jmc

For each positive integer $n$, let $S_n=\{k:1\le k\le n\}$, and let $c_n$ be the number of spacy subsets of $S_n$. Then $c_1=2$, $c_2=3$, and $c_3=4$. For $n\ge 4$, the spacy subsets of $S_n$ can be partitioned into two types: those that contain $n$ and those that do not. Those that do not contain $n$ are precisely the spacy subsets of $S_{n-1}$. Those that contain $n$ do not contain either $n-1$ or $n-2$ and are therefore in one-to-one correspondence with the spacy subsets of $S_{n-3}$. It follows that $c_n=c_{n-3}+c_{n-1}$. Thus the first twelve terms in the sequence $\left(c_n\right)$ are $2$, $3$, $4$, $6$, $9$, $13$, $19$, $28$, $41$, $60$, $88$, $129$, and there are $c_{12}=\boxed{129}$ spacy subsets of $S_{12}$.