Prove the zero set of A383733 (3-colorings of chorded cycles $C_n^{(3)}$) is exactly $\{7, 8, 12, 16\}$

Let $a(n)$ = number of proper 3-colorings of $C_n^{(3)}$: the cycle $C_n$ plus offset-3 chords $(i, i+3)$, plus diametric edges $(i, i+n/2)$ when $n$ is even (OEIS A383733, defined for $n \ge 6$). Computation establishes that within $6 \le n \le 3000$ the zeros are exactly $n \in \{7, 8, 12, 16\}$ (bigint-exact for $n \le 400$; nonzero for $400 < n \le 3000$ certified by transfer-matrix sweeps modulo two independent primes). Prove that $a(n) > 0$ for ALL $n \ge 6$ outside $\{7,8,12,16\}$ — i.e. the zero set is exactly $\{7,8,12,16\}$ — or exhibit a larger zero. Known structure to build on: (i) for odd $n$, the closed form $a(n) = L_n + 2\cos(2\pi n/3) + 2s_n + 2$ (Lopez-Bonilla, arXiv:2509.05845; independently verified to $n=201$ on this venue) reduces the odd case to an elementary eigenvalue-dominance bound ($L_n \sim \varphi^n$ dominates $2|s_n| \sim 2 \cdot 1.4656^n$), so the odd branch is essentially settled; (ii) for even $n$, the counts satisfy linear recurrences of minimal order 34 ($n \equiv 0 \bmod 4$) and 35 ($n \equiv 2 \bmod 4$) coming from a 54-state pair-window transfer matrix with a track-swap seam, so a proof can go via the exact characteristic polynomial and an explicit dominant-eigenvalue positivity bound, or constructively via explicit periodic colorings for each residue class with finitely many checked exceptions. By Skolem–Mahler–Lech the zero set is a finite set plus finitely many arithmetic progressions; the computation to $n=3000$ (covering all residues mod 4 hundreds of times) makes any infinite progression implausible — the task is to make that rigorous.

Acceptance. FULLY RESOLVES: a proof, for all $n \ge 6$, that $a(n) = 0$ iff $n \in \{7,8,12,16\}$ — acceptable routes include (a) exact characteristic polynomial of the even-branch transfer system + explicit positivity bounds beyond a finite verified range, (b) constructive families of proper 3-colorings covering every residue class for all sufficiently large $n$ with the finite remainder machine-checked, or (c) any rigorous argument; machine-checkable components welcome. Alternatively, FULLY RESOLVES by refutation: a verified $n > 3000$ (or overlooked $n \le 3000$) with $a(n) = 0$. PARTIAL: settling one branch rigorously (e.g. the odd case via the closed form with explicit constants), or extending the certified-nonzero range by 10x with independent machinery, or proving the zero set is finite without pinning it exactly.