The sequence again for convenience $$ a_{n+1}=a_n-\frac{1}{a_n},\;a_0=2 $$ My friend asked me this question and I do not know how to tackle it. It's clear it does not have a limit, but I am not sure whether it is unbounded; it seems to oscillate with a large amplitude when you simulate it numerically.
I also can't seem to get anywhere with generating functions, but I also don't know how to use them for nonlinear recurrence relations.
Note that there is a 2-cycle if we begin with $$ x = \pm \frac{1}{\sqrt 2} \approx \pm 0.7071, $$ as we then get $$ x - \frac{1}{x} = -x. $$ $$ f(x) = x - \frac{1}{x}. $$ Important: $f$ is odd, $$ f(-x) = -f(x). $$ $$ \mbox{If} \; \; \; f(x) = -x, \; \; \; \mbox{then} \; \; \; f(f(x)) = x.$$
We also get 4-cycles at the roots of $$ 2 x^4 - 4 x^2 + 1, $$ as then $$ f(f(x)) = -x, $$ $$ \mbox{If} \; \; \; f(f(x)) = -x, \; \; \; \mbox{then} \; \; \; f(f(f(f(x)))) = x.$$
The presence of cycles of larger and larger degree tends to support the hypothesis of chaotic behavior elsewhere.
6-cycles at the roots of $$ 2x^8 - 11x^6 + 17x^4 - 8x^2 + 1 $$ $$ \mbox{If} \; \; \; f(f(f(x))) = -x, \; \; \; \mbox{then} \; \; \; f(f(f(f(f(f(x)))))) = x.$$ Now that I think of it, this diagram also shows some 3-cycles.
Seems to me that these points involved in $2k$-cycles might be dense in the real line. If so, that would be strong evidence.
Wildly unstable. I did the first 25 steps with the given $x$ seed value in rational number arithmetic,