GPT 6 Luna

openai/gpt-6-luna·anthropic/claude-opus-5.5

Let $P_n$ be the Pell numbers and, for $\ell\ge1$, let $\rho_\ell(n)$ be the coefficient of $x^n$ in $G_\ell(x)=\prod_{j=2}^{\ell+1}(1+x^{P_j}+x^{2P_j})$, that is, the number of words $(d_2,\dots,d_{\ell+1})\in\{0,1,2\}^\ell$ with $\sum_j d_jP_j=n$. We prove that $B_\ell=\max_n\rho_\ell(n)=2^{\lfloor(\ell-1)/2\rfloor}$ for every $\ell\ge1$, and we determine the set $M_\ell$ of all maximizers for every length. For odd $\ell\ge3$ it consists of the four integers $u_\ell,\ u_\ell+2,\ u_\ell+P_{\ell+1},\ u_\ell+P_{\ell+1}+2$ with $u_\ell=(P_{\ell+2}-5)/2$, and $M_1=\{0,2,4\}$. For even lengths $|M_2|=9$ and $|M_{2k}|=16k-8$ for $k\ge2$; moreover $M_{2k}$ is the union of two explicit six-element cores and four explicit chains of two-element blocks. All proofs rest on one structural fact: when two consecutive Pell weights are adjoined, the nine shifted copies of the previous coefficient sequence overlap at most two at a time. As an application, let $\delta=1+\sqrt2$ and let $\nu_m(t)$ be the number of strings $(d_k)_{k=-2m}^{2m}\in\{0,1,2\}^{4m+1}$ with $\sum_kd_k\delta^k=t$. We prove the exact identity $\nu_m(t)=\rho_{4m+1}(tP_{2m+2})$ for every integer $t$. Consequently $\max_{t\in\mathbb Z}\nu_m(t)=4^m$, attained only at $t=\sum_{j=-m}^{m}\delta^{2j}=P_{2m+1}+P_{2m}$ when $m\ge1$ and at $t\in\{0,1,2\}$ when $m=0$, and every fibre over $\mathbb Z[\sqrt2]$ in the same window contains at most $4^m$ strings. The results and proofs constitute human-directed, AI-authored mathematical research; they have been checked only by AI-model audits; no independent human proof review has taken place, and novelty has not been established.