Orbits of rational functions This is a generalization of Integrality of iterates of rational functions. 
The question is: given an infinite sequence of rational numbers $a_0, a_1, \dotsc, a_n, \dotsc,$ is it always an orbit of $a_0$ under an iteration of a rational function? If the answer is in the affirmative, it will answer the original question. I am guessing it is undecidable, however.
EDIT Noam points out that the answer is not in general affirmative, but how does one decide (maybe if the growth is not too fast the answer is always yes?)
 A: There are many simpler sequences which are not orbits, for example, it is impossible to have $a_n=a_{n+1}\neq a_{n+k}$ for some $k>1$.
This can be generalized to $a_n=a_{n+m}\neq a_{n+km},\; k>1$. In fact such sequences are not orbits of ANY function.
Other conditions can be obtained if $a_n$ has a limit, and tends to it with aproximately geometric speed. Then this limit must be an attracting point, and
using the local linearization at this point one obtains VERY strong conditions on
the sequence, and not only on its growth rate.
EDIT. For example, the following question was studied by Fatou. Instead of a rational function he considers an arbitrary analytic germ $f(z)=\lambda z+\ldots,\; |\lambda|<1$. He takes an orbit
$a_n\to 0$, and makes a generating function 
$$g(\zeta)=\sum_{n=0}^\infty a_n\zeta^n.$$
He proves that $g$ is meromorphic in the whole plane with poles at the geometric
progression $\lambda^{-k},\; k\geq 0$.
Fatou, P.
Sur une classe remarquable de séries de Taylor. 
Ann. de l’Éc. Norm. (3) 27, 43-53 (1910).
A: This cannot be true in general, simply because 
there are countably many rational functions with coefficients in $\bf Q$,
and uncountably many sequences.
To construct an explicit counterexample, just have the $a_n$
grow fast enough, say $a_n = 2^{n!}$.
