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 $0\ldots010$ or $0\ldots0110$. 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 $\boxed{\text{\bf{(C)}  } 65}$.

Contributors:
~Original Author
~solasky
~BakedPotato66
~[CrazyVideoGamez](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez)
How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$