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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.

Dan Brumleve
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    Why the statement is not $P(0) \land P(1) \land (\forall x: P(3 x + 2) \implies P(2 x + 1)) \land (\forall x: P(\mathbf{2 x}) \implies P(\mathbf{x}))$ ? Is it equivalent? For the distance, it seems to me that for most (not for all) elements $x$ : $P^x(x) = 1$ or $2$. In this graph you can see some exceptions: http://www.opimedia.be/3nP1/T_100.pdf . (I made this few years ago, so I don't remember details.) – Olivier Pirson Aug 12 '15 at 22:05
  • I think you are misunderstanding a few different things. If you reverse the implication arrow in the last clause, it is not equivalent. $P(x)$ denotes an arbitrary predicate, not the Collatz function, so it doesn't make sense to say $P^{x}(x)$. And I am not concerned here about bounds on stopping time anyway, I am just looking for an example of of a statement that can be proved this way. – Dan Brumleve Aug 13 '15 at 21:34
  • Yes, I don't understand. Have you references to this equivalence result? – Olivier Pirson Aug 13 '15 at 22:14
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    It is like ordinary induction but instead of starting at $1$ and counting up, start at $1$ and follow the Collatz function backwards. If the Collatz conjecture is true then we can reach every number that way, and if the truth of the predicate at each step is implied by the the previous step then the predicate is universally true. I'm not sure how to make it any more clear. – Dan Brumleve Aug 13 '15 at 22:21
  • For the other direction, note that if the Collatz conjecture fails we can construct a not-universally-true predicate that still satisfies the induction conditions (just define it to be false on a non-trivial cycle or non-terminating chain and true elsewhere). But that doesn't mean that _every_ statement provable this way implies the CC. – Dan Brumleve Aug 13 '15 at 22:41
  • What is the meaning of $P$? $P(x)$ means that $x$ verifies the conjecture. No? $P(0)$, $P(1)$, $\forall x: P(3 x + 2) \implies P(2 x + 1)$ and $\forall x: P(x) \implies P(2 x)$ are trivially true. So your formulation is **not** equivalent to the conjecture. No? Moreover, why $P(0)$? – Olivier Pirson Aug 14 '15 at 12:04
  • I think there is a mistake in your condition, shouldn't it be $P(3 (2x+1) +1) \Rightarrow P(2x+1)$ in the first part? As for an answer, it is probably possible to prove some simple proof by induction exercise using this, like $\sum_{i=0}^n i = n(n+1)/2$, but that will probably fall under trivial result. (I just tried, it works, but it is a really overly complicated approach) Also it does not really use any structure of the conjecture. – mlk Aug 16 '15 at 19:12
  • @mlk that version works just as well (since $P(3 \cdot x + 2) \implies P(6 \cdot x + 4)$ by the other condition and $3 \cdot (2 \cdot x + 1) + 1 = 6 \cdot x + 4$). I'm hoping for something where the proof is even simpler than with ordinary induction, even if the result itself is not very interesting. One idea I've been playing with since posting this is "any set of $x$ numbers can be ordered into a sequence $y_n$ such that if $a, b, c$ are in arithmetic progression then $y_a, y_b, y_c$ are not in arithmetic progression.". With this statement we can show $P(x) \implies P(2 \cdot x)$ [cont] – Dan Brumleve Aug 17 '15 at 04:53
  • ... (and ordinarily we would finish the proof with $P(x) \implies P(2 \cdot x + 1)$) but there doesn't seem to be any way to make it work with the other condition. If we change the statement to the stronger (and dubious) "any set of $x$ numbers can be ordered into a sequence $y_n$ such that if $a \lt b \lt c$ then $y_a, y_b, y_c$ are not in arithmetic progression" then $P(3 \cdot x + 2) \implies P(2 \cdot x + 1)$ is straightforward but $P(x) \implies P(2 \cdot x)$ becomes difficult. I wonder if there is some in-between statement that can work? – Dan Brumleve Aug 17 '15 at 04:54

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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.

Dan Brumleve
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