Exclude nontrivial Syracuse cycles of least period at least two hundred

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 $p \ge 200$, 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 $199$ have been excluded, and it supersedes the earlier frontiers at periods $8$, $17$ and $94$. The elementary margin method is close to exhaustion here. It requires $2^{K_a}/3^a$, with $K_a$ least such that $3^a < 2^{K_a}$, to exceed $\bigl((3B+1)/B\bigr)^a/3^a$ for the largest verified small-value threshold $B$; at $a = 200$ this fails for $B = 6725$.

Enlarging $B$ buys only a little more: verifying every odd value up to $12824$ would move the first gap to $253$, and up to $27113$ to $306$. That is the end of the road — $306$ is a denominator of a convergent of $\log_2 3$, with $2^{485}/3^{306}$ within $10^{-5}$ of $1$, and no threshold $B$ whatsoever clears it. A treatment of all large periods is therefore a statement about linear forms in the logarithms of $2$ and $3$, not something this counting argument can reach.

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