Results: the barrier law survives the grid; the constant is what moves

2026-09-10. Grade: measured (floating LP, HiGHS dual simplex), with every optimum recomputed from the returned coefficients through the constraint definition. No asymptotic statement is established and nothing here bears on RH (docs/08).

Landed on main 2026-10-10 as Hunt #130 (it was #121 on its branch, claude/repo-hunt-content-2smbid), with its front door at docs/42 (numbered 39 there). Every script was re-run on 2026-10-10 (RUNS.md, the landing manifest) and test_quotient_exponent.py pins the numbers on this page, most of them by re-solving: the LP values came back to 1e-12 or better. Four stated numbers did not reproduce and are marked where they are stated (sections 2, 6 and 6b); one sentence in RUNS.md was false and is corrected there; and the audit's exact witnesses for the zeros at y*, whose scripts are not in the tree, were re-run by an independent route and hold (section 6, note). The page's grade stays measured. Recorded text is left as written.

This page was rewritten after an independent adversarial audit. The audit is in AUDIT.md: fifteen attacks, four of which landed, and one of the four took the first version's headline down. That version said the conjectured shape E >= c N / sqrt(y) "is not the shape of T*". It is. What the grid shows is that the constant in front of it depends on the support exponent, which BARRIER.md had already said in its own words, and which the first version's model comparison hid by pitting a three-parameter free fit against a one-parameter reading of the conjecture.

1. The verdict

reproduced, four published diagonal values to 14-15 significant digits new diagonal rows N = 3 x 10^6 and 10^7 T*(sqrt N, N) - psi(N) at 10^7 30297.736768, 0.1704 N^{3/4} published reading "(0.18 to 0.23) N^{3/4}" the conjectured inequality on all 20 grid rows E >= 0.1704 N/sqrt(y), holds the published constant for T* 0.2, which is 15% too large at the 10^7 diagonal point zero excess first reached at y* = 63, 173, 589, 1938 for N = 10^3 .. 10^6 alpha* = log y* / log N 0.5998, 0.5595, 0.5540, 0.5479

Three things this hunt adds, and one it withdraws.

Withdrawn: that the conjectured shape is wrong. Section 4.

The ladder, extended two decades, with the first constant that falls outside the published band.

The excess is a staircase in y, not a curve, with long plateaus and a collapse to exactly zero at a support this hunt locates for four cutoffs.

What keeps the excess positive is the cells with no prime mass at all. Only 40, 99 and 275 of the 61, 198 and 630 attainable cells carry weight at N = 10^3, 10^4, 10^5, and W = 1 on those alone is satisfiable far below y*. The binding requirement is W >= 1 at the zero-weight cells.

2. The reformulation, and what it is not

The Chebyshev identity gives exactly

B_c(N) - psi(N) = sum_{q in Q_N} w_q e_q, w_q >= 0, e_q = W_c(q) - 1 >= 0,

so solving in (c, e) with objective sum_q w_q e_q makes the objective the excess itself. The identities that make the two programmes the same are measured at every cutoff before the solve is trusted:

Not reproduced as printed, 2026-10-10. The 3 x 10^6 row's first column is 1.6e-16 in artifacts/lp_large.json and in the re-run, not 0. Every other entry reproduces.

The first version claimed this reformulation was necessary, and that is false. It argued that the factorial objective asks a float solver for a quantity near 1e4 as a difference of quantities near 1e8, and that this one does not. The arithmetic is right and the consequence is not: solved in the factorial form, the same optimum comes back at

HiGHS absorbs the cancellation. The reformulation is better conditioned in principle, it is what these numbers were computed with, and it is not the reason 10^7 was reachable. The reason is stated in door 2 and it is memory.

3. The diagonal, extended

The first four are the published rows. They agree with the source to 14 and 15 significant digits against the exact-rational pins in its DUAL_WITNESS.md (41.28216944295942 here against 41.28216944295939183...), which is float agreement and not reproduction "to every printed digit", as the first version of this page said. The last two rows are new.

