The problem reads:
Let the integers $1,2,\dots,n$ be arranged consecutively around a circle, and let $g(n)$ be the number of ways of choosing a subset, no two consecutive on the circle.
In a previous exercise, I have the same problem but with the numbers arranged in a line. The answer to this previous problem is $f(n)=F_{n+2}$, where $F$ is a Fibonacci number.
So I have two cases:
$n$ is not counted in the $g(n-1)$ subsets, in which case, the number of subsets it's just $F_{n-1+2} = F_{n+1}$. (In other words, taking away $n$ I can arrange the remaining numbers on the line and then just close it in a necklace. No problem here I guess.
Suppose $n$ is counted in the subsets. So I have to take away the $1$ and the $n-1$ and then do the arrangement with $\lbrace 2,3,\dots,n-1,n\rbrace$. In this case I have $n-2$ objects which I can arrange on the line and then form a necklace, I would have $F_{n-2+2}$ ways to do so. However, my book says it should be $F_{n-1}$ ways for this case. Am I misisng something?
Let $S$ be a subset containing $n$; then $S\setminus\{n\}$ can be any subset of the linear array $\{2,\ldots,n-2\}$, and there are $F_{\big((n-2)-1\big)+2}=F_{n-1}$ of those.
That is, you need to remove all three of the elements $n-1,n$, and $1$ from consideration and look at subsets of the remaining $n-3$ elements that have no adjacent elements in the linear arrangement.