Find the number of partitions of the set $\{1, 2, \cdots, n\}$ into three subsets $A_1,A_2,A_3$, some of which may be empty, such that the following conditions are satisfied:

$(i)$ After the elements of every subset have been put in ascending order, every two consecutive elements of any subset have different parity.

$(ii)$ If $A_1,A_2,A_3$ are all nonempty, then in exactly one of them the minimal number is even .

[i]Proposed by Poland.[/i]
To solve the problem, we need to find the number of partitions of the set $\{1, 2, \ldots, n\}$ into three subsets $A_1, A_2, A_3$ such that:
1. Every two consecutive elements of any subset have different parity.
2. If $A_1, A_2, A_3$ are all nonempty, then exactly one of them has an even minimal number.

We will use a combinatorial approach to count the valid partitions.

1. **Partitioning into Consecutive Strings**:
   We start by noting that the condition on parity implies that each subset must alternate between odd and even numbers. This means that each subset can be represented as a sequence of consecutive integers. For example, a valid subset could be $\{1, 2, 3, 4\}$, but not $\{1, 3, 5\}$.

2. **Bijection with Consecutive Strings**:
   We claim that valid partitions are in bijection with partitions of $\{1, \ldots, n\}$ into consecutive strings of integers. For example:
   - $\{1, 2, 3, 4, 5, 6\} \to \{1, 2, 3, 4, 5, 6\}, \varnothing, \varnothing$
   - $\{1\}, \{2, 3, 4, 5, 6\}, \varnothing$
   - $\{1, 2\}, \{3, 4, 5, 6\}, \varnothing$
   - $\{1\}, \{2\}, \{3, 4, 5, 6\}$

3. **Counting the Partitions**:
   To count the number of valid partitions, we need to consider the number of ways to partition $\{1, \ldots, n\}$ into up to three non-overlapping consecutive strings. This is equivalent to counting the number of ways to place two dividers in the sequence $\{1, 2, \ldots, n\}$ to create up to three parts.

4. **Inclusion-Exclusion Principle**:
   We use the inclusion-exclusion principle to count the number of partitions:
   - Total number of ways to partition $\{1, \ldots, n\}$ into three parts (some of which may be empty) is given by the number of ways to place two dividers in $n$ elements, which is $\binom{n+2}{2}$.
   - We need to subtract the cases where all three subsets are nonempty and do not satisfy the condition that exactly one subset has an even minimal number.

5. **Final Calculation**:
   The number of valid partitions is given by:
   \[
   \text{Number of valid partitions} = \binom{n+2}{2} - \text{(invalid partitions)}
   \]
   where invalid partitions are those where all three subsets are nonempty and do not satisfy the condition on the minimal even number.

The final answer is $\boxed{\binom{n+2}{2} - \text{(invalid partitions)}}$.