Problem 1650

[Proof] It is sufficient to prove the case for $k \geqslant 1$. From the recursive formula, we have
$$a_{n} \equiv a_{n-2}(\bmod 2), n=2,3,4, \cdots$$

Given $a_{0}=0, a_{1}=1$, we have
$$2\left|a_{n} \Leftrightarrow 2\right| n,$$

which means the conclusion holds for $k=1$. Next, we prove the case for $k>1$. From the recursive formula and $a_{0}=0, a_{1}=1$, we can easily obtain the general term formula
$$a_{n}=\frac{\sqrt{2}}{4}\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right],$$

Thus, $\quad a_{2^{m} n}=\frac{\sqrt{2}}{4}\left[(1+\sqrt{2})^{2^{\prime \prime \prime} n}-\left(1-\sqrt{2}^{2^{2 m} n}\right]\right.$.
Let $\quad(1+\sqrt{2})^{2 m}+(1-\sqrt{2})^{2^{m}}=2 \sum c_{2}^{k m} 2^{\frac{k}{2}}=2 b_{2}{ }^{m \prime}$,
where the sum is over all even numbers in $\left[0,2^{m}\right]$, and
$$(1+\sqrt{2})^{2^{\prime \prime \prime}} \cdot(1-\sqrt{2})^{2^{m}}=\left\{\begin{array}{r} -1, m=0 \\ 1, m>0 \end{array}\right.$$

Therefore, $\left[\lambda-(1+\sqrt{2})^{2^{m}}\right]\left[\lambda-(1-\sqrt{2})^{2^{\prime \prime}}\right]=\left\{\begin{array}{l}\lambda^{2}-2 b_{2}{ }^{m} \lambda-1, m=0, \\ \lambda^{2}-2 b_{2}{ }^{m} \lambda+1, m>0 .\end{array}\right.$ From this, we can derive the recursive formula for the sequence $\left\{a_{2}{ }^{m}{ }_{n}\right\}, n=0,1,2, \cdots$ as
$$a_{2}{ }^{m}{ }_{n}=\left\{\begin{array}{l} 2 b_{2}{ }^{m} a_{2}{ }^{\prime \prime \prime}(n-1)+a_{2}{ }^{\prime \prime \prime}(n-2), m=0, \\ 2 b_{2}{ }^{\prime \prime} a_{2}{ }^{m \prime \prime}(n-1)-a_{2}{ }^{m \prime}(n-2), m>0 . \end{array}\right.$$

Consider the sequence $\quad b_{n}=\frac{1}{2}\left[(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}\right], n=0,1,2, \cdots$
It is easy to see that $b_{0}=b_{1}=1, b_{n}=2 b_{n-1}+b_{n-2}$, for any $n \geqslant 2$.
Thus, for any non-negative integer $n, b_{n}$ is odd. From the recursive formula, we can see that when $m \geqslant 1$, we have

Thus, $\quad a_{2}{ }^{m}=2^{m} b_{2^{m-1}} b_{2^{m}}{ }^{m} \cdots b_{1}$.
Since $b_{1}, b_{2}, \cdots, b_{2}{ }^{m-1}$ are all odd, we have
$$2^{m}|| a_{2^{m}}$$

which means $2^{m} \mid a_{2}{ }^{m}$ but $2^{m+1} \mid a_{2}{ }^{m}$. From the recursive formula, we can also get
$$2^{m} \mid a_{2^{m}}{ }_{n} \text { and } a_{2}{ }^{m}(n+2) \equiv a_{2} m_{n}\left(\bmod 2^{m+1}\right), n=0,1,2 \cdots \text { . }$$

From $2^{m+1} \mid a_{0}=0,2^{m+1} \times a_{2} m$, we can conclude that for any $m \geqslant 1$,
$$2^{m+1}\left|a_{2^{m}}{ }_{n} \Leftrightarrow 2\right| n .$$

For any natural number $k>1$. If $2^{k} \mid n$, then $n=2^{k-1} l$, where $l$ is even. From the above result, we have $2^{k} \mid a_{n}$. Conversely, if $2^{k} \times n$, then there exists $0 \leqslant m \leqslant k-1$ such that $n=2^{m} l$, where $l$ is odd. Thus, $2^{m+1} \times a_{n}$, and since $k \geqslant m+1$, we have $2^{k} \times a_{n}$. The above shows that for $k>1$, the conclusion also holds.