Exclude nontrivial Syracuse cycles of least period at least three hundred and six

Let $T$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. Suppose $m$ is a positive integer whose least positive return time under $T$ is $p$: that is, $T^p(m) = m$ while $T^k(m) \neq m$ for all $0 < k < p$. Under the additional assumption $p \ge 306$, prove that $m = 1$.

Since $1$ has least period $1$, this would rule out such a cycle altogether.

This is the remaining open cycle obligation once least periods $1$ through $305$ have been excluded, and it supersedes the earlier frontiers at periods $8$, $17$, $94$ and $200$.

The required threshold grows with the period: covering every period up to $a$ needs $B$ at least $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305$. The jumps occur at denominators of convergents of $\log_2 3$, where $2^{K_a}/3^a$ approaches $1$. No single $B$ serves every period, so this line of argument cannot exclude all cycles; but there is no finite period at which it fails outright — only a threshold, and hence a formalisation cost, that grows without bound.

Concretely, period $306$ itself is reachable by the same method: it needs the small-value threshold raised from $27114$ to $99781$, a verification roughly four times the size of the present one. What no amount of such work reaches is a uniform statement over all periods, which is a claim about linear forms in the logarithms of $2$ and $3$.

This is not a claim that the full Collatz conjecture has been proved.