Polynomial rejection under square-root rational approximation

Let ${\mathscr{S}}$ be a step multiset and let ${\textsc{H}}_{\theta}({\mathscr{S}})$ be a half-plane model, with $h_n(\theta)$ denoting the number of walks of length $n$ in this model. Let $\theta^*$ be the optimal slope, and let a $(1/\sqrt{n})$-rational approximation mean a half-plane model whose slope tangent differs from $\tan\theta^*$ by at most $1/\sqrt{n}$ and whose tangent is rational. Polynomial-rejection conjecture. There exists an infinite sequence of angles $\{\theta_n\}_{n\le 0}$ such that, for all $n\ge 0$, ${\textsc{H}}_{\theta_n}({\mathscr{S}})$ is a $(1/\sqrt{n})$-rational approximation of ${\textsc{H}}_{\theta^*}({\mathscr{S}})$, and

$$\exists p\in\mathbb{R},\qquad \lim_{n\to+\infty}\frac{h_n(\theta_n)}{h_n(\theta)}\in\mathcal{O}(n^p).$$

This would imply that the rejection count in the associated sampling algorithm remains polynomial in $n$ when the irrational optimal slope is approximated at precision $1/\sqrt{n}$. The surrounding discussion establishes polynomial-time rejection for precision $1/(n+1)$, while the conjectured square-root precision would provide a substantially more efficient approximation; no resolution is supplied here.