E / N^{3/4} at 10^7 is 0.1704, below the 0.18 to 0.23 band the source reports over its four decades. That is the one number here that contradicts a published reading, and it does so by 15 percent at one point.

4. The grid, and the headline that was wrong

E sqrt(y) / N is what the conjectured shape says is constant:

It is not constant. It falls by a factor of 2.1 to 2.8 as y grows inside the conjecture's own range.

The first version read that as the shape being wrong. That reading does not survive a like-for-like comparison. BARRIER.md section 3 states the conjecture with a moving constant, in its own words: away from y = sqrt(N) the constant moves but the exponent does not. Fitted that way, on the same 20 rows with alpha <= 0.5:

The conjectured exponents fit better than the free ones, with one parameter fewer. And once the constant is allowed to move, the free exponents come back to a = 1.042, b = 0.609, near (1, 0.5) rather than the (1.159, 0.874) the first version reported. The 2.7 times worse figure it quoted compared a three-parameter model against a one-parameter one.

The conjecture is a lower bound, and every measured point satisfies it. E >= 0.1704 N / sqrt(y) holds at all 20 rows. The only thing the measurement touches is the published constant 0.2, at the single new 10^7 diagonal point, by 15 percent. That is the measurement's actual bite and the first version never stated it.

What remains true, and is worth keeping: on the diagonal alone none of these models can be told apart, because the diagonal only ever constrains a - b/2. Separating them needs the grid. The grid then says the exponents are where the conjecture puts them and the constant is not.

5. The staircase

Every integer support at N = 10^3: 33 distinct optima across 76 supports.

y 42..47 excess 10.346570 y 48..50 excess 10.283916 y 51..55 excess 10.247100 y 56..62 excess 2.426015 y 63..77 excess 0

Fourteen supports from 42 to 55 buy 1% of the excess. The next seven buy a factor of 4.2. The next one buys the rest. At N = 10^4 the value 9.406483 holds unchanged across sixteen consecutive supports, y = 157 to 172, then zero at 173.

hunts/quotient_certificate/RESULTS.md names the support bound as its one frozen constant "with trade shape". Measured, the trade is not a curve. A plateau means a block of added columns changed no optimum, which is a rank statement about floor(q/j), and it says the door is not "raise y" but "find which columns did the work".

6. Where it reaches zero, and what holds it up

alpha* falls monotonically, by about 0.007 per decade.

Not reproduced, 2026-10-10. The least-squares slope of alpha* against log10 N over the four rows is -0.0161 per decade; 0.0070 is the slope per unit of natural log N, which is how AUDIT.md attack 13 states it. The monotone fall reproduces.

The zeros at y*, re-checked exactly, 2026-10-10. AUDIT.md attack 3 reports exact rational witnesses for zero excess at y* = 63, 173, 589, with scripts that are not in the tree. Re-run by a different route (the minimum-mass vertex of the zero-excess feasibility programme, rounded with limit_denominator and checked with Python fractions on every attainable cell): witnesses exist at all three, with denominators 4, 7 and 9350 (the auditor's were 2, 60 and 10^6; witnesses are not unique). So zero excess at those three supports is an exact finite fact on this tree's own evidence (test_zero_excess_at_y_star_has_an_exact_rational_witness). Attack 4's Farkas certificates for strict positivity at y* - 1 were not re-run; on this tree that half is float-measured.

hunts/quotient_certificate refuted a dimension count that inferred zero excess from having more columns than constrained cells, with an exact N = 27, y = 9 example whose minimum excess is log 2, so the count is not sufficient. Two of these four rows have the zero arriving while the columns are still outnumbered, so it is not necessary either.

But |Q_N| is the wrong denominator. Only 40, 99 and 275 of the 61, 198 and 630 attainable cells carry any prime mass, and W = 1 on those alone is satisfiable well below y*. What keeps the excess positive up to y* is the requirement W >= 1 at the cells with no weight at all: cells that contribute nothing to the objective and constrain the feasible set anyway. That is a sharper statement than a ratio against a mixed denominator, and it is the door section 7 ranks first.

