1.27 Divide the set $\{1,2, \cdots, n\}$ into 3 subsets $A_{1}, A_{2}, A_{3}$, where empty sets are allowed, and find the number of ways to do this that satisfy the following conditions:
（1）When the numbers in each subset are arranged in increasing order, the parity of any two adjacent numbers is different;
(2) If $A_{1}, A_{2}, A_{3}$ are all non-empty, then exactly one of the subsets has an even number as its smallest element.
[Solution] Let $1 \in A_{1}$, and the smallest element in $A_{2}$ is less than the smallest element in $A_{3}$. We construct $A_{1}, A_{2}, A_{3}$ by sequentially placing $2, 3, \cdots, n$ into these three sets according to the requirements. First, 2 has two placement options: place it in $A_{1}$ or $A_{2}$. Assume that for all natural numbers less than $k$, there are two methods and they have been placed properly. Now, consider the placement of $k$.
(1) Neither $A_{2}$ nor $A_{3}$ has any elements placed in them yet. In this case, $k$ can be placed in $A_{1}$ or $A_{2}$, but not in $A_{3}$, giving a total of two placement options.
(2) $A_{2}$ already has elements, but $A_{3}$ has not yet received any elements.
(i) $k$ and the smallest element in $A_{2}$ are both odd. Since the $k-1$ numbers already placed in sets $A_{1}$ and $A_{2}$ are evenly split between odd and even, the largest elements in these two sets are both even. Thus, $k$ can be placed in $A_{1}$ or $A_{2}$, but not in $A_{3}$, giving a total of two placement options. Similarly, when $k$ and the smallest element in $A_{2}$ are both even, there are also two placement options.
(ii) $k$ and the smallest element in $A_{2}$ have different parities. Suppose $k$ is odd, then the $k-1$ numbers already placed in $A_{1}$ and $A_{2}$ are evenly split between odd and even. Thus, the largest elements in $A_{1}$ and $A_{2}$ have different parities. This means that the odd number $k$ can be placed in either $A_{1}$ or $A_{2}$, and also in $A_{3}$, giving a total of two placement options. Similarly, when $k$ is even, there are also two placement options.
(3) Both $A_{2}$ and $A_{3}$ already have elements. Before this, $k-1$ had two placement options. Assume $k-1$ could be placed in both $A_{1}$ and $A_{2}$, but was actually placed in $A_{1}$. Then, $k$ can be placed in $A_{1}$ or $A_{3}$, but not in $A_{2}$, giving a total of two placement options.

In summary, each number from 2 to $n$ has two different placement options. Therefore, the number of partitions that meet the requirements is $2^{n-1}$.