10. Find the number of ways to divide the set $\{1,2, \cdots, n\}$ into three subsets $A_{1}, A_{2}, A_{3}$. The subsets can be empty, and must satisfy the following conditions:
(1) After arranging the elements of each subset in ascending order, the parity of adjacent elements is different;
(2) If the sets $A_{1}, A_{2}, A_{3}$ are all non-empty, then one of the sets must have the smallest element as an even number.

Hint: The partition of the set is determined by a family of sets $A_{1}$, $A_{2}$, $A_{3}$, which satisfy:
$$
A_{1} \cup A_{2} \cup A_{3} = \{1,2, \cdots, n\}_{3}
$$
$$
A_{1} \cap A_{2}=A_{2} \cap A_{3}=A_{3} \cap A_{1}=\phi_{3}
$$

Another arrangement of the sets, for example, $\dot{A}_{2}, A_{3}, A_{1}$ and $A_{1}, A_{2}, A_{1}$, is considered the same partition.
10. Let $A_{1}$ be a set containing the element 1, and let $A_{2}, A$ be two sets such that the smallest element of $A_{2}$ is less than the smallest element of $A_{3}$. Now, numbers are sequentially allocated to subsets that always satisfy certain conditions to construct a partition that meets condition (2). The number 1 must be an element of $A_{1}$, and it will be proven that the subsequent numbers have exactly two possibilities. If $A_{2}$ and $A_{3}$ are both empty, then the next number can either enter $A_{1}$ or $A_{2}$. After a number is allocated to $A_{2}$, conditions (1) and the parity of the number are considered. In fact: if $A_{3}$ is empty, then according to (2), the first number to enter it must have a different parity from the smallest number in $A_{2}$; for non-empty subsets, the parity of the next number is determined by (1). Finally, the proof is completed by induction. Once the first element is allocated to $A_{2}$, there are two possibilities for the next number to enter $A_{2}$ or $A_{3}$. Assuming at some step, the next number can enter $A_{i_{1}}$ or $A_{i_{2}}$, and it indeed enters $A_{i_{1}}$ (where $i_{1}$, $i_{2}$, $i_{3}$ are a permutation of the set $\{1,2,3\}$), because the next number has a different parity, $A_{i_{3}}$ becomes accessible, while $A_{i_{2}}$ cannot be entered. Entering $A_{1}$ is also possible. Thus, the next step still has two possibilities, which provides the inductive step. This results in $2^{n-1}$ ways to partition the numbers.