k >= 2: what the budget's superadditivity does and does not buy

Date: 2026-08-13. Reads: gram_form.py (the three-term form), kpair_identity.py (the identity and the joint search), PROOF-LEDGER.md defect #20.

0. The state, corrected

PROOF-LEDGER.md carries an entry reading "k >= 2: OPEN on the BUDGET side ... short by a factor 1.99". That entry is superseded a few lines below it by defect #20, and a reader who stops at the first entry will price this problem wrongly. The 1.99 came from multi_pair_requirement, which charges damage per pair against a joint budget floor; the identity shows the two sides are maximised by different configurations, so the factor does not bound the truth.

What the object itself measures (joint annealing over atoms, centres and depths, n <= 24, 60 restarts):

No downward trend in k. The statement looks true with about a third of relative margin in hand, and the planted-fault ladder agrees with that number independently (a worst margin of 0.343 predicts first violation at damage inflation 1.522; the ladder first goes negative at 1.5).

So k >= 2 is missing a proof, not missing plausibility.

1. The reduction that is already in hand

gram_form.py puts the slack in three terms,

slack_k = B(T,y) + Cross(X,T,y) + R(X)/400,

with B >= 0 provable in one line from Retention.energy_F_ge (already kernel-checked) and R >= 0 a sum of squares. So the whole question is

(Q) B(T,y) + R(X)/400 >= sum_{a,p} D(y, x_a - t_p).

2. The budget is exactly superadditive, and by how much

Writing |That(w)|^2 = k + sum_{p != q} cos(tau_pq w) and integrating term by term against c2(w) cosh^2(y w) dw gives an identity that is not written down anywhere else in this hunt:

B(T, y) - sum_p B({t_p}, y) = sum_{p != q} Psi(tau_pq, y),

Psi(tau, y) = -(1/2) [ D(0, tau) + D(2y, tau) ].

Checked numerically over 200 random configurations (k in 2..6, y in [0.05, 0.5], centres in [-40, 40]): worst residual 1.78e-15. It is the same cross-pair term the kpair_identity slack already carries, arrived at from the budget side instead.

3. What that kills

The obvious route to k >= 2 is to apply the closed k = 1 inequality once per pair and add. Doing so needs, exactly,

(S) sum_{p != q} Psi(tau_pq, y) >= (k - 1) * R(X)/400,

because each of the k copies consumes the whole repulsion R, and only one R is available. (S) is false in general, and the identity says why in one line: its left side depends only on the centre gaps tau, its right side only on the atoms X. Spread the centres and D(., tau) -> 0, so the left side goes to zero while R(X) is untouched and can be large.

The escape "use only a fraction lambda of R per pair" needs lambda <= 1/k, i.e. a k = 1 bound that is asymptotically repulsion-free, and PROOF-LEDGER.md already records that the repulsion-free route is "arithmetically dead, not merely lossy".

So the per-pair decomposition is dead for a structural reason, not for want of sharper constants. That agrees with the ledger's verdict and now has a one-line proof rather than a measurement behind it.

4. Where that leaves the search

Any proof of (Q) has to be genuinely joint in X and T. The two adversary routes both pay, and a proof has to make that quantitative in one argument:

• Spread the centres so every pair sees atoms at a damage peak, but only two positions carry the top peak (+/- 6.517) and the rest decay like 1/s^2, so damage/gain falls (measured 1.0420 at k = 2, 0.2483 at k = 12).
• Stack the centres to keep damage/k maximal, but coincident centres contribute D(2y,0) + D(0,0), which enters with a minus sign and is a gain of 1.7556 per ordered pair.

The quantity that interpolates them is the centre packing: tight packing makes Psi large, loose packing limits how many pairs one atom can damage at once (counting_bound.py's "at most one pair per damage window" is the combinatorial half of this). A proof of (Q) most plausibly comes from bounding sum_{a,p} D(y, x_a - t_p) by an incidence count controlled by the minimum centre gap, and playing that against Psi summed over the same gaps.

