Mysteries of Wolfram's rule 18 [Unfortunately, I made some mistakes in my original question. I tacitly corrected them wherever I found them.]

Wolfram's rule 18 gives rise to fractal patterns, but when started with two black cells on $\mathbb{Z}/k$ the pattern necessarily evolves into a limit cycle.
The limit cycle for given $k$ is determined by

*

*the number $\tau$ of steps it takes to reach it, i.e. to arrive for the first time at a pattern $\pi$ that will repeat,


*the number $\lambda$ of steps it takes until $\pi$ appears for the second time, and


*a number $\sigma$ of cells by which $\pi$ is possibly shifted after $\lambda$ steps.
The length of the limit cycle is given by $L = k / \text{gcd}(k,\sigma) \cdot \lambda$.
Because rule 18 is totalistic, i.e. left-right symmetric, $\sigma$ can only be $0$ (necessarily when $k$ is odd) or $k/2$. I determined $\sigma$, $\lambda$, and $\tau$ for all $k \leq 70$ and some selected ones up to $526$, making up a set $K$, and made the following observations:

*

*$(\forall k\in K)\ \  \sigma(k) = 0 \implies k \text { is odd } \vee (\exists n)\ k = 2\cdot(2^n-1)$


*$(\forall k\in K)\ \ k \text { is even } \implies \tau(k) = \tau(k+1)$


*$(\forall k\in K)\ \ k \text { is even } \implies L(k) = L(k+1)$


*$(\forall k\in K) \ (\exists n)\ (\exists m)\ \  \lambda(k) = 2^n(2^m - 1)$


*$(\forall k\in K)\ \ k = 2^{n+1} \implies \lambda(k) = 2^{n} - 1$


*$(\forall k\in K)\ \ k = 2^{n+1} - 4 \implies \lambda(k) = \lambda(k) = 2^{n} - 1$
Beyond these parameter-free findings I found something more (as a straight line in the logarithmic $\sigma$-$\lambda$ plot):

*

*$(\forall k\in K) \ \ k = 4\cdot n \implies \lambda(k) = 2^{n} - 1$for $n\in\{2,3,5,6,9,11,14,\dots\}$
Is there a general approach to prove the parameter-free statements for arbitrary $k$, or has each of the statements to be proved by own means? If there is a general approach: Could you please sketch it?
And what about the parameterized statement: Is there a chance to prove it (in case it can be generalized)? I have no idea how the sequence would proceed. At least there are some OEIS sequences containing $2,3,5,6,9,11,14,…$
The straight line mentioned above together with some pairs $k=2^{n}$ and $k=2^{n}-4$ can be seen in the $\sigma$-$\lambda$ plot:

