No nontrivial Syracuse cycle of period at most 4295

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 $1 \le a \le 4295$, then $m = 1$.

Equivalently, every Syracuse cycle of length at most $4295$ is the trivial cycle $\{1\}$; the return time need not be minimal.

The proof is uniform across the $4295$ 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 $1505449$ is available because no nontrivial cycle contains a value below it.

Period $4296$ is the first not covered. Writing $B(a)$ for the least threshold covering every period up to $a$, the sequence reads $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781, 330750, 583288, 860564, 1166400, 1505449$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305, 970, 1635, 2300, 2965, 3630, 4295$, the jumps landing on denominators of convergents of $\log_2 3$. Since $B(a) \to \infty$, no single threshold covers every period, so a finite verification of this kind can never settle the cycle question; equally, no finite period defeats the method — only the certificate size grows without bound.