smc

Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem). We know our valid sequence must end in a $0$. Then, since we cannot have two consecutive $0$s, it must end in a $10$. Now, we only have two cases: it ends with $010$, or it ends with $110$ which is equivalent to $0110$. Thus, our sequence must be of the forms $0\dots 010$ or $0\dots 0110$. In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$. In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$. Therefore, it must be the case that $f(n)=f(n-3)+f(n-2)$, with $n\ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions). It is easy to find $f(3)=1$ since the only possible valid sequence is $010$. $f(4)=1$ since the only possible valid sequence is $0110$. $f(5)=1$ since the only possible valid sequence is $01010$. The recursive sequence is then as follows: $$f(3)=1$$ $$f(4)=1$$ $$f(5)=1$$ $$f(6)=1+1=2$$ $$f(7)=1+1=2$$ $$f(8)=1+2=3$$ $$f(9)=2+2=4$$ $$f(10)=2+3=5$$ $$f(11)=3+4=7$$ $$f(12)=4+5=9$$ $$f(13)=5+7=12$$ $$f(14)=7+9=16$$ $$f(15)=9+12=21$$ $$f(16)=12+16=28$$ $$f(17)=16+21=37$$ $$f(18)=21+28=49$$ $$f(19)=28+37=65$$ So, our answer is $\boxed{65}$.