# Collatz $2x + 1$ conjecture?

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?

Examples:

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

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

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