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About the Collatz Conjecture:

Every body looks at it from "leaves to root" - to use the tree analogy.

I have another approach. My approach is to look at it from the root - the number 1 - and see if, by means of the inverse algorithm functions, I can generate all Counting Numbers.

Here I will wait for a first kind response to this: why is it t that the simple fact that lines 2n and (n-1)/3 - the inverse functions of the algorithm - are obviously continuous and all integers (I mean (n, m) points with m,n integers) belong to one or another isn't enough as a proof?

functions: all integers belong to one or another

(note that the generated numbers, by 2n function,: 2,4,6,8,10... plus division generates all naturals: 2,6/2,4,10/2,...)

all integers in those lines are hailstorm numbers. : their big bang like origin - a single point - guarantees that.

So, what I seek is to understand what is to prove!

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  • $\begingroup$ as you said it at the beginning of your post, you have to prove that if you begin with a random integer you will finally end at 1. $\endgroup$ – fonfonx May 26 '17 at 2:33
  • $\begingroup$ There's a lot of people who looked at the collatz-problem this way. A somehow prominent one (Peter Schorer) thinks that he even has proven the conjecture using the reverse tree and offers a handful of dollars if you find and point to some error/flaw. You can also look at my early fiddlings at go.helms-net.de/math/collatz/aboutloop/collloopintro_main.htm and especially open the subpage about "textual" and "graphical" trees. $\endgroup$ – Gottfried Helms May 26 '17 at 5:50
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Yes, a line like $2n$ is contiguous, but that's not what we're doing.

We're applying the mapping $n \rightarrow 2n$. It's important to realize that the input domain is $\mathbb{N}$. We can never get $3$ with this mapping for example, as there is no integer that we can double to get 3.

Neither mapping $n \rightarrow 2n$ or $n \rightarrow \frac{n-1}{3}$ covers every integer by itself. The interesting part of the Collatz conjecture is that we can combine and alternate these two mappings to construct integers. But it's incredibly hard to prove that this combination can reach all integers.

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Let's define a Collatz sequence recursively, take $a_0$ to be any positive integer as a starting point and say

$$a_{n+1} = \begin{cases} \dfrac{3a_n+1}{2} & \text{$a_n$ odd}\\ \dfrac{a_n}{2} & \text{$a_n$ even} \end{cases} $$ If $a_n\equiv 0,1 \mod 3$, there is no problem in traversing the Collatz sequence in reverse. Take $a_{n-1} = 2a_n$. The problem is when $a_n\equiv 2 \mod 3$. We don't know whether to take $a_{n-1} = 2a_n$ or whether we should take $a_{n-1}=\dfrac{2a_{n-1}-1}{3}$. Either choice is possible, so how does one tell which choice is needed to reach any particular integer.

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  • $\begingroup$ There are redundancy in the generating algorithm. I think that corresponds (somehow!) to the numbers that have equal steps. $\endgroup$ – Gil Costa May 26 '17 at 23:02
  • $\begingroup$ That possible redundancy is what could ruin any effort to prove the Collatz conjecture by using an inverse sequence from 1 to n, since there is that ambiguity you can't be certain the algorithm in the usual direction will really lead to 1. If the Collatz conjecture were false, there could be an $n$ which was reachable from 1 traversing in the reverse direction but which did not lead to 1 by applying the usual algorithm. $\endgroup$ – sharding4 May 27 '17 at 0:07

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