This comment reads:
Let $g(x) = x 2^{-v_2(x)}3^{-v_3(x)}$ and $F(x)=x+c$ or any function then $f(x)=g(F(g(x))$ can be seen as a function $\Bbb{Q^*/<2,3> \to Q^*/<2,3>}$. There are no other ways. With $F(x)=x+c$ then replacing $f$ by $c^{-1}f(cx)$ is the same as replacing $c$ by $1$ which is easily understood : the iterates converges when the input is of the form $n/6^r$. Try replacing $6$ by $2$ to see how (with $c=1$) then $f(n/6^r)= \lfloor (n+1)/2\rfloor$. When $x$ is not of the form $n/6^r$ then the binary expansion of $x$ and hence the sequence of iterates of $f$ is periodic.
Can you please rework this for $c=\frac13$ and for the equivalence class or coset $\langle2\rangle$ instead of the one generated by both primes $2$ and $3$, and explain a little better?
- I understand $gFg:\Bbb Q^\times/\langle2,3\rangle\to\Bbb Q^\times/\langle2,3\rangle$ fine.
- I get how $c^{-1}f(cx)=c^{-1}gFgc(x+1)$ so that's what's meant by it's like having $c=1$
- I don't get why this makes iterates converge when the input is of the form $n/6^r$
- However from what I do understand, I tentatively think the method applies directly to the Collatz conjecture, and reworking with the above quotient and value of $c$ will indicate which elements of $\Bbb Q$ converge.
If the $p_j$ are distinct primes and $n$ is a non-negative integer and $$f(n) = g(n+1),\qquad F(n)=h(n+1),\qquad g(n) = n\prod_{j\le J} p_j^{-v_{p_j}(n)}$$ $$ N=\prod_{j\le J} p_j,\quad h(n) = n N^{-v_N(n)},\quad v_N(n) = \max\{ a,N^a|n\}$$ with $f^{k+1}=f\circ f^k$ $$|f^N(n)|\le |F^N(n)| \le \frac{|n|+N}{N}$$ Thus for $m \ge N\ v_N(n)$ $$f^m(n)=1$$ If $n$ is strictly negative then $f^l(n)=0$ for some $l < N \ v_N(n)$ and $g(0)$ is undefined.
When looking at rational input $x$ then the question is if the denominator is a power of $N$, if it is then $f(x)$ is an integer, if it is not then the $N$-decimal expansion of $f(x)$ is periodic and $(f^m(x))_{m\ge 1}$ is periodic.