No nontrivial Syracuse cycle of length $6$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. Its only positive point of period dividing $6$ is $1$:

Equivalently, the Collatz map has no nontrivial cycle whose accelerated form has length $6$.

Mathematical role. With the shorter lengths this closes every accelerated cycle length up to $6$. Multiplying the six step relations $2^{v_i}x_{i+1}=3x_i+1$ around the orbit gives

with $P$ the product of the orbit and $e_j$ its elementary symmetric functions. Positivity of the right side forces $2^{K}>729$, hence $2^{K}\ge 1024$ and a margin of $295$ — a generous one, so the comparison already fails once every orbit point is at least $7$.

This case is notably easier than length $5$, where the margin was only $13$ and the threshold rose to $33$. The margin is governed by how far the least power of two above $3^{a}$ has to reach, and at $a=6$ it reaches from $729$ to $1024$.

On the residual step. Earlier exclusions each finished with a hand check that no small value is periodic of that particular length. That is no longer needed: it is now known that no nontrivial $T$-cycle contains an element at most $33$, whatever its length, so the threshold $7$ obtained here is immediately enough. The exclusion therefore consists of the symmetric-function comparison alone.

Formalization note. No oddness hypothesis is needed: a point of period dividing $6$ under $T$ is automatically odd. The statement covers lengths $1$, $2$, $3$ and $6$ simultaneously, since each divides $6$.