Brief: refute a barrier claim

You are auditing a mathematical claim that the person who wrote this believes is TRUE and wants you to attack. Your job is refutation, not confirmation. A refutation is worth far more than an agreement, so do not soften anything.

Implement everything yourself from the definitions below. Do not look for prior implementations. Do not trust any number stated here: every constant is a claim to be recomputed.

Definitions

Kernel, for real x:

K(x) = ∫_{-1/2}^{1/2} cos(√2 · t) · cos(2π · x · t) dt k(x) = K(x) / K(0) w(x) = k(x)²

K has a closed form as a combination of two sinc terms. Derive it yourself.

For an integer n ≥ 3 (this is n points, so k = n − 1 gaps), a pressure p > 0, and a vector of nonnegative gaps g = (g_1, …, g_k):

F_{n,p}(g) = (1/p) · Σ_{i=1..k} g_i

• Σ_{s=1..n−1} (2/(n−s)) · Σ_{i=1..n−s} w(g_i + g_{i+1} + … + g_{i+s−1})

The inner sum runs over the n − s windows of s consecutive gaps. Write S(g) = Σ g_i for the total length, and W(g) = F_{n,p}(g) − S(g)/p for the pressure-free part, which does not depend on p.

Constant:

H = 3/2 − (1/√2) · cot(1/√2)

A triple (c, m, p) is admissible for n when c > 0, m and p are integers with m ≥ n and p > 0, the cap condition c · (m − (n−1)) ≤ 1 holds, and the certificate condition holds:

c ≤ F_{n,p}(g) for EVERY g with all g_i ≥ 0

For an admissible triple define

Φ_n(c, m, p) = (H − (n−1)(m−1)/(p·m)) / (1 − c·(m − (n−1))/m)

The value 0.6818286874638 comes from a separate argument. Treat it here purely as a target number.

The claim to attack

CLAIM. The supremum of Φ_n(c, m, p) over all n ≥ 3 and all admissible (c, m, p) is at most 0.6751676068, and in particular is strictly below 0.6818286874638. No matter how many points are used, this family cannot reach the target.

The supporting argument has two halves.

Algebraic half: verify or break each step

1. Φ_n = (H·m − (n−1)(m−1)/p) / (m − c(m−(n−1)))
2. the cap c(m−(n−1)) ≤ 1 makes the denominator ≥ m−1, so Φ_n ≤ H·m/(m−1) − (n−1)/p
3. with m at its largest useful value (n−1) + ⌊1/c⌋, we have m−1 ≥ 1/c when n ≥ 3, so Φ_n ≤ H + H·c − (n−1)/p
4. c is a floor over ALL nonnegative g, so for ANY single witness g, c ≤ F_{n,p}(g) = S(g)/p + W(g), hence Φ_n ≤ H + H·W(g) + (H·S(g) − (n−1))/p
5. therefore any witness g with S(g) ≤ (n−1)/H gives Φ_n ≤ H·(1 + W(g)) for EVERY p

State plainly whether each step is valid, including sign conditions, and any case where a numerator could be negative or a denominator could vanish or change sign.

Numerical half

The claim needs, for every n, a witness g with S(g) ≤ (n−1)/H whose energy W(g) is small. Reaching 0.6818286874638 would require W ≥ 0.0138706 for every admissible witness simultaneously.

What would refute the claim: hunt for these

(a) A step in 1–5 that is invalid, or valid only under an unstated condition.

(b) An n for which NO witness g exists with S(g) ≤ (n−1)/H and H(1 + W(g)) ≤ 0.6751676068. This is the main target. Search over arbitrary nonnegative gap vectors. In particular do not restrict to regular or periodic patterns: try irregular gaps, clustered gaps, gaps far from any zero of k, and whatever else your own analysis of w suggests. Report the n with the largest minimum-W you find.

(c) A concrete admissible triple (c, m, p) and an n with Φ_n(c, m, p) > 0.6751676068, where you have real evidence that c is a genuine floor, a serious global minimisation of F_{n,p} over nonnegative g that does not go below c.

(d) Any reason the supremum over n could fail to be controlled as n → ∞.

What to compute and report

1. Your own values of H and K(0), to 16 digits.
2. For n = 7, 8, 9, 10, 12, 14, 16, 20, 30, 56, 100: the smallest W(g) you can find subject to S(g) ≤ (n−1)/H, the witness attaining it, and the bound H(1 + W(g)). Say which optimiser you used and how you guarded against local minima.
3. Whether any of those bounds exceeds 0.6818286874638 (breaking the barrier at that n) or 0.6751676068 (breaking the stated supremum).
4. For n = 7 and p = 3000: the global minimum of F_{7,3000} over nonnegative g, to 15 digits, with the minimiser.
5. Your step-by-step verdict on the algebraic half.
6. Interval arithmetic wherever you can manage it. The witness direction is the unsafe one: an underestimated W would make a false barrier look true. If you can bound W rigorously from above for your witnesses, do it and say so. If not, say that plainly.

Rules

• Implement from the definitions above only.
• Report every disagreement with a stated number loudly, with your value beside it.
• If you cannot break the claim, say exactly what you tried hardest to break and where the argument is most fragile. "I could not refute it" plus a map of the weak points is a complete and valuable answer.
• Do not conclude that the claim is correct because it looks correct. Compute.

Write your report to results/REPORT.md and leave your scripts in this directory.