Call a set of integers "spacy" if it contains no more than one out of any three consecutive integers.  How many subsets of $\{1, 2,
3, \dots, 12\}$, including the empty set, are spacy?
One possible approach is to use a recursive method, where we try to find a relation between the number of spacy subsets of $\{1, 2, 3, \dots, n\}$ and the number of spacy subsets of $\{1, 2, 3, \dots, n-1\}$ or smaller sets. Let's call the number of spacy subsets of $\{1, 2, 3, \dots, n\}$ by $S(n)$. Then, we can notice that any spacy subset of $\{1, 2, 3, \dots, n\}$ either does not contain $n$, or it contains $n$ but not $n-1$ or $n-2$. This means that $S(n) = S(n-1) + S(n-3)$, because we can either add $n$ to any spacy subset of $\{1, 2, 3, \dots, n-3\}$, or we can just take any spacy subset of $\{1, 2, 3, \dots, n-1\}$ that does not contain $n$. This is a recursive relation that we can use to find $S(n)$ for any positive integer $n$. We just need to find some initial values of $S(n)$ for small $n$. For $n = 1$, we have $S(1) = 2$, because the only spacy subsets of $\{1\}$ are the empty set and the whole set. For $n = 2$, we have $S(2) = 3$, because the spacy subsets of $\{1, 2\}$ are the empty set, $\{1\}$, and $\{2\}$. For $n = 3$, we have $S(3) = 4$, because the spacy subsets of $\{1, 2, 3\}$ are the empty set, $\{1\}$, $\{2\}$, and $\{3\}$. Now, we can use the recursive relation to find $S(4)$, $S(5)$, and so on, until we reach $S(12)$. We have $S(4) = S(3) + S(1) = 4 + 2 = 6$. We have $S(5) = S(4) + S(2) = 6 + 3 = 9$. We have $S(6) = S(5) + S(3) = 9 + 4 = 13$. We have $S(7) = S(6) + S(4) = 13 + 6 = 19$. We have $S(8) = S(7) + S(5) = 19 + 9 = 28$. We have $S(9) = S(8) + S(6) = 28 + 13 = 41$. We have $S(10) = S(9) + S(7) = 41 + 19 = 60$. We have $S(11) = S(10) + S(8) = 60 + 28 = 88$. We have $S(12) = S(11) + S(9) = 88 + 41 = 129$. Therefore, the final answer is 129. 