We first introduce several functions motivated by Kolakoski's sequence. The conjecture itself can be stated independently of Kolakoski's sequence. You can skip straight to the formulation of the conjecture below the line. Kolakoski's sequence is the unique sequence of 1's and 2's which describes its own sequence of block lengths and whose first term is 1. The first few terms are 1, 2, 2, 1, 1, 2, 1....

Notation: we'll write sequences as strings. We use a sequence of length 1 and a single term interchangeably. The concatenation of sequences $s$ and $t$ is $st.$

Let $m, n$ be different positive integers. $\{m, n\}^\omega$ is the set of finite nonempty sequences whose terms are all $m$ or $n.$ We define $E_{m, n}: \{m, n\}^\omega \times \{m, n\}^\omega \rightarrow \{m, n\}^\omega$ recursively as follows.

If $t$ has length $1$, then $E_{m, n}(s, t)$ is the unique finite sequence whose terms are all $m$ or $n$, whose first term is the term in $t$, and whose sequence of block lengths is $s.$ For example, $E_{2, 3}(233, 2)=22333222.$

If $t$ has length greater than $1,$ then suppose that $t=yu,$ where $y$ is a single term (the first term), $u$ is the rest of the sequence, and $yu$ is the concatenation. Then $E_{m, n}(s, yu) = E_{m, n}(E_{m, n}(s, y), u).$

One can show that for finite sequences $s_1, s_2, t_1,$ there exists a unique sequence $t_2$ of the same length as $t_1$ (all in $\{m, n\}^\omega$) such that $E_{m, n}(s_1s_2, t_1) = E_{m, n}(s_1, t_1)E_{m, n}(s_2, t_2).$ Furthermore, $t_2$ depends only on $s_1, t_1.$ We define the function $C_{m, n}$ such that $C_{m, n}(s_1, t_1) = t_2.$

Another way to define $C_{m, n}: \{m, n\} \times \{m, n\}^\omega \rightarrow \{m, n\}^\omega$, where $m, n$ are positive integers.

We've restricted the first argument to $\{m, n\}$ because that's all that's necessary for the conjecture. Aside: in general, it makes sense for the first argument to be a finite sequence. See the above paragraph. One can show that $C_{m, n}(s_1 s_2, t_1) = C_{m, n}(s_2,C_{m, n}(s_1, t_1)).$

For $x, t$ in $\{m, n\},$ we define $C_{m, n}(x, t)$ to be $\bar{t},$ where $\bar{t}$ is the value in $\{m, n\}$ not equal to $t.$

For $x, y$ in $\{m, n\}$ and $u$ in $\{m, n\}^\omega,$ we define $f(s):=C_{m, n}(y, s)$ and $C_{m, n}(x, yu):=\bar{y}f^x(u),$ where the superscript denotes iteration.

For example, $C_{1, 2}(1, 222) = 1 C_{1, 2}(2, 22) = 11\left(C_{1, 2}(2, C_{1, 2}(2, 2)) \right) = 112.$

Some more examples: $C_{1, 2}(1, -)$ sends $111, 112, 121, 122, 211, 212, 221, 222$ to $222, 221, 212, 211, 121, 122, 111, 112$ respectively, and $C_{1, 2}(2, -)$ sends $111, 112, 121, 122, 211, 212, 221, 222$ to $211, 212, 221, 222, 111, 112, 121, 122$ respectively.

Observe that $C_{m, n}(x, -),$ where $x$ is fixed, is bijective on $\{m, n\}^\omega,$ and it fixes lengths. Here is the conjecture:

Conjecture. Let $n, j$ be positive integers with $n$ even. Let $1^j$ be the sequence of $j$ terms which are all $1.$ Then the size of the orbit of $1^j$ under the function $C_{m, n}(1, -)$ equals $2^{\lfloor(j+1)/2\rfloor}.$

One can show that the following conjecture is equivalent: Let $n, j$ be positive integers with $n$ with $j$ even. Then the sequence $E_{1, n}(1^{2^j}, 1^{2j})$ has odd length. $(*)$

We can also define $C_{m, n}: \{m, n\} \times \{m, n\}^\omega \rightarrow \{m, n\}^\omega$ where $m, n$ are any integers. Well, we define it exactly as above. A negative power means involves taking the inverse of the function. To do this, we need the functions $C_{m, n}(x, -)$ to be bijective and length preserving. These properties are easy to show.

Here is another conjecture.

Conjecture. Let $n, j$ be positive integers with $n$ even. Let $s_1, s_2$ be the two sequences of length $j$ which terms alternate between $-1$ and $n.$ Then the size of the orbits of $s_1$ and $s_2$ under the function $C_{-1, n}(-1,-)$ both equal $2^{\lfloor (k+1)/2 \rfloor}.$

Empirical evidence: Using C++ and a fast routine to exponentiate maps as well as some elementary observations, I proved both conjectures for all $j \le 26$ and all even integers $n.$ Using the equivalent form $(*),$ I proved the conjecture for the cases $n=2$ and $j \le 46.$

Remarks. One can easily show that the sizes of the orbit of an arbitrary string of length $j$ under the function $C_{m, n}(s, -)$ for an arbitrary string $s$ is at most $2^{\lfloor(j+1)/2\rfloor}.$ Thus the sizes of the orbits which we've stated are in some sense the maximum possible.

Note that this conjecture is not true when replacing the strings $1^j, s_1, s_2$ with arbitrary strings. For arbitrary strings, I've empirically found that the lengths of strings related to $E_{1, n}(1^{2^j}, 1^{2j})$ seem to be randomly even or odd, whereas they are always odd for the cases in the conjecture. Because I don't understand this conjecture well, it sort of looks like there are several hundred coin flips landing the same way. I also suspect that understanding these conjectures will involve fields outside combinatorics, but I have no idea which.

This was first observed by Yongyi Chen