1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=u, a_{n+1}=\frac{v}{a_{n}} + 1\left(u \in \mathbf{R}, v \geqslant-\frac{1}{4}\right)$. If the set formed by the sequence $\left\{a_{n}\right\}$ is a finite set, then $u, v$ should satisfy $\qquad$ .
1. $u^{2}=u+v$.
Since the sequence $\left\{a_{n}\right\}$ forms a finite set, there exist $m, n \in \mathbf{N}_{+}$ such that $a_{m}=a_{n}$. Also, $a_{n+1}$ is uniquely determined by $a_{n}$, so from the $n$-th or $m$-th term onward, the sequence $\left\{a_{n}\right\}$ is periodic.

Since $a_{n+1}-\lambda=\frac{(1-\lambda)\left(a_{n}-\lambda\right)}{a_{n}}$ (here $\lambda^{2}-\lambda-v=0$ has real solutions), when $v=-\frac{1}{4}$, there is only $\lambda_{1}=\frac{1}{2}$, hence
$$
\frac{1}{a_{n+1}-\frac{1}{2}}=\frac{1}{a_{n}-\frac{1}{2}}+2\left(a_{n} \neq \frac{1}{2}\right) \text {. }
$$

In this case, $\left\{a_{n}\right\}$ cannot be a periodic sequence.
Also, $a_{n+1}=\frac{1}{2} \Leftrightarrow a_{n}=\frac{1}{2}$, so $a_{1}=u=\frac{1}{2}=a_{n}$, which satisfies $u^{2}=u+v$.
When $v>-\frac{1}{4}$, there are $\lambda_{1,2}=\frac{1 \pm \sqrt{1+4 v}}{2}$.
Similarly, $a_{n+1}=\lambda_{i} (i=1,2) \Leftrightarrow a_{n}=\lambda_{i}$.
Thus, $u=\lambda_{i}$, meaning that when $u^{2}=u+v$, $a_{n}=\lambda_{i}$ is a constant sequence.
When $a_{n} \neq \lambda_{i}$, by
$$
\left(\frac{a_{n+1}-\lambda_{1}}{a_{n+1}-\lambda_{2}}\right)=\left(\frac{1-\lambda_{1}}{1-\lambda_{2}}\right)\left(\frac{a_{n}-\lambda_{1}}{a_{n}-\lambda_{2}}\right)
$$

and the fact that there are two equal terms in $\left\{a_{n}\right\}$, we know there exists $k \in \mathbf{N}$ such that $\left(\frac{1-\lambda_{1}}{1-\lambda_{2}}\right)^{k}=1$, which is a contradiction.