Here’s another argument that almost no parity sequences can correspond to Collatz counterexample cycles.
There are about 2^k/k rational cycles of length k.
Let’s show at most 1.3^k of them can be integer cycles, a vanishing fraction as k increases.
Every parity vector v of length k corresponds to some rational number m that returns to itself after following the even-odd sequence v.
A cycle is a set of parity vectors that are rotations of each other.
There are about 2^k/k cycles (actually a tiny, tiny bit less, due to periodic vectors).
Previously, if two vectors v_1 and v_2 in the same cycle have a common prefix longer than 0.38\,k, then their associated cycle members m_1 and m_2 can’t both be integers, so the cycle can’t be a Collatz counterexample.
We can apply the same idea across cycles, to show that two distinct cycles can’t both be Collatz counter-examples (even if one of them might).
Build a graph with a node for each cycle. Connect any pair of nodes A and B if there’s some parity vector (rotation) v_1 in the A cycle, and some parity vector (rotation) v_2 in the B cycle, such that v_1 and v_2 have a shared prefix longer than 0.38\,k.
Question: What is the largest set S of nodes where none are pairwise connected? All cycles in this set can (conceivably) be Collatz counter-examples simultaneously. This is the maximum independent set of the graph.
In this case, we can show that the set S can’t be bigger than 1.3^k.
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The total number of available, distinct blocks of length 0.38k is 2^{0.38k}. (Omitting floors and ceilings for simplicity.)
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Every cycle has some parity vector with some block of length 0.38k.
So, if S contains more than 2^{0.38k} cycles, some block will appear twice, meaning two of the cycles will be connected, and therefore S isn’t an independent set.
That means the maximum number of pairwise unconnected nodes is 2^{0.38k} \approx 1.3^k, so there are at most that many conceivable Collatz counterexamples.
As usual, even though almost no parity sequences can form integer cycles, we don’t have much of a grasp on which ones might, or might not …
PS. Similar arguments show the maximum clique T in the graph is also smaller than 1.3^k. Among the cycles in T, at most one can be a Collatz counterexample.