1
$\begingroup$

Remark : I've found a rather trivial answer for this question and so very likely the premise of paralleling it with the Zsigmondy-theorem is wrong, so this question might better be retracted. I'll give it a certain time loking at reactions on the discussion in the MO-meta site.
Following a hint in META I added a comment where you can vote for deletion.


I'm (recreationally) studying properties of the generalized Collatz-problem ("gCp" for shortness) for $3x+\rho$ and look for help in the question of the existence of cycles in gCp's with composite $\rho$ as parameter in relation to that in gCp's with the primefactors of $\rho$ as parameters.

To explain the problem formally, I'd like to use here the compact "Syracuse"-notation of the gCp in the following way: $$ T_\rho(a;[A]) := {3a+\rho \over 2^A} \qquad \qquad \text{where } A=v_2(3a+\rho) $$ with odd positive integer in the parameter $\rho$. By that definition the transformation $b=T(a;[A])$ works on odd numbers $a,b$ only.
For an iterated transformation of $N$ steps I write simply $a_{N+1}=T_\rho(a_1;[A_1,A_2,...,A_N]) $. I denote $S$ as the sum of exponents $A_k$ and to make the following formulae short, I define $E_{N,S} = [A_1,A_2,...,A_N]$ as the vector for a fixed set of exponents $A_k$ having length $N$ and sum $S$.

I'm interested in the question on existence of cycles in $T_\rho()$ for some composite $\rho$ in relation to that in $T_\varphi()$ where $\varphi$ is a primefactor of $\rho$ .

For example let $\varphi, \sigma \in \Bbb P$ being prime and $\rho=\varphi \cdot \sigma$ being composite. Let some vector $E_1 = E_{N_1,S_1}$ be such that $a_{N+1} = T_\varphi(a_1;E_1)=a_1$ is a cycle.
It is easy to derive that then also $ b_1 = T_\rho(b_1;E_1)$ with $b_1= \rho/\varphi \cdot a_1$ is a cycle in $T_\rho()$.
Of course the analogue is true for a cycle in $T_\sigma()$ in relation to $T_\rho()$ with another set of exponents $E_2 := E_{N_2,S_2}$

  • a) So we have first -by some simple analysis-, that a generalized Collatz-problem gCp with $T_\rho()$ with a composite parameter $\rho$ defines the same cycles as the gCp's with the primefactors of $\rho$ as parameters (where only its elements $a_k$ are rescaled by some factor as said above).

The two interesting parts are the following which I'd like to understand/to prove.

  • b) The gCp of a composite parameter $\rho$ seems to have in general always additional cycles besides that of the gCp's defined by $\rho$'s primefactors.
    This reminds me of the theorem of Zsigmondy about the primefactorization of Mersenne-numbers (where he proves the existence of the then so called "primitive primefactors"). Possibly it can be proven the same way, but I could not follow his proof. Let's call that additional cycles analoguously "primitive cycles".

  • c) Different from the "general" case: when we have the primefactor $\sigma=3$ involved, the composition $\rho=\varphi \cdot 3$ with that primefactor $T_{\varphi \cdot 3}()$ seems to not to lead to such additional "primitive cycles" - again in analogy to Zsigmondy's observation on the Mersenne-number $M_6 = M_{2 \cdot 3}$ which has no "primitive" factors.
    Of course I do such a statement only based on heuristics and the basic assumption that the Collatz-conjecture for $T_1()$ is true and that it has only the "trivial cycle" $1=T_1(1;[2,2,2,...,2])$.

So my question asks for help in proving the observations in b) and c) (under assumption of truth of the Collatz-conjecture). Perhaps it would be sufficient to only get Zsigmondy's derivations transparent enough.


*Remark: for some examples and possibly for a better explanation of the elementary analysis of this you might look at the short treatize [at my webspace][3] (instead of the term "primitive cycles" I've used "unexplained cycles" there)*
$\endgroup$
2
  • $\begingroup$ Poll for opinions: vote this comment if I should delete that question. $\endgroup$ Commented Sep 7, 2017 at 12:44
  • $\begingroup$ On a note related to $3x+p$ you might want to look into convergence of the function $3x+p\cdot2^{v_2(x)}$, when viewed projected down to $\mathbb{N}_2^{\times}$ in the 2-adic space. You will see that the numbers which converge to a cycle of order $1$, in this alternative view of the problem, they are convergent in the 2-adic space while those converging on a cycle of order greater than one, they contain multiple convergent trajectories. Those which rise indefinitely are divergent, and their image in $\mathbb{N}_2^{\times}$ must be divergent too. $\endgroup$ Commented Sep 20, 2017 at 17:49

1 Answer 1

1
$\begingroup$

The answers for b) and c) came out to be trivial and have likely nothing to to with Zsigmondy, so possibly I should retract my question.


For the definition of a cycle for some $a_1=T_\rho(a_1;E_{N,S})$ we have the formula (6.b) in the referred article: $$ a_1 = \rho \cdot { Q(E_{N,S}) \over 2^S - 3^N} \tag 1 $$ Let $\frac uv$ with $v>1$ denote the fully cancelled fraction in (1) then by $$ a_1 = \rho \cdot \frac uv \tag 2 $$ it is obvious that we can have a cycle at $a_1$ only if $v | \rho$. Now let $v$ be prime, then setting $\rho=v$ makes this having an integral solution and thus having a cycle at $a_1$. But of course any composite $\rho = v\cdot t$ makes this having an integral solution (and having a cycle at $t \cdot a_1$) as well. So a gCp with composite $\rho$ has cycles at the same exponentsvector $E_{N,S}$ as all gCp's with parameters being primefactor of $\rho$.

At question b): the "unexplained" (or "primitive") cycles at some $E_{N,S}$ occur if in (2) the denominator is composite and not prime - because in the gCp's with prime parameter the denominator $v$ is not completely cancelled, and thus that gCp's cannot have a cycle at that same $E_{N,S}$
This explains question b).

The question c) is simply explained by the additional fact, that the denominator in (2) cannot have the primefactor $3$ - so for any composite $\rho $ the possession of primefactor $3$ does not introduce any new ("unexplained","primitive") cycle.

So this seems to have nothing to do with Zsigmondy's theorem on mersenne-numbers.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .