(SW) at (-1,-1);(W) at (-1,0); (NW) at (-1,1); (B1) at (1.2,1.2); (B2) at (-1.2,-1.2);

(B1) –++(0,-2.4); (B1) –++ (-2.4,0); (B2) –++(0,2.4); (B2) –++ (2.4,0);

& $\displaystyle \frac{8}{\pi\sqrt{3 \cdot 3}} \cdot n^{-1} \cdot 8^n = \frac{8}{3\pi} \cdot \frac{8^n}n$\

\

Main results
------------

We present two main results in this work. The first appears as Theorem \[thm:asm\].

Let ${\ensuremath{\mathcal{S}}}\subseteq \{-1,0,1\}^d \setminus \{\mathbf{0}\}$ be a set of unit steps in dimension $d$. If ${\ensuremath{\mathcal{S}}}$ is symmetric with respect to each axis, and ${\ensuremath{\mathcal{S}}}$ takes a positive step in each direction, then the number of walks of length $n$ taking steps in ${\ensuremath{\mathcal{S}}}$, beginning at the origin, and never leaving the positive orthant has asymptotic expansion $$s_n  = \left[ \left(s^{(1)} \cdots s^{(d)}\right)^{-1/2} \pi^{-d/2} |\mathcal{S}|^{d/2}\right] \cdot n^{-d/2} \cdot |\mathcal{S}|^n + O\left( n^{-(d+1)/2} \cdot |\mathcal{S}|^n \right),$$ where $s^{(k)}$ denotes the number of steps in ${\ensuremath{\mathcal{S}}}$ which have $k^\text{th}$ coordinate 1.

This formula is easy to apply to any given model, and for certain infinite families as well.

When $d=2$ there are four non-isomorphic highly symmetric walks in the quarter plane, listed in Table \[tab:verif\]. Applying Theorem \[thm:asm\] verifies the asymptotic results guessed previously by [@BoKa09].

Let ${\ensuremath{\mathcal{S}}}= \{-1,0,1\}^d \setminus \{\mathbf{0}\}$, the full set of possible steps. 