Drop a comment if you want to verify the findings, I can provide data.
 A: Hopefully I guessed correctly that your two $1$s are next to each other, and your initial pattern is $110^{k-2}$ (considered periodically).
I'll do the "kink-elimination", to reduce the analysis of your initial configurations to problems about spacetime diagrams of
ECA $90$, which is linear with local rule $F'(a, b, c) = a \oplus c$. The initial configurations are roughly as simple, namely $101$ surrounded by zeroes, and I increase $k$ by $1$ or $2$.
The point is that questions about linear rules should be tractable, even if tedious. I did not attempt to solve the specific ones that appear, and I did not look too hard at your specific findings. Some look like they are obvious from this analysis, some less so.
Linear means linear with respect to the natural $\mathbb{F}_2$-vector space structure of $\mathbb{F}_2^{\mathbb{Z}}$. The cellular automaton $90$ multiplies the formal Laurent series $\sum_{k = -\infty}^\infty a_k x^k$ ($a_k \in \mathbb{F}_2$) by $x + 1/x$. For cyclic configurations, you can directly write a matrix over $\mathbb{F}_2$ for it.
Rule $18$, which I'll call $f$, has local rule
$$
F(a, b, c) = \left\{ \begin{array}{ll}
0 & \mbox{ if } b = 1 \\
a \oplus c & \mbox{ if } b = 0 \\
\end{array} \right. $$
where $\oplus$ is addition mod $2$. I'll refer to Rule $90$ as $g$.
Suppose first that $k$ is odd. Then an easy induction shows that the $f$-evolution of
$A = 10^{k-2}1$ on $\mathbb{Z}/k$ is the same as the $f$-evolution of
$B = 10^{k-2}10$ on $\mathbb{Z}/(k+1)$, in the sense that apart from the last cell of $B$, every cell contains the same bit, and the last cell in $B$ will always contain $0$.
($A$ is your initial configuration shifted, I have split the $11$ just for indexing purposes.)
The fact that the last bit stays $0$ is immediate from left-right symmetry of the
configuration around this cell, and the fact $18$ maps left-right symmetric patterns to $0$
and preserves left-right symmetry.
Suppose then that $f^i(A) = abuba$ and $f^i(B) = abuba0$ (they look like this again by symmetry).
If $a = 1$, then $f^{i+1}(A) = 0cvc0$ for some $c \in \{0,1\}, v \in \{0,1\}^{k-4}$, and $f^{i+1}(B) = 0cvc00$.
If $a = 0$, then $f^{i+1}(A) = bcvcb$ for some $c \in \{0,1\}, v \in \{0,1\}^{k-4}$, and $f^{i+1}(B) = bcvcb0$
concluding the induction.
Now we observe that that both gaps in $B = 10^{k-2}10$ have odd length. We say a kink is a word of the form $10^{2n}1$. It is easy to show by induction that kinks are never introduced by $f$, namely if a configuration (considered on $\mathbb{Z}$) contains no kinks, then its $1$s lie on at most one $2$-periodic arithmetic progression, and in the image, they have to lie in the other one.
It is also immediate that when there are no kinks, $f$ behaves exactly like $g$. Namely, the only situation where $f$ and $g$ differ is when $g$ maps a cell with $1$ to $1$. This can only happen in the presense of the specific kink $11$.
Thus, $f^i(B) = g^i(B)$, and $f^i(A)$ is obtained by removing its last bit. Thus, we have completely reduced the behavior for odd $k$ into the behavior of the linear Rule 90.
Suppose then that $k$ is even. Now $A = 10^{k-2}1$ has two kinks, one at the boundary, one at the middle. Now we can eliminate the kinks by letting $B = 10^{k-1}10$. Note that $B$ has $0$ at $k/2$ ($0$-indexing the positions) and at the end, and $A$ is obtained from $B$ by dropping these $0$s. This will hold by induction: the bits stay $0$ by symmetry of the configuration around them, and otherwise the calculation is the same as above.
All in all, using a bit of symmetry, we have obtained that for odd $k$,
$$ f^i(10^{k-2}1) = D_0(g^i(010^{k-2}1)) $$
for all $i$, where $D_j$ is the obvious ``drop $i$th symbol'' operator (and the dropped symbol is always $0$), and
for even $k$,
$$ f^i(10^{k-2}1) = D_{k/2}(D_0(g^i(010^{k-1}1))) $$
for all $i$ (and the dropped symbols are always $0$).
Next, we observe that $\tau$ is not modified when stepping from the analysis of $f^i(A)$ to the analysis of
$g^i(B)$, since the definition of $\tau$ is the same whether or not we consider configurations up to a shift
(and since we only ever drop cells that contain $0$).
Now let's analyze $\lambda$ and $\sigma$. If $k$ is odd, as you mention $f$ cannot give nontrivial $\sigma$.
We may see nontrivial $\sigma$ in the $g$-evolution, for example for $k = 5$ we see.
$g(101000) = 000101$ so $\tau = 0, \lambda = 1, \sigma = 3$.
So when analyzing odd $k$, we can just remember that for $f$ we will have $\sigma = 0$, and $\lambda$ is determined by the time we have actual recurrence in $g^i(B)$, not just up to a shift.
If $k$ is even, a simple symmetry consideration (the dropped pair of zero-cells are distance $k/2+1$ apart, and the configuration stays symmetric around them) shows that we will see $\sigma = k/2+1$ for $g$ if and only if
we see $\sigma = k/2$.
So as claimed, we may analyze all of $\tau, \lambda, \sigma$ by analyzing spacetime diagrams of Rule $90$.
