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"Rule 150" is a fascinating one-dimensional and simple cellular automaton giving raise to some quite chaotic behaviour. This is the starting point of this question.

Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to ${\sf XOR}$ in computer science).

Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i) + x(i-1) + x(i+1) \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.

Question. For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?

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    $\begingroup$ This cellular automaton simply amounts to multiplying the formal Laurent series $\sum_{k=\infty}^\infty a_kx^k$ (over the field $F_2$)by $1+x+1/x.$ We observe that $(1+x+1/x)^{2^n}=1+x^{2^n}+x^{-2^n}$. $\endgroup$ Commented Aug 29, 2022 at 22:17
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    $\begingroup$ This is not rule 30. $\endgroup$
    – Ville Salo
    Commented Aug 30, 2022 at 6:23
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    $\begingroup$ (To save you the trouble: you are talking about rule 150.) $\endgroup$
    – Ville Salo
    Commented Aug 30, 2022 at 6:56
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    $\begingroup$ Wolfram notation is rubbish. $\endgroup$ Commented Aug 30, 2022 at 16:59
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    $\begingroup$ I will write a new question with the correct cellular automaton $\endgroup$ Commented Aug 31, 2022 at 14:05

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As already mentioned, this cellular automaton is Rule 150.

Rule 150 is an example of a bipermutive cellular automaton, meaning that it is both left permutive (the value of $f(x)(i)$ can be permuted by permuting the value of $x(i-1)$) and right permutive (the value of $f(x)(i)$ can be permuted by permuting the value of $x(i+1)$). Due to bipermutivity, by Theorem 6.3 in https://core.ac.uk/download/pdf/236376428.pdf Rule 150 is conjugate to the shift map $\sigma$ on $\{0,1,2,3\}^\mathbb{N}$ defined by $\sigma(x)(i)=x(i+1)$ for $x\in\{0,1,2,3\}^\mathbb{N}$, $i\in\mathbb{N}$ (conjugacy means that there is a bijection $\phi:\{0,1\}^\mathbb{Z}\to \{0,1,2,3\}^\mathbb{N}$ such that $\phi\circ f = \sigma\circ \phi$). For all integers there is a sequence $x\in\{0,1,2,3\}^\mathbb{N}$ such that $\sigma^{(n)}(x)=x$ but $\sigma^{(k)}(x)\neq x$ for $k<n$, so due to conjugacy the same holds for the function $f$.

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