Where the products lose

A cell of this chart FAILS when its least height h is strictly below the least height over pure products — products of factors x^d − 1 with at least J of them and degree below M — which happens at 83 of the 695. On this chart every failing cell sits at lattice rank between 5 and 18 — the rank being width minus depth — and 325 cells outside that window fail zero times, 221 of them with height 2 or more and so able to fail at all — a cell whose h is 1 cannot fail whatever else is true of it, since the pure bound is 1 there as well and no height is below 1. Call its two edges the LOW WALL, rank 4 and below, and the HIGH WALL, rank 19 and above. But that window is not a law about the rank. It is this chart's own corner crossing a threshold in the DEPTH. Scan each rank's own column in the depth, past the corner, and record the first depth at which the cell fails; then the ranks able to fail on a chart cut at width 40 and depth 30 are exactly those whose first failing depth is at most the corner allows — and that set is 5 through 18, the census's failing ranks, rank for rank.

Where that agreement is evidence has to be said, because inside the window it is not. For a rank between 5 and 18, “its first failing depth sits on the chart” and “it fails on the chart” are the same sentence; the comparison restates one fact there and checks the arithmetic rather than showing anything. THE CONTENT IS AT THE RANKS THE WINDOW EXCLUDES — which is to say, it is the two walls, and they turn out to be different things. The scan reaches ranks 1 to 22: the whole of the low wall, every failing rank, and the first four ranks of the high wall, 19 to 22 — not the whole of that wall, which runs to 38. At ranks 23 to 38 the chart's own depths are at most 17, inside the census's own rectangle, so their exclusion is inherited from that census and not measured here.

The mechanism is the vanishing order at −1. The vectors of width M flattened to depth J form a lattice — the integer multiples of (x−1)^J cut at degree below M, of rank M − J — so every one of them is a cofactor q times (x−1)^J with q of degree below the rank, and the pure products land in the same variable. The coefficients of (x−1)^J are the alternating binomial row, a bell of width about the square root of the depth, and a factor (1+x) of the cofactor differences that row, costing a factor of about one over that square root in the height. So to leading order in the depth only the vanishing order at −1 matters, and the largest such order a cofactor of that degree can carry is one less than the rank, attained by the integer multiples of (1+x) raised to that power and by no other cofactor, ±(1+x) to that power being the two of least height among them — call that cofactor the CHAMPION of its rank. That cofactor IS a pure product — a power of x²−1 times a power of x−1 — whenever the depth is at least one less than the rank, which is what it takes for that second power to exist at all; below that the champion is not in the pure family, and the chart's own corner has such cells, a rank-38 one there having depth 2. So wherever the mechanism binds the pure family wins by construction, and the least height has a CLOSED FORM with no search of any kind — it is the height of (1+x)^r−1(x−1)^J, one polynomial written down from the rank and the depth — at every depth past that rank’s own threshold the sweep reached, and there only. Below it the argument says nothing; at the 83 failing cells the closed form is false by the definition of failing; and above it what is checked is eight depths at ranks 5 to 8, where the scan stops at a crossing, while at ranks 3 and 4 — which fail at no depth swept — the closed form attains the minimum from depths 7 and 13 continuously through depth 140. Both are stopping rules and neither is a proof that it never reverses. Rank 2 is the one place where the prediction is not a stopping rule at all: it is proved there, at every depth and with no computation at any value. The mechanism itself is not a claim here and carries no tier: it is a heuristic asymptotic argument, unproved at every rank, and what it is good for is having PREDICTED the closed form the paragraphs below then measure — and, at rank 2, prove by a route that owes it nothing.

