Some of my initial constraints were looser than necessary (though still true), so I have tightened them up below and can get all the way to 2 now. Here's my pro

Some of my initial constraints were looser than necessary (though still true), so I have tightened them up below and can get all the way to 2 now. Here's my proof of the Collatz conjecture. I hope someone will call me on it if there's any problem. Given the function: f(n)={n/2 if n=0 mod 2, 3n+1 if n=1 mod 2}, for integers n >= 1 And the set of sequences: a_i={n if i=0, f(a_[i-1]) if i>0}, for integers i >= 0 and integers n >= 1 Lemma: In any given sequence, any given odd must be followed by an even. Proof: a_[k-1]=2n-1 (clearly odd), for k >= 1 a_[k]=f(2n-1)=3(2n-1)+1=2(3n-1) (clearly even) Corrollary: The elements of both the domain and range of f (namely, the elements of the positive integers) are trivially either even or odd, so every non-initial odd in a sequence must also have been immediately preceded by at least one even. We can now exhaustively describe all the shortest sub-sequences that start and end on evens. These can be found simply by assuming an even start and fully permuting the the next two iterations of f to find the next two numbers in a sequence. Specifically, there are 4 cases: 1) f(f(f(n:even):even):even), which contains a repeated sub-sequence: f(f(n:even):even)=(n/2)/2=n/4 Which is less than the first (n:even) for n>0. 2) f(f(f(n:even):even):odd), which ends with an odd and can only be followed by an even (Lemma), and so describes the same sub-sequence as: f(f(f(f(n:even):even):odd):even)=(3((n/2)/2)+1)/2=((3/4)n+1)/2=(3/8)n+1/2 Which is less than the first (n:even) for n>4/5. 3) f(f(f(n:even):odd):even)=(3(n/2)+1)/2=(3/4)n+1/2 Which is less than the first (n:even) for n>2. 4) f(f(f(n:even):odd):odd) Which is impossible (Lemma) as it contains 2 consecutive odds. I have included this case merely to illustrate the completeness of the permutations. The 3 valid cases above (cases 1 through 3) show that for any initial even greater than 2 there exists an even that is both strictly lower than the original even and is within 4 iterations of f (within a sub-sequence of 5 numbers, beginning with the first even) beyond the first even. Now consider any full sequence beginning with an even. By the above exhaustion, all such full-length sequences begin with one of the above sub-sequences. The end of that sub-sequence is even and, when taken again as the first even in one of the above sub-sequences, as many times as necessary, shows that the evens, with certainty, reach some smallest even integer greater than 2, after only a finite number of steps. This smallest even integer greater 2 is followed by exactly 1 odd (Lemma and Corrollary) and is, by definition, less than or equal to 2. All inputs less than or equal to 2 terminate trivially in the well-known 1-cycle (2, 1, 4, 2, 1, ...), therefore all sequences beginning with an even number end in this 1-cycle. Furthermore, all odd inputs are followed by an even after at most 1 iteration (Lemma), so sequences beginning with odd numbers also terminate in this 1-cycle. Q.E.D(ammit). This argument is not qualitatively different from the one in my original comment. I have only attempted to connect the intermediary steps more clearly, in addition to tightening up the constraints to lower the original boundary condition. I'm not sure if anybody's still reading this, but I would welcome any further analysis. (For example, I'm sure it could be written more professionally, and with less English.)