Consider the following iterations :
$x_0 = z$
Where $z$ is complex.
$x_n = \frac{ x_{n-1}^2 - 1}{n}$
It is well known that for real $z > 3$ the sequence grows double exponentially. It is known that for $z = 3$ the sequence grows linear ; in fact like $3,4,5,6,7,...$.
In fact When considering positive sqrt :
$$ 3 = \sqrt{1 + 2 \sqrt{ 1 + 3 \sqrt ...}} $$
Is true.
Since $(-3)^2 = 3^2 $ this explains the behaviour of $x_n$ for real $z$. For real $z$ with $z^2 < 9$ , $x_n$ Goes to zero.
So What remains are the nonreal $z$.
Clearly $z$ behaves like $-z$ and $conj(z)$. Also When $abs > 3$ the sequence goes to complex infinity at double exponential speed.
Im not sure if i may use the term julia set , but something similar is going on here. There is a connected region where the iterations remain bounded.
Are there values that do not grow double exponential nor linear and yet go to infinity ?? ( probably not )
We know ${-3,3}$ are on the boundary of this " Julia set " , So I wonder If all points on the boundary go to infinity ? And do so at linear speed ?
Does the boundary have a Fractal structure ?
Do all interior points go to $0$ ?
Are there attracting points apart from $0$ ?
A plot would be nice too.
Can we compute other values on the boundary apart from $ {-3,3} $ and some trivial ones ?
By trivial I mean for instance $z= \sqrt{-7}$ Because
$$ (( ( z^2 - 1)/2 )^2 - 1 )/3 = 5 $$
Has solutions ${-3,3,-\sqrt{-7},\sqrt{-7}}$.
It would be nice to know the boundary point with $arg(z) = \frac{5}{4}$ for instance.
What is the area of this Julia set ? Is there a parametric description of the boundary ?
Many questions.
This is not an answer, but it is not practical to put it in a comment.
Your iterative scheme can by written on the form $$F(w,z)=\left(w+1,\frac{z^2-1}{w}\right)$$ which is a rational map of $\mathbb C^2$ called a skew-product (because the $w$- coordinate depends only on $w$). You are asking about the dynamical behaviour of the slice $\{w=1\}$. There is a well-defined notion of Julia set in this context; it will indeed be the boundary of those $z$ such that the second component of $F^n(1,z)$ escapes to infinity. That should enable you to draw some pictures. It will probably be a fractal set, but needs not be self-similar in quite the same way as a Julia set.