## Collatz Question [closed]

I'd like some help with my proof of the Collatz Conjecture. I believe I have taken a new tact, and can't find a place to get the help I want. (I have tried e-mailing professors, and commenting on open problem garden is too limited) The general gist is:

I use the following assumptions: Given:

Operation 1, f_n+1 (x) = f_n (x) / 2, when f_n (x) is even

Operation 2, f_n+1(x) = (f_n (x) * 3 + 1) / 2, when f_n(x) is odd

1. All interesting odd numbers are reducible to 2b+1

2. All uninteresting numbers (even) eventually become 2b+1 through operation 2

3. Every step in the series must remain whole

To end: 1. The process must cycle indefinitely between two numbers. 2. The process must become 1. 3. The process must grow indefinitely.

What I see after basic iteration (and since basic iteration can be applied multiple times recursively, meaning after a given sequence has been shown to increase factor requirements a certain amount, chaining them won't remove these factors, simply increase their size) is that the factors of b required to maintain a whole f_n+1 (x) grow far faster than the coefficient of ab+c. (the constant originally described by 2b+1)

Examples:

Process 2, 1: 1.5b + 1

To remain whole: b = 2y

Process 2: 3b + 2

To remain whole: b = y

Process 2, 1, 2: 4.5b +2

To remain whole: b = 2y

Process 2, 2: 4.5b + 3.5

To remain whole: b = 2y + 1

Process 2, 2, 1: 2.25b + 1.75

To remain whole: b = 4y + 1

Process 2, 2, 2: 6.75b + 5.75

To remain whole: b = 4y + 3

Process 2, 2, 2, 1: 3.375b + 2.875

To remain whole: b = 8y + 3

Process 2, 2, 2, 2: 10.175b +9.125

To remain whole: b = 40y + 5

Process 2, 2, 2, 2, 1: 5.0875b + 4.5625

To remain whole: b = 80y + 5

Process 2, 2, 2, 2, 2: 15.7625b + 14.1875

To remain whole: b = 80y + 5

So, because the coefficient of the factors (the a in ab+c) is growing so much faster than the coefficient describing the number multiplied by a constant, the conjecture must be true; the end resolution can not have factors bigger than itself and remain whole.

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This is not really the place for this. Meanwhile, en.wikipedia.org/wiki/Tack_%28sailing%29#Position – Will Jagy Jan 3 2012 at 21:42
MO is not the place to promote proofs of well-known, difficult open problems. Please see the FAQ. This question will be closed. – David Roberts Jan 3 2012 at 21:43
You could perhaps try posting your question at math.stackexchange.com, but it will likely quickly closed here. If you haven't read the FAQ I'd like to suggest that. – Ryan Budney Jan 3 2012 at 21:44
Just regarding idiom, I think Will is right here. You cannot "take a new tact", any more than you can "take a new politeness" – Yemon Choi Jan 4 2012 at 2:37
Dear Noobermath: I'm afraid it is quite common for experts to read claims by outsiders of solutions to famous open problems, and almost unheard of for them to be correct. (That said, look up the Stark-Heegner theorem.) You might get a better reception if you regard your argument with extraordinary skepticism: "Almost certainly I have overlooked something, but if not then this argument seems to imply the Collatz conjecture. If you have time, would you please look at this and tell me if you see some mistake which I have missed?" Also, it is not clear exactly what your question is. Good luck! – Frank Thorne Jan 4 2012 at 3:42