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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.

  • $\begingroup$ sorry, I now see you made six definitions. It's clear now. Thanks for this. $\endgroup$ – samerivertwice Dec 26 '19 at 20:11
  • $\begingroup$ $h$ just factors out the same powers of every prime $\leq J$ until at least one of the primes is not a factor, right? $\endgroup$ – samerivertwice Dec 27 '19 at 6:58
  • $\begingroup$ I'm still digesting this by the way. $\endgroup$ – samerivertwice Dec 27 '19 at 16:40
  • $\begingroup$ Is $N=p_J\#$ or are you permitting incomplete sets of primes less than or equal to $J$? $\endgroup$ – samerivertwice Dec 28 '19 at 14:01
  • $\begingroup$ 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? $\endgroup$ – samerivertwice Dec 28 '19 at 16:03

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