Let $a_n$ denote the number of subsets that can be selected from $n$ pots given that at least one is taken and no two consecutive pots are taken.
Number the pots from $1$ to $n$ so that we are considering subsets of $\{1, 2, \ldots, n\}$.
$a_1 = 1$ since the only possible selection is $\{1\}$
$a_2 = 2$ since the only possible selections are the singleton sets $\{1\}, \{2\}$.
$a_3 = 4$ since the only possible selections are $\{1\}, \{2\}, \{3\}, \{1, 3\}$.
$a_4 = 6$ since the only possible selections are $\{1\}, \{2\}, \{3\}, \{4\}, \{1, 3\}, \{1, 4\}, \{2, 4\}$.
Note that when $n = 4$, there are three types of selections that can be made:
- the allowable subsets of $\{1, 2, 3\}$, of which there are $a_3$;
- the union of the set $\{4\}$ with each of the allowable subsets of $\{1, 2\}$, of which there are $a_2$;
- the singleton set $\{4\}$.
In general, the allowable subsets of $\{1, 2, \ldots, n\}$, where $n \geq 3$, consist of
- the allowable subsets of $\{1, 2, \ldots, n - 1\}$, of which there are $a_{n - 1}$;
- the union of the set $\{n\}$ with each of the allowable subsets of $\{1, 2, \ldots, n - 2\}$, of which there are $a_{n - 2}$ since $n$ can only be paired with elements that are less than $n - 1$;
- the singleton set $\{n\}$.
Hence,
\begin{align*}
a_1 & = 1\\
a_2 & = 2\\
a_n & = a_{n - 1} + a_{n - 2} + 1, n \geq 3
\end{align*}
Since $a_1 = F_3 - 1$ and $a_2 = F_4 - 1$, where $F_n$ denotes the $n$th Fibonacci number, this suggests that
$$a_n = F_{n + 1} - 1 + F_{n} - 1 + 1 = F_{n + 2} - 1$$
which can be proved by induction using the observations above.