When considering Collatz for the first time, most people will notice that if convergence is proven for odd numbers only, then even numbers will automatically follow suit. The same argument can be extended to limit our attention to only those odd numbers with the quality of being "origin" numbers. All numbers with this quality are of the form 3 mod 6 or 6k + 3. Or in other words, if, 3, 9, 15, 21, … all converge, then so too will any other number.
Consider the following four sequences.
3, 5, 8, 4
9, 14, 7, 11, 17, 26, 13, 20, 10
15, 23, 35, 53, 80, 40
21, 32, 16
Notice how we depict 6k + 3 numbers 3, 9, 15, and 21 ending at 6k + 4 numbers 4, 10, 40, and 16. This is a convention that I have established for the purposes of this proof. We can look at 6k + 4 numbers as the end points of arrows (i.e. their "pointy end") emerging from 6k + 3 origin numbers, traveling along for a short while before becoming "the sharp end of the 6k + 4 tip".
This convention is not without meaning. The reason I have chosen to define a Collatz chain as terminating at 6x + 4 is because it is the first occurrence of two even numbers in a row from the starting point. This is significant because it indicates a collision with another sequence, or an "origin swap".
As we traverse a trajectory, if we constantly point to the number and ask where it naturally comes from, we will notice sometimes that the "origins" of our numbers change. An "origin" is a conventional term I am using to describe the starting point or common ancestor of our numbers.
We say, for example, that origin-9 (3 mod 6) proceeds to end-point-10 (4 mod 6), eight places further down than 9, where it ends abruptly. At this point, we collide with a "perpendicular" sequence coming from another origin. Therefore, 9 has a "stopping time" of eight and an end-point of 10. End of story. We're going to stop the "9-train" at 10, right before it encounters 5.
But what about the other numbers 5 through 1? Long story short...they don't belong to 9.
With this convention in place, I can prove, using this fact,
sum of all Fibonacci_n / 2^n = 1
that the Collatz Conjecture is true.
The argument goes like this. Consider an infinite sequence of FINITE numbers greater than 2 which are to represent the lengths of our arrows.
...,3, 4, 5, 5, 3, 6, 6, 6, 3, 4, 7, 7, 3, 7, 7, 7, 3, 4, 5, 5, 3, 8, 8, 8, 3, 4, 8, 8, 3, 8, 8, 8,...
Lemma : We see 3, 4, 5, 6, 7, and 8, etc. We see 3 occurs once every 4 places, 4 occurs once every 8 places, 5 occurs twice every 16 places, 6 occurs three times every 32 places, 7 occurs five times every 64 places, 8 occurs eight times every 128 places, and if we had enough room to include it, 9 would occur 13 times every 256 places, etc. These are the Fibonacci numbers and the powers of 2.
Thus we see the relevance of:
F_n / 2^n
Now, a keen observer would point out that this above sequence of numbers is merely simulated. In reality, our numbers might not look exactly like this. However, this is what our numbers actually look like:
...,4, 9, 6, 3, 26, 8, 8, 3, 4, 10, 8, 3, 6, 5, 5, 3, 4,...
We can clearly see the patterns of 3 and 4. By expanding the domain, we will also inevitably see the pattern for 5, 6, 7, 8, and so on.
...,4, 9, 6, 3, 26, 8, 8, 3, 4, 10, 8, 3, 6, 5, 5, 3, 4, 8, 16, 3, 9, 11, 7, 3, 4, 6, 13, 3, 10, 5, 5, 3, 4, 7, 6, 3, 8, 15, 11, 3, 4, 9, 10, 3, 6, 5, 5, 3, 4, 9, 14, 3, 7, 7, 20, 3, 4, 6, 7, 3, 9, 5, 5, 3, 4, 10, 6, 3, 9, 18, 10, 3, 4, 8, 12, 3, 6, 5, 5, 3, 4, 11, 24, 3, 10, 12, 7, 3, 4, 6, 11, 3, 8, 5, 5, 3, 4, 7, 6, 3, 16, 9, 10, 3, 4, 10, 18, 3, 6, 5, 5, 3, 4, 17, 8, 3, 4...
Therefore, our premise is correct.
Now let's repeat our Lemma.
Lemma : We see 3, 4, 5, 6, 7, and 8, etc. We see 3 occurs once every 4 places, 4 occurs once every 8 places, 5 occurs twice every 16 places, 6 occurs three times every 32 places, 7 occurs five times every 64 places, 8 occurs eight times every 128 places, and if we had enough room to include it, 9 would occur 13 times every 256 places, etc. These are the Fibonacci numbers and the powers of 2.
From here, it is a routine calculation to show that:
1/4 + 1/8 + 2/16 + 3/32 + 5/64 + 8/128 + 13/256 + ... + F_n / 2^n = 1
This represents the probability that a Collatz chain has a finite length.
We are therefore, led to the following proof by contradiction:
Let's suppose we have an INFINITE number, X. Let's suppose X occurs somewhere on our number line. This would be a contradiction because every number on this line is finite. Therefore X is not infinite.
And finally, since this number, X, represents the length of our arrows, we are led to conclude that all Collatz chains are finite.
QED
