A Snake-y Problem

Let us consider a sequence ${ a }_{ n }$ which denotes the number of ways to reach the ${n}^{th}$ stair. Given the restriction to take steps of $1$ or $2$ , the recursion ${ a }_{ n }\ =\ { a }_{ n-1 }+{ a }_{ n-2 }$ is pretty obvious; as there are only $2$ ways to get to the ${n}^{th}$ stair - by taking a single step from the ${n-1}^{th}$ stair or taking a double step from the ${n-2}^{th}$ stair. Also, it is clear that ${ a }_{ 1 }=1$ and ${ a }_{ 2}=2$ . Using the recursion, we find that ${ a }_{ 6 }=13$ .

At the ${6}^{th}$ stair, the boy has only one choice, i.e., to jump directly to the ${8}^{th}$ stair, hence ${ a }_{ 8 }=13$ . Again, to get to the ${9}^{th}$ stair, he has only one choice, i.e., take a single step and hence ${ a }_{ 9 }=13$ .

Edit 1- It is mandatory for the boy to land on the ${6}^{th}$ stair because of the fact that he can only take steps of $1$ or $2$ . In order for him to reach the ${9}^{th}$ stair, he must either take $2$ steps from the ${7}^{th}$ stair or $1$ from the ${8}^{th}$ . Since, he cannot be on the ${7}^{th}$ stair at any point of time, it must be from the ${8}^{th}$ stair. Moreover, to reach the ${8}^{th}$ stair, again, he must do it from either the ${7}^{th}$ or the ${6}^{th}$ stair. Since the ${7}^{th}$ stair is out of question, it must be the ${6}^{th}$ stair.