hexagon.md

--- title: The hexagon moire lead: The moire of the stacked diagonal slices: the exact cut-ink laws, the doubling sign law, the quarter-line bands, and the ghost star, whose cell-frame decay is a closed form at every band width, -1/4 at the arm and -1/8 in the limit, every layer-pair constant an exact rational, and no frame-free coefficient. figure: research-hexagon slug: hexagon ---

Cut the solid cube of odd side n through its centre, perpendicular to the main diagonal, and the section is a regular hexagon; the slices page counts that mesh and the fills that live on it. Do it at every odd side at once, hold the layers up to the light, and the hexagons interfere. This page is the arithmetic of that interference: which lines the stack lights, which pairs of layers agree, how the agreement decays, and which constants survive to the limit. The layers are the four historical families of slices - carpet, net, tree and void - and everything below is either an exact statement about one layer or an exact statement about a pair.

Every claim carries a tag. Proved means a proof is given or restated here; Verified means recomputed from scratch by the study named; Conjecture means a fit with no derivation. One generator prints every number on this page, lab/rs/hexagon-moire, which runs in under a minute and whose sections are the sections below; a number it does not print is not on this page, and the one place a shelf paper is the source instead says so. The volume demo stacks a cube design into a solid field and cuts it on any plane, the central diagonal cut being this hexagon, and the slices demo draws one layer triangle by triangle.

Three frames run through the page and they are not the same frame. The ideal frame is the cut plane's own coordinate square: a point is (X, Z) in [0,1]^2, the layer's cell is x = floor(4nX), z = 2*floor(2nZ), y = 6n - 2 - x - z, the point is on the layer when 0 <= y < 4n, and every layer is resampled onto one common grid. The lattice frame is the rendered raster of the cube's own cells. The cell frame is the cube's own cell lattice read one layer at a time, nothing resampled and no raster, every quantity a ratio of exact cell counts. The plane x + y + z = 6n - 2 sits half a cell off the true centre, so each layer's hexagon is displaced by 1/(2n) from the last, and that displacement is real geometry rather than an artifact: a scan raster that ignores it misreads the crosshair strengths, and one quantity below - the star's decay coefficient - is a different number in each of the three.

The cut ink of one layer

The cut ink laws of all four families are exact closed forms. Proved. With chi = (-1)^((3n-1)/2), the inked share of the hexagon at odd side n is

carpet  1/2 + chi/8 + 1/(2n) - chi/(8n^2)
net     1 - carpet
tree    1/4 + (1/3 - chi/12)/n + (1 - chi)/(6n^2)
void    1/4 - chi/(4n) + 1/(2n^2)

and the generator checks each against the counted hexagon in exact rational arithmetic: 28 of 28 layers match for all four families at every odd n <= 55. The character chi is the same chi_4 the Walsh spectrometer proves the quasipolynomial by, and the 1/n and 1/n^2 orders of that law are proved on slices under the slice ink. The wider range n <= 101 has no generator, so nothing is claimed there.

The centre cell is the first place the families separate. Over the same 28 layers the carpet's centre cell is ink in 14, the net's in 14, the tree's in 14, and the void's in 28 of 28 - the void's centre is ink at every odd n, by a two-line parity argument on the three coordinates. That single cell is the seed of the two star sections below: the carpet's centre agrees with the background and fades, the void's never does. Verified.

Layer pairs and the doubling law

Correlate two layers on the full hexagon, exactly, cell by cell. The seven pairs the generator reads blind are

(Verified.) The coprime pairs are not zero, and that is the whole point: on the square stack they are exactly zero, and the reason the hexagon loses it is a theorem of its own, stated under WHERE THE EXACTNESS STOPS.

The corrected law on the hexagon is dyadic. Proved. What breaks coprime independence is a hidden half-cell-shifted overtone at doubled frequency, chi_4(n) s(2nX + 1/2)/8, that the plane constraint forces into every carpet slice, plus the hexagon's non-product tent marginal; together they couple layer m to layers 2m +- 1 and m +- 2 regardless of gcd. The sign is forced: sign r(m, 2m +- 1) = -chi_4(m) chi_4(2m +- 1), and the generator holds it on 18 of 18 pairs from (3, 5) to (601, 1201) across all four residue branches. Two of those rows, the largest computed, read (501, 1001) at r = -0.11732362 and (601, 1201) at r = -0.11729091.

