How does $3n+1$ change the proximity of $n$ to a power of two?

Question

This is part of an attempt to prove Collatz's conjecture. I proved a modification of Collatz's conjecture, where instead of $3n+1$ if $n$ is odd, you do $n+1$. In Collatz's conjecture, if you get to a power of two, you're going to get to one. Let $p$ represent the proximity to a power of two. We shall keep the value of $p$ as $⌈\log_2(n)⌉-\log_2(n)$. Where $p=0$, $n$ must be a power of two. With my $n+1$ version, $p$ decreases if $n$ is odd, obviously and will stay the same if $n$ divides by two. However, is there a way to explain how $p$ decreases if $n$ becomes $3n+1$?

You might consider studying how the $3n+1$ map behaves modulo $2^k$ for each $k$ and each odd residue class. — Justin Benfield, Jun 26 '16 at 10:38
Please could you explain that using more noob-friendly maths? I'm only fourteen. :-D — 0WJYxW9FMN, Jun 26 '16 at 10:41
Are you familiar with modular arithmetic? (e.g. https://en.wikipedia.org/wiki/Modular_arithmetic). — Justin Benfield, Jun 26 '16 at 10:42
The remainders are also known as residue classes, so consider what happens to $1,3,5,7,$,etc. when you apply $3n+1$ and compute the remainder modulo $2^k$. For instance, if your number was congruent to $7 \operatorname{mod} 8$, then after applying $3n+1$ is will be congruent to $3(7)+1=21+1=22\equiv 6 \operatorname{mod} 8$. — Justin Benfield, Jun 26 '16 at 10:50
So is $k$ any integer? I'm not sure how what you're saying would help. — 0WJYxW9FMN, Jun 26 '16 at 10:53
Any natural number (negatives (and 0) don't make sense here). The idea is to try and determine how many times you will be able to divide by 2 after taking the $3n+1$ step. — Justin Benfield, Jun 26 '16 at 10:54
Ah. Thanks - that's a great idea! How would that help work out a value for $p$, though? (Or is $p$ too 'chaotic' to work out in this case?) I'll give you credit if I ever publish a proof of Collatz's conjecture based on this. — 0WJYxW9FMN, Jun 26 '16 at 10:58
I am, are you wanting to chat further about this? (chat would be better than loading up comments here) — Justin Benfield, Jun 26 '16 at 11:13
Not sure about the "proximity" approach, but in comparing the $3n + 1$ and $n+1$ algorithms, you've hit upon a major reason the Collatz conjecture is so elusive to prove. $\frac{n+1}{2} < n$ for all natural $n > 1$, so the "pseudo-Collatz" with $n+1$ is trivially easy to show by strong induction. In contrast, $\frac{3n+1}{2} > n$ for all natural $n$. So the odd numbers keep mapping to higher even numbers, which makes strong induction (from this approach) impossible. Anyway, the Collatz is a nice and accessible problem for enthusiastic amateurs, I wish you the best of luck. — Deepak, Jun 26 '16 at 11:17
@JustinBenfield Yeah, that would be a good idea - if you read Deepak's comment, he could be right, so I'm trying to find a different approach. — 0WJYxW9FMN, Jun 26 '16 at 11:26
@JustinBenfield I'll have to go in a minute, but if you're online later, I'll move this discussion to the chat. — 0WJYxW9FMN, Jun 26 '16 at 11:27
Did you want to write $\log_2$ instead of $\log2$? (You typeset the first as \log_2.) — Martin Sleziak, Jun 26 '16 at 11:34
You may want to use $\min{\lceil\log_2n\rceil-\log_2n,\log_2n-\lfloor\log_2n\rfloor}$ instead since it could happen that you are closer to the next lower power of $2$ rather than next highest. — Justin Benfield, Jun 26 '16 at 11:50
@J843136028 You'll find this thread interesting, I think: http://math.stackexchange.com/questions/14569/the-5n1-problem In particular, the first answer gives an illuminating heuristic. Also see the linked thread in the comment for that answer. — Deepak, Jun 26 '16 at 13:02

How does $3n+1$ change the proximity of $n$ to a power of two?

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