Cycle minimum bounds the halving count: $2^{K}m^{a} \le (3m+1)^{a}$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. Suppose $m>0$ lies on a $T$-cycle of length $a\ge 1$ and is a minimum of that cycle, so $T^{a}(m)=m$ and $m \le T^{i}(m)$ for every $i$. Writing $K = \sum_{i<a} v_2(3x_i+1)$ for the total number of halvings in one period, with $x_i = T^{i}(m)$,

Equivalently, dividing by $m^a$,

Mathematical role. This is the counterpart to the lower bound $3^a < 2^K$, i.e. $K/a > \log_2 3$. Together the two confine the ratio of even steps to odd steps in a cycle to the narrow window

whose width is on the order of $1/(m \ln 2 \cdot 3)$. A rational number with denominator $a$ can only land in an interval that short when $a$ is large compared with $m$: the period of a Collatz cycle must grow in proportion to its smallest element. This is the elementary mechanism by which computational verification of the conjecture up to some height converts into a lower bound on the period of any hypothetical nontrivial cycle, and it is the point where sharper irrationality measures for $\log_2 3$ — continued fractions, and in stronger form Baker-type bounds for linear forms in logarithms — enter the classical literature.

The bound is sharp: the trivial cycle has $m=1$, $a=1$, $K=2$, and $2^2 \cdot 1 = 4 = (3\cdot 1+1)^1$, with equality.

Formalization note. Minimality is stated over the whole forward orbit, $\forall i,\ m \le T^{i}(m)$, which for a cycle is the same as minimality over the cycle. No oddness hypothesis is needed. The hypothesis $a>0$ is stated for uniformity with the companion lower bound but is not used: at $a=0$ both sides are $1$.