The doubling magnitude reads between 0.11711630 and 0.11715991. Verified. Richardson extrapolation r = r_inf + a/m + b/m^2 on sliding triples of m = [157, 201, 301, 401, 501, 601] returns r_inf = -0.11715991, -0.11711630, -0.11714296, -0.11711630, and the two branch extrapolations in 1/m land on 0.1171270 at (501, 601) and 0.1171274 at (103, 403). That kills the rival 19/162 = 0.11728395, which sits above every reading in that spread and above both branch extrapolations; the closed form below kills it outright.

The exact doubling constant is 253/2160, and the smear integral behind it is closed. Proved. Write layer N's cut cell at hexagon point (X, Z) as x = floor(4NX), z = 2 floor(2NZ), y = 6N - 2 - x - z. Its block parities are s_x = floor(NX) mod 2, s_z = floor(NZ) mod 2 and, exactly, s_y = s_x + s_z + w mod 2 with w = floor((6N - 2 - p - 2q)/4) mod 2 for p = floor(4 frac(NX)) and q = floor(2 frac(NZ)). That w is the half-cell overtone in closed form: it is 1 at exactly the two phase cells (p, q) = (0, 0) and (3, 1) when N = 1 mod 4 and its complement when N = 3 mod 4, so the plane constraint is the entire coupling and the third coordinate is never free. For n = 2m + 1 the identity nX = 2 mX + X puts layer n's phase at (2 frac(mX) + X) mod 2: the two layers' phases lie on one closed geodesic of the torus instead of filling it, which is why coprimality buys nothing on this cut, and as m grows (mX mod 2, mZ mod 2) equidistributes over the hexagon at rate O(1/m), since the region is a fixed polygon and every nonzero frequency integrates to O(1/m). What is left is a piecewise-constant integral over (X, Z) and the phase, with all breakpoints rational, and it evaluates in exact rational arithmetic to covariance 253/9216. Every layer's variance is 15/64, so r = 253/2160 at all four branches, carrying the sign law's sign. The generator reads -0.11745304 at (301, 601) and -0.11729091 at (601, 1201) against -253/2160 = -0.11712963, the gap times m holding at -0.0974 and -0.0969; those are both m = 1 mod 4, and the other class converges to the same magnitude about twice as slowly, (103, 205) reading +0.11914004 and (203, 405) reading +0.11814528 against +253/2160 with the gap times m at +0.207 and +0.206. The rival 19/162 is dead.

The other pair families

The breakage is rule-specific in the limit. Proved. Run the same exact correlation across the four families:

So the persistent doubling coupling is carpet and net only - their doubling covariance is +-253/9216 and the tree's and the void's are exactly 0 - the tree keeps a neighbour coupling at (m, m+2) and no doubling, and the void is nearly independent at both. The gcd echo survives in all four. The controls agree that this is signal and not raster noise: at the coprime pairs (101, 173) the three families read -0.00248023, +0.00033407 and +0.00018183, and at (97, 251) they read +0.00212976, +0.00022204 and +0.00008585.

Each pair family on this page is one integral with a different phase map, and each limit is rational. Proved. The claim is about a fixed affine phase map n = a m + c with m growing inside one class mod 4, measured on the mask this page always uses, the full hexagon of the common cut with exact area weighting; a general pair (m, n) has no limit theorem here. The map is n = a m + c, which sends layer m's phase alpha = mX mod 2 to layer n's (a alpha + c X) mod 2; the doubling pairs are a = 2, c = +-1, the adjacent pair is a = 1, c = 2, and the gcd echo is a = 3, c = 0, where the phase map forgets (X, Z) entirely and the hexagon's shape drops out. The same exact integral run on the four families' corner tables gives

with covariances +-253/9216, -11/576, +29/576 for carpet and net, 0, -61/4608, +1/36 for the tree and 0, +7/4608, +1/72 for the void. The tree's and the void's doubling zeros are exact and not small: the integral returns their joint ink as 1/16 on the nose against a background of 1/4 each, so the doubling coupling is a carpet and net theorem and nothing wider. The measured rows above sit O(1/m) from these: (249, 251) reads -0.08150224 against -11/135 and (99, 297) reads +0.21476417 against +29/135, gaps of 2.1e-5 and 5.1e-5 with gap * m at -0.0052 and -0.0050.

