$d$-dimensional lattice model restricted to the first orthant, and let $F({\ensuremath{\mathbf{z}}},t)$ be the generating function for this model, counting the number of walks of length $n$ with marked endpoint. Let $V=\{1, \dots, d\}$, so that it is the set of coordinates $j$ for which there is at least one step in ${\ensuremath{\mathcal{S}}}$ with $-1$ in the $j$-th coordinate (this is the full set of indices by our assumptions). Then, by translating the combinatorial recurrence described above, we see that $F({\ensuremath{\mathbf{z}}},t)$ satisfies the functional equation $$\begin{aligned}
\begin{split}
(z_1\cdots z_d)F({\ensuremath{\mathbf{z}}},t) = (z_1\cdots z_d) &+ t(z_1\cdots z_d)S({\ensuremath{\mathbf{z}}}) F({\ensuremath{\mathbf{z}}},t)\\
& - t\sum_{V'\subseteq V} (-1)^{|V'|} \left[(z_1\cdots z_d)S({\ensuremath{\mathbf{z}}}) F({\ensuremath{\mathbf{z}}},t)\right]_{\{z_j=0: j\in V'\}}.
\end{split}\end{aligned}$$ Basic manipulations then give the following result.

Let $F({\ensuremath{\mathbf{z}}},t)$ be the multivariate generating function described above. Then $$(z_1\cdots z_d)\left(1-tS({\ensuremath{\mathbf{z}}})\right)F({\ensuremath{\mathbf{z}}},t) = (z_1\cdots z_d) + \sum_{k=1}^d A_k(z_1,\dots,z_{k-1},z_{k+1}\dots,z_d,t), \label{eq:kernel}$$ for some $A_k \in \mathbb{Q}[z_1,\dots,z_{k-1},z_{k+1},\dots,z_d][\![t]\!]$.

Set ${\ensuremath{\mathcal{S}}}=\{e_1, -e_1, \dots, e_d, -e_d\}$. In this case $S({\ensuremath{\mathbf{z}}})=\sum_{j=1}^d (z_j+{\ensuremath{\overline{z}}}_j)$, so $(z_1\cdots z_d)S({\ensuremath{\mathbf{z}}})$ vanishes when at least two of the $z_j$ are zero, and the generating function satisfies $$(z_1\cdots z_d)\left(1-tS({\ensuremath{\mathbf{z}}})\right)F({\ensuremath{\mathbf{z}}},t) = (z_1\cdots z_d) + \sum_{k=1}^d t(z_1\dots z_{k-1} z_{k+1} \dots z_d) F(z_1, \dots, z_{j-1}, 0, z_{j+1}\dots, z_d).$$

The Orbit Sum Method {#sec:orbit}
--------------------

The orbit sum method, when it applies, has three main 