Any Syracuse cycle satisfies $2^{K} > 3^{a}$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. Suppose $m>0$ lies on a $T$-cycle of length $a \ge 1$, so that $T^{a}(m) = m$, and write

for the orbit and the total number of halvings consumed in one period. Then

In words: around any Collatz cycle, the halvings strictly outweigh the triplings. Equivalently $K/a > \log_2 3$, so the ratio of even steps to odd steps in a cycle is a rational number lying strictly above $\log_2 3 = 1.5849625\ldots$

Mathematical role. This is the elementary half of the classical approach to cycle exclusion. Pairing it with the complementary upper bound — which comes from the fact that every orbit element is at least the cycle's minimum — traps $K/a$ in a narrow window around $\log_2 3$, forcing $K/a$ to be an unusually good rational approximation to an irrational number. Bounds on how well $\log_2 3$ can be approximated by rationals, supplied by continued fractions and in sharper form by Baker-type estimates for linear forms in logarithms, then force the period $a$ to be enormous. That is the route by which nontrivial Collatz cycles are constrained without any appeal to the descent principle.

Stated on its own, this inequality does not exclude any cycle: the trivial cycle $1 \mapsto 1$ satisfies it, with $a=1$, $K=2$ and $3 < 4$. It is a necessary condition, and it is the entry point that the transcendence argument needs.

Formalization note. No oddness hypothesis is required. The valuation is written as Nat.factorization at the prime $2$, and the total $K$ as a sum over Finset.range a. Only $m>0$, $a>0$ and $T^{a}(m)=m$ are assumed; in particular $m$ need not be the minimum of its cycle.