The quarter line and the crosshairs

Stack the carpet's layers in the ideal frame at N = 55, 3601 samples per axis, and read one-sided bands of width 0.018 against the hexagon mean 0.543621:

The quarter-line law. Verified. The strongest interior lines of the stacked hexagram sit at quarter-cell coordinates a/4, generally a/(4b) with b odd, and the step across such a line is an exact one-sided +-1/8 that every layer votes for identically, because the overtone's chi_4 sign meets the layer's own chi_4 and squares away. Read straight at the line the plateau converges: mean magnitude 0.12210, 0.12340, 0.12412, 0.12446 at N = 151, 301, 601, 1201 against the limit 1/8, the one-sided pair at N = 1201 reading +0.12320 and -0.12573. The odd-fraction crosshairs at (-1)^a/(4q), q odd, follow behind. The X, Z and X+Z profiles are numerically identical on the render by the slice's permutation symmetry, so "in five directions but never horizontal" is false, and the missing-Z-overtone statement holds only in the rectangle-cell frame.

The three 60-degree crosshair families obey a limit law. Verified. The line at coordinate a/q carries strength (-1)^a/(4q) for odd q and nothing for even q. In the coarse cut model, 200003 samples along a line, the excesses converge:

The N = 55 column is the whole lesson about frames: 1/7 reads -0.058065 where the N = 5555 reading is -0.036036 against a predicted -0.035714, and the even denominators read -0.014353 and -0.023394 where the prediction is zero. That gap is the per-layer registration drift of 1/(2n), and it is what a drifted scan raster misreads. In registration-correct frames the rays are visible in the real render as well, the X + Z = 1.25 line at N = 55 reading -0.045236 and +0.028988 and the X = 1/3 one-sided bands reading +0.020969 and -0.058390. A null claiming no rays at all was about the scan geometry and not about the object.

The void star

The void slice stack keeps its star forever. Verified. In the ideal frame at N = 55, 3601 samples per axis, centred bands of half-width 0.004, against a hexagon mean of 0.278670 whose limit is 1/4:

The law behind the table is six central lines at plateau ink 1/2 against a background of 1/4, ratio 2, every layer voting on all six, with the model-frame Z arm weaker at 3/8 - the Z = 0.50 row - and a centre dot that is ink at every odd n, the 28/28 of the first section. Nothing about this decays. The carpet's star, by contrast, is a finite-layer artifact and fades; the two snowflake stacks differ by a theorem, not by a constant.

Void and carpet have complementary line spectra on the cut. Verified. The void's lines sit at even-denominator twisted positions X = a/(2b) with b odd, exactly where the carpet is silent, and the void is silent at the carpet's odd rationals. The tree carries the only untwisted crosshair family plus a permanent ratio-2 line at 2X + Z = 3/2, and ignores its free axis. The net is the exact pixelwise complement of the carpet, since "at most one odd" and "at least two odd" exhaust the cases - the same partition that makes carpet and net tile one hexagon in slices.

The twist that kills the ray family

The chi_4 twist kills the pair-resonance ray family. Verified. Average chi_4(n) T(nx) over odd n <= N with T the layer's tent, and the average falls like 1/N at every x, rational or not:

The last row is irrational and behaves like the rest. So the snowflake stack has no analogue of the carpet stack's bright main diagonal: in the coarse cut model the A = C excess goes -0.003580 at N = 55 to -0.000437 at N = 555 to -0.000052 at N = 5555, with the background falling 0.550865, 0.507266, 0.500942 to its limit 1/2. What is left is only the three single-wave crosshair families parallel to the hexagon's edge directions, one per lattice axis, at odd-denominator rational coordinates - the families the previous section measures.

The ghost star

The ghost star at the hexagon's centre is a finite-layer artifact. Verified. Each layer's centre is entirely ink or entirely paper, flipping with n mod 4, so 28 layers give exactly 1/2, which is also the limiting background; the star is the difference between the two, and it goes to zero. On a 1200 by 2399 raster of the carpet's cut layers at odd n <= 111, with 2147316 hexagon pixels of which 25440 are star and 1981584 background:

The decay is (ln L)/L for L the number of layers stacked, and the coefficient is where the frame bites. In the ideal frame, main diagonal X = Z, band half-width 0.01, 2801 samples per axis, excess * L runs -0.7779, -0.9942, -1.1194, -1.2519 at L = 28, 100, 200, 400, with slopes against ln L of -0.1807 from 100 to 200 and -0.1911 from 200 to 400. In the lattice frame, the band |x - y| <= 0.01 of the cube side read against the exact ink law layer by layer, excess * L runs -0.9337, -1.0212, -1.1387, -1.1923, -1.2791, -1.3645 at L = 14, 28, 56, 100, 200, 400, with slopes -0.1252 from 100 to 200 and -0.1233 from 200 to 400. The ghost star demo stacks the hexagonal cuts one per odd side and draws the arm, whose ink is exactly 1/2 + chi_8(n)/(2n) and whose decay is exactly -1/4 on the cube's own cells and a different number in every frame that resamples or widens it.

