6 Let the sequence of real numbers $\left\{x_{n}\right\}$ satisfy: $x_{0}=0, x_{2}=\sqrt[3]{2} x_{1}, x_{3}$ is a positive integer, and $x_{n+1}=\frac{1}{\sqrt[3]{4}} x_{n}+\sqrt[3]{4} x_{n-1}+\frac{1}{2} x_{n-2}(n \geqslant 2)$. Question: What is the minimum number of integer terms in such a sequence? (Provided by Huang Yumin)
Let $n \geqslant 2$, then
$$
x_{n+1}-\sqrt[3]{2} x_{n}-\frac{1}{\sqrt[3]{2}} x_{n-1}=\frac{1}{\sqrt[3]{4} x_{n}}-\sqrt[3]{2} x_{n}+\sqrt[3]{4} x_{n-1}-\frac{1}{\sqrt[3]{2}} x_{n-1}+\frac{1}{2} x_{n-2}
$$
$$
\begin{array}{l}
=-\frac{\sqrt[3]{2}}{2} x_{n}+\frac{\sqrt[3]{4}}{2} x_{n-1}+\frac{1}{2} x_{n-2} \\
=-\frac{\sqrt[3]{2}}{2}\left(x_{n}-\sqrt[3]{2} x_{n-1}-\frac{1}{\sqrt[3]{2}} x_{n-2}\right) .
\end{array}
$$

Since $x_{2}-\sqrt[3]{2} x_{1}-\frac{1}{\sqrt[3]{2}} x_{0}=0$, we have
$$
x_{n+1}=\sqrt[3]{2} x_{n}+\frac{1}{\sqrt[3]{2}} x_{n-1}(\forall n \geqslant 1) .
$$
The characteristic equation of (1) is $\lambda^{2}=\sqrt[3]{2} \lambda+\frac{1}{\sqrt[3]{2}}$, solving it yields the roots
$$
\lambda=\frac{\sqrt[3]{2}}{2} \pm \sqrt{\frac{\sqrt[3]{4}}{4}+\frac{1}{\sqrt[3]{2}}}=\frac{\sqrt[3]{2}}{2}(1 \pm \sqrt{3}),
$$

Using $x_{0}=0$, we get $x_{n}=A\left(\frac{\sqrt[3]{2}}{2}\right)^{n}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right]$,
$$
\begin{array}{l}
\text { Thus } x_{3}=\frac{A}{4}\left[(1+\sqrt{3})^{3}-(1-\sqrt{3})^{3}\right]=3 \sqrt{3} A, \\
\text { So } \quad A=\frac{x_{3}}{3 \sqrt{3}}, \\
\text { Therefore } x_{n}=\frac{x_{3}}{3 \sqrt{3}}\left(\frac{\sqrt[3]{2}}{2}\right)^{n}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right] . \\
\text { Let } a_{n}=\frac{1}{\sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right] \text {, clearly }\left\{a_{n}\right\} \text { is an even sequence and by } \\
x_{3} \text { being a positive integer and (2), the necessary condition for } x_{n} \text { to be an integer is } 3 \mid n \text {. Let } b_{n}=(1+ \\
\sqrt{3})^{n}+(1-\sqrt{3})^{n}(n=0,1,2, \cdots) \text {, then }\left\{b_{n}\right\rangle \text { is also an even sequence, and it is easy to see that for } \\
\text { any non-negative integers } m, n \text {, we have } \\
\left\{\begin{array}{l}
a_{n+m}=\frac{1}{2}\left(a_{n} b_{m}+a_{m} b_{n}\right), \\
b_{n+m}=\frac{1}{2}\left(b_{n} b_{m}+3 a_{n} a_{m}\right) .
\end{array}\right. \\
\text { Setting } m=n \text { in (3), we get } \\
\left\{\begin{array}{l}
a_{2 n}=a_{n} b_{n}, \\
b_{2 n}=\frac{1}{2}\left(b_{n}^{2}+3 a_{n}^{2}\right) .
\end{array}\right. \\
\text { Suppose } a_{n}=2^{k}, p_{n}, b_{n}=2^{l}, q_{n} \text {, where } n \text { is a positive integer, } k_{n}, l_{n} \text { are positive integers, } \\
p_{n}, q_{n} \text { are odd numbers, and since } a_{1}=b_{1}=2 \text {, i.e., } k_{1}=l_{1}=1 \text {, using (4) we can see that } \\
k_{2}=2, l_{2}=3 ; k_{4}=5, l_{4}=3 ; k_{8}=8, l_{8}=5 \text {. } \\
\text { Generally, by induction, we get } \\
k_{2}=\left\{\begin{array}{l}
1, \quad m=0, \\
2, \quad m=1, \\
2^{m-1}+m+1, \quad m \geqslant 2 ;
\end{array}\right. \\
l_{2^{m}}=\left\{\begin{array}{ll}
1, & m=0, \\
2, & m=1, \\
2^{m-1}+1, & m \geqslant 2 .
\end{array}\right. \\
\end{array}
$$

For any $m_{1}>m_{2} \geqslant 2$, by (3) we get
$$
\begin{array}{l}
\text { Since } a_{3}=12 \text {, so } 31 a_{3} \text {. From (4), we know } 3 \mid a_{3} \cdot 2^{r} \text {, where } r \text { is a non-negative } \\
k_{n}-\frac{2}{3} n \leqslant 1-\frac{n}{6} \text { (6). } \\
\text { If } n \equiv 0(\bmod 4) \text {, since } 3 \mid n \text {, we can set } n=2 r^{\prime} 3^{k} \text {, where } r \geq 2 \text {, } \\
k \geqslant 1, q \text { does not contain the factor } 3 \text {, from (5) we know } k_{n}=2^{r-1} 3^{k} q+r+1 \text {, thus, } \\
k_{n}-\frac{2}{3} n=2^{r-1} 3^{k} q+r+1-2^{r+1} 3^{k-1} q \\
=r+1-2^{-1} 3^{k-1} q \\
\leqslant r+1-2^{r-1}, \\
\text { equality holds if and only if } k=q=1 \text {. Since when } r>3 \text {, } 2^{-1}=(1+1)^{-1}> \\
r+1 \text {, thus when } r>3 \text {, or when } 2 \leqslant r \leqslant 3 \text {, but } k, q \text { are not both } \\
1 \text {, we have } \\
k_{n}-\frac{2}{3} n<0 \text {. } \\
\text { From (1) and (8), if } x_{3}=3 \text {, then }\left\{x_{n}\right\} \text { has only } x_{0}, x_{3}, x_{6}, x_{12}, x_{24} \\
\text { as integers. } \\
\end{array}
$$
```
Integers.
k24}=16=\frac{2}{3}\cdot24, \text{ and } x_{3}, x_{6}, x_{12}, x_{24} \text{ are all integers.}
n}2+1, \text{ so
```
In summary, this type of sequence has at least 5 integer terms.