polynomials in n = (b+1)/2 on each class, exact from b = 11, isolates the real roots of the two sign polynomials of THE COMPOSITION's side and of the case guard, s_01 s_10, or s_00 on a class where s_01 vanishes, all at most n = 3, the guard's only one being code 128 at b = 5, and checks every odd base up to 15 exactly. The empty slices are the seven codes whose patterns all have weight 2, at b = 3, and the code of the single pattern 111, at b = 5.
Code 23 is one of the 93 above at 3 mod 4, and THE CLAIM's mod-4 side is this law at e = 1, o = 3. Within parity designs the side is set by sgn(o - e) and b mod 4 alone, which answers, for them, the OPEN QUESTIONS bullet "The split is a property of the rule as well as the base."; the middle-digit solid is not a parity design.

What stays open

Closed forms for C_1 and the four constants of (3, 5, ..).
The sign law along dim at base 3, Conjecture S of cuts, is untouched: everything here runs along the base.

WHERE THE REST LIVES

The dimension axis at fixed base 3: the ceil(dim/2) order law, the product formula over 3-adic angle towers, and the unconditional pinning |rho_dim - fill/3| <= 2(dim-1)/3 live in the shelf paper slice-recurrence-order; the order law at every odd base is also cuts.
The sign law in every even dimension at bases 3 and 5, the certificate machines, the transient constant ln(R)/4, the tent rank law and the layer-2 window law live in the shelf paper slice-sign-even-half.
The layer-2 window itself - its generator g_D, its ceiling C_D, the family shift law and the corrector law behind them - is regenerated by lab/py/smith-window.
So are the two statements that close the corrector law. The tent identity is proved above as arithmetic in b, e, k, R among Law E's own closed forms, so only the reach law reach = R - jmax = 3 min(p, N - 1 - p) + 2 [e even] + [k odd](1 + p mod 2) is still read off a sweep, with one row per odd octave escaping it at D = 4^m + 3.
Off those escaping rows floor(reach/3) = C_D - deg g_D + [k odd and e even]; on them it reads 1 against C_D - deg g_D = K - deg g_D = 0, so min(K - deg g_D, floor(reach/3)) = C_D - deg g_D at every row and the corrector law's statement reads off (b, e, k, R) with no span test in it. The deduction behind it still carries one.
The carry matrix M_even is defined once, in cuts, and is not redefined here.
The hexagon-triangle substitution, the carpet slice census row on CLAIMS, and the carry automaton: cuts. The hexagon mesh itself: slices. The fill polynomial of bang dim 3, code 23: method.
Every finding on a tagged line: CLAIMS. Every source resolved: REFS.