Frogs spinning in a pot In the classic jumping frogs game, we have 15 frogs on a circular dish, each in a pot. The dish is spinning and every few seconds it stops abruptly, so the frogs jump and land again either in their original pot or in one of the two neighboring.
In how many ways can they be rearranged after they fall? All frogs are different.

I have calculated it to be $610+(377+1)*2$ but it doesn't seem correct to me.

Any help is much appreciated.
 A: In the linear case we get Fibonacci numbers because each of the $n$ frogs they either stay put or exchange position with one of their neighbors. One can think of this as tiling some $1\times n$ space with $1\times 1$ dominos (corresponding to the frogs that stay put) and $1\times 2$ dominos (corresponding to the pair of frogs that exchange positions). Denote $g_n$ to be this number of linear arrangements for $n$ frogs. Then $g_1 = 1$ and $g_2 = 2$, and for each $n\ge 3$, we have $g_n = g_{n-1} + g_{n-2}$, because if the first frog stays put, then we have $g_{n-1}$ many ways to arrange the rest, and if the first frog exchanges with the second frog, then we have $g_{n-2}$ many ways. Hence the recurrence, which is just the Fibonacci numbers (shifted). Here we have $g_{13} = 377$, $g_{14} = 610$.
For the circular case, there are several interpretations of what you may want.
In the simplest situation where the pots are all fixed in place (so we will ignore circular symmetry for now), and assuming the frogs will either stay in place or exchange with one of the neighbors. Denote $a_n$ to be the number of outcomes for $n$ many frogs. Then looking at frog number 1, it can either
(1) stay in place, so the rest of the $n-1$ frogs are in a line, giving $g_{n-1}$ many ways;
(2) exchange with frog 2, giving $g_{n-2}$ many ways;
(3) exchange with frog $n$, giving $g_{n-2}$ many ways.
So this gives $a_n = g_{n-1} + 2g_{n-2}$.
Now, there is also the situation where all the frogs jump to the left, or all the frogs jump to the right. Then in this case we have $a_n = g_{n-1} + 2g_{n-2} +2 $. (However if you take rotational symmetry into consideration, these two extra configurations are just rotationally the same no frog moved.) With $n = 15$, it seems to agree with what you got.
However, as the circular dish is spinning, it seems to me that we should take rotational symmetry into consideration, then we will get $a_n = g_{n-1} + 2g_{n-2}$.
If there are any other missing ways of how the frogs can move, then we will need to account for those. Otherwise we have our result.
A: There are two special cases- if frog $1$ moves clockwise and frog $2$ moves clockwise, then they all must move clockwise. The same holds for counterclockwise. In all other cases, each frog must either stay in place, or switch places with a neighbor. First, we think of the frogs as being in a line. When a frog stays in place, we write down $1$. When two frogs switch places, we write a $2$. These numbers will add up to $15$, so we're asking how many integer compositions of $15$ there are with no part greater than $2$. This is counted by the Fibonacci numbers. The $n$-th Fibonacci number $F_n$ counts the number of integer composition of $n-1$ with no part greater than $2$. These rearrangements are counted by $F_{16}=987$.
However, there's one problem with my argument above. By treating the frogs as being in a line instead of a circle, we forget that we can also allow frog $1$ and frog $15$ to switch places. Hence, we also need to count the rearrangements where this happens. Assuming $1$ and $15$ swap positions, we can then treat frogs $2$ through $14$ as being in a line, and do the same thing above. Since there are $13$ frogs left whose positions have not been decided, we count these rearrangements using $F_{14}=377$. This time, there's no problem treating the remaining frogs as being in a line, as frogs $2$ and $14$ are not neighbors.
The total number of these rearrangements is $987+377+2$, which is the same as your answer.
In general, if there are $n$ frogs with $n\geq 3$, then this is counted by $F_{n+1}+F_{n-1}+2$
