Syracuse descent for odd numbers between 267823 and 271823

Let $T$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. For every odd $m$ with $267823 \le m \le 271823$, some iterate of $T$ falls strictly below $m$:

This is a descent statement, not a convergence statement: nothing is claimed about where the orbit eventually goes, only that it dips below its starting point.

Descent is exactly what is needed to exclude cycles. The minimum of a nontrivial cycle can never descend, so any $m$ satisfying the above is disqualified from being a cycle minimum. This is cheaper to certify than convergence, because each orbit need only be followed until its first drop below $267823$ — which is automatically below $m$ — rather than all the way to $1$.

Half the range is handled uniformly rather than by certificate: if $m \equiv 1 \pmod 4$ then $4 \mid 3m+1$, so $T(m) \le (3m+1)/4 < m$ for $m > 1$, a single step. Only the residues $m \equiv 3 \pmod 4$ need explicit orbit data, here $5995$ step facts.