Collatz descent for $n \equiv 3 \pmod 4$

Let $C$ denote the Collatz step map. The assertion is that every starting value congruent to $3$ modulo $4$ has a finite stopping time: for each $n$ with $n \equiv 3 \pmod 4$ there is an $m \in \mathbb{N}$ with

This is the one residue class modulo $4$ that elementary parity bookkeeping does not settle. For even $n$ a single step suffices, and for $n \equiv 1 \pmod 4$ with $n > 1$ three steps suffice, because $n = 4k+1$ is carried to $3k+1$. For $n \equiv 3 \pmod 4$ the first two steps produce $(3n+1)/2 > n$, and no bounded number of steps works uniformly: the length of the initial ascending run is governed by the $2$-adic behaviour of $n$ and is unbounded over the class.

The statement is open. It is the substantive content of the general descent principle for the Collatz map, and by the residue-class decomposition it is equivalent to that principle; the descent principle in turn implies the Collatz conjecture by strong induction. Tao's theorem gives descent for almost all $n$ in the sense of logarithmic density, which does not cover any specified starting value, and finite verification up to $2^{68}$ does not constrain larger $n$.

Formalization Note. No positivity hypothesis is needed: $n \equiv 3 \pmod 4$ already forces $n \ge 3$. The index $m$ is unconstrained, but $m = 0$ can never witness the conclusion since $n < n$ is false, so the statement genuinely asserts descent after a positive number of steps.