Let a uniform tiling be defined by a vertex configuration $(n_1.n_2.\cdots.n_k)^m$, which is either spherical, Euclidean or hyperbolic. Assume that the tiling is vertex-transitive, especially that each vertex has the same number of vertices at all distances $d=1,2,3,\dots$ (not the geometrical distance but the number of edges between two vertices).
Let the distance spectrum $\delta$ of such a tiling be the vector with $\delta_d$ = number of vertices at distance $d$ (which is by assumption the same for all vertices).
For the spherical uniform tilings, i.e. the Platonic polyhedra, the distance spectrum is finite:
tetrahedron $(3)^3 \rightarrow$ $\delta = \langle 3 \rangle$
cube $(4)^3 \rightarrow \delta = \langle 3,3,1 \rangle$
octahedron $(3)^4 \rightarrow \delta = \langle 4,1 \rangle$
dodecahedron $(5)^3 \rightarrow \delta = \langle 3,6,6,3,1 \rangle$
icosahedron $(3)^5 \rightarrow \delta = \langle 5,5,1 \rangle$
I am more interested in uniform tilings of the plane (Euclidean or hyperbolic), which of course have infinite distance spectra which have to be expressed by a function $\delta(d)$.
For the three regular tilings of the Euclidean plane this function can be given easily:
triangular $(3)^6 \rightarrow \delta(d) = 6\cdot d$
square $(4)^4 \rightarrow \delta(d) = 4\cdot d$
hexagonal $(6)^3 \rightarrow \delta(d) = 3\cdot d$
which seems to depend on $k$ and $m$ in $(n_1.n_2.\cdots.n_k)^m$, but not on the specific value of $n_1$. The general pattern is $\delta(d) = k\cdot m \cdot d$.
But what about arbitrary vertex configurations?
For which (especially hyperbolic) vertex configurations is $\delta(k)$ known in explicit form – or is there even a chance to derive it systematically from $(n_1.n_2.\cdots.n_k)^m$?
I especially wonder if generating functions may be of help.
For my favorite configuration $(3.4)^3$ a manual count gives $\delta(2) = 21$, but it's hard to determine even $\delta(3)$, even with this picture at hand:
At least some kind of pattern looms: $\delta(2) = 21 = \color{red}{\mathbf{3}} + \color{green}{\mathbf{6}} \cdot \color{blue}{\mathbf{3}}$ which is
$\color{red}{\mathbf{3}} = m \cdot \min(n_2 - 3, 2)$, i.e. the number of quadrilaterals $m = 3$ times $1$ (would be $2$ if $n_2 - 3 > 1$).
$2$ and $3$ are fixed numbers and don't depend on the vertex configuration.$\color{green}{\mathbf{6}} = m \cdot k$ which is the degree of the vertices
$\color{blue}{\mathbf{3}} = m\cdot k - 3$
$3$ is a fixed number and doesn't depend on the vertex configuration.
To sum it up (adding the term $\min(n_1 - 3, 2) = 0$):
$$\delta(2) = m \big(m k^2 - 3 k + \min(n_1 - 3, 2) + \min(n_2 - 3, 2) \big)$$
In general – i.e. for arbitrary configurations $(n_1.n_2.\cdots.n_k)^m$ with $k>1$ – it should hold that
$$\delta(2) = m \big(m k^2 - 3 k + \sum_{i=1}^k \min(n_i - 3, 2)\big)$$
Among the non-hyperbolic tilings above, the formula holds for all but
$(3)^5$ where it yields $10$ instead of $5$
$(3)^6$ where it yields $18$ instead of $12$
$(6)^3$ where it yields $15$ instead of $6$
How does the formula for $\delta(2)$ have to be adjusted to cover also these cases?