The cell frame, defined. Registration: the cube's own cell lattice, one layer at a time, nothing resampled onto a common grid, so the half-cell displacement 1/(2n) between consecutive layers is carried rather than averaged away. Band normalisation: the star intensity is the ink share of the exact arm x = y of the cut, a diameter of 2n cells with no width to choose, and the background intensity is the layer's whole-hexagon ink share, the Proved closed form of the first section; the excess is star minus background, and the L-layer excess is its mean over the first L odd layers.

The star arm's ink is exactly 1/2 + chi_8(n)/(2n). Proved. The arm x = y at odd n is the 2n cells with z = 6n - 2 - 2x, and 0 <= z < 4n pins n <= x <= 3n - 1. On the arm the two block parities agree, so "at most one of three odd" collapses to floor(x/4) even, and with f(m) = 4 floor(m/8) + min(m mod 8, 4) the inked count is f(3n) - f(n) = n + chi_8(n), where chi_8 is the real character mod 8 of Q(sqrt 2), +1 at n = 1, 7 and -1 at n = 3, 5. The generator checks that rational against the counted arm at every odd n <= 2001, 1001 of 1001, and the three arms x = y, y = z, z = x return the same rational at every odd n <= 221, 111 of 111, by the slice's permutation symmetry.

In the cell frame at even L the decay coefficient is exactly -1/4, with an exact constant beside it. Proved. Writing excess_L for the L-layer excess, at even L

L * excess_L = -(ln L)/4 + C + O(1/L^2)
C = ln(1 + sqrt 2)/(2 sqrt 2) - G/8 - gamma/4 - (ln 2)/2 = -0.2937605857

The proof is one subtraction of closed forms. The per-layer excess is -chi/8 + (chi_8(n) - 1)/(2n) + chi/(8 n^2); summed over the first L odd n the first term is 0 at even L and +1/8 at odd L, the character part converges to L(1, chi_8)/2 = ln(1 + sqrt 2)/(2 sqrt 2) by the class number formula for Q(sqrt 2), the third converges to -G/8 with G Catalan's constant, and -sum 1/(2n) over the odd n below 2L is -(ln L)/4 - gamma/4 - (ln 2)/2 + O(1/L^2), the 1/(4L) terms of the two harmonic sums cancelling exactly. Every limb of C is a closed form that subtraction hands over, so C is derived and not a decimal recognised after the fact, and the whole of the ghost star's decay is the 1/(2n) of the hexagon's ink law read against an arm that has none. The generator prints L(1, chi_8) = 0.62322524 from its own series against ln(1 + sqrt 2)/sqrt 2 = 0.62322524.

Even L is a hypothesis and not decoration. Proved. The -chi/8 term is the whole of the difference: at odd L it sums to +1/8, so the law reads C + 1/8 in place of C, and what is left is +-1/(4L) rather than O(1/L^2). The -1/4 coefficient is untouched by parity. The generator's odd ladder, columns L * excess_L + (ln L)/4 - C and that miss less 1/8 times L:

The 1/L^2 term is exact and reads L mod 4. Proved. Three tails carry it: the character tail -(1/2) sum chi_8(n)/n over odd n > 2L is -1/(8L^2) at L = 0 mod 4 and +1/(8L^2) at L = 2 mod 4, since the sign pattern +--+ on the four odd residues starts at n = 2L + 1 and flips with that residue; the Catalan tail contributes +1/(64 L^2) and the harmonic remainder -1/(96 L^2), both blind to the residue. So the coefficient is -1/8 + 1/64 - 1/96 = -23/192 at L = 0 mod 4 and +1/8 + 1/64 - 1/96 = +25/192 at L = 2 mod 4. The residual column above is every row at L = 0 mod 4 and falls by 4.00 at all six doublings from L = 100 onto -23/192 = -0.11979167, the L = 6400 row overshooting at the floor of double precision where the residual itself is 3e-9; the generator's L = 2 mod 4 ladder reads +0.13020778, +0.13020819, +0.13020825 at L = 802, 1602, 3202 against +25/192 = +0.13020833.

