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
All interesting odd numbers are reducible to 2b+1
All uninteresting numbers (even) eventually become 2b+1 through operation 2
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.
