Examples of "eventually reaches y under iteration" other than the Collatz problem

An alternative is $$ x_{k+1} = \left\{ \begin{array} {} x_k /2 & \text{if } x_k \equiv 0 \pmod 2 \\ 3 x_k+1 & \text{if } x_k \equiv 1 \pmod 4 \\ 3 x_k-1 & \text{if } x_k \equiv 3 \pmod 4 \\ \end{array} \right. $$

This is -more than the Collatz-transformation- even guaranteed to run into the trivial cycle for every positive starting number $x_0$

If we write the trajectories displaying only the odd natural numbers and always beginning at odd multiples of 3 (because multiples of 3 cannot occur in the tail of a trajectory) we get $$ \begin{array} {} 3 &\to 1 \\ 9 &\to 7 &\to 5 & \to 1 \\ 15 &\to 11 &\to 1 \\ 21 &\to 1 \\ 27 &\to 5 &\to 1 \\ 33 &\to 25 &\to 19 &\to 7 &\to \text{(see above)}\\ ... \\ \end{array}$$

Tightly related are version like $5x+1$ where the following division is made by $2$ or $3$ depending on the modular residue class, where then apparently everything runs into the trivial cycle. (But well, these are not really qualitatively different ideas)

Let the set $X$ the positive integers which are divisible by $9$. Let the transformation $\sigma$ be the operation of computing the digit sum.
Then taking any element from $X$ and applying repeatedly $\sigma$ gives $y=9$.

A similar recursive algorithm can be based on “good numbers”. Given any Integer n, the sum of the digits of n^2 is the “good number" of n. The relation is surjective (many Integers have the same good number)

    For example:
    2713^2 = 7360369, 
    7+3+6+0+3+6+9 = 34, 
    34 is the good number of 2713.

We can recursively calculate the good number of 34,

    34^2 = 1156,      
    1+1+5+6 = 13,    
    13 is the good number of 34, 

We carry on

    13^2 = 169
    1+6+9 = 16
    16 is the good number of 13

and so on.

So doing we see Integers connected into 3 convergent trees, respectively converging towards three “attractors” which are 1, 9, and the couple 13-16. See graph of the 3 converging trees here

COMPLEMENTARY NOTE

Given an Integer n we know that the sum of its digits S(n) is congruent to n modulo 9

    n ≡ S(n) (modulo 9)  
   (n and S(n) have the same remainder when divided by 9)

So any Integer can be written as n = 9k + r and its square n^2 = (9k + r)^2
or (81k^2 + 18kr + r^2) or (9k’ + r^2)

    The table   r   1   2   3   4   5   6   7   8
              r^2   1   4   9   16  25  36  49  64
        r^2 mod 9   1   4   0   7   7   0   4   1   

    shows that good numbers can only be of the form 
       9k,      9k + 1,     9k + 4,     9k + 7

 The first few are given below:
            9k,         9k + 1,     9k + 4,     9k + 7
    k=0      0           1           4           7
    k=1      9          10          13          16
    k=2     18          19          22          25
    k=3     27          28          31          34
    k=4     36          37          40          43