No nontrivial Syracuse cycle of period at most nine hundred and seventy

Let $T$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. If $m$ is a positive integer with $T^a(m) = m$ for some $a$ with $1 \le a \le 970$, then $m = 1$.

Equivalently, every Syracuse cycle of length at most $970$ is the trivial cycle $\{1\}$. The return time $a$ need not be minimal.

The proof is uniform across the $970$ periods. For each $a$ one checks the single numerical inequality

where $K_a$ is the least exponent with $3^a < 2^{K_a}$, and feeds it to the margin criterion; the threshold $99781$ is available because no nontrivial cycle contains a value below $99781$. At $a = 970$ these compare integers of roughly $4900$ digits.

Period $971$ is the first not covered, and needs the threshold raised to $330750$. 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.