Exclude nontrivial Syracuse cycles of least period at least nine hundred and seventy-one

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 $p \ge 971$, 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 $970$ have been excluded, and it supersedes the earlier frontiers at periods $8$, $17$, $94$, $200$, $253$ and $306$.

The least threshold $B(a)$ covering every period up to $a$ runs $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781, 330750$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305, 970$, with the jumps landing on denominators of convergents of $\log_2 3$, where $2^{K_a}/3^a$ returns close to $1$. Since $B(a) \to \infty$, no single threshold serves every period, and this argument cannot settle the cycle question outright — but no finite period defeats it either. Only the certificate size, and hence the formalisation cost, grows without limit.

Concretely, the next step is again routine rather than blocked: raising the verified small-value threshold from $99781$ to $330750$ covers every period up to the following gap. What is not routine — and what this statement really stands for — is a uniform treatment of all large periods, which is a claim about linear forms in the logarithms of $2$ and $3$ rather than anything a finite verification can reach.

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