I know ultrafilters are considered powerful by more-learned mathematicians than I. I cannot profess to understand the reasons how and why although I can see the power of Zorn's Lemma and the axiom of choice in there.
I'm interested in whether the power of ultrafilters adds anything to this problem:
Let $X=\Bbb Z[\frac1{ a\cdot b\cdot p}]^+$ be the non-negative half of a rational ring extension of $\Bbb Z$ - making it a semiring.
Let an affine prime-quotient function (for the purposes of this question) be any function of the form $f(x)=ax+b\cdot p^{\nu_p(x)}$ where $p$ is some prime and $a,b$ are positive integers coprime with $p$.
For the avoidance of ambiguity I let $f(0)=0$
Then we can define affine prime-quotient filters on $X$ by saying the filter at $x$, denoted $F_x$, is the set of all the numbers $y$ for whom the orbit of $f$ first passes through $y$ and subesquently passes through $x$ (including the case $x=y$).
The non-intersecting ultrafilters of these filters are in 1:1 correspondence with the orbits of $f$ or equivalently, the sets on which $f$ acts transitively.
Question
How many affine prime-quotient ultrafilters does a rational semiring have?
This question can be posed in a few similar ways...
For which $a,b,p$ does $f$ act transitively?
If $f$ does not act transitively, how many orbits does it have? Are they finite in number?
How many non-intersecting ultrafilters are there on $X$?
A slightly weaker question, which is a little bit stronger than Lagarias' periodicity conjecture is: For which $a,b,p$, is every orbit of $f$ eventually periodic?
A special case
A special case of this question is the Collatz conjecture, which requires that the affine prime-quotient ultrafilters for $a=3, b=1, c=2$ fall into three noninstersecting ultrafilters:
- $F_x: x=0,$
- $F_x: x=2^p:p\equiv1\pmod2,$
- $F_x: x=2^p:p\equiv0\pmod2$