There is no frame-free decay coefficient. Refuted. Widen the star from the exact arm to a band of half-width W cells about it, change nothing else, and the coefficient walks over a family:

Every entry is the exact rational the width family below hands over, and the measurement converges on it: recomputed to L = 3200, four times the sweep, W = 0 through 12 land on -1/4, -1/6, -1/10, -3/28, -5/36, -3/22, -3/26 within 5.1e-8, every gap falling by four per doubling of L from L = 800, where the sweep's own gap is 8.1e-7. Proved. Beside them a band whose half-width is a fixed fraction of the cube side rather than a fixed number of cells reads -0.1252 and -0.1233, and the ideal frame, which resamples every layer onto one 2801-point grid, reads -0.1807 and -0.1911 and is still moving at L = 400. Resample the layers or do not, and pick the star's width: more than a dozen coefficients come out of one star, so the coefficient is a property of the measurement and not of the stack. What survives the frame change is the (ln L)/L order, the sign, and the largest coefficient of them all, the cell frame's -1/4, which is the family's first row and the only one a band of zero width can read.

The width family is exact at every half-width, and the arm is its first row. Proved. Widen the star from the arm to the band |x - y| <= W cells and write K = floor(W/2): x - y is even on the cut, so W and W - 1 read one point set at odd W and K is the only width there is. With b = 1 when floor(K/2) is even and 0 when it is odd, and with E(K) = #{|j| <= K : j = 3, 4, 5 mod 8} + floor((K + 2)/4) - K, which is the 8-periodic run 0, -1, -1, 0, 1, 2, 2, 1 at K = 0..7 mod 8, the band's excess over the hexagon's ink law at every odd side n >= K is exactly

excess_W(n) = kappa chi + (m + q chi_8(n))/n + chi/(8 n^2)
kappa = -(-1)^K/(8(2K + 1))
m     = -(K + b)/(2(2K + 1))
q     = (1 - 2 E(K))/(2(2K + 1))

with chi = (-1)^((3n-1)/2) the ink law's character, -1 at n = 1 mod 4, and chi_8 the arm's, the real character mod 8 of Q(sqrt 2). Below n = K the band is clipped by the hexagon's edge and the identity is false, W = 6 at n = 1 missing by 2/7. At W = 0 it is K = 0, b = 1, kappa = -1/8, m = -1/2, q = 1/2, which is the arm's proved excess -chi/8 + (chi_8(n) - 1)/(2n) + chi/(8 n^2) term for term, so the arm is not a separate theorem. Summed over the first L odd layers the chi term vanishes at even L and the 1/n term hands the decay coefficient m/2 = -(K + b)/(4(2K + 1)), which at even W is the -(W + 2b)/(8(W + 1)) the sweep predicted, at odd W is -(W - 1 + 2b)/(8W) - -1/4 at W = 1 and -1/6 at W = 3, where the even-W formula was false - and tends to -1/8 as W grows, the value the fixed-fraction band of the lattice frame measures.

The proof is a block-tail sum. On the cut plane the band's cells are x = B + j, y = B - j, z = 6n - 2 - 2B with B running over the 2n integers of [n, 3n) and |j| <= K, and nothing is clipped exactly when K <= n. The space rule is s(x) + s(y) + s(z) <= 1 for the block parity s(c) = floor(c/4) mod 2, so a row is ink at 2K + 1 - U(B) cells when s(z) = 0 and at P(B) cells when s(z) = 1, where U(B) counts the j with s(B + j) = s(B - j) = 1 and P(B) the j with both 0; writing D = 2K + 1 - U - P for the j where the two block parities disagree, the band's ink is sum_B (2K + 1 - U(B)) - sum_B s(z_B) D(B). The second sum collapses to one product. s(z_B) = 1 exactly on B = 0, 3 mod 4 when n = 1 mod 4 and exactly on B = 1, 2 mod 4 when n = 3 mod 4, while D takes only two values, D_a = 2(K - floor(K/4)) on B = 0, 3 mod 4 and D_b = 2 floor((K + 2)/4) on B = 1, 2 mod 4, because the disagreement pattern in j is the 8-periodic 0, 1, 1, 1, 0, 1, 1, 1 in the first case and 0, 0, 1, 0, 0, 0, 1, 0 in the second. The cut's character and the band's parity are the same partition of B mod 4, so s(z) D is a constant times an indicator and sum_B s(z_B) D(B) = (n - 1) D_chi exactly, with D_chi = D_a at chi = -1 and D_b at chi = +1. The first sum is a block tail: U is 8-periodic in B, and for any 8-periodic h the tail sum_{B=n}^{3n-1} h(B) - 2n mean(h) equals psi_h(3n mod 8) - psi_h(n mod 8) for psi_h the periodised partial sum, which flips sign between n = 1 and n = 3 mod 8 and between n = 5 and n = 7, so it carries no trivial and no chi_-8 component and is a combination of chi_4 and chi_8 alone. Three identities close it. (D_a + D_b)/2 = A(K) for A(K) = sum_{j=1}^{K} w(j mod 4) with w = 0, 1, 2, 1, and A(K) = K + 1 - b by the four cases of K mod 4, which is b's definition read backwards and is where the -(K + b) comes from. 2 mean(U) + A(K) = 2K + 1, because w and the mean pattern 2, 1, 0, 1 sum to 2 at every residue, which is why the constant term is kappa chi with no drift. The tail then splits into its two characters by one shift. s(c + 4) = 1 - s(c) turns P into U four steps along, P(B) = U(B + 4), so U(B) + U(B + 4) = 2K + 1 - D(B). Its chi_4 weight is (e_1 + e_5)/2 for e_r = 2(U(r) - mean U), while D_chi carries (D_a - D_b)/2; since e_1 + e_5 = D_a - D_b = K + (K mod 2) both weights are (K + (K mod 2))/2 and cancel identically at every K, which is why the 1/n term has no chi_4. Its chi_8 weight is (e_1 - e_5)/2 = U(1) - U(5), and U(1) reads straight off the pattern as the count of j in [-K, K] with j = 3, 4, 5 mod 8, so U(1) - U(5) = 2 U(1) - (2K + 1) + D_b and q = (U(5) - U(1))/(2(2K + 1)) = (1 - 2E(K))/(2(2K + 1)). E is 8-periodic because a block of eight adds 6 + 2 - 8 = 0, and the generator holds it against the run at K = 0..200, 201 of 201. Nothing in the family is fitted.

