The Lay of the Land, by n

The Lay of the Land, by `n`

Where the program has spent effort, and what came of it.

$n$	Status	Standing best	Role here	What has been done
5	proved, $2 + \frac{1}{2} \sqrt{2}$	$2.70710678 \dots$	positive control	`sqsearch --selftest` recovers it on every run. exp-007: the bracketing quench refines annealer output to $2.22 \times 10^{- 15}$ —the analytic value to machine precision
8	proved, $3$	$3$	census kill line	The $n$ at which H-011’s discovery curve must plateau, or enumeration is abandoned. No rounds
10	proved, $3 + \frac{1}{2} \sqrt{2}$	$3.70710678 \dots$	positive control	Five rounds. The annealer stops $4.19 \times 10^{- 4}$ short (exp-002); exp-008 closes it to $1.33 \times 10^{- 15}$ ; exp-031 returns all four source perturbations within $2.221 \times 10^{- 15}$
11	proved: $s (11) = T = 3.87708359 \dots$ (T-060)	$3.87708359 \dots$ (Trump 1979)	settled target	The former open-case account is `n = 11`, End to End. Exact verification over $ℚ (u)$ (T-1); the cell decomposition (T-2), corner (T-3), and repaired lower-bound certificate (T-4); nine rounds. Search remains $\approx 6 \times 10^{- 2}$ short, exp-013 proves Trump’s exact pose locally isolated, exp-016 rejects Stromquist’s printed proof, and exp-017 independently restores its numerical bound
12	open; $4$ believed optimal	$4$	open-case calibration	Two rounds. Returns exactly $4.0$ on all five seeds, which is baseline evidence rather than a known-answer guard. Also where the search and proof lanes are planned to meet
16	proved, $4$	$4$	proved not-below control	The valid replacement for the old $n = 12$ guard: any reported side below $4$ is known to be invalid
17	open	$4.67553009 \dots$ (Bidwell 1998)	mechanism-matched calibration	The nearest case whose record uses genuinely oblique structure—tilts of $0^{\circ}$ , $+ {39.80496}^{\circ}$ , and $- {36.62379}^{\circ}$ . One round: exp-011 reports $5.0$ , the trivial $5 \times 5$ grid, on all five binary64 screening seeds
97	proved, $10$	$10$ (grid)	opportunistic slot, closed	Proved on 2026-10-03 as the case $m = 10$ of Evan Daniel’s $s (m^{2} - 3) = m$ (T-064), his `Valid7` checker replayed in full and his Lean reduction built here; $n = 61$ and $n = 78$ , the other two of the slot, were proved on 2026-10-02 by replayed mixed covers. The registered analytic Cleemann-style attempt at $\arctan (3 / 4)$ was never made
1–324	77 proved, 247 open	—	the corpus	One schema-validated artifact per case in `frontier/`; see the Frontier corpus summary for the current aggregate lower-bound counts

Three facts about this table drive the strategy.

Every proved control in the ladder is a 45° mechanism. $n = 5$ and $n = 10$ are symmetric arrangements that blind search reaches without help. $n = 11$ needs an oblique core at an irrational angle, which neither control exercises, so the ladder validates machinery, not strategy. T-060’s later proof of $s (11)$ is a lower-bound argument, not a search, and does not change this.

The ladder now discriminates sharply, and the target does not move. The bracketing quench takes $n = 5$ and $n = 10$ to machine precision and leaves $n = 11$ essentially where the annealer put it. That is the cleanest statement of where the difficulty lives: the refiner is not the problem.

$n = 17$ adds one mechanism-matched negative result. It was the only registered instance cell testing record-finding rather than machinery, and it was the last one never run. exp-011 ran it: the annealer reports $5.0$ on all five binary64 screening seeds—the trivial $5 \times 5$ grid—against Bidwell’s $4.67553$ , a gap of $+ 0.324$ . The retained final states do not leave the grid basin.

That scopes one failure at a second cell: this implementation, five seeds, and the registered $10^{8}$ moves per chain did not reach Bidwell’s oblique record at $n = 17$ . The retained final best does not show which orientations the trajectory visited, and a single budget cannot establish that no larger budget or related proposer can reach oblique records as a class (H-020).