Syracuse descent for $n \equiv 15 \pmod{16}$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. The assertion is that every $n \equiv 15 \pmod{16}$ has a finite accelerated stopping time:

This statement is open, and it is the most resistant of the classes left by the refinement of the Collatz descent problem modulo $32$: for $n = 16j+15$ the classical map ascends for four of its first eight steps, accumulating a multiplier $3^{4}/2^{4} = 81/16$. In accelerated form the same obstruction appears as a run of $T$-steps with $v_2(3n+1) = 1$, each multiplying by $3/2$.

The statement covers the classes $15$ and $31$ modulo $32$, and implies its classical counterpart because the $T$-orbit of an odd $n$ is a subsequence of its $C$-orbit.

Formalization Note. No positivity or oddness hypothesis is needed: $n \equiv 15 \pmod{16}$ forces $n \ge 15$ and $n$ odd. The index $t$ is unconstrained, but $t = 0$ can never witness the conclusion since $n < n$ is false.