Syracuse descent for $n \equiv 27 \pmod{32}$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. The assertion is that every $n \equiv 27 \pmod{32}$ has a finite accelerated stopping time:

This statement is open. It is the accelerated form of one of the three residue classes left open by the refinement of the Collatz descent problem modulo $32$, and it implies its classical counterpart because the $T$-orbit of an odd $n$ is a subsequence of its $C$-orbit.

Each $T$-step multiplies by $3/2^{v}$ with $v \ge 1$, so the accelerated map contracts on average while the classical map alternates growth and contraction; this is why the quantitative theory, from Terras's density theorem to Tao's almost-all result, is developed for $T$.

Formalization Note. No positivity or oddness hypothesis is needed: $n \equiv 27 \pmod{32}$ forces $n \ge 27$ and $n$ odd. The index $t$ is unconstrained, but $t = 0$ can never witness the conclusion since $n < n$ is false.