So the low wall is a real absence, and it is far wider than the chart: at ranks 1, 2, 3 and 4 the cell is clean at every depth from 2 to 45, a completed scan reaching width 49 and so past this chart at all four — and at ranks 2, 3 and 4 much further still, since the second scan takes those three to the depth ceiling and finds no first failure at all, 139 depths each over depths 2 to 140, width out to 144. At rank 2 that is now the weaker statement, the theorem below making the cell clean at EVERY depth rather than at every depth scanned. Rank 1 alone stops at 45, and it is the one that needs no scan. Its case is a proof rather than a count. At rank 1 the lattice is generated by (x−1)^M−1 alone, so its shortest vector is that generator, and the only pure product with M−1 factors fitting degree M−1 is the same polynomial, every part being forced to 1; the two bounds coincide by construction.

And the high wall is not a wall. Rank 19 fails first at depth 26, rank 20 at 23, rank 21 at 25 and rank 22 at 30 — cells of width 45, 43, 46 and 52. Every rank above the window that this scan reached does fail — ranks 19 to 22, and not the whole of the high wall, since 23 to 38 were never taken past the corner; what is shown is that the wall's first four ranks are not clean, not that none of the twenty is. They fail just outside the corner: a rank-19 cell here has depth at most 21, five short of where its failures start. Rank 19 and above never failing was a true statement about a chart and a false one about the lattice.

The failing depths at a fixed rank are CONTAINED in a band and do not fill it. Where the sweep reached the far end, rank 5 is a clean interval, thirteen depths for thirteen, while rank 6 is clean at depth 23 inside 22 to 33, rank 7 at 55 and 57 inside 26 to 58, and rank 8 at seven depths inside 19 to 60 — the holes clustering at the ends. The low edge is ragged too: the first failing depth falls with rising rank seven times over ranks 5 to 22, so no formula reads it off on this evidence, and the derivation above never needed one — only each rank's own comparison against the corner. What confirms the far end is a second and independent threshold: the depth from which the closed form attains the minimum is 31, 34, 59 and 61 at ranks 5, 6, 7 and 8, and at all four that is exactly one past the last failing depth. One threshold is read off a comparison with the whole pure family, the other off a single polynomial, and they meet at every rank where both were reached.

Above the threshold the minimiser is unique up to sign. The collecting pass is a second one, holding its search radius at the winning height instead of shrinking it on every improvement, so it meets every minimiser of a cell and not only the first. Over ranks 2, 3 and 4 it closes 86 cells, and at sixty-nine of them exactly two minimisers stand and both are plus and minus that closed form: every one of those 86 from depth 2 at rank 2, from 9 at rank 3 and from 13 at rank 4. At rank 2 the theorem below gives that same uniqueness at every depth and not only at the 29 this column collects. Those three depths are a STRICTER threshold than the attainment one above, and at rank 3 a different number, since the closed form attains the minimum from depth 7 there and is the whole minimiser set only from 9, the two depths between being cells where it ties without being alone — with vanishing order at −1 equal to 1, 2 and 3 there — one less than the rank at every one, the mechanism's own signature read off the answer. Below those depths the set is larger, reaching eight minimisers at rank 3 and depth 3, carrying three different vanishing orders.

Inside the window, on this chart, the condition is far from sufficient — 83 of 370 cells, 22.4% raw and 25.9% over the 321 that can fail at all — and the depth then drives the rate: 1.9%, 15.2% and 60.6% raw across the three bands, 2.9%, 15.2% and 60.6% conditioned, the deeper two unmoved because no cell in either has height 1. Outside the window the rate is zero at every depth on either reading. Nothing else selects. Eight cheap per-cell quantities were scored against the failure flag — the largest repeated factor in the best pure product, the number of divisors of J, a ball-counting heuristic at two radii, two measures of what the pure bound loses when width is taken from it, the rank and the depth — and seven of the eight have no cut better than declaring the whole chart clean, while the eighth buys 7 of the 83 failures at the price of one false alarm. That scoring is ONE-SIDED, and the limit matters here rather than being a technicality: a cut declares failure on one side of a number, and a window has two sides, so the rank scores as a null in that table for a reason about the scoring and not about the rank. The seven nulls are nulls for one-sided cuts alone. No quantity but the rank was tried as a window, and whether any of the other seven carries one is untested, not answered. What does select, once the rank is held fixed, is the depth — through that rank's own first failing depth, which is what the derivation above rests on and which no single cut over the whole chart can express.

