@Collag3n It’s a very interesting idea that proving l(v) = \lceil s(v) \log_2 3 \rceil for an integer cycle is a “way station” on the road to proving v can’t be an integer cycle at all.
For example, above you’ve got a condition where there are \log_2 (s(v)/3 + 1) ones at the start of a_0, the smallest member of v … and that’s enough to require l(v) = \lceil s(v) \log_2 3 \rceil.
If we strengthen that to 16 \log_2 s(v) initial ones, we can rule out the integer cycle altogether.
Let v = 1^y w be a parity vector of length l(v) with s(v) > 13 and y > 16 \log_2 s(v).
Let a_0 be the rational start number that, following v, returns to a_0.
Critically, a_0 must also be the smallest member of its cycle.
Then, a_0 is not an integer.
\beta(v) = 3^{s(w)}(3^y-2^y) + 2^y \beta(w)
= 3^{s(v)} - 2^y 3^{s(w)} + 2^y \beta(w)
If a_0 = \cfrac{\beta(v)}{2^{l(v)}-3^{s(v)}} is an integer, then so is
\cfrac{\beta(v)}{2^{l(v)}-3^{s(v)}} + 1
= \cfrac{\beta(v)+2^{l(v)}-3^{s(v)}}{2^{l(v)}-3^{s(v)}}
= \cfrac{2^{l(v)} - 2^y 3^{s(w)} + 2^y \beta(w)}{2^{l(v)}-3^{s(v)}}
= \cfrac{2^y \left(2^{l(w)} - 3^{s(w)} + \beta(w)\right)}{2^{l(v)}-3^{s(v)}}
and therefore so is
\cfrac{2^{l(w)} - 3^{s(w)} + \beta(w)}{2^{l(v)}-3^{s(v)}}
However, this quantity will be less than 1, provided the just-canceled 2^y is large enough.
\cfrac{2^{l(w)} - 3^{s(w)} + \beta(w)}{2^{l(v)}-3^{s(v)}}
= \cfrac{\frac{\beta(v)}{2^{l(v)}-3^{s(v)}}+1}{2^y}\ \ \ \text{(from equality above)}
= \cfrac{1}{2^y} \left[ \cfrac{\beta(v)}{2^{l(v)}-3^{s(v)}}+1 \right]
< \cfrac{1}{2^y} \left[ \cfrac{3^{s(v)} \frac{s(v)}{3}}{2^{l(v)}-3^{s(v)}}+1 \right] (from this bound)
<\cfrac{1}{2^y} \left[ \cfrac{3^{s(v)} \frac{s(v)}{3}}{\frac{3^{s(v)}}{250\ s(v)^{13.3}}}+1 \right] \text{(from Rhin's bound)}
= \cfrac{1}{2^y} \left[ 84\ s(v)^{14.3} + 1 \right]
< \cfrac{1}{2^{16 \log s(v)}} \left[ 84\ s(v)^{14.3} + 1 \right] \ \text{because}\ \ y > 16 \log s(v)
= \cfrac{1}{s(v)^{16}} \left[ 84\ s(v)^{14.3} + 1 \right]
= 84\ s(v)^{-1.7} + s(v)^{-16}
< 1 … for all s(v)>13, ruling out an integer cycle.
At least, I think that’s right.