Just to finish this line off … any rational cycle with all members greater than 1 needs to have k < 2x.
The smallest member of any rational cycle v is less than that of the high cycle.
Easy case, let k=2x.
f(v) \leq f(v_h) = \cfrac{\sum_{i=0}^{x-1} 3^{x-i-1} 2^{\lfloor \frac{k}{x}\,i \rfloor}}{2^k-3^x}
= \cfrac{\sum_{i=0}^{x-1} 3^{x-i-1} 2^{2i}}{2^{2x}-3^x}
= \cfrac{4^x-3^x}{4^x-3^x} = 1
Actually, the high cycle’s smallest member is 1 when k=2x because its parity sequence (10)^* follows the trivial cycle.
Here’s a general account of the high cycle’s smallest member:
f(v) \leq f(v_h) = \cfrac{\sum_{i=0}^{x-1} 3^{x-i-1} 2^{\lfloor \frac{k}{x}\,i \rfloor}}{2^k-3^x}
\leq \cfrac{\sum_{i=0}^{x-1} 3^{x-i-1} 2^{\frac{k}{x}\,i}}{2^k-3^x}
= \cfrac{2^k-3^x}{2^{k/x}-3}\cdot\cfrac{1}{2^k-3^x}
= \cfrac{1}{2^{k/x}-3}
So for a rational cycle to have all members greater than 1, we require 2^{k/x}-3 < 1, or k < 2x.
From an integer cycle perspective, this bound is very weak, but rational high cycles with k = 2x -1 manage to have all members above 1.