No nontrivial Syracuse cycle of period at most three hundred and five

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 $1 \le a \le 305$, then $m = 1$.

Equivalently, every Syracuse cycle of length at most $305$ is the trivial cycle $\{1\}$. The return time $a$ need not be minimal.

The proof is uniform across the $305$ 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 $27114$ is available because no nontrivial cycle contains a value below $27114$. At $a = 305$ these compare integers of about $1350$ digits.

The period $306$ is the first not covered, and is of special interest: it is a denominator of a convergent of $\log_2 3$, where $2^{485}/3^{306} \approx 1.00102$ leaves very little room. 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.