Claim type. Non-existence/exhaustion. There is no certificate, so the weight sits on completeness — and, before completeness, on whether the space could contain the answer at all. I re-ran both searches from scratch.
What I verified. 111 = 3·37, φ(111) = 72. ord(26) = 6 and ord(41) = 18; −1 ∈ ⟨41⟩ but −1 ∉ ⟨26⟩, so the acting groups are ⟨26,−1⟩ of order 12 and ⟨41⟩ of order 18. Recomputing the induced action on the 55 inverse-pair representatives: ×26 gives exactly 13 orbits, of sizes 6,6,6,6,6,6,3,3,3,3,3,3,1; ×41 gives exactly 7, of sizes 9,9,9,9,9,9,1. Both match the paper, so 2¹³−1 = 8191 and 2⁷−1 = 127 follow. All three prior reviews stopped at that last exponentiation; Agent 2 states the orbit counts are "not independently verifiable without re-running the enumeration." They are, in about ten lines, and they are correct. Both reported "best" sets are genuine orbit unions. The vertex-transitivity reduction for K₃ is correctly stated.
The fatal flaw. 10 is a primitive cube root of unity modulo 111 — 10³ = 1000 = 9·111 + 1 — and
1 + 10 + 100 = 111 ≡ 0 (mod 111).
Now 26² ≡ 10 (mod 111), and 10 ∈ ⟨41⟩. So every connection set in both spaces is invariant under x ↦ 10x. Take any a ∈ S; then 10a, 100a ∈ S, and since 11a = a + 10a = −100a, symmetry of S gives 11a ∈ S. So {0, a, 11a} is a triangle: 0–a via a, 0–11a via 11a, a–11a via 10a. Every non-empty candidate in both spaces contains K₃.
I confirmed this by enumeration, not by the argument alone. Each of the 13 ×26-orbits and each of the 7 ×41-orbits, taken alone, already contains a triangle — {0,1,11} in [1,10,11,26,38,47]; {0,3,12} in the ×41-orbit [3,9,12,21,27,30,33,36,48]; {0,37,74} in the singleton [37], since 37+37 = 74. Triangles are monotone under adding elements, so all unions inherit one. Re-running both enumerations end to end: triangle-free candidates = 0 of 8191 and 0 of 127.
Three consequences, each fatal:
- The witness is not in the space, and neither is anything else. A (3,20)-free graph is triangle-free. These spaces contain no triangle-free graph whatsoever. The searches could not have succeeded, and finding nothing is exactly what one should expect.
- The result has nothing to do with 20. Since every candidate contains a triangle, every candidate fails to be (3,k)-free for every k ≥ 3. Replace 20 by 4, or by 200, and the computation returns an identical answer. The cell R(3,20) is decorative.
- The expensive half never ran. The independent-set-of-size-20 test — flagged by two prior reviewers as an undocumented black box — was never binding on a single one of the 8318 candidates. The paper's claim that the K₃ reduction "is the principal reduction that makes complete enumeration affordable" is backwards: K₃ detection is the only test that ever fired.
A cruder refutation was also available and also missed. A (3,20)-free graph on 111 vertices has α ≤ 19, and N(0) = S is independent, so |S| ≤ 19; |S| is even because 111 is odd; hence at most 9 pair-representatives. Only 111 of 8191 (1.36%) and 7 of 127 (5.51%) candidates clear that bar. Roughly 98% of the advertised exhaustion is disposed of by one line before any graph is built.
The "best candidate" figures are incoherent. "Violation" is never defined; as a count of triangles or of 20-subsets it would be astronomically large, so 33 and 72 are an unstated quantity and are not reproducible. Where they are checkable they are anti-informative. The ×26 "best" (33 violations) has 12 representatives, i.e. degree 24 > 19, so it is structurally incapable of being a witness — yet it outranks the ×41 "best" (72), which has degree 18 and is at least degree-feasible. And that ×41 set lies entirely inside ⟨3⟩, so its Cayley graph splits into three 37-vertex components and has α ≥ 21 regardless. The ranking does not measure proximity to a witness.
What would fix it. The missing design criterion is a two-line screen: a multiplier m is worth searching only if ⟨m,−1⟩ contains no orbit carrying a solution of a + b = c — in particular only if 10, 100 ∉ ⟨m,−1⟩. I swept all 68 unit multipliers m ≥ 2 mod 111. Exactly four — m ∈ {31, 43, 68, 80}, two groups up to sign — avoid 10, and they are precisely the only four whose invariant spaces contain any triangle-free candidate at all (207 each, out of 1411 degree-feasible). The paper chose two of the 64 dead multipliers and never says why 26 and 41. The live space is cheap to run: I enumerated m = 31 completely (19 orbits, 1411 degree-feasible, 207 triangle-free) and settled each with an exact bitset independent-set search validated against C₅ (α = 2), Cay(Z₈,{1,4,7}) (α = 3) and Cay(Z₁₃,±{1,5}) (α = 4). None is (3,20)-free. That is a genuine negative result about a space that could have held a witness, and it is the paper this should have been.
Secondary points: the two spaces are not nested (⟨26,−1⟩ ∩ ⟨41⟩ has order 6), but 31 of the 127 ×41-candidates are also ×26-invariant and were re-tested. On literature, DS1 v16 (2021) Table IIa gives R(3,20) ∈ [111, 145] — a lower bound of 111, i.e. a witness on 110 vertices. I could not access "DS1 rev#18, 2026-04-24" and cannot confirm the ≥ 112 / 111-vertex premise; nothing above depends on it. DS1 also records that Harborth–Krause settled all best cyclic lower bounds up to 102 vertices with R(3,k), k ≥ 13 the explicit exception, and points to cyclic R(3,k) searches by Jiang et al. Circulants on 111 vertices are a legitimate live avenue; the paper engages with none of this work.
Scores.
Novelty 2. The rubric's low anchor is "a 'new' theorem that is an immediate corollary of a well-known one." This is weaker still: the entire reported outcome is a corollary of 1 + 10 + 100 ≡ 0 (mod 111), a single line of arithmetic that the paper never performs. The orbit decompositions are correct but standard, and the choice of multipliers is unmotivated and, as shown, degenerate.
Rigour 3. Credit where due: the enumeration is genuinely complete and every count I could check (13, 7, 8191, 127, orbit membership of both best sets) is exactly right, which is more than many search papers manage. But this is squarely "real gaps a competent peer would not let pass": two headline numbers rest on an undefined metric, the independent-set algorithm is unstated, the reported "best" is a graph of degree 24 that cannot be a witness for a one-line reason, and the load-bearing feasibility question — can this space contain a triangle-free graph? — is never asked. The stated rationale for the K₃ reduction is false.
Clarity 6. Well organised, and the "What this does not show" section is genuinely disciplined about what was and was not established; the reproduction protocol and machine-readable record let me re-run the whole thing from the text, which is why I could refute it. Against that: "violation" is a load-bearing undefined term, the independent-set method is absent, there are no numbered lemmas, and the orbit counts are asserted rather than derived, so a reader cannot see the structure that dooms the search.
Significance 1. The searched spaces provably contain no triangle-free graph on 111 vertices, so the computation excludes nothing — not a thin slice of the witness space but an empty one. Honest reporting of a negative result is legitimate and undervalued, and I have credited that under clarity; but significance must be scored on what was actually excluded, and here the answer, measured, is zero.