Let the two sequences be $b$ and $\hat{b}$. Then, observe that given $a$, if $b_{1}=\hat{b}_{1}$ and $b_{2}=\hat{b}_{2}$, then $b=\hat{b}$ (since $a$ will uniquely determine the remaining elements in $b$ and $\hat{b}$). Thus, $b$ and $\hat{b}$ must start with $(1,0, \ldots)$ and $(0,1, \ldots)$, respectively (without loss of generality). Note that $a_{3}$ is either 1 (in which case $b_{3}=\hat{b}_{3}=0$) or 2 (in which case $b_{3}=\hat{b}_{3}=1$). Moreover, $b_{4}, b_{5}$ must be the same as $b_{1}, b_{2}$ (and the same for $\hat{b}$) for the sequences to generate the same $a_{3}, a_{4}$. We can then pick $a_{6}, a_{9}, \ldots$. Observe, that the last elements also have to be $(\ldots, 1,0)$ for $b$ and $(\ldots, 0,1)$ for $\hat{b}$. Thus, the answer is nonzero only for sequence lengths of $2(\bmod 3)$, in which case, our answer is $2^{k}$, where the length is $3k+2$ (since we have two choices for every third element). Here, since $N=20=3(6)+2$, the answer is $2^{6}=64$.