(14) (50 points) The sequence $\left\{x_{n}\right\}$ is defined as $x_{1}=3, x_{n+1}=\left[\sqrt{2} x_{n}\right]\left(n \in \mathbf{N}^{*}\right)$. Find all $n$ such that $x_{n}, x_{n+1}, x_{n+2}$ form an arithmetic sequence. (Here $[x]$ denotes the greatest integer not exceeding the real number $x$)
14 Because $x_{n}, x_{n+1}, x_{n+2}$ form an arithmetic sequence, we have $x_{n}+x_{n+2}=2 x_{n+2}$, which means
$$
2\left[\sqrt{2} x_{n}\right]=x_{n}+\left[\sqrt{2}\left[\sqrt{2} x_{n}\right]\right] .
$$

Since $x-1x_{n}+\sqrt{2}\left[\sqrt{2} x_{n}\right]-1 \\
& >x_{n}+\sqrt{2}\left(\sqrt{2} x_{n}-1\right)-1 \\
& =3 x_{n}-\sqrt{2}-1 .
\end{aligned}
$$

Combining (1) we get $3 x_{n}-\sqrt{2}-1x_{1}$.
$$
\begin{aligned}
x_{n+1} & =\left[\sqrt{2} x_{n}\right]>\sqrt{2} x_{n}-1 \\
& =x_{n}+x_{n}(\sqrt{2}-1)-1 \\
& \geqslant x_{n}+3(\sqrt{2}-1)-1>x_{n} .
\end{aligned}
$$

By calculation we get
$$
x_{1}=3, x_{2}=4, x_{3}=5, x_{4}=7, x_{5}=9, x_{6}=12, x_{7}=16 \text {. }
$$

Therefore, when $n \geqslant 7$, $x_{n}>15$, so the required $n \leqslant 6$.
Thus, from (2) we know that the $n$ satisfying the condition is 1 or 3.