Example 13 Given $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)(n \geqslant 1)$, find $a_{n}$.
Let $b_{n}=\sqrt{1+24 a_{n}}$, i.e., $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$. The original recurrence relation becomes
$\left(2 b_{n+1}+1\right)^{2}=\left(b_{n}+3\right)^{2}$.
Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$, i.e., $b_{n+1}=\frac{1}{2} b_{n}+\frac{3}{2}$, and $b_{1}=\sqrt{1+24 a_{1}}=5$. The fixed point of $f(x)=\frac{1}{2} x+\frac{3}{2}$ is 3, so $b_{n}=\left(\frac{1}{2}\right)^{n-1}\left(b_{1}-3\right)+3=2^{2-n}+3$, hence $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)=\left(2^{2 n-1}+3 \cdot 2^{n-1}+1\right) /\left(3 \cdot 2^{2 n-1}\right), n \geqslant 1$.
$a_{n}=\frac{\beta A-\alpha}{A-1}$, where $A=r^{1+k+\cdots+k^{n-1}} \cdot\left(\frac{a_{1}-\alpha}{a_{1}-\beta}\right)^{n^{n}}$.
In particular, we have
(1) If $f(x)=\frac{a x+b}{c x+d}(c \neq 0, a d-b c \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then: (i) $\left\{\frac{a_{n}-\alpha}{a_{n}-\beta}\right\}$ is a geometric sequence with common ratio $r=\frac{a-c \alpha}{a-c \beta}$ if and only if $\alpha, \beta$ are two distinct fixed points of $f(x)$; (ii) $\left\{\frac{1}{a_{n}-\alpha}\right\}$ is an arithmetic sequence with common difference $\frac{2 c}{a+d}$ if and only if $\alpha=\beta$.
(2) If $f(x)=\frac{a x^{2}+b x+c}{d x^{2}+e x+f}\left(d\right.$ and $e$ are not both zero), the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1}=f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then $\frac{a_{n}-\alpha}{a_{n}-\beta}=\frac{a-d \alpha}{a-d \beta} \cdot\left(\frac{a_{n-1}-\alpha}{a_{n-1}-\beta}\right)^{2}(n \geqslant 2)$ if and only if $b e=4 d c, e(e-2 a)=2 d(2 f-$ $b), \alpha$ and $\beta$ are two distinct fixed points of $f(x)$, and when $d \neq 0$, the third fixed point of $f(x)$ is $-\frac{e}{2 d}$.
In particular, (i) If $f(x)=\frac{a x^{2}+b x+c}{e x+f}(a e \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then $\left\{\ln \left(\frac{a_{n}-\alpha}{a_{n}-\beta}\right)\right\}$ is a geometric sequence with common ratio 2, or
$$
\frac{a_{n}-\alpha}{a_{n}-\beta}=\left(\frac{a_{n-1}-\alpha}{a_{n-1}-\beta}\right)^{2} \text {. }
$$
(ii) If $f(x)=\frac{a x^{2}+b x}{d x^{2}+e x+f}(b d \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then
$$
\frac{a_{n}-\alpha}{a_{n}-\beta}=\frac{a-d \alpha}{a-d \beta} \cdot\left(\frac{a_{n-1}-\alpha}{a_{n-1}-\beta}\right)^{2} \quad(n \geqslant 2, \alpha \neq \beta, \alpha \neq 0, \beta \neq 0) .
$$
(iii) If $f(x)=\frac{a x^{2}}{d x^{2}+e x+f}(a d f \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then
$$
\frac{a_{n}-\alpha}{a_{n}}=\frac{a-d \alpha}{a} \cdot\left(\frac{a_{n-1}-\alpha}{a_{n-1}}\right)^{2} \quad(n \geqslant 2, \alpha \neq 0) .
$$
(3) If $f(x)=\frac{a x^{3}+b x^{2}+c x+d}{e x^{2}+f x+g}(a e \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then $\frac{a_{n}-\alpha}{a_{n}-\beta}=\left(\frac{a_{n-1}-\alpha}{a_{n-1}-\beta}\right)^{3}(n \geqslant 2, \alpha \neq \beta)$, or the sequence $\left\{\ln \left(\frac{a_{n}-\alpha}{a_{n}-\beta}\right)\right\}$ is a geometric sequence with common ratio 3, if and only if $b=0, e=3 a, c f=9 a d, f^{2}=$ $3 a(3 g-c), \alpha$ and $\beta$ are two distinct fixed points of $f(x)$, and the third fixed point of $f(x)$ is $-\frac{f}{6 d}$.
In particular, if $f(x)=\frac{a x^{3}+b x^{2}+c x}{e x^{2}+f x+g}(a e \neq 0, c \neq 0)$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq$ $f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then $\frac{a_{n}-\alpha}{a_{n}-\beta}=\left(\frac{a_{n-1}-\alpha}{a_{n-1}-\beta}\right)^{3}(n \geqslant 2, \alpha \neq$ $\beta, \alpha \neq 0, \beta \neq 0)$, or the sequence $\left\{\left.\ln \left(\frac{a_{n}-\alpha}{a_{n}-\beta}\right) \right\rvert\,\right.$ is a geometric sequence with common ratio 3, if and only if $b=f=0, e=3 a, c=$ $3 g, \alpha$ and $\beta$ are two distinct non-zero fixed points of $f(x)$.
(4) If $f(x)=\frac{\alpha(x-\beta)^{k}-\beta(x-\alpha)^{k}}{(x-\beta)^{k}-(x-\alpha)^{k}}(\alpha \neq \beta, k \in \mathbf{N})$, the sequence $\left\{a_{n}\right\}$ is determined by the initial condition $a_{1} \neq f\left(a_{1}\right)$ and the recurrence relation $a_{n}=f\left(a_{n-1}\right)(n \geqslant 2)$, then the sequence $\left\{\ln \left(\frac{a_{n}-\alpha}{a_{n}-\beta}\right)\right\}$ is a geometric sequence with common ratio $k$ (this is the case when $r=1$ in Theorem 11).