the number of arrangements of N players around a round table, where each player can sit on one of 3 contiguous chairs
Consider the variant of the problem in which the people are arranged in a segment instead of circle and let $S(n)$ be the number feasible arrangements for $n$ people.
Now lets go back to your original problem, and let's name $C(n)$ the number of feasible arrangements of $n$ people.
Clearly $C(0)=C(1)=1$ and $C(2)=2$, so we will henceforth suppose that $n \ge 3$.
Let's name the people $p_1, \dots, p_n$ where the assigned position of $p_i$ is $i$.
If $p_1$ is sitting in position 1 then the number of possible arrangements of $p_2, \dots, p_n$ is exactly $S(n-1)$.
Otherwise, $p_1$ is sitting either in position $2$ or in position $n \neq 2$. We will only consider the former case since the latter one is symmetric.
Clearly $p_2$ cannot sit in position $2$, so he/she must either sit in position $1$ or in position $3$. If $p_2$ is sitting in position $1$, then the number of possible arrangements of $p_3, \dots, p_n$ is exactly $S(n-2)$.
Otherwise, $p_2$ must be sitting in position $3$, which means that $p_3$ must be sitting in position $4$ (since position $2$ and $3$ are occupied) and, in general, $p_i$ must be sitting in position $(i \bmod n) + 1$. Since this completely determines everyone's position, this case only contributes $1$ to the total number of configurations.
To summarize, we have that:
$$
C(n) = S(n-1) + 2(S(n-2)+1) = S(n-1) + 2S(n-2) + 2.
$$
We are left with figuring out what $S(n)$ is. Clearly $S(0) = S(1) = 1$, so we consider $n \ge 2$.
In this case $p_1$ can only sit in position $1$ or $2$.
If $p_1$ is sitting in position $1$ then there are $S(n-1)$ possible arrangements of $p_2, \dots, p_n$. If $p_1$ is sitting in position $2$, then $p_2$ must be sitting in position $1$ (as it is the only other person that can sit there) and there are $S(n-2)$ possible arrangements for $p_3, \dots, p_n$.
We hence have:
$$S(n) = \begin{cases} 1 &amp; \mbox{if } n \in \{0,1\} \\ S(n-1) + S(n-2) &amp; \mbox{otherwise} \end{cases} = \mathcal{F}_{n+1},$$
where $\mathcal{F}_{i}$ denotes the $i$-th Fibonacci number. Substituting back, we obtain:
$$
C(n) =
\begin{cases}
1 &amp; \mbox{if } n \in \{0,1\} \\
2 &amp; \mbox{if } n = 2 \\ 
\mathcal{F}_n + 2 \mathcal{F}_{n-1} + 2 &amp; \mbox{otherwise}
\end{cases}.
$$