# Tightening the Level Ledger **2026-08-17. Script: `research/level-ledger-tight.js` (30.7 s, all asserts passed, full output pasted in the file). R\*(19) from `--deep`, 2717.1 s, and folded in.** `research/FOLD-PROFILE.md` §2 proves, for the twin-slot tile T_x of width W = x# and census D, and for every prime p > x and every residue a mod p, $$\bigl| h(a) - D/p \bigr| \;\le\; 2\cdot 3^{\pi(x)-1}, \qquad h(a) = \#\{\, r \in T_x : r \equiv a \ (\mathrm{mod}\ p)\,\}.$$ `research/discrepancy-two-class.md` §5 then showed the bound has the right base and twice the right exponent. This note asks what can actually be taken back. **Lead with what is hard.** The sup-versus-variance gap is not closed here, and §4 shows it cannot be closed by any L¹ Fourier method at all: such a bound is already ≫ 2^{π(x)} from its own j = 1 term, which sits strictly above √3^{π(x)}. What replaces the exponent is an exact reduction of the Level Ledger to a single finite quantity, that quantity computed by exhaustion up to y = 19, and a proven **constant** factor of 81.0 uniform in x, p and a. The 81.0 needs the y = 19 exhaustion; the y = 17 seed, which costs seconds against the deep run's 2717 s, gives 53.9. The factor does not grow. There is a second, genuinely growing unconditional route (Erdős–Turán with the exact Fourier coefficients, ×1.18 per fold), and it is never the best bound available in the computable range. Saying otherwise would be the interesting result, and it is not the one that came out. --- ## 1. Custody Everything downstream is checked against the published numbers first, through code written from scratch for this file. | target | source | reproduced | |---|---|---| | max_a\|h−D/p\| = 1.36, 1.62, 3.35, 3.63, 6.13, 16.9 | FOLD-PROFILE §3 | to the digit (16.93) | | K(0), min K, max K, sd(K), six rows | FOLD-PROFILE §3 | every cell | | bound/mean = 5.56, 1.24, 0.266, 0.048 | FOLD-PROFILE §2 | 5.5636, 1.2436, 0.2657, 0.0479 | | ΔΦ₂ max/min, 0.3000/−1.1000 .. 26.9038/−27.8325 | discrepancy-two-class §2 | every cell | | looseness ratios 25.7, 36.0, 54.4, 124, 186, 256, 479 | discrepancy-two-class §5 | 25.7 .. 479.5 | **"Right base, twice the exponent" VERIFIED.** The looseness ratio grows by **1.6284** per fold on the interval object (6 folds, x = 5→23) and **1.8127** on the arithmetic-progression object (5 folds, x = 7→23). √3 = 1.7321 sits between them. The claim survives independent checking. **The scope of the first custody row.** FOLD-PROFILE §3's sequence is quoted at the *ladder* prime, the smallest admissible p at every level, and the quantity is not uniform in p. Maximising over p instead: | x | §3 value (ladder p) | max over p | at p | range of p searched | |---|---|---|---|---| | 11 | 1.62 | 2.87 | 47 | ≤ 5000 | | 13 | 3.35 | 5.97 | 149 | ≤ 5000 | | 17 | 3.63 | 12.03 | 37 | ≤ 2000 | So the diagonal sequence moves p and x together and is not a statement about x alone. The bound is uniform in p, the measurement it is compared against is not, and no exponential base should be fitted to either. FOLD-PROFILE §3 carries the three maxima; the primes attaining them and the ranges searched are recorded here because it does not. --- ## 2. The reduction: the Level Ledger is a dilated-interval discrepancy Write, for a set S mod W of density ρ, $$G_S(u) = \#\bigl(S \cap [0,u)\bigr) - \rho u, \qquad \operatorname{range}(S) = \max_u G_S - \min_u G_S ,$$ and define $$R^*(x) \;=\; \max_{\alpha \in (\mathbb{Z}/W)^{\times}} \operatorname{range}(\alpha \cdot T_x).$$ **Theorem 1 (Dilation Reduction, PROVEN).** For every prime p > x and every a, $$\bigl|h(a) - D/p\bigr| \;\le\; R^*(x) + D/W \;<\; R^*(x) + \tfrac12 .$$ *Proof.* r ≡ a (mod p) with r ∈ [0,W) means r = a + ps for 0 ≤ s < L = (W−a)/p, and r ∈ T_x means s ∈ S := p^{-1}(T_x − a) mod W, which is an admissible dilate of T_x. So h(a) = ρL + G_S(L) = D/p − ρa/p + G_S(L). The family of dilates is closed under translation, and the supremum over translates of sup_u |G| is exactly the range; ρa/p < ρ = D/W. ∎ Nothing is discarded but that D/W. **The arithmetic-progression discrepancy of the tile *is* the worst dilated-interval discrepancy of the tile**, so the Level Ledger and Holt's ΔΦ₂ channel are the same object seen twice, and any bound on one transfers instantly to the other. That settles `discrepancy-two-class.md` §9's loose end: the natal ledger deviation is a sup of a k = 2 discrepancy, and its base should be √3, not 2. **Theorem 2 (Transfer, PROVEN).** For y ≤ x, R\*(x) ≤ 3^{π(x)−π(y)} · R\*(y). *Proof.* Expand the conditions only at the primes in (y, x]: the indicator factorises into 3^{π(x)−π(y)} signed terms, one for each squarefree d | W₂ = ∏_{y x, every a, $$\bigl|h(a) - D/p\bigr| \;\le\; 53.972817 \cdot 3^{\pi(x)-8} + \tfrac12 ,$$ which beats 2·3^{π(x)−1} by **at least 80.98** uniformly in x, p and a (80.98 at x = 19, rising to 81.04 once the +½ is negligible). The y = 17 seed alone gives 53.9 with no overnight run. The same constants apply to the interval channel with the +½ dropped. **The parametrisation is checked independently.