The family's constant and both its 1/L^2 branches are closed forms at every width. Proved. At even L

L * excess_L = (m/2) ln L + C_W + O(1/L^2)
C_W = m (ln 2 + gamma/2) + q L(1, chi_8) - G/8 + Delta_W

with L(1, chi_8) = ln(1 + sqrt 2)/sqrt 2 by the class number formula for Q(sqrt 2), G Catalan's constant, and Delta_W = sum over odd n < K of (counted excess - the identity), the exact rational the finitely many clipped layers contribute, 0 through W = 5 and 2/7 at W = 6. The chi_4 cancellation is at the 1/n order only, so L(1, chi_4) = pi/4 is absent at every width while L(2, chi_4) = G sits in every one: C_W carries gamma, ln 2, L(1, chi_8) and Catalan's G, and widening the band moves only m and q. Three tails carry the 1/L^2 term. The character tail -q sum chi_8(n)/n over odd n > 2L is -q/(4L^2) at L = 0 mod 4 and +q/(4L^2) at L = 2 mod 4, since the sign pattern +--+ on the four odd residues starts at n = 2L + 1 and flips with that residue; the Catalan tail of the background's chi/(8 n^2) gives +1/(64 L^2) at every width, blind to the residue; and the harmonic remainder of m (H_{2L} - H_L/2) gives +m/(48 L^2), the 1/(4L) parts of the two harmonic sums cancelling exactly. So the coefficient is -q/4 + 1/64 + m/48 at L = 0 mod 4 and +q/4 + 1/64 + m/48 at L = 2 mod 4, which at W = 0 is -23/192 and +25/192; at odd L the constant is C_W - kappa with an O(1/L) error, C + 1/8 at W = 0. The sliding-window slope the sweep reads has a residue trap of its own: the window from L/2 to L cancels the chi term only at L = 0 mod 4, and at L = 2 mod 4 it converges to m/2 + kappa/ln 2 instead, -0.4303 in place of -1/4 at W = 0 and -0.1066 in place of -1/6 at W = 2, so a sweep that never leaves one class of L cannot see its own offset. The generator checks the per-layer identity against the counted band in exact rational arithmetic at 14 distinct half-widths from W = 0 to W = 64, every odd n from K to 201, and reports the four residue classes n = 1, 3, 5, 7 mod 8 separately: 1354 of 1354, no class short. The sweep also prints nine odd widths, which repeat their even neighbours cell for cell and are not counted twice. Seven summed ladders confirm the constant to 1e-9 at L = 1600 and both 1/L^2 branches to 1e-7; W = 0 returns C = -0.2937605857, -23/192 and +25/192 unchanged, and the sliding window reads -0.24999980 at L = 1600 against -0.43078703 at L = 1602, whose limit is -0.43033688, the 4.5e-4 gap being the odd-L error at L/2 = 801.

