, 2017

Solution:

For $i=1,2, \ldots, 6$, let $b_{i}=a_{i} a_{i+1}$. From the problem condition each of $b_{1}, b_{2}, \ldots, b_{6}$ can only be $-1,0$, or $1$. Since the sum of these six numbers is $4$, either there are five $1$s and a $-1$ or there are four $1$s and two $0$s.

In the first case, there are $6$ ways to choose $i$ such that $b_{i}=-1$. Once that is fixed, determining the value of $a_{1}$ (one of $1$ and $-1$) will determine the value of all the remaining $a_{i}$'s, so there are $6 \cdot 2=12$ possible ways in this case.

In the second case, since if one of $b_{2}, b_{3}, b_{4}, b_{5}$ is zero, then one of the adjacent terms to this zero term must also be zero. Therefore the two zeroes must be next to each other or be $b_{1}$ and $b_{6}$.

If $b_{1}=b_{2}=0$, then $a_{2}$ must be zero. $a_{1}$'s value doesn't matter, and $a_{3}, a_{4}, \ldots, a_{7}$ must have the same sign. The same goes for $b_{5}=b_{6}=0$, giving $3 \cdot 2 \cdot 2=12$ possibilities in these two cases.

If $b_{i}=b_{i+1}=0$ for $i=2,3,4$, then $a_{i+1}$ must be zero. Moreover, $a_{1}, a_{2}, \ldots, a_{i}$ must have the same sign, and so do $a_{i+2}, \ldots, a_{7}$. This gives $2 \cdot 2 \cdot 3=12$ possibilities in these three cases.

If $b_{1}=b_{6}=0$, then $a_{1}=a_{7}=0$. Also, $a_{2}, a_{3}, \ldots, a_{6}$ must have the same sign so there are $2$ possibilities. Combining these cases gives $12+12+12+2=38$ possible sequences in total.