Problem 1562

To solve this problem, we need to find the number of sets of subsets $T = \{A, B, C\}$ such that:

1. $A, B, C \subseteq S$
2. $A \cup B \cup C = S$
3. $(A \cap C) \cup (B \cap C) = \emptyset$
4. No subset contains two consecutive integers.

Let's break down the problem step by step.

1. Condition Analysis:
- $A \cup B \cup C = S$ means every element in $S$ must be in at least one of $A$, $B$, or $C$.
- $(A \cap C) \cup (B \cap C) = \emptyset$ means $A$ and $B$ are disjoint from $C$. Therefore, $C$ must be disjoint from both $A$ and $B$.
- No subset contains two consecutive integers.

2. Disjoint Subsets:
- Since $A$ and $B$ are disjoint from $C$, we can consider $C$ as a subset of $S$ that does not contain consecutive integers.
- The remaining elements in $S \setminus C$ must be partitioned into $A$ and $B$ such that $A$ and $B$ do not contain consecutive integers.

3. Counting Non-Consecutive Subsets:
- We need to count the number of ways to partition $S$ into three subsets $A$, $B$, and $C$ such that no subset contains consecutive integers.
- Let's denote the number of ways to partition $S$ into three such subsets as $f(n)$, where $n$ is the number of elements in $S$.

4. Recursive Approach:
- We can use a recursive approach to count the number of valid partitions.
- For each element $i$ in $S$, we have three choices: put $i$ in $A$, $B$, or $C$.
- However, we need to ensure that no subset contains consecutive integers. This means if $i$ is in $A$, then $i+1$ cannot be in $A$, and similarly for $B$ and $C$.

5. Base Cases:
- For $n = 1$, there are 3 ways to partition $\{1\}$ into $A$, $B$, and $C$.
- For $n = 2$, there are 6 ways to partition $\{1, 2\}$ into $A$, $B$, and $C$ such that no subset contains consecutive integers.

6. General Case:
- For $n \geq 3$, we can use the following recurrence relation:
$$f(n) = 3 \cdot f(n-1) - 2 \cdot f(n-2)$$
- This recurrence relation accounts for the fact that each element can be placed in one of three subsets, but we need to subtract the cases where consecutive integers are placed in the same subset.

7. Calculation:
- Using the recurrence relation, we can calculate $f(7)$:
$$f(1) = 3$$
$$f(2) = 6$$
$$f(3) = 3 \cdot f(2) - 2 \cdot f(1) = 3 \cdot 6 - 2 \cdot 3 = 18 - 6 = 12$$
$$f(4) = 3 \cdot f(3) - 2 \cdot f(2) = 3 \cdot 12 - 2 \cdot 6 = 36 - 12 = 24$$
$$f(5) = 3 \cdot f(4) - 2 \cdot f(3) = 3 \cdot 24 - 2 \cdot 12 = 72 - 24 = 48$$
$$f(6) = 3 \cdot f(5) - 2 \cdot f(4) = 3 \cdot 48 - 2 \cdot 24 = 144 - 48 = 96$$
$$f(7) = 3 \cdot f(6) - 2 \cdot f(5) = 3 \cdot 96 - 2 \cdot 48 = 288 - 96 = 192$$