Every positive Collatz cycle contains an odd number

Let $C$ be the classical Collatz map, $C(n)=n/2$ for even $n$ and $C(n)=3n+1$ for odd $n$. Every positive periodic point of $C$ has an odd number somewhere in its orbit:

Mathematical role. This is the bridge between the classical map and its accelerated form. The Syracuse map $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ is defined on odd inputs and collapses each ascending step together with the ensuing run of halvings; results about cycles are far easier to state and prove for $T$, since a $T$-cycle carries the arithmetic of the problem without the intervening halvings. To transfer any such result back to the classical map one must first know that a classical cycle meets the odd numbers at all — otherwise there is no point at which to start the accelerated dynamics.

The reason is immediate once stated: an orbit consisting entirely of even numbers halves at every step, so it strictly decreases, and a strictly decreasing orbit cannot return to its starting point. The hypothesis $x>0$ is essential, since $0$ is a fixed point whose orbit is entirely even.

Formalization note. Oddness is written as $\neg\,\mathrm{Even}$ rather than $\mathrm{Odd}$; over $\mathbb{N}$ the two are equivalent. The index $i$ is unconstrained, though it may always be taken below $p$.