# Branching Densities of Cube-Free and Square-Free Words
## 3 Positions Fixed by Big Cubes
Then we study \(\mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1)\) depending on \(\delta\): \[\delta\geq\gamma :\ \mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1) =\tfrac{16}{5}\gamma-2\delta+4>\tfrac{16}{5}>\delta+\gamma 3\] \[\tfrac{5}{3}\leq\delta\leq\gamma :\ \mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1) =\tfrac{1}{5}\gamma+\delta+4>\delta+\gamma 3\] \[\tfrac{3}{5}\gamma\leq\delta\leq\tfrac{5}{3} :\ \mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1) =\tfrac{15}{5}\gamma+\delta+4\geq 2\gamma-2\delta+4\geq\tfrac{22}{3}-2\delta\] \[\delta\leq\tfrac{5}{3}\gamma :\ \mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1) =\tfrac{16}{5}\gamma-4\delta+4\geq 2\gamma-2\delta+4\geq\tfrac{22}{3}-2\delta.\] In all cases, \(\mathsf{span}(\delta,\gamma,\frac{3}{5}\gamma,1)\geq\mathsf{span}(\delta,1, \frac{5}{3},1)\). Thus, we have one more correct transformation rule which extends the canonical suffix:  
* given a sequence \((q_{0},\ldots,\delta,\gamma,\frac{3}{5}\gamma,q_{i},\ldots,q_{s})\) with a canonical suffix \((q_{i},\ldots,q_{s})\), replace \((\gamma,\frac{3}{5}\gamma)\) by \((1,\frac{5}{3})\). Starting with an arbitrary sequence \(\vec{q}\), we apply T1 and/or T2 to get a sequence with a nonempty canonical suffix. For any sequence with such a suffix preceded by three or more numbers, at least one of the transformations T3-T5 is applicable. Note that T3 either increases the length of the canonical suffix or makes one of T4, T5 applicable, while each of T4 and T5 increases this length. Thus, we eventually arrive at the situation where the canonical suffix of the current sequence \(\vec{r}=(r_{0},\ldots,r_{s})\) is preceded by 0, 1, or 2 numbers, so that no other transformations are possible. If this suffix begins with \(r_{2}\), then T3 and/or T4 was already applied, and then either \(r_{1}=\frac{5}{3}\) or \(r_{1}=\frac{3}{3}r_{0}\). In particular, \(\vec{r}\) is a good sequence. To find a good sequence of minimum span, we note that \(\mathsf{span}(1,1,1)=4>\frac{10}{3}=\mathsf{span}(1,\frac{5}{3},1)\). Hence, a unique canonical sequence of odd length and minimum span is \((1,\frac{5}{3},1,\frac{5}{3},\ldots,1)\) and one of the canonical sequences of even length and minimum span is \((1,1,\frac{5}{3},1,\frac{5}{3},\ldots,1)\). Now it is easy to find, using (6), good sequences of minimum span. Namely, for even (resp., odd) \(s\), we have \(\vec{r}=(\beta,\frac{3}{5}\beta,1,\frac{5}{3},1,\frac{5}{3},\ldots,1)\) (resp., \(\vec{r}=(\beta,\frac{3}{5}\beta,1,1,\frac{5}{3},1,\frac{5}{3},\ldots,1)\)), where \(\beta=2-\varepsilon\) for \(\varepsilon\) as small as possible. Thus, in the case of even (resp., odd) \(s\), one has \(\mathsf{span}(\vec{r})=\frac{5}{3}s-\frac{14}{15}+\frac{4}{5}\varepsilon\) (resp., \(\mathsf{span}(\vec{r})=\frac{5}{3}s-\frac{3}{5}+\frac{4}{5}\varepsilon\)). Thus, \(\mathsf{span}(\vec{q})>\frac{5}{3}s-1\) for any sequence \(\vec{q}=(q_{0},\ldots,q_{s})\) such that \(q_{i}\in[1,2)\) for all \(i\). Returning to the problem (\(\star\)) we are solving, recall that \(\mathsf{span}(p\vec{q})=p\cdot\mathsf{span}(\vec{q})\), Thus, we have the lower bound \(l_{s}-l_{0}>(\frac{5}{3}s-1)p\). This means, at most, \(s+1\) letters fixed by cubes with periods in \([p..2p-1]\) among each \(l=(\frac{5}{5}sp-p+1)\) consecutive positions of \(w\). Now, easy arithmetic gives \(s+1<\frac{8}{5}+\frac{3l}{5p}\), as required. We split the range from \(h\) to infinity into disjoint finite ranges such that the \(k\)th range is \([2^{k-1}h..2^{k}h-1]\). Consider the density of positions in a cube-free word \(\mathbf{w}\), fixed by \(p\)-cubes with \(p\) in the \(k\)th range. By Lemma 3 and the definition of density, it is upper-bounded by \(\frac{3}{5\cdot 2^{k-1}h}\). Summing up the geometric series of all these upper bounds, we get the required bound \(\frac{6}{5h}\).