6a. The door, tested: consecutive support is what the barrier is about

Doors item 2 below names "a support chosen for rank rather than for being an interval" as the obvious untested variation. A scouting pass in the same session reported that freeing the positions reaches zero excess. It does, and the interesting version of that is not the obvious one.

Freeing the positions with no limit on how many says nothing. Allow every j <= N/2 and the excess is zero, but with 5000 columns available at N = 10^4 that is unsurprising: the threshold run above already reaches zero at 173 consecutive columns.

At a comparable support size it says a lot. Excess is sum_q w_q e_q with every term non-negative, so zero excess is exactly W(q) = 1 at every cell carrying prime mass and W(q) >= 1 at the rest. Minimising the coefficient mass under that and counting the positions the optimum uses:

Freely chosen positions reach zero excess on roughly half the support that consecutive ones need, and at N = 10^5 on fewer than sqrt(N) of them. Against the consecutive family at y = sqrt(N), where the excess is 41.28, 226.83 and 1035.23, the same count of positions chosen freely gives zero.

At N = 10^3 this is exact. The float solution rounded to rationals over 10^9 was re-checked with Python integers on all 61 attainable cells: no cell has W(q) < 1, and W(q) = 1 at all 40 prime-mass cells, so the excess is exactly zero. The 10^4 and 10^5 rows are float LP values whose reported minimum slack is slightly negative (-8e-13, -6e-12) and are measured only; they have no exact witness here.

What it costs, and why this is not a free win. The positions are few but they reach far: max j is 201, 1251, 14286, about N^{0.77}, N^{0.77} and N^{0.83}. The source's own criterion asks for excess and mass both under N^{1/2+eps}, and both hold here (mass ~ N^{0.52} at 10^5), but it says nothing about how far the support may reach, and evaluating such a certificate costs max j, not the count.

So the barrier this hunt measured is a property of consecutive support. It is not a property of the floor-sum family, and the conjecture as stated does not speak to the family with free positions, which is a different object and not one these three points settle.

6b. The last cell standing, and why the plateau values are logarithms

Section 6 says the zero-weight cells are what the support has to grow past. This is what that looks like at the end, and it is sharper than expected.

Just before the excess reaches zero, E(N, y) sits on a plateau: a run of consecutive y on which the value does not move at all. The runs are wide and they widen with N.

Not reproduced, 2026-10-10. The 10^5 row is the scan window, not the plateau: closed_form.py searched y = 545..580 and the value filled it. Re-solved on every support from 537 to 589, the plateau is 538 .. 588, width 51 (E = 87.28 at 537, zero at 589, as threshold.json already implied with the same value at 588). The widths still grow with N: 7, 16, 51.

Those three numbers are not arbitrary reals. To every digit the float solver carries they are a small rational multiple of the logarithm of a single prime:

The reason is structural, and nothing here was fitted. The objective is sum_q w_q e_q, and w_q sums Lambda(d) over the integers d in (N/(q+1), N/q]. Lambda vanishes off the prime powers, so a cell whose interval contains no prime power is arithmetically empty: it costs the objective nothing however much excess sits in it, and the solver parks excess there for free. On the plateau the optimum carries positive excess in dozens of cells and exactly one of them has nonzero weight, at every N and at every y across the plateau:

Note, 2026-10-10. closed_form.json sampled three supports per plateau. Re-run at every support of all three plateaus (7, 16 and 51 solves), exactly one excess-carrying cell is weighted at every one of them, and it is always the same carrier, so "at every y across the plateau" now holds as checked. The counts of cells carrying excess widen over the full plateaus: 14 to 15 at 10^3 (as stated), 50 to 57 at 10^4 and 165 to 182 at 10^5, not 52 to 57 and 167 to 181.

So the objective collapses to a single term, e_q * Lambda(p^k), and the prime is the unique prime power in that one cell's interval. The rational is read out of the solution rather than fitted to the value: 7/2, 3 and 6 are the excesses e_q themselves, recovered by Fraction(...).limit_denominator(64) and only then multiplied by log p and compared. closed_form.py does that and writes artifacts/closed_form.json.

