I was thinking about odd integers when viewing the collatz conjecture in reverse.
In another question, I presented an argument to show that if $x$ is an odd integer, then it resolves in $n$ steps if and only if $4x+1$ resolves in $n$ steps.
It seems to me that if this is true, then there is an argument to be made that there cannot be a finite number of nontrivial cycles
Let:
- $\nu_2(x)$ be the 2-adic valuation of $x$
- $x_1, x_2, \dots, x_n$ be $n$ distinct odd integers such that:
- $x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$
- $x_1, x_2, \dots, x_n$ be a nontrivial cycle if all $x_i > 1$ and $x_1 = \dfrac{3x_n+1}{2^{\nu_2(3x_n+1)}}$
- $x_1$ resolves in $n$ steps if $\dfrac{3x_n+1}{2^{\nu_2(3x_n + 1)}} = 1$
Here's the argument:
(1) Assume that there are a finite number of nontrivial cycles.
(2) So, let $x_1, x_2, \dots, x_n$ be all the odd integers that are part of any of these nontrivial cycles. So, it may not be assumed that any 2 are part of the same cycle.
(3) Let $a$ be the number of $x_i$ where $x_i \equiv 1 \pmod 4$ and $b$ the number of $x_j$ where $x_j \equiv 3 \pmod 4$
(4) It follows from step (2) that $n = a + b$
(5) Each $4x_k + 1$ must necessarily be part of a nontrivial cycle (otherwise, $x_k$ cannot be in a nontrivial cycle, by the assumption in the previous question)
(6) Then $b=0$ since each odd integer has a 1-to-1 mapping with a unique $4x_j+1$ which means that $n = a$.
(7) But now we have a contradiction because a nontrivial cycle requires at least $1$ odd integer where $x_k \equiv 3 \pmod 4$.
This follows since every odd integer congruent to $1$ modulo $4$ is followed by a smaller odd integer and every odd integer congruent to $3$ modulo $4$ is followed by a larger odd integer. A nontrivial cycle requires both.
Is my reasoning sound? Did I make a mistake?
Edit: Step (5) is incorrect. $4x+1$ will end up at $x$ but without being part of a cycle.