** Dilating T₁₇ by 4000 random α ∈ (Z/510510)^× and taking the range directly, never touching the e-tuples, reaches **27.019392 exactly** (at α = 473,611) and never exceeds it. The enumerated maximum is attained by a genuine dilation. **Checked, not assumed.** The nine-term expansion at (x, y, p) = (13, 7, 17) was run term by term: it reproduces h(a) for every a, no single term exceeds R\*(7) + D₁/W₁ = 3.0714 (worst 1.5294), and the assembled bound 27.64 brackets the truth 3.3529 against the ledger's 486. Theorem 1 was then stress-tested against the actual maximum over **every** p ≤ 5000 at x = 11, 13 and every p ≤ 2000 at x = 17: 2.8723 ≤ 6.6429, 5.9664 ≤ 14.4341, 12.0270 ≤ 27.0630. **The target shape, and the deep run breaking it.** Up to y = 17 the fit looked exact: R\*(y) grew by **1.73226** per fold in geometric mean from y = 2, against √3 = 1.73205, with R\*(17) = 27.0194 against 3^{(π(17)−1)/2} = 27. The caveat attached at the time was that this is a two-endpoint statement, and the caveat was right. **R\*(19) = 53.9728 breaks it**: the step is 1.9976, essentially 2, the seven-fold geometric mean rises to **1.76788**, and 3^{(π(19)−1)/2} = 46.77 is 15% low. So R\* runs *above* the √3 shape. That is expected and is not evidence against √3 for the twin pattern itself. R\* is a maximum over ∏(q−1) patterns, a count that grows with y, so it carries a √(2 ln #patterns) extreme-value factor that √3^{π(y)} does not. The individual steps run 1.000, 1.800, 1.667, 2.195, 2.185, 1.878, 1.998: no law, and the "exact" hit at y = 17 was a coincidence. --- ## 4. Why the exponent does not move The gain is 2·3^{π(y)−1}/R\*(y). It grows by about ×1.4 for each extra level of y that can be enumerated, and y is capped by compute: y = 17 costs 92,160 patterns and 11 s, y = 19 costs 1,658,880 patterns and about 50 minutes, y = 23 costs 36,495,360 patterns on a 2.2 × 10^8 tile and is out of reach. **A growing gain needs R\*(y) for unbounded y, which is the √3 statement itself.** The transfer is lossless in base and costs exactly ×3 per fold above the seed, so seeding low and transferring high cannot recover the exponent: the gap is not a small-level artefact. Four routes to the exponent were tried. One gives a growing gain and is still never the best bound available; the other three do not survive. **(a) Second moment plus extreme value.** Parseval on the p-point group gives Σ_a (h(a)−D/p)² = (1/p)Σ_{t≠0}|F(t)|², and sup ≤ √(Σ_a (h−D/p)²), so a second-moment bound *would* give a sup bound with no extreme-value step at all. The second moment has an exact identity: $$\sum_a h(a)^2 \;=\; D + 2\sum_{m\ge 1} N(mp), \qquad N(h) = \#\{r : r,\ r+h \in T_x,\ r+h < W\},$$ because p | r − r′ is a condition on *integers*, not on residues mod W. Verified exactly at (11,13), (11,29), (13,17), (13,41). But that turns the problem into Hardy–Littlewood pair correlation: Σ_m N(mp) must be known to absolute precision 3^{π(x)/2}, i.e. the singular series must equidistribute over h ≡ 0 (mod p) to that precision. That is the same wall, not an easier one. The measured p·Var is 4.1e−2, 3.3e−1, 4.9e−2, 1.0e−1 of the binomial D(1−1/p) at those four cells and **2.52e−4 at (23, 29)**, a fifth cell the script now prints (added 2026-08-20; the ratio there is one O(D) residue histogram, while the identity check at W = 223,092,870 would cost ~6e13 operations and is not run — the four small cells are what certify the identity). So every general inequality that stops at the binomial line is vacuous against this object, the large sieve included: it gives Σ_{a≠0}|F(a)|² ≤ (p²+W)D and hence nothing below binomial. **(b) A better per-fold recursion.** The one-step identity for a new prime q is Δ_{new}(u) = Δ_S(u) − Δ_{S₀}(u/q) − Δ_{S_e}((u−e)/q), and S_e is a *translate* of S₀ by q^{-1}e mod W, which is essentially arbitrary. Rewriting the third term through that translate gives four evaluations instead of three, so the direct ×3 is the better of the two. Beating ×3 needs the three arguments not to be extreme together, which is an equidistribution statement of exactly the kind being sought. **(c) Erdős–Turán with the exact Fourier coefficients, and the L¹ ceiling.** This is the one route that *does* give a growing unconditional gain, and it still cannot reach √3. The coefficients of a 2-class pattern factorise exactly: with removed classes {0, e_q} mod q, $$S(j) = \prod_q S_q(j), \qquad S_q(j) = q-2 \ \text{if } q \mid j, \quad -1 - e(j e_q/q) \ \text{otherwise},$$ so |S_q(j)| = 2|cos(π j e_q/q)| off the divisors. Since |Σ_{n