# Branching Densities of Cube-Free and Square-Free Words
## 3 Positions Fixed by Big Cubes
Then Theorem 2 implies the following inequalities related to the boundaries of the polygon in Figure 2 (\(\beta\) and \(\alpha\) play the roles of \(p\) and \(q\), respectively): \[l \geq 4\alpha-4\beta \text{if}\quad\alpha\geq\tfrac{5}{3}\beta\] (5a) \[l \geq\alpha+\beta \text{if}\quad\beta\leq\alpha\leq\tfrac{5}{3}\beta\] (5b) \[l \geq 4\alpha-2\beta \text{if}\quad\tfrac{3}{5}\beta\leq\alpha\leq\beta\] (5c) \[l \geq\beta-\alpha \text{if}\quad\alpha\leq\tfrac{3}{5}\beta.\] (5d) Assume that \[\vec{q}=(q_{0},\ldots,q_{s})\] is a sequence of positive rational numbers such that \[\max_{i\in[0..s]}q_{i}<2\min_{i\in[0..s]}q_{i}.\] We define its \[\text{span}(\vec{q})\] as the lower bound for the difference Figure 5:0 for the sequence Figure 5:1 of periods. Precisely, Figure 5:2 is the minimum number such that there exists a sequence Figure 5:3 satisfying, for each Figure 5:4, the property \"the point Figure 5:5 is on the border of the red polygon in Figure 2, in which Figure 5:6 is substituted for Figure 5:7 \". Thus, Figure 5:8, where the minimum is taken over all sequences of length Figure 5:9 in the given range Figure 6:0, is the lower bound sought in ( Figure 6:1  
any sequence \(\vec{q}\) in the given range. Precise definitions are as follows. We call a sequence \((r_{0},\ldots,r_{t})\)_canonical_ if it contains only numbers \(1\) and \(\frac{5}{3}\) in a way that no two \((\frac{5}{3})\)'s are consecutive and, in addition, \(r_{0}=r_{t}=1\). A sequence \(\vec{q}=(q_{0},\ldots,q_{s})\) is _good_ if it has a nonempty canonical suffix beginning at \(q_{0},q_{1}\), or \(q_{2}\). We transform an arbitrary sequence into a good one with local transformations changing either one element or two consecutive elements of a sequence. Note that if we change, say, \(q_{i}\) and \(q_{i+1}\), this affects \(\mathsf{span}(q_{i-1},q_{i},q_{i+1},q_{i+2})\) but preserves \(\mathsf{span}(q_{1},\ldots,q_{i-1})\) and \(\mathsf{span}(q_{i+2},\ldots,q_{s})\). By \((*)\) and \((6)\) one has  
Figure 6:2 Figure 6:3  
These inequalities justify the first transformation rule:  
* given a sequence \((q_{0},\ldots,\beta,\alpha)\), replace \(\alpha\) by \(1\) if \(\beta\leq\frac{5}{3}\) and by \(\frac{3}{5}\beta\) otherwise. Next, we consider the span of a triple \((\gamma,\beta,\frac{3}{5}\beta)\) as a function of \(\beta\). Here, \(\beta\geq\frac{5}{3}\), so \(\mathsf{span}(\gamma,\beta)\geq\mathsf{span}(\gamma,\frac{5}{3})\) by \((*)\). Since \(\mathsf{span}(\beta,\frac{3}{5}\beta)=\frac{2}{5}\beta\geq\frac{2}{5}=\mathsf{ span}(\frac{5}{3},1)\), we have  
Figure 6:4  
Further, compare \(\mathsf{span}(\gamma,\frac{5}{3},1)\) to \(\mathsf{span}(\gamma,1,1)\). For \(\gamma\geq\frac{5}{3}\) we obtain, by \((6)\),  
Figure 6:5  
For \(\gamma\leq\frac{5}{3}\), \((6)\) gives us \(\mathsf{span}(\gamma,\frac{5}{3},1)=\gamma+\frac{7}{3}\) and \(\mathsf{span}(\gamma,1,1)=6-2\gamma\). The first number is bigger (resp., smaller) if \(\gamma\) is bigger (resp., smaller) than \(\frac{11}{9}\). Therefore, we justified the second transformation rule:  
* given a sequence \((q_{0},\ldots,\gamma,\beta,\frac{3}{5}\beta)\), replace \((\beta,\frac{3}{5}\beta)\) by \((1,1)\) if \(\gamma\geq\frac{11}{9}\) and by \((\frac{5}{3},1)\) otherwise. Rules T1 and/or T2 replace the last number in the processed sequence \(\vec{q}\) by \(1\) and serve as the base case in transforming \(\vec{q}\) into a good sequence. Now we describe the general case, assuming that \(\vec{q}\) has a nonempty canonical suffix \((q_{i},\ldots,q_{s})\). The subsequent transformations preserve the numbers \(q_{i},\ldots,q_{s}\) and aim at extending the canonical suffix. Consider the span of a triple \((\gamma,\beta,1)\) as a function of \(\beta\). By \((6)\), for \(\gamma\geq\frac{5}{3}\), we have  
\[\mathsf{span}(\gamma,\beta)= \begin{cases}\gamma+\beta,\text{ if }\beta\geq \gamma;\\ 4\beta-2\gamma,\text{ if }\frac{3}{5}\gamma\leq\beta\leq\gamma;\\ \gamma-\beta,\text{ if }\beta\leq\frac{3}{5}\gamma;\end{cases} \quad\mathsf{ span}(\beta,1)=