SpectraSpectra

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 on this page is regenerated by lab/py/odd-base-slice-grammar, a raster-free digit recursion cross-checked against a direct layer census. What stays Conjecture is the grammar itself: a 2x2 map reproducing cell counts is weaker than a substitution acting on tiles, and only base 3 has the latter from a source.

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 are the same object is Proved, by the hexagram bijection in cuts, and 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

  • Conjecture. For every odd base, the centroid diagonal slice of the parity solid bang dim 3, code 23 is a graph-directed set on exactly two tiles, whose 2x2 integer substitution matrix is a fixed rational function of base within each class of base mod 4, and whose slice dimension sits above log(fill) / log(base) - 1 when base = 3 mod 4 and below it when base = 1 mod 4.
  • 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 claim 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. Counts obeying a fixed 2x2 map is weaker than a substitution acting on tiles, and only base 3 has the latter from a source.
  • The closed form. Verified at base 3..21 by lab/py/odd-base-slice-grammar; every odd base is the claim above. 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. Any odd base whose slice needs a third tile symbol; any census breaking its own two-term recurrence past the fitted levels; any base where the mod-4 side is wrong.
  • 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.

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.
  • Does the mod-4 split have a mechanism, or is it numerology? Conjecture. It holds at every odd base up to 401 by exact rational comparison in lab/py/odd-base-slice-grammar, so it is not four data points. The candidate mechanism - the middle diagonal layer sits at coordinate sum 3*(base-1)/2, odd exactly when base = 3 mod 4, forcing a different parity of cell into the middle layer in each class - is unwritten and unchecked.
  • 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. Any statement of it must pin base, dimension and digit rule before it means anything.
  • Is the two-tile grammar geometric, or only arithmetic? Reproducing cell counts is weaker than a substitution acting on tiles, and no source supplies the latter past base 3.
  • Does the base axis factor at all? 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, carry contraction, palindromic symmetry - is what the two-tile claim has never been given, and it is exactly the kind of statement that would decide the mod-4 split.

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.

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 are the carlomitchener/research/slice-recurrence-order lane.
  • 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 are the carlomitchener/research/slice-sign-even-half lane.
  • 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 hexagram bijection and the mesh census: cuts. The hexagon mesh itself: slices. The fill polynomial of bang dim 3, code 23: method.
  • Every finding on a tagged line: DISCOVERIES. Every source resolved: REFS.