18. Let $B_{n}$ be the set of sequences with the stated property $\left(S_{n}=\left|B_{n}\right|\right)$. We shall prove by induction on $n$ that $S_{n} \geq \frac{3}{2} S_{n-1}$ for every $n$. Suppose that for every $i \leq n, S_{i} \geq \frac{3}{2} S_{i-1}$, and consequently $S_{i} \leq \left(\frac{2}{3}\right)^{n-i} S_{n}$. Let us consider the $2 S_{n}$ sequences obtained by putting 0 or 1 at the end of any sequence from $B_{n}$. If some sequence among them does not belong to $B_{n+1}$, then for some $k \geq 1$ it can be obtained by extending some sequence from $B_{n+1-6 k}$ by a sequence of $k$ terms repeated six times. The number of such sequences is $2^{k} S_{n+1-6 k}$. Hence the number of sequences not satisfying our condition is not greater than
$$\sum_{k \geq 1} 2^{k} S_{n+1-6 k} \leq \sum_{k \geq 1} 2^{k}\left(\frac{2}{3}\right)^{6 k-1} S_{n}=\frac{3}{2} S_{n} \frac{2(2 / 3)^{6}}{1-2(2 / 3)^{6}}=\frac{192}{601} S_{n}<\frac{1}{2} S_{n}$$
Therefore $S_{n+1}$ is not smaller than $2 S_{n}-\frac{1}{2} S_{n}=\frac{3}{2} S_{n}$. Thus we have $S_{n} \geq\left(\frac{3}{2}\right)^{n}$.