Re: Proof of Collatz Conjecture?

You wrote:
But this doesn't even begin to show that there
aren't any cycles of length bigger than 3, or
that there aren't any numbers that never wind
up in a cycle.

I believe that what I had established, by the arbitrariness of n and a (or b) in my argument, is that every standard Collatz sequence Cn has a length-3 cycle because there is always one solution with b = 1 to the expression
Cn = <n, m, l, ..., (i, h, …, d, c, b)>
for any natural number n, i in { n, m, l, …, d, c, b } and, hence, every Cn is of the form
Cn = <n, m, l, ..., (4, 2, 1)>.
I stopped at length-3 cycle because a length-3 cycle could not be included in a larger cycle with length greater than 3 -- although two input terms to the standard Collatz function could give the same output term, there is always only one succeeding term to any term in any sequence or the cycle. Thus, all of your above concerns are answered -- all of them due to the arbitrariness of n and b in my argument.

If it is necessary to show that there are no cycles with length greater than 3, then one could simply proceed as follows:
8. Suppose there exists a
Cn = <n, m, l, ..., (e, d, c, b)>
— that is, a length-4 standard Collatz cycle — for some
starting natural number n with
b = min(e,d,c,b).
Then, we simply rehash our earlier argument:
e = f(b) = 3b + 1
d = f(e) = (3b+1)/2 = 4b yields b = -1/5
c = f(d) = 3[(3b+1)/2] + 1 = 2b yields b = -1
-- therefore, there is no valid solution so that the cycle
(e,d,c,b) does not exist.
Again, suppose there exists a
Cn = <n, m, l, ..., (f*,e,d,c,b)>
-- that is, a length-5 standard Collatz cycle -- for some
starting natural number n with
b = min(f*,e,d,c,b).
Then, we simply rehash our earlier argument:
f* = f(b) = 3b + 1
e = f(f*) = (3b+1)/2
d = f(e) = 3[(3b+1)/2] + 1 = (9b+5)/2 = 4b
yields b = -5
c = f(d) = 3[(9b+5)/2] + 1 = (27b+17)/2 = 2b
yields b = -17/23
-- therefore, there is no valid solution so that the
cycle (f*,e,d,c,b) does not exist.

I think that there is a pattern here (I did not explore this further because I didn’t think it is necessary since I believe that the arbitrariness of n in my earlier argument already covers these concerns) that we could easily generalize by mathematical induction to prove that there are no cycles with length greater than 3 (hence, there is also no divergent sequence).
.

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