Exclude nontrivial Syracuse cycles of least period at least 4961

Let $T$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. Suppose $m$ is a positive integer whose least positive return time under $T$ is $p$: that is, $T^p(m) = m$ while $T^k(m) \neq m$ for all $0 < k < p$. Under the additional assumption $p \ge 4961$, prove that $m = 1$.

Since $1$ has least period $1$, this would rule out such a cycle altogether.

This is the remaining open cycle obligation once least periods $1$ through $4960$ have been excluded, superseding the earlier frontiers at $8$, $17$, $94$, $200$, $253$, $306$, $971$, $1636$, $2301$, $2966$, $3631$ and $4296$.

Writing $B(a)$ for the least small-value threshold covering every period up to $a$, the sequence runs $34, 147, 387, 1193, 3343, 6725, 12825, 27114, 99781, 330750, 583288, 860564, 1166400, 1505449, 1883432$ as $a$ passes $16, 28, 40, 93, 146, 199, 252, 305, 970, 1635, 2300, 2965, 3630, 4295, 4960$, with jumps at denominators of convergents of $\log_2 3$. Because $B(a) \to \infty$, no single threshold covers every period, so a finite verification of this kind can never settle the cycle question — yet no finite period defeats the method either; only the certificate size grows without bound.

The certificates are descent statements rather than convergence statements: a cycle minimum cannot descend, so it suffices to exhibit some strictly smaller iterate, and orbits need only be followed to their first drop. A uniform treatment of all large periods remains a claim about linear forms in the logarithms of $2$ and $3$.

This is not a claim that the full Collatz conjecture has been proved.