Call a set of integers [i]spacy[/i] if it contains no more than one out of any three consecutive integers. How many subsets of $ \{1,2,3,\ldots,12\},$ including the empty set, are spacy?
	
$ \textbf{(A)}\ 121 \qquad \textbf{(B)}\ 123 \qquad \textbf{(C)}\ 125 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 129$
1. **Base Cases:**
   - For \( n = 1 \), the spacy subsets are \(\{\}, \{1\}\). Thus, \( a_1 = 2 \).
   - For \( n = 2 \), the spacy subsets are \(\{\}, \{1\}, \{2\}, \{1, 2\}\). However, \(\{1, 2\}\) is not spacy. Thus, \( a_2 = 3 \).
   - For \( n = 3 \), the spacy subsets are \(\{\}, \{1\}, \{2\}, \{3\}, \{1, 3\}\). Thus, \( a_3 = 4 \).

2. **Recurrence Relation:**
   - If \( n \) is in the subset, then neither \( n-1 \) nor \( n-2 \) can be in the subset. Therefore, the number of spacy subsets that include \( n \) is \( a_{n-3} \).
   - If \( n \) is not in the subset, then the number of spacy subsets is \( a_{n-1} \).
   - Combining these cases, we get the recurrence relation:
     \[
     a_n = a_{n-3} + a_{n-1}
     \]

3. **Calculate \( a_n \) for \( n = 4 \) to \( n = 12 \):**
   - Using the recurrence relation and the base cases:
     \[
     \begin{align*}
     a_4 &= a_1 + a_3 = 2 + 4 = 6, \\
     a_5 &= a_2 + a_4 = 3 + 6 = 9, \\
     a_6 &= a_3 + a_5 = 4 + 9 = 13, \\
     a_7 &= a_4 + a_6 = 6 + 13 = 19, \\
     a_8 &= a_5 + a_7 = 9 + 19 = 28, \\
     a_9 &= a_6 + a_8 = 13 + 28 = 41, \\
     a_{10} &= a_7 + a_9 = 19 + 41 = 60, \\
     a_{11} &= a_8 + a_{10} = 28 + 60 = 88, \\
     a_{12} &= a_9 + a_{11} = 41 + 88 = 129.
     \end{align*}
     \]

Thus, the number of spacy subsets of \(\{1, 2, 3, \ldots, 12\}\) is \( \boxed{129} \).