The punchline is at N = 10^6, where there is no closed form because there is nothing left to close. At y = 1995 the excess is exactly 0.0, and 497 cells still carry positive excess. Every one of them is weightless. The LP never drove the constraint violation to zero. It drove the weighted violation to zero by moving all of it into the cells where Lambda vanishes, and the plateau is the last stage of that process, the interval on which one prime power is still in the way. The certificate stops improving not because the excess is gone but because it has all been pushed somewhere the objective cannot see.

Note, 2026-10-10. The 10^6 plateau was not scanned, as stated, but its value is in this tree: the bisection in threshold.log hit E = 123.8331 at y = 1812, 1921, 1934 and 1937, and threshold.json stores 123.83310675500282 at 1937 (RUNS.md's note says no run in the tree carries it, and is corrected there). Since E is non-increasing in y (both staircases), the plateau spans at least 1812 .. 1937, 126 supports. Re-solved at y = 1937 the value reproduces, with 513 cells carrying excess of which two are weighted, so it has no single-carrier closed form of the kind above.

Grade: measured. One route, float LP, agreement at 1e-14 to 1e-18. The rationals are exact objects and the primes are exact, but the excesses come from a float solver, so the identity E = (7/2) log 2 is an observation about a numerical solution and not yet a proved property of the programme. Making it a proved one is an exact rational LP at these three (N, y), which is the same door section 7 already names for the zero-excess witnesses, and it would settle whether 7/2 is exactly 7/2.

7. The doors

1. Active constraints at the optimum. Positivity binds on 31, 103, 321, 1011, 1746 and 3193 cells at the six diagonal cutoffs, which is 0.504 to 0.520 of the cells every time. That looks like a regularity and is not: a vertex has as many active constraints as variables, there are y of those, and the diagonal fixes y = floor(sqrt N) while the cell count is 2 floor(sqrt N) - 1. Half is arithmetic. What is not forced is the small excess over y: binding/y runs 1.000, 1.030, 1.016, 1.011, 1.008, 1.010. The binding constraint that is not arithmetic is the zero-weight cells, per section 6: they cost nothing in the objective, they are more than half the constraint set, and they are what the support has to grow past. Section 6b prices that exactly at the end of the descent: on the plateau the optimum carries excess in dozens of cells and exactly ONE of them is weighted, so the whole remaining objective is one prime power's Lambda times one rational. The last thing binding this family is a single cell, and which cell it is depends only on where the prime powers fall.
2. Frozen-constant inventory.
3. The zero-weight cells, newly identified as the binding thing. The trade: they are exactly the cells floor(N/d) for d composite, and dropping them is not free, because the Chebyshev identity needs W >= 1 wherever Lambda(d) > 0, which is where they are not. The question with real shape is how much of the excess survives if positivity is required only on a sublattice, and it is one LP per choice.
4. The support bound y, now measured as a staircase. Most of its range is inert and two points are worth everything. "Find which columns matter" is a rank question about floor(q/j) and is answerable exactly.
5. Consecutive denominators j = 1..y, now tested (section 6a). Freely chosen positions reach zero excess on roughly half the support consecutive ones need, exactly at N = 10^3 and measured at 10^4 and 10^5. The trade is reach: the positions run out to about N^{0.8}, and evaluation cost follows the largest one rather than the count. The next question is the one this does not answer, whether a support bounded in BOTH count and reach still beats the interval.
6. The 10^7 ceiling, which is memory. The constraint block is dense 2 sqrt(N) x y, which at N = 10^8 is 1.6 GB before the solver's own copy, and lp.py builds a CSR copy of it, which doubles that. Column generation, which the source already uses for V*, is the way past it.
7. Information class. Everything stays inside the floor-sum span and reads no prime locations. The doors that read more are the source's own: prime awareness below sqrt N, and an exactly evaluable dictionary whose fluctuation is not sawtooth-shaped. This hunt adds one that reads less: the staircase and the zero-weight cells both say the family's information is concentrated in a small part of its own constraint set, so how small is a compression question about the same data.