The Collatz conjecture is equivalent to the following "induction principle":
If $P(0) \land P(1) \land (\forall{x} P(3 \cdot x + 2) \implies P(2 \cdot x + 1)) \land (\forall x P(x) \implies P(2 \cdot x))$,
then $\forall x P(x)$.
I am wondering if there are any statements that can be proved using this principle that are not obvious and are not obviously equivalent to the Collatz conjecture itself? I'm not so much interested in open problems (implications of the Collatz conjecture), rather something that may be easily provable a different way but also has a simple proof using "Collatz induction".
I have tried some statements $P(x)$ like "there exists a number of the form $2^a \cdot 3^b$ within a distance $f(x)$ from $x$" but I can't quite make this work.
Here is the best I'm able to do so far, following up on my idea in the comments:
Let $P(x)$ be the claim that any set of $x$ integers can be ordered in a sequence $s_i$ with $1 \le i \le x$ such that if $a \lt b \lt c$, then $s_a + s_b \ne s_c$.
Suppose $S$ is a set containing $2 \cdot x$ integers. We have $S = V \cup W$ where $V$ contains the $x$ smallest elements of $S$ and $W$ contains the $x$ largest elements of $S$. If $V$ and $W$ each correspond to a sequence with the property then we can make a sequence for $S$ by starting with $W$'s sequence and appending $V$'s sequence to it. So $\forall x P(x) \implies P(2 \cdot x)$.
If a set can be ordered in such a way so can all of its subsets: to find a sequence for a set after having removed some elements, simply remove those same elements from the original set's sequence. So if $y < x$ then $P(x) \implies P(y)$, and in particular this gives $\forall x P(3 \cdot x + 2) \implies P(2 \cdot x + 1)$.
Therefore if the Collatz conjecture is true, $\forall x P(x)$.
Now this is disappointing because it is just a long-winded way of saying sort the set from highest to lowest, also it proves much more than $P(3 \cdot x + 2) \implies P(2 \cdot x + 1)$. Maybe a starting point for a better example.