September 20th, 2017, 07:47 PM  #1 
Newbie Joined: Sep 2017 From: Canada Posts: 1 Thanks: 0  Collatz Conjecture Idea
Writing from a phone, so please excuse any poor formatting. So obviously with the Collatz conjecture all odd numbers don't need to be checked. That being said would it be just as hard to prove that over time all numbers put through the function have a tendency to decrease in size? I mean obviously proving a tendency doesn't prove the conjecture but if it can be shown that the numbers will always eventually decrease it would be a way to build to the proof. Since we know the first large chunk are guaranteed to go to one by brute force. Obviously, I'm probably stating the obvious here since I'm new to the problem, but it seems like you just need to show all numbers will eventually decrease. Which is probably much harder than expected. Perhaps this is because you can only ever have one increase in a row Ie 3n+1 × will always end in a positive integer which will than be divided by two. So as such you're very likely to get multiple even numbers in a row and decrease it quicker than it increases since you can only get one odd number at a time. Or in other words you have to show that all numbers will have more even numbers decreasing the number than odd numbers can consistently increase it. I guess that's probably just as hard to show though. Just a thought journal lmao. Last edited by skipjack; September 20th, 2017 at 11:28 PM. 
September 21st, 2017, 09:09 AM  #2 
Newbie Joined: Sep 2017 From: Belgium Posts: 4 Thanks: 2 
You probably know the direct children (Collatz tree) of 1 which are $1,5,21,85,341,...$ or $\frac{2^l1}{3}$ From them you can build any odd $n$ of these forms: $8i+1$ which as a direct parent $6i+$1 $32i+5$ which as a direct parent $6i+1$ $128i+21$ which as a direct parent $6i+1$ $512i+85$ which as a direct parent $6i+1$ ... Parents (which is the number obtained by applying the condensed Collatz function $\frac{3n+1}{2^m}$) are always smaller. The direct children of 5 which are $3,13,53,213,...$ or $\frac{5\cdot2^l1}{3}$ help you build any odd $n$ of these forms: $4i+3$ which as a direct parent $6i+$5 $16i+13$ which as a direct parent $6i+5$ $64i+53$ which as a direct parent $6i+5$ $256i+213$ which as a direct parent $6i+5$ ... Now, same thing: parents are always smaller except for the first representation $4i+3$ Note: The above forms cover all natural numbers. e.g $17$, which is $8\cdot2+1$, have $6\cdot2+1 = 13$ as parent, which is smaller. $7$ or $4\cdot1+3$, has parent $6\cdot1+5 = 11$ which is greater. $245$ or $64\cdot3+53$, has a smaller parent $6\cdot3+5 = 23$ But don't get me wrong, $4i+3$ is still half of the odd numbers, so it is not so easy to go that way. Last edited by skipjack; September 21st, 2017 at 10:32 AM. 
September 29th, 2017, 06:09 AM  #3  
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  Quote:
I'll be lazy and quote Wikipedia: If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average 3/4 of the previous one.[16] (More precisely, the geometric mean of the ratios of outcomes is 3/4.) This yields a heuristic argument that every Hailstone sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2adic extension of the Collatz process has two division steps for every multiplication step for almost all 2adic starting values.) And even if the probabilistic reasoning were rigorous, this would still imply only that the conjecture is almost surely true for any given integer, which does not necessarily imply that it is true for all integers. It's not even great evidence against divergence, because a small number of chains that kept rising infinitely would be such a small proportion of the number chain as to have no significant impact on the mean outcome.  

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