Do we know of any Collatz theorem involving similar functions. For example what do we know about iterations of:

$$ f(x) = \begin{cases} \dfrac{x + 1}{2} \text{, if } x \text{ is odd}. \\ 2x + 1, \text{ if } x \text{ is even}.\end{cases} $$

Since this one doesn't involve $3$ maybe it is easier to solve? Why isn't this one the lowest hanging fruit rather than the $3x +1$ problem?


$$ f(1) = 1 \\ f(2) = 5 \to 3 \to 2 \to \dots \\ f(4) = 9 \to 5 \dots \\ f(6) = 13 \to 7 \to 4 \to \dots $$

So one of the finite number of conjectured cycles would be $5 \to 3 \to 2$.

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    $\begingroup$ I think this can be solved by analysing the binary expansion...? $\endgroup$ – Trebor Mar 26 at 2:08
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    $\begingroup$ Why $\pm$? If it’s just minus, the two things are mutually inverse, and all numbers pair off. I guess you just want +? $\endgroup$ – rschwieb Mar 26 at 2:10
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    $\begingroup$ You might want to check your definition here. The $-$ sign obviously oscillates (e.g. 4, 9, 4, 9, ...) and the $+$ sign just increases by 1 each time (e.g. 4, 9, 5, 11, ...). $\endgroup$ – eyeballfrog Mar 26 at 2:10
  • $\begingroup$ @eyballfrog, it's $+$ yes. It has finite cycles just like Collatz. $4, 9, 5, 11, 6, 13, 7, 4$. 🍌😹 $\endgroup$ – Shine On You Crazy Diamond Mar 26 at 2:12
  • $\begingroup$ Perhaps this answer of mine (image) is interesting: mathoverflow.net/a/200126/7710 It looks at the generalization of $3x+1$ to $mx+1$ $\endgroup$ – Gottfried Helms Mar 26 at 8:30

The Collatz conjecture is interesting because it is hard. If you use the $-1$ version, every even $x$ goes to $2x+1$, which goes back to $x$, so we can just say every pair $(2k, 4k+1)$ is a cycle including $(0,1)$ and be done.

Even for the $+1$ version, every doubling is followed by a divide by $2$, which might be followed by another divide by $2$, so numbers can't grow much beyond twice the start. There is a cycle $(2,5,3)$ which everything except $1$ seems to fall into. We can prove this by considering what happens to numbers of the form $4k$ and $4k+2$. A number of the form $4k$ goes to $8k+1$, which then goes to $4k+1$. A number of the form $4k+1$ goes to $2k+1$ which then goes to $k+1$, so any large number gets divided by $2$ in a few steps. We can expand this to show every number except $1$ goes to the $(2,5,3)$ cycle.

The multiplication by $3$ in the Collatz conjecture is just right to balance the up steps and the down steps if you consider the chances of odd numbers in a row. If you increase the multiplier most numbers will go off to infinity. If you decrease it, all numbers fall into a small cycle and we can prove that. Chaos lives on the boundary between two or more well behaved regions.


If $x$ is even, then $f(x)=2x+1$ is odd. So $$f(f(x))=\frac{(2x+1)+1}{2}= x+1,$$ which is also odd. So $$ f(f(f(x))) = \frac{(x+1)+1}{2} =\frac{x}{2}+1 \leq x.$$ Thus no matter wheter $x$ is even or odd at the beginning, the number will always become smaller after a couple of iterations. Thus there is no number $x$ for which the series tends to infinity.


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