The constants

The slice stacks' surviving constants are Leibniz, odd Basel and Catalan. Proved. Every character series below runs over the exact ink laws of the first section, so the limits are theorems about those closed forms and not fits; the carpet split converges at balanced layer counts, and N = 55 is 28 layers.

The third and fourth rows are the reason the void's constant has to be stated to eight places: (pi + pi^2)/16 = 0.8131998159 and pi^2 ln 2/(7 zeta(3)) = 0.8130217042 collide in the fourth decimal and are different numbers, the second belonging to the flat square stack and not to this one.

One law prints every row of that table, and the layer count's parity is the only residue it reads. Proved. Write a family's ink law of the first section as I(n) = A + B chi + (c + d chi)/n + (e + f chi)/n^2 with chi = -chi_4(n), let mean be the average of I over the first M odd sides n = 1, 3, ..., 2M - 1, and let eps be the average of 1/n over the same layers. Subtracting A and c eps removes exactly the two limbs that do not converge, and what is left is termwise three character sums and one zeta tail, so at every M, exactly,

M (mean - A - c eps) = -B S - d s_1 + e s_3 - f s_2
S = sum chi_4(n)   s_1 = sum chi_4(n)/n   s_2 = sum chi_4(n)/n^2   s_3 = sum 1/n^2

all four sums over those layers, s_3 the trivial character's tail and the other three chi_4's. chi_4 alternates on the odd numbers, so S = 0 at even M and 1 at odd M, and the other three walk to their L-values, L(1, chi_4) = pi/4, sum 1/n^2 = pi^2/8 over odd n, and L(2, chi_4) = G:

M (mean - A - c eps) -> -B [M odd] - d pi/4 + e pi^2/8 - f G

The leading term is the same phase-map integral the pair limits run, taken at the identity map a = 1, c = 0: over the hexagon polygon the cut cell law s_y = s_x + s_z + w returns A + B chi at each residue class of n mod 4, the carpet's 3/8 and 5/8 and the tree's and the void's 1/4, which the generator prints as the E fields of every pair limit. So A and B are the integral's, B chi being the half-cell overtone w and nothing else, and c, d, e and f are the finite-n orders no phase integral sees - which is where pi and G come from and why they cannot reach a pair limit, every one of which is rational.

The hypothesis carries two limbs: the object is the cut ink of a 3D parity design, whose Walsh quasipolynomial carries only chi_4 and terminates at the 1/n^2 order, proved in the Walsh spectrometer lane, and the quantity is the recentred M (mean - A - c eps). Inside it, pi reaches a one-layer constant only through the 1/n order of the ink law, G only through the 1/n^2 order, the residue class only through B, and no other character and no other constant can appear - which is why L(2, chi_-3) is absent from the table by a theorem and not by a search. Both limbs carry weight: the ghost star's width family above is a one-layer object of this same page whose constant C_W = m (ln 2 + gamma/2) + q L(1, chi_8) - G/8 + Delta_W holds gamma, ln 2 and L(1, chi_8), because its character is mod 8 and its 1/n limb is not subtracted. Since N = 2M - 1, even M is N = 3 mod 4 and odd M is N = 1 mod 4, and the modulus is 4 and no finer: the generator holds the summed identity against the counted hexagons in exact rational arithmetic at every layer count to N = 55, all four families, the four classes n = 1, 3, 5, 7 mod 8 counted apart, 7 of 7 in each. The one row of the table that is not an instance is the carpet split, M (I1 - I3 + 1/4) = s_1 + s_3/4 -> pi/4 + pi^2/32, and it needs even M: at odd M the two classes hold (M + 1)/2 and (M - 1)/2 layers and the difference of their means is not a character sum at all.

The Catalan statement, at both residue classes. Proved. The carpet is B = 1/8, d = e = 0, f = -1/8, so M (mean ink - 1/2 - eps/2) -> G/8 = 0.1144956993 along N = 3 mod 4, measured 0.1144757884 at N = 55, and -> G/8 - 1/8 = -0.0105043007 along N = 1 mod 4, measured -0.0104828892 at N = 53, with G = 0.9159655942 from its own series. The 1/8 step is the ink law's own chi averaged over an odd number of layers, the same parity term the ghost star's even-L hypothesis carries, and Catalan enters only as L(2, chi_4), one order below the pi the tree and the void collect. Nothing in it is fitted.