The band is measured and not proved. Each rank's first failing depth is read off a completed upward scan, so it is exact; each rank's LAST one is read off a stopping rule — four consecutive depths at which the closed form attains the minimum — and a stopping rule is not a proof, so the far end holds within the swept range and no further. That rule replaced a weaker one, four consecutive CLEAN depths, which declared rank 8 finished at depth 26 when its failures run to 60: a cell can be clean because some other pure product wins, and that state is transient where the closed form's win is the one the mechanism predicts is permanent. The mechanism itself is an asymptotic argument in the depth at fixed rank, and it is proved at no rank — including rank 2, where what is proved is its CONCLUSION and by a different argument, the order-counting that predicted it playing no part. It says nothing about where each rank's band BEGINS. The eight scored quantities are named above; a best cut is chosen for each by minimising its two error counts, and the two controls the scoring runs against — a quantity that must separate perfectly and one that must not — behave as they must before any verdict is read.

verifiers: explore_flatten_band.py, explore_flatten_select.py (the census whose counts of failing and clean cells this block quotes)

At rank 2 the lattice is exactly the products (a + bx)(x−1)^J over integer pairs, so the closed form there is a statement about binomial coefficients and nothing else. Write c_k = C(J, k), ZERO OUTSIDE 0 ≤ k ≤ J — the convention is not decoration: the two indices it makes meaningful, k = 0 and k = J+1, are where the target's own bound is checked, and they are what confines the enumeration in the scope below to a finite region. The coefficient of x^k in that product is ±(a c_k − b c_k−1), so its HEIGHT is the maximum of |a c_k − b c_k−1| over k and h at this cell is the least value that maximum takes over nonzero pairs — which is the whole of what has to be minimised. Then for every J ≥ 2 and every integer pair (a, b) other than (0, 0), the maximum over all integers k obeys max |a c_k − b c_k−1| ≥ max |c_k − c_k−1|, with equality only at (a, b) = ±(1, 1). So h(J+2, J) is that difference of consecutive binomial coefficients and the minimiser is ±(1+x)(x−1)^J and nothing else, at every depth, with no computation at any value — the rank-2 instance of the champion above, at the LOWEST rank where the question has content. Rank 1 needs no threshold either, but there the lattice has a single generator and there is nothing to minimise over.

THE PROOF TURNS ON THE PALINDROME. Write N for the largest coefficient of the row, C(J, ⌊J/2⌋), and D for the target. Two facts about the row do all the work, and a third — that the endpoint indices give |a| and |b|, so the maximum is at least the larger of them — is not one of them, though it is what makes the enumeration below finite. The row reads the same backwards, so the substitution k ↦ J+1−k carries the expression to minus its own swap: the lattice is invariant under reversal and reversal SWAPS a and b, which collapses the positive quadrant to a ≥ b and is the whole reason the proof is four lines. Then read the peak at m = ⌊J/2⌋ — the FIRST index attaining the maximum, where the row is still strictly rising, so 1 ≤ c_m−1 < c_m = N. A pair with a > b ≥ 1 already pays more than (a−b)N ≥ N there, while the target is at most N − 1, two unequal positive integers differing by less than the larger. A pair with a = b = t pays tD outright, by homogeneity, which is D at t = 1 and at least 2D otherwise — and that is the case the uniqueness clause rests on, so it is the one a case list cannot leave to the reader. A zero coordinate pays a multiple of N and opposite signs pay at least N + 1, on the same line. The one hypothesis J ≥ 2 enters at exactly one place, N ≥ 2, and at J = 1 the conclusion is false — the row is 1, 1 and (1, 0) ties with (1, 1).

