No. If you could prove that you'll always reach a previously seen number, then you could simply say "Let n be the smallest number for which the Collatz Conjecture is false, we have shown that we'll always reach a number < n for which the Collatz Conjecture is true, hence n does not exist."
I don't follow this either. ((3n+1) / 2) is always greater than n. If all Collatz sequences are periodic (or periodic after a finite prefix), how does that prove that they all achieve a value less than some constant k?
(Or even that they all achieve a value which is less than their own first element?)
(3n+1)/2 is always greater than n, yes, but you'd need to find a loop where you're always alternating 3n+1 and n/2 to reach a rate of increase which could not be overcome.
Which I take to mean that if n/2 occurs > 50% of the time, will eventually take the series to a number lower than your starting value and eventually to 1.
Maybe the threshold isn't 50%, but the proof then becomes finding that threshold
I am satisfied that, under the assumption that all Collatz sequences are periodic[1], they must all achieve a value less than their own initial value. (Unless that initial value is 1.)
Because the sequence is periodic[1], the ratio of (3n+1) applications to (n/2) applications must be constant[1]. This implies that the value of a term of the sequence will look like ((3^pn)k / (2^qn)), where p/q is approximately equal to the ratio of (3n+1) applications to (n/2) applications. Since no integer greater than 1 is simultaneously a power of 2 and a power of 3, this must trend toward positive infinity or toward zero as n increases. But for the sequence to increase without bound violates the assumption that it is periodic[2] -- so the terms of the sequence must instead trend toward zero.
[1] I'll use the simplifying assumption that all Collatz sequences are periodic after a prefix of 0 terms. For sequences that are periodic after a prefix of more than 0 terms, just treat them as equivalent to the same sequence with the prefix removed.
[2] A periodic sequence can take on only a finite number of values (equal to the period). But if the sequence increases without bound, it must take on an infinite number of distinct values.