The approach to every one of those limits is a closed form. Proved. Write sigma = +1 at even M and -1 at odd M, and a = 2M + 1 for the first layer past the ladder. Each tail solves its own one-line recursion, T(a) + T(a + 2) = a^-s twisted and T(a) - T(a + 2) = a^-s untwisted, in powers of 1/a:

sum chi_4(n)/n   over odd n >= a = sigma (1/(2a) + 1/(2a^2)) + O(a^-4)    = sigma/(4M) + O(M^-3)
sum chi_4(n)/n^2 over odd n >= a = sigma (1/(2a^2) + 1/a^3) + O(a^-5)     = sigma/(8M^2) + O(M^-4)
sum 1/n^2        over odd n >= a = 1/(2a) + 1/(2a^2) + 1/(3a^3) + O(a^-5) = 1/(4M) + O(M^-3)

the 1/M^2 limb of each cancelling on the way from a to M, so the gap to the limit is (sigma d - e)/(4M) + sigma f/(8 M^2) + O(1/M^3):

The carpet's 1/M term is absent because d = e = 0, which is why its ladder is the sharpest of the five; the tree's second limb is the only place sigma f/(8 M^2) is separately visible, and it carries the last three digits of both tree readings. The generator prints all four classes of M mod 4, so all four of N mod 8, at M = 400, 1600 and 3200, and the two parities are the only split in them.

Eisenstein is absent from the base-2 slice stack. Verified. L(2, chi_-3) = 0.7813024129, the L-value the bases page finds under base 3, appears nowhere in the list above; the two characters this slice generates are chi_4, from the ink law, and the mod-8 character of Q(sqrt 2), from the star arm above, and the hexagonal geometry contributes rational tent integrals rather than an Eisenstein L-function. The hexagon in this page is a shape, not an Eisenstein lattice.

Where the exactness stops

The flat analogue is exact and proved on the shelf. Proved. For the square parity carpets at odd scales, moire-correlation-laws proves that the Pearson correlation of two layers is a closed form in gcd(m, n) which vanishes if and only if the scales are coprime, for each of the four fields the plane rule generates. The same paper proves why none of that reaches this page: the plane rule "ink unless both indices are odd" expands as 3/4 + (sigma_1 + sigma_2)/4 - sigma_1 sigma_2 /4, and the pair term is the entire mechanism, while the space rule "fill if at most one of three indices is odd" expands as 1/2 + (sigma_1 + sigma_2 + sigma_3)/4 - sigma_1 sigma_2 sigma_3/4, whose three pairwise coefficients are all exactly zero. A triple product takes the pair term's place, and a planar section of the space rule inherits nothing.

What replaces the exactness is the dyadic overtone of the second section, and the numbers depend on the mask. The paper's own hexagonal remark reports measurements on rendered sections, mean -0.037 over a range [-0.205, +0.147], and labels them measurements on rendered images rather than theorems, the extreme values mask-dependent. Conjecture. The exact full-hexagon computation on this page reads (5, 9) = -0.14179450 and (5, 7) = -0.08542646 for the same two pairs, so any number published for a hexagon layer pair must pin the mask convention it was measured under. What is robust across masks is the shape of the failure: coprime pairs correlate, the gcd echo survives, and the sign is the dyadic law's.

Where the numbers live

lab/rs/hexagon-moire prints every number above in one run: the exact cut ink of every odd n <= 55 against the four closed forms in rational arithmetic, the exact full-hexagon Pearson correlations and the doubling sign law, the pair limit as an exact rational for all four families and all three phase maps with the finite-layer gap beside it, the ideal-frame stack at 3601 samples per axis for the quarter line and the void star, the twisted averages and the coarse crosshair model at 200003 samples, the rendered ghost star on its 1200 by 2399 raster, its decay in the ideal, lattice and cell frames with the arm ink law in exact rationals to n = 2001, the cell frame's ladders at L = 0 mod 4, L = 2 mod 4 and odd L, the swept band widths with the odd-W rows, the recompute to L = 3200 and the four held-out widths, the width family's per-layer identity in exact rationals at 23 half-widths broken out by n mod 8, its seven summed ladders at L = 0 mod 4, L = 2 mod 4 and odd L with the sliding window in both even classes, the constants from the mrlynum series with the partial character sums at N = 53 and N = 55, and the one-layer law: the six coefficients of each family read off its ink law, the summed identity against the counted hexagons in exact rationals at every layer count to N = 55 with n mod 8 broken out, and the approach ladders at M = 400, 1600 and 3200 in all four classes of M mod 4. Every domain is the source's own. The mesh those layers are counted on, and the fills that live on it, are slices; the ink law's quasipolynomial is proved in the Walsh spectrometer lane; the flat stack this one is measured against is the moire-correlation-laws lane; and the L-value it does not contain is bases.