Internal mathematical review

September 9, 2026. Originating-agent review; no separate-agent or human review.

Universal proof obligations

1. Definition. The missing set excludes zero and uses all nonempty subsequences. The subgroup may be arbitrary, but the coset must exclude zero. Both primary conjecture statements explicitly impose this condition.
2. Length. The sequence contains 2n-3 copies of f and four circuit terms, hence has length 2n+1. Baayen's established upper bound and the standard lower example give d(G)=2n+2.
3. Zero-sum freeness. The elementary-two projection of the four exceptional terms has kernel exactly the empty and full subsets. In the full-subset case the cyclic sum is 2+k, where 0<=k<=2n-3. No such sum vanishes modulo 2n. The empty-subset case is equally explicit.
4. All holes. Each elementary-two coordinate has exactly two complementary subset representations. Their cyclic sum intervals each omit two residues. Computing the intersection shows that only subsets of size two contribute holes. The congruence reduction divides by two only after reducing the modulus from 2n to n; it uses gcd(q,n)=1. Since n>=3 is odd, k=2 is the unique solution among 0,1,2,3,4.
5. Coset obstruction. A containing coset forces f into its subgroup from an adjacent pair. A second difference puts u+v into the subgroup. A displayed hole then lies in that subgroup, contradicting the nonzero-coset requirement. This argument covers arbitrary subgroups, including composite-index subgroups.
6. Exact invariant. Failure at length d-1 proves nu>=d. Every zero-sum-free sequence of length d has no holes, since any hole would furnish a further term. Hence nu<=d.
7. Extensions. The chosen first extra term is its own negative and is a hole, making the extension zero-sum free. Its negative total sum gives the last term of an atom. The length and maximality assertions follow directly.
8. Parity statement. Within the displayed constant-lift shape, zero-sum freeness is equivalent to 4t being 1 or 2 modulo 2n. The first residue is impossible by parity, and the remaining congruence has a solution precisely for odd n. This statement does not assert a classification of all sequences for even n.

Independent finite checks

Run python3 check.py. For all 50 odd n from 3 through 101, it verifies zero-sum freeness, the exact six holes, and containment of a hole in the subgroup generated by hole differences. The subgroup has order 8n, so the obstruction is also visible without a character test. It then verifies the explicit extremal extension and longest atom. A separate brute-force base-case routine enumerates all 127 nonempty subsequences and finds 41 distinct nonzero sums.

The hole computation uses a generic coordinate subset-sum dynamic program, not the interval formula in the proof. The subgroup computation uses a generic breadth-first closure under generators, not the three-line coset argument. Minimality of the final atom is certified by zero sum and zero-sum freeness after deletion of its last term; any proper zero-sum subsequence containing that term would leave a nonempty zero-sum complement.

An exploratory exhaustive C++ search was used to discover the example but is not a premise of the paper. In particular, a failure of the index-two condition alone was not accepted as a failure of the original conjecture. Prime-order characters alone were also not used as a general arbitrary-coset criterion.

Artifact review

The final LaTeX, full Markdown, PDF, mathematical checks, source assessment, and machine-readable artifact checks are included. PDF layout and Markdown equation checks are recorded in the adjacent JSON files. Outside mathematical correctness and priority remain provisional.