Please let me first describe the general background.
The state of the system at time $t$ will be described by a scalar or phase $u=u^t$. Both $t$ and $u$ are discrete. $u$ take values from the fixed set $\mathfrak{S}=\left\{0,1,2,\ldots,e,e+1,e+2,\ldots,e+r\right\}$ of nonnegative integers and $t\geq 0$. Elements of the subset $\left\{1,2,\ldots,e\right\}$ will be identified as the "excited" states, elements of the subset $\left\{e+1,e+2,\ldots,e+r\right\}$ as the "refractory" states, and the singleton $\left\{0\right\}$ as the "rest" or "equilibrium" state. Assume throughout that $r\geq e$.
The dynamics of the model are specified by the rule $$ u^{t+1}=\mathfrak{E}(u^t), $$ where $$ \mathfrak{E}(k)=k+1,~1\leq k\leq e+r-1 $$ and $$ \mathfrak{E}(e+r)=\mathfrak{E}(0)=0. $$ Now we want to put the neighbour cells into consideration. In one-dimension this is done as follows: $$ u_k^{t+1}=\mathfrak{E}(u_k^t)+D(u_k^t; u_{k+1}^t, u_{k-1}^t),~~~(*) $$ where $$ D(u;v_1,v_2)=\begin{cases}1, & \text{if }u=0\text{ and }1\leq v_i\leq e\text{ for some }i=1,2\\0, & \text{otherwise}\end{cases}. $$ That is, if a cell is in rest and at least one of its two neighbors is excited, then the cell becomes excited.
--- Now consider the following special case.
Let $r+2\leqslant L\leqslant e+r$ and consider the initial conditions $$ u_k^0=k,~~0\leqslant k\leqslant L-1~~~(**) $$ with periodic boundary conditions, that is, we take the neighbours of cell $0$ to be cell $1$ and cell $L-1$ and the neighbours of cell $L-1$ to be cell $0$ and cell $L-2$. Apply the rule $(*)$ on this.
(The solution $u_k^t$, corresponding to the initial data $(**)$ has been dubbed a "caterpillar-wave" because the widths of the excited and refractory regions alternately expand and contract.)
Now the question is, which topological entropy this example has.
I looked at an example, namely $L=5, e=r=3$. What I get for the evolution is $$ 01234\\ 12345\\ 23456\\ 34560\\ 45601\\ 56012\\ 60123\\ 01234\\ $$ and so on, i.e. period $7$.
My idea therefore is to use that topological entropy is an invariant of conjugation.
Consider $$ Y\subset\left\{0,1,2,\ldots,e+r\right\}^{\mathbb{Z}}, $$ consisting of the sequences where
- k has k+1 to its right for $0\leq k\leq e+r-1$
- e+r has 0 to its right.
On $Y$ consider the left-shift $\sigma$.
As far as I see there is a homeomorphism $\varphi$ from $Y$ to the configurations $u_k^0, u_k^1,\ldots$ on $[0,\ldots L-1]$.
So that the topological entropy that I am searching for can be computed as the topological entropy of $(Y,\sigma)$.
But this is known to be $\ln\lambda$, where $\lambda$ here is the biggest (in absolute value) eigenvalue of the $(e+r+1)\times (e+r+1)$-dimensional adjacency-matrix. And if I am not wrong, it is $\lambda=1$.
So my question is, if I am right? Do the "caterpillar waves" indeed have topological entropy $$ \ln 1=0? $$
Am I right or do I think much too easy? I am very unsure!
With greetings
This is not a complete answer, I am writing here because SE does not like long lines of comments.
If you consider the dynamics on the space of admissible sequences of configurations i.e. on the space of all possible (one-sided) trajectories of your $\mathfrak{C}+D$ map on the space of configurations, then it is indeed corresponds to the shift on a certain subspace $A$ of $\{0,\ldots, e+r \}^\mathbb{N}$.
When you consider only special initial conditions you consider another subspace $B\subset A$. Then indeed you have to compute the adjacency matrix and (if the shift is irreducible) get information about its largest e.v.
So finally after all I think your idea is right but you need to consider one-sided sequences instead of two-sided. If you did your computation of the matrix and its largest e.v. (which I did not check) then this gives your the right answer.