Among the Collatz conjecture we have other "similar" problems that are solved and have repeating cycles.
$5n+1$ has the repeating cycle $13, 66, 33, 166, 83, 416, 208, 104, 52, 26$, with a length of $10$
$3n-1$ has the repeating cycle $5,14,7,20,10$
If there is a repeating cycle, must there its smallest length be aprox. of size $2k$, where the algorithm is $kn\pm1$ for odd numbers?
Also, are the following diagrams meaningful?
First row of "Collatz" number diagram can be seen as N, second as 2N-1, third as 3N-1
If you look at diagram a), you can see that positional value y betwen 2x in forth row and x in third row is given by formula 4y-1
2 in column 1 => 4 in column 3
5 in column 3 => 10 in column 11
.
.
.
14 in column 5 => 28 in column 19
Another very naive, probably stupid but do not care subquestion? Finding some solution for $(3n-1)/2$ is the same as solving system of equations with 2 unknowns.
Look at last diagram. Can there be some generality written?
Best Answer
If you take $k=2^m-1$, then the $kn+1$ problem maps $$1 \mapsto 2^m \mapsto 2^{m-1} \mapsto \cdots \mapsto 1.$$ In this case, the cycle length is $O(\log k)$. This is significantly different to $2k$. (A similar construction works for the $kn-1$ problem.)