No nontrivial Syracuse cycle of period at most 1635

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 $1 \le a \le 1635$, then $m = 1$.

Equivalently, every Syracuse cycle of length at most $1635$ is the trivial cycle $\{1\}$. The return time need not be minimal.

The proof is uniform across the $1635$ 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 $330750$ is available because no nontrivial cycle contains a value below it. At $a = 1635$ one has $K_a = 2592$ and the comparison is between integers of about $9000$ digits.

Period $1636$ is the first not covered. Writing $B(a)$ for the least threshold covering every period up to $a$, the sequence runs $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781, 330750$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305, 970, 1635$, with jumps at denominators of convergents of $\log_2 3$. Since $B(a) \to \infty$, no single threshold covers all periods, so this argument cannot settle the cycle question — but no finite period defeats it either; only the certificate size grows without bound.