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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.

Larry Freeman
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    '(5): Each $4x_k+1$ must necessarily be part of a nontrivial cycle' seems wrong to me. $4x_k+1$ must lead to a nontrivial cycle, but that doesn't imply that it's part of a cycle; it could be on the 'tail of the rho' as it were, part of the chain leading into the ultimate cycle. – Steven Stadnicki Feb 21 '21 at 22:12
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    @StevenStadnicki In fact, not only could that happen, it must happen. By definition of cycle, we have $$x_k=c_1\mapsto c_2\mapsto\ldots\mapsto c_r\mapsto c_1,$$ where all $c_i$ are distinct. But also $$4x_k+1\mapsto c_2\mapsto\ldots\mapsto c_r\mapsto c_1\mapsto c_2\mapsto\ldots.$$ So $4x_k+1$ leads to a cycle but is not in a cycle. For example, the trivial cycle is $1\mapsto1\mapsto1\mapsto\ldots$. We also have $5\mapsto1$, $21\mapsto1$, $85\mapsto1$, etc., but none of $5,\ 21,\ 85,\ \ldots$ are in a cycle. – Will Orrick Feb 22 '21 at 00:36
  • @WillOrrick I am not clear. If $x$ is part of a nontrivial cycle, then it does not resolve in any steps. If $4x+1$ is not part of any nontrivial cycle and it does not resolve in any steps, then does it follow that it must be part of a never-ending sequence? Is that the point? That's a great point. I hadn't considered never-ending sequences. Did I understand correctly? – Larry Freeman Feb 22 '21 at 02:00
  • @StevenStadnicki I am not clear what you mean by 'tail of the rho' or 'the ultimate cycle'. Could you add more details? – Larry Freeman Feb 22 '21 at 02:01
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    Not quite. Whether $x$ resolves or not is not so important: if $x$ reaches $1$ in $n$ steps then $4x+1$ reaches $1$ in $n$ steps; if $x$ reaches $c_r$ (an element of a nontrivial cycle) in $n$ steps then $4x+1$ reaches $c_r$ in $n$ steps; if $x$ reaches $z$ (an element of a never-ending sequence) in $n$ steps than $4x+1$ reaches $z$ in $n$ steps. The point is that you assumed $x_k$ to be part of a nontrivial cycle. If $x_k$ comes back to $x_k$ after $r$ steps then $4x_k+1$ reaches $x_k$ after $r$ steps. Importantly, $4x_k+1$ never comes back to itself; it is not in a cycle. – Will Orrick Feb 22 '21 at 02:19
  • @WillOrrick Thanks! Got it. These are great points! I suspected that there was a mistake in the reasoning. – Larry Freeman Feb 22 '21 at 02:48

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