Consider the quadratic function $f(x)=x^{2}-4 x$, its fixed points are $f(0)=0$ and $f(5)=5$. Let its second-order fixed points be $f(\alpha)=\beta, f(\beta)=\alpha$, then we have $\left\{\begin{array}{l}\alpha^{2}-4 \alpha=\beta, \\ \beta^{2}-4 \beta=\alpha\end{array} \Rightarrow \alpha+\beta=3\right.$,
Substituting $\alpha^{2}-4 \alpha=3-\alpha \Rightarrow \alpha^{2}-3 \alpha-3=0 \Rightarrow \alpha=\frac{3+\sqrt{21}}{2}, \beta=\frac{3-\sqrt{21}}{2}$.
Construct a sequence with a period of 2: $x_{2 k-1}=\alpha, x_{2 k}=\beta$, then all terms in the sequence $\leqslant \alpha \Rightarrow b \leqslant \alpha$.
Below, we prove that there must be some term $\geqslant \alpha$ in the sequence. By contradiction, assume $\forall k \in \mathbf{N}^{*}, x_{k}\beta=f(\alpha)=f(4-\alpha) \Rightarrow x_{k-1}\alpha$ (discard), thus $\forall k \in \mathbf{N}^{*}, \beta\alpha>3 \Rightarrow\left|x_{k}\right|>3^{k-1}\left|x_{1}\right|$ for any positive integer $k$ $\Rightarrow x_{1}=0$, contradiction. In summary, $b=\alpha \Rightarrow(2 b-3)^{2}=21$.
9. Let the recursive sequence $\left\{x_{n}\right\}$ satisfy: $x_{n+1}=x_{n}^{2}-4 x_{n}, n=1,2, \cdots$, If for any initial term $x_{1} \in \mathbf{R}$ and $x_{1} \neq 0$, there exists some term $x_{k} \geqslant b$ in the sequence, then the largest integer not exceeding $(2 b-3)^{2}$ is $\qquad$.