# Rework this statement to the case $c=\frac13$ and $\sim=\langle2\rangle$

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.
• $h$ just factors out the same powers of every prime $\leq J$ until at least one of the primes is not a factor, right? – samerivertwice Dec 27 '19 at 6:58
• Is $N=p_J\#$ or are you permitting incomplete sets of primes less than or equal to $J$? – samerivertwice Dec 28 '19 at 14:01
• You start saying $n$ to be non-negative, then use $\lvert n\rvert$. So I'm just checking $\lvert n\rvert= n$ if $n$ is positive and $-n$ if $n$ is negative - is that correct? – samerivertwice Dec 28 '19 at 16:03