Rotation divisibility is one test, not many
Plain English: Rotating a proposed odd-cycle exponent word does not provide a new integrality obstruction.
Setup
Choose a finite word of positive integers \(a_0,\ldots,a_{k-1}\). Write \(A=\sum a_i\), \(S_j=\sum_{i<j}a_i\), and
\[D=2^A-3^k,\qquad C=\sum_{j=0}^{k-1}3^{k-1-j}2^{S_j}.\]
Composing the affine steps \(x\mapsto(3x+1)/2^{a_i}\) gives \(x\mapsto(3^kx+C)/2^A\). Hence its rational fixed point is \(C/D\). The denominator is nonzero because a positive power of two cannot equal a positive power of three.
Let \(C'\) be the numerator for the left-rotated word \((a_1,\ldots,a_{k-1},a_0)\). Direct expansion yields
\[2^{a_0}C'=3C+D.\]
Indeed, multiplying the rotated numerator by \(2^{a_0}\) produces all terms of \(3C\) except its first term \(3^k\), and adds \(2^A\).
Exact consequence
The integer \(D\) is odd and is not divisible by three. Therefore both \(2^{a_0}\) and three are invertible modulo \(|D|\). The displayed identity implies
\[D\mid C'\quad\Longleftrightarrow\quad D\mid C.\]
Apply this equivalence successively to every rotation. All rotated fixed points are integral together, or none are. Testing their divisibility separately cannot strengthen the original test.
What this does and does not establish
If \(D>0\) and \(D\mid C\), each rotated fixed point is a positive integer. Each numerator is odd, so these integers are odd. The identity then gives the required adjacent relation \(2^{a_i}n_{i+1}=3n_i+1\); since \(n_{i+1}\) is odd, the selected division exponent is exact. This supplies a positive odd-cycle certificate for that exponent word, not necessarily a cycle of least period \(k\).
The obstruction that remains is to rule out all exponent words other than repetitions of the trivial odd cycle. Rotation redundancy does not do that. Different orderings not related by rotation can still have different numerators.
Ledger decision
Retire: treating cyclic rotations as independent divisibility filters.
Keep open: non-rotational ordering constraints and rigorous ways to eliminate families of exponent words.
This page records an algebraic derivation, not a literature novelty claim or a proof of the Collatz conjecture. A machine verifier is a separate deliverable; this update does not claim one was run.