Consider sequences that consist entirely of $ A$'s and $ B$'s and that have the property that every run of consecutive $ A$'s has even length, and every run of consecutive $ B$'s has odd length. Examples of such sequences are $ AA$, $ B$, and $ AABAA$, while $ BBAB$ is not such a sequence. How many such sequences have length 14?
1. **Understanding the Problem:**
   We need to count the number of sequences of length 14 consisting of 'A's and 'B's such that every run of consecutive 'A's has even length, and every run of consecutive 'B's has odd length.

2. **Generalizing the Problem:**
   Let's generalize the problem for a sequence of length \(2n\). We need to find sequences of length \(2n\) with the given properties.

3. **Analyzing the Structure:**
   - Every run of 'B's must be odd.
   - Every run of 'A's must be even.
   - We must have an even number of chunks of 'B's because of parity issues.

4. **Forming the Sequence:**
   Each chunk of 'B's, except possibly the last one, must be followed by at least two 'A's to separate it from the next chunk of 'B's. Thus, the sequence must look like:
   \[
   \boxed{\text{A's}} \boxed{\text{B's}} BAA \boxed{\text{A's}} \cdots \boxed{\text{B's}} BAA \boxed{\text{A's}} \boxed{\text{B's}} B \boxed{\text{A's}}
   \]
   where each box holds an even number of letters (possibly zero).

5. **Counting the Letters:**
   If we have \(2k\) chunks of 'B's, then we have \(2n - 6k + 2\) letters left to fill the boxes. Since each box must have an even number of letters, we can think of each box as containing pairs of letters. Thus, we have \((n - 3k + 1)\) pairs of letters to distribute into \(4k + 1\) boxes.

6. **Using Combinatorics:**
   By the classic balls-and-bins argument, the number of ways to distribute \((n - 3k + 1)\) pairs into \(4k + 1\) boxes is given by:
   \[
   \binom{(n - 3k + 1) + (4k + 1) - 1}{(4k + 1) - 1} = \binom{n + k + 1}{4k + 1}
   \]

7. **Summing Over All Possible \(k\):**
   The total number of sequences is the sum over all possible values of \(k\):
   \[
   \sum_{k} \binom{n + k + 1}{4k + 1}
   \]

8. **Evaluating for \(n = 7\):**
   For \(n = 7\), we need to evaluate:
   \[
   \sum_{k} \binom{8 + k}{4k + 1}
   \]
   The possible values of \(k\) are such that \(4k + 1 \leq 14\), i.e., \(k \leq 3\).

   - For \(k = 0\):
     \[
     \binom{8}{1} = 8
     \]
   - For \(k = 1\):
     \[
     \binom{9}{5} = 126
     \]
   - For \(k = 2\):
     \[
     \binom{10}{9} = 10
     \]
   - For \(k = 3\):
     \[
     \binom{11}{13} = 0
     \]

   Summing these values:
   \[
   8 + 126 + 10 = 144
   \]