## Appendix A Review on the Conversions between the Integers and the Corresponding Binary Sequences  
First, consider the case when a non-negative integer is represented using the octave integer base. Denote mod \((\cdot,\cdot)\) and \([\cdot]\) as the modulo operator and the operator taking the smallest integer of a real number, respectively. Define the following nonlinear map: \(\mathfrak{H}:Z^{+}\cup\{0\}\rightarrow\{-1,1\}\times\{-1,1\}\times\cdots\), such that \(\mathbf{x}^{\prime}=\mathfrak{H}(n)\) and the \(k\)th element of \(\mathbf{x}^{{}^{\prime}}\) is equal to \(2\text{mod}\Big{(}\big{\lfloor}\frac{n}{2^{k-1}}\big{\rfloor},2\Big{)}-1\). This nonlinear map is to convert a non-negative decimal number \(n\) to a vector of a binary sequence \(\mathbf{x}^{{}^{\prime}}\), in which the elements in \(\mathbf{x}^{{}^{\prime}}\) are either \(1\) or \(-1\). For example, if \(n=11\), then \(\mathbf{x}^{{}^{\prime}}=\big{[} \begin{array}{cccccc}1&1&-1&1&-1&-1&\cdots \end{array} \Big{]}^{T}\). Define a vector \(\xi\), in which the \(k\)th element of \(\xi\) is equal to \(2^{k-1}\) and denote the vector \(1\) as the vector with all its elements being equal to \(1\). Then, we have \(\xi^{T}\frac{\mathbf{x}^{{}^{\prime}}+1}{2}=n\). This equation is to convert the vector of the binary sequence \(\mathbf{x}^{{}^{\prime}}\) back to a non-negative decimal number \(n\). From here, we can see that \(\mathfrak{H}\) is invertible. For the completeness, consider the case when a non-negative integer is represented using the non-octave integer bases. Similarly, define the following nonlinear map: \(\mathfrak{H}^{\prime}:Z^{+}\cup\{0\}\rightarrow\{0,\cdots,\lambda^{\prime}-1 \}\times\{0,1,\cdots,\lambda^{\prime}-1\}\times\cdots\), such that \(\mathbf{x}^{{}^{\prime}}=\mathfrak{H}^{\prime}(n)\) and the \(k\)th element of \(\mathbf{x}^{{}^{\prime}}\) is equal to mod\(\Big{(}\big{\lfloor}\frac{n}{\lambda^{k-1}}\big{\rfloor},\lambda^{\prime}\Big{)}\). This nonlinear map is to convert a non-negative decimal number \(n\) to a vector of an integer sequence \(\mathbf{x}^{{}^{\prime}}\) in which the elements in \(\mathbf{x}^{{}^{\prime}}\) are in \(\{0,1,\cdots,\lambda^{\prime}-1\}\). Define a vector \(\xi^{{}^{\prime}}\) in which the \(k\)th element of \(\xi^{{}^{\prime}}\) is equal to \(\lambda^{\prime}{}^{k-1}\). Then, we have \(\xi^{{}^{\prime}T}\mathbf{x}^{{}^{\prime}}=n\). This equation is to convert the vector of the integer sequence \(\mathbf{x}^{{}^{\prime}}\) back to a non-negative decimal number \(n\). From here, we can also see that \(\mathfrak{H}^{\prime}\) is invertible. Appendix B Review on Both the Fractal and the Chaotic Behaviors of the Dynamical System with Its State Vectors Being the Corresponding Binary Sequences of the Consecutive Integers  
Now, denote \(\lambda\) as a complex number with its modulus greater than \(1\). Define \(\Gamma=\big{\{}\lambda^{-(k-1)}:k\geq 1\big{\}}\). Let \(c_{k}\in\{1,-1\}\) for \(k\geq 1\) be a binary combinational coefficient. Define a vector \(\psi\) in which the \(k\)th element of \(\psi\) is \(\lambda^{-(k-1)}\). Let \(y(n)\) be a complex number that can be represented as a binary combination of the elements in \(\Gamma\). That is, \(y(n)=\sum\limits_{k\geq 1}c_{k}\lambda^{-k}\). Denote \(Y\equiv\{y(n):n\geq 0\}\). In this section, we will characterize \(Y\). First, the integer \(0\) is converted to the corresponding binary sequence \(\mathfrak{H}(0)\). Then, we can compute the corresponding complex number \(y(0)\). Next, increment the integer \(0\) to the integer \(1\) and convert the integer \(1\) to the corresponding binary sequence \(\mathfrak{H}(1)\) and compute \(y(1)\). Eventually, we obtain \(Y\). More precisely, for a given binary sequence \(\mathbf{x}(n)\), as \(\mathfrak{H}\) is invertible, the corresponding integer is \(\xi^{T}\Big{(}\frac{\mathbf{x}(n)+1}{2}\Big{)}\).