No nontrivial Syracuse cycle of period at most two hundred and fifty-two

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 $1 \le a \le 252$, then $m = 1$.

Equivalently, every Syracuse cycle of length at most $252$ is the trivial cycle $\{1\}$. The return time $a$ need not be minimal.

The proof is uniform across the $252$ 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 $12825$ is available because no nontrivial cycle contains a value below $12825$. At $a = 252$ these compare integers of about $1060$ digits.

The least threshold covering every period up to $a$ grows without bound: it is $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305$, with the jumps landing on denominators of convergents of $\log_2 3$, where $2^{K_a}/3^a$ returns close to $1$. No single threshold serves every period, so 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.