(Here I discuss the Collatz problem only for positive integers.)

It is possible, by computation, to find all cycles in the Collatz iteration of a fixed length.

It is clear that an increase must be followed by a decrease (for if $n$ is odd, then $3n+1$ is even) and a decrease can be followed by either an increase or decrease (for if $n$ is even, $\frac{n}{2}$ may be odd or even).

Using this, it is easy, for example, to show $4, 2, 1$ is the only cycle of length $3$:

Over the course of a cycle, we must have both increases and decreases. Position an increase at the beginning of the cycle. Then the changes of the cycle must be $IDD$ (where $I$ stands for increase and $D$ stands for decrease), for there is no other way to include an increase and avoid two consecutive increases. Then the cycle consists of $a, 3a+1, \frac{3a+1}{2}$, so that $a = \frac{3a+1}{4}$ and $a = 1$, leading to the familiar $1, 4, 2 = 4, 2, 1$ cycle.

For length $4$, there are two possibilities for a pattern of increase and decrease along a cycle which start with an increase: $IDID$ and $IDDD$.

The $IDID$ possibility leads to $a = \frac{9a+5}{4}$ so $a = -1$.

The $IDDD$ possibility leads to $a = \frac{3a+1}{8}$ so $a = \frac{1}{5}$. Neither of these values of $a$ is a positive integer, so there are no cycles of length $4$.

Likewise, for length $5$, there are only $3$ possibilities: $IDDDD$, $IDIDD$, and $IDDID$. Since last two are equivalent, this leads to $IDDDD$ and $IDIDD$.

$IDDDD$ leads to $a = \frac{3a+1}{16}$ so $a = \frac{1}{13}$.

$IDIDD$ leads to $a = \frac{9a+5}{8}$ so $a = -5$. So there are no cycles of length $5$.

There is more to be said for this way of considering cycles in the Collatz iteration. If a sequence of increases and decreases leads to the equation $a = Ta + U$, then $T$ is positive and $T \neq 1$, since $T = \frac{3^{m}}{2^{n}}$ for some positive integers $m, n$. The positivity of $T$ and the fact that $T \neq 1$ ensure that the solution of $a = Ta + U$ also solves (and therefore is the only solution to) $a = T(Ta + U) + U$ (or the equation for fixed points of higher-multiplicity iterates of the function $f(a) = Ta + U$), so that only cycles that form primitive circular words need to be considered (and $IDID$ was unnecessary, given how it reduces to $ID$). With this noted, it becomes very easy to handle cycles of length $6$:

The number of increases in a cycle is $1$, $2$, or $3$, since no two of them can be consecutive (and since there must be at least one increase). $3$ increases in a cycle can only be realized by $IDIDID$, which is not a primitive circular word. $2$ increases in a cycle can be realized by $IDIDDD$, $IDDIDD$, or $IDDDID$. $IDIDDD$ and $IDDDID$ are equivalent, while $IDDIDD$ is not a primitive circular word. So the only cases to consider are $IDDDDD$ and $IDIDDD$.

$IDDDDD$ leads to $a = \frac{3a+1}{32}$ so $a = \frac{1}{29}$.

$IDIDDD$ leads to $a = \frac{9a+5}{16}$ so $a = \frac{5}{7}$.

So there are no cycles of length $6$.

This is natural and simple enough that someone must have considered it before. Who has, and in what paper(s)? Also, has this been used to settle the question of whether or not there is a cycle in the positive integers larger (both in length, and in member-wise comparison) than the $4, 2, 1$ cycle?

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