This is a direction, not a schedule. Nothing above is a proof of k >= 2, and none of it moves the constant. Nothing here is evidence about RH.

5. The anatomy of the binding configurations

Decomposing slack_k into its four terms at the worst configuration found for each k (fixed-n annealer, 20 000 iterations):

Three things this says, none of them obvious from the identity:

1. The cross term is a cost, not a gain, at every binding configuration, and by k = 12 it is comparable to the damage (-0.102 against 0.142). The +1.7556-per-ordered-pair gain from coincident centres is real but the adversary does not take it, stacking buys relief it does not want.
2. Incidence is not limited. One atom sits inside the damage window of 10 of the 12 pairs at k = 12. Whatever "at most one pair per damage window" bounds, it is not this, so the damage side scales with k essentially in full.
3. Everything scales linearly in k. cross/k runs 0.0092, 0.0072, 0.0065, 0.0085; gain and damage likewise. That is why the relative margin stabilises instead of collapsing, and it means an argument has to be linear in k too, anything that loses a factor of k cannot work.

6. Repulsion cannot be dropped (confirming the ledger independently)

The table above shows repulsion contributing under 20% of the gain at the binding configurations, which invites the thought that it could be discarded and the shared-repulsion difficulty with it. It cannot. Minimising the relative margin with the repulsion term deleted:

Negative already at k = 1, decisively. (The k = 12 positive is search weakness on a harder landscape, not a reversal.) The small repulsion values in §5 are an artefact of reading them off configurations chosen to bind the full slack; an adversary allowed to ignore repulsion immediately picks different ones.

So PROOF-LEDGER.md's "the route that drops the repulsion term is arithmetically dead" holds, reached here from a different direction. Repulsion is load-bearing, it is paid once however large k is, and any proof must carry that asymmetry rather than route around it.

7. The extremal configuration lives on a 2*pi lattice

The saved k = 12 witness (k_trend.WITNESS_K12) has centre gaps

37.553 131.918 44.043 12.632 6.225 6.236 12.803 6.235 6.238 12.904 18.789

and dividing by 2*pi gives 5.977, 20.995, 7.010, 2.010, 0.991, 0.992, 2.038, 0.992, 0.993, 2.054, 2.990, every one within 2..5% of an integer. The atoms show the same thing, in clusters of near-coincident points separated by multiples of the same step.

That is not decoration. The nine damage windows sit at 6.5167, 12.6988, 18.9400, ..., whose successive gaps run 6.182, 6.241, 6.260, 6.270, 6.274, 6.277, 6.279, 6.279, converging to 2*pi. The adversary needs its atoms inside the damage windows of as many pairs at once as possible, and the window spacing is the lattice step.

The restriction is essentially free. Minimising with centres and atoms confined to c0 + 2*pi*Z and x0 + 2*pi*Z (two real offsets, integer sites, depths free):

At k = 8 the restricted search finds a worse configuration than the free one, and by k = 16 the two agree to 0.007. Restricting costs nothing and searches better, because the space is smaller.

This is a genuine reduction: the continuum of centre and atom positions collapses to integers plus two offsets. It is not yet a finite problem, since k and the occupancy pattern remain.

8. But the occupancy is not periodic, so there is no small cell

The obvious next hope, if the sites are a lattice, take the occupancy periodic and let k -> infinity on one cell, fails. Occupying every p-th site with centres and every q-th with atoms, at depth 1/2, offsets optimised on a 24 x 24 grid:

Every uniform periodic pattern is benign, the best is +0.694 against a binding ~0.24, a factor of three of slack. The margins do converge in k (p=q=2 runs 0.8100, 0.7514, 0.7154, 0.6940), so the limits exist; they are simply not where the adversary lives.

What the binding configurations use is the lattice with irregular occupancy and mixed depths, at k = 12, ten of twelve depths on an endpoint and a gap pattern 1,1,2,1,1,2,3,... rather than a constant stride. So the k -> infinity problem does not reduce to a small periodic cell, and the combinatorics of which sites are occupied is carrying the difficulty.