A Syracuse-periodic point that reaches $1$ equals $1$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. A periodic point of $T$ whose orbit ever reaches $1$ must itself be $1$:

Mathematical role. Cycles and convergence to $1$ are mutually exclusive, and this states it in the form actually needed. Every exclusion of a specific accelerated cycle length ends the same way: the argument bounds the cycle below by some explicit threshold $B$, and one is left checking that no small value is periodic. That final check is a statement about each individual residual value and has to be redone for each period.

This lemma removes that repetition. Because $1$ is fixed by $T$, an orbit which reaches $1$ stays there; a periodic orbit, on the other hand, returns to its starting point at every multiple of its length. Running far enough forward that both apply forces the starting point to equal $1$. The conclusion holds for every period $a$ at once, so it replaces the residual step uniformly: once one knows that the odd numbers below a threshold all reach $1$ under $T$, no nontrivial cycle can have its minimum among them, whatever its length.

Combined with the bound $m \lesssim a\,3^{a-1}$ on the minimum of an accelerated cycle, this reduces the entire cycle problem to convergence for small values: a nontrivial $T$-cycle of length $a$ would have to have its minimum among the odd numbers up to roughly $a\,3^{a-1}$, none of which may reach $1$.

Formalization note. The hypothesis is $T^{a}(z)=z$ for some positive $a$, not that $a$ is minimal. No positivity or oddness assumption on $z$ is needed; reaching $1$ supplies both.