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Topic: Cope-atz Conjecture (Read 4763 times) |
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aicoped
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Cope-atz Conjecture
« on: Nov 6th, 2011, 7:02am » |
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ok briefly the collatz conjecture states for any positive integer if it is even you divide it by two else you multiply it by 3 and add 1. repeat until you get a cycle. It has been shown that all numbers under 57 bajillion or some such high number are true but there is no proof. So I decided to check to see if it is true for other variations.
Here is my new claim If you leave the divide by 2 rule alone, and only mess with the second half of the form ax+y, then for all 3x+y it will eventually reach a cycle no matter how large y is. And for any prime"a" that is higher than 3, it will not cycle and will have exceptions that get higher than any bound(ie infinite) regardless of the value of y.
Can anyone prove this conjecture wrong(or right)?
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aicoped
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Re: Cope-atz Conjecture
« Reply #1 on: Nov 6th, 2011, 7:08am » |
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Just for some extra clarification, there is already another stricter collatz conjecture for numbers of form 5x+1 and 7x+1, but they add extra rules. So for 5x+1 the rules are as follows: if even divide by 2, if not divide by 3 if it is divisible by three, else multiply by 5x+1. That has been verified to cycle to fairly large numbers although not proven. Ditto the 7x+1.
So working with everything we know so far it seems like colltaz is true if the average move is less than one. I don't know a way to word that better. In other words on the 3x+1 we are moving on an average of 3/4 on each move if we count the divide by 2 step after the 3x+1 to be a part of the step since you know you will always divide by 2 at least 1 time after a 3x+1 move. Additionally it appears that the 5x+1 will result in a 5/6 average and so on.
So using "my version" of the new conjecture where we just either divide by 2 else multiply by 5x+y, we get a trend of 5/4 per move which will have certain numbers that do not cycle.
I hope i made myself clear, and would appreciate any feedback.
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| « Last Edit: Nov 6th, 2011, 7:09am by aicoped » |
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towr
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Re: Cope-atz Conjecture
« Reply #2 on: Nov 6th, 2011, 8:24am » |
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on Nov 6th, 2011, 7:02am, aicoped wrote:| Here is my new claim If you leave the divide by 2 rule alone, and only mess with the second half of the form ax+y, then for all 3x+y it will eventually reach a cycle no matter how large y is. |
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Try picking y a multiple of three.
That way you keep introducing a factor of three which you can never get rid of.
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| « Last Edit: Nov 6th, 2011, 9:13am by towr » |
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aicoped
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Re: Cope-atz Conjecture
« Reply #3 on: Nov 6th, 2011, 9:26am » |
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it will still cycle, just not at 1 any more, to use 11 as an example.
11,36,18,9,30,15,48,24,12,6,3,12,6,3,etc
13,42,21,66,33,102,51,166,83,252,126,63,192,96,48,24,12,6,3 etc.
With the normal collatz the cycle is centered around 1. when you change the y it simple has the potential to change the cycle point, but it still cycles to the best i can figure.
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towr
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Re: Cope-atz Conjecture
« Reply #4 on: Nov 6th, 2011, 10:42am » |
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Hmm, yeah, I made a mistake there, it only works the way I imagined for y=0.
On the other hand, just picking y even means you never get to divide by two anyway; so it'd be fair enough to at least require y be odd.
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aicoped
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Re: Cope-atz Conjecture
« Reply #5 on: Nov 6th, 2011, 11:01am » |
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sorry yes y must be odd. I did leave that out of my statement, but in practice i was only looking at the values where it was odd. So with the time you have put in, would you personally agree it appears to be true or false?
What I did was made a spreadsheet in excel with an if statement then copied it for the next 1000 cells and I found that when the second part was 3x+y the numbers would eventually cycle, but if i made it so that it was 5x+y or any odd number higher, they would never cycle. I have not seen this on any other math site and I spent about 6 hours perusing the literature, so I thought i would get it posted as a conjecture and see if it has any obvious holes from all of the smart people here.
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