We define a sequence with $a_{1}=850$ and

$$
a_{n+1}=\frac{a_{n}^{2}}{a_{n}-1}
$$

for $n \geq 1$. Determine all values of $n$ for which $\left\lfloor a_{n}\right\rfloor=2024$.
Here, the floor $\lfloor a\rfloor$ of a real number $a$ denotes the greatest integer less than or equal to $a$.
Answer: the only value that satisfies is $n=1175$.
First, we note that we can rewrite the recursion as

$$
a_{n+1}=\frac{a_{n}^{2}-1+1}{a_{n}-1}=\frac{a_{n}^{2}-1}{a_{n}-1}+\frac{1}{a_{n}-1}=a_{n}+1+\frac{1}{a_{n}-1} .
$$

Since the difference $a_{n+1}-a_{n}>1$, there is at most one value of $n$ that satisfies. Now we will show that $n=1175$ indeed satisfies. First, the above gives us

$$
\begin{aligned}
a_{1175} & =a_{1}+1174+\frac{1}{a_{1}-1}+\frac{1}{a_{2}-1}+\cdots+\frac{1}{a_{1174}-1} \\
& >850+1174 \\
& =2024
\end{aligned}
$$

More generally, it follows from (??) analogously that $a_{n} \geq 849+n$. From this, we deduce that

$$
\begin{aligned}
\frac{1}{a_{1}-1}+\cdots+\frac{1}{a_{1174}-1} & \leq \frac{1}{849}+\frac{1}{850}+\cdots+\frac{1}{2022} \\
& <51 \cdot \frac{1}{849}+100 \cdot \frac{1}{900}+200 \cdot \frac{1}{1000}+400 \cdot \frac{1}{1200}+400 \cdot \frac{1}{1600}+23 \cdot \frac{1}{2000} \\
& =\frac{17}{283}+\frac{1}{9}+\frac{1}{5}+\frac{1}{3}+\frac{1}{4}+\frac{23}{2000} \\
& =\frac{17}{283}+\frac{20+36+60+45}{180}+\frac{23}{2000} \\
& <\frac{11}{180}+\frac{161}{180}+\frac{3}{180} \\
& <1
\end{aligned}
$$

With this, we conclude that indeed

$$
a_{1175} \leq 850+1174+\frac{1}{849}+\frac{1}{850}+\cdots+\frac{1}{2022}<2024+1
$$