THE SAME CASE LIST GIVES THE SECOND SUCCESSIVE MINIMUM AND ITS ATTAINING SET, at no extra cost and with no reduction and no search of any kind. Every pair NOT parallel to the minimiser is one of the three the list bounds below by N rather than by D — a zero coordinate, an opposite-signed pair, or one with a > b ≥ 1 after the collapse — and among those the bound is an equality only at ±(1, 0) and ±(0, 1). So in the largest-coefficient norm the two successive minima are D and C(J, ⌊J/2⌋) exactly, the second attained — AMONG VECTORS INDEPENDENT OF THE MINIMISER — at those four and nowhere else, a scope that is load-bearing rather than pedantry, since a MULTIPLE of the minimiser also reaches height exactly N whenever D divides N, which happens at depths 2, 4, 5 and 13 in the first thirteen alone. So a rank-2 lattice here has NO second short vector: its two minima are separated by the whole of the gain differencing buys, with ratio about (√e/2)√(J+1), 2.41% high at depth 10 and 0.01% at depth 2000. And the choice of ⌊J/2⌋ over ⌈J/2⌉ is what decides whether the case list gives the attaining SET or only the value: at odd depths the two central entries are equal and every pair (t+1, t) would tie at N instead of losing. One index apart is the difference between a value and a uniqueness.

What carries this from the lattice to the CELL — to h against the pure bound, which is what a failure on this page is measured by — is that the rank-2 pure family offers exactly three cofactors up to sign: 1, x−1 and 1+x. The cofactor 1 gives (x−1)^J, of height exactly N, so the champion beats it by the same bound that proves the theorem — D ≤ N − 1 — and x−1 gives (x−1)^J+1, of height C(J+1, ⌊(J+1)/2⌋), which Pascal's rule puts above N. So the pure bound equals h at rank 2 at every depth, and that rank leaves the failing set for a reason rather than for want of a counterexample. Rank 1 is clean for a reason too, and a cheaper one — one generator, one candidate — so what rank 2 is alone in is being clean by an ARGUMENT rather than by having no choices: ranks 3 and 4 are clean over the 139 depths the column scan above reaches and no further, that count being that scan's and not this proof's, and every argument here rests on the cofactor being linear.

Scope. Theorem: every J ≥ 2, no computation at any value, the hypothesis used once. The instrument CHECKS the proof rather than supplying it — the lemmas at every depth from 2 to 200, and at depths 2 through 14 a complete enumeration over all of the integer plane, which reproduces both readings independently at each of those thirteen depths rather than sampling them. That enumeration is complete for a reason worth stating, since the endpoints alone are not enough: they bound a box only up to that box's own side, which settles the minimum and is blind to every pair whose coordinates fall between D and the second minimum. Adding the peak inequality cuts the plane to a strip of about two candidates per first coordinate, and six, eight or ten pairs in all of it carry height at most that second minimum — the count being 4 plus twice the number of multiples of the minimiser that fit under it, so it reads ⌊N/D⌋ off directly, which is 2 at ten of the thirteen depths and 3 at the deepest three. Controls run before any verdict: the same machinery at J = 1, where the statement is FALSE, must report the uniqueness break; two mutated targets must be rejected at every depth; and the strip must equal a brute-force box exactly at depths 2 through 9. The RATIO's asymptotic is the one figure here that is measured and not proved — a saddle-point estimate whose residual oscillates rather than decaying, the argmax being an integer where the saddle point is not. AND THE THEOREM SAYS NOTHING ABOUT WHERE THE PRODUCTS LOSE, which on this page is the thing to be careful of: rank 2 is a rank that never fails, so what is proved here is that one column of the LOW WALL is real rather than an artefact of where the chart stops. It does not reach rank 3, where the cofactor gains a quadratic term and the peak comparison alone is not enough; it does not reach the failing ranks 5 to 18, where h < the pure bound is the whole point; and it does not reach the high wall.

verifier: explore_flatten_theorem.py