When are infinitely many points in the orbit of a polynomial integers? This question is inspired by a riddle in math.stackexchange.
Let $P$ be a polynomial, and $O = \{P^{(n)}(0) : n \geq 0\}$ be its orbit under zero (viewed as a set). Suppose that $O$ contains infinitely many integers. Is it true that for some $n$, $P^{(n)}$ is a polynomial with integral coefficients?
We can ask the same question replacing integers with rationals.
EDIT: Nick and David gave simple counterexamples for the first question.
Still open:


*

*In the setting of the original question, is it true that some composition power of $P$ takes integers to integers?

*The original question with rationals.

 A: $P(x)= \frac{x(x+1)}{2} +1$. 
It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.
But I think (didn't check it, might be one of these facts which are obvious but wrong) that 
$$P^{(n)}(x) = \frac{1}{2^{m}} x^{2^n}+....\notin \mathbb{Z} $$
where $m$ is probably $m=2^n+1$.
The right question to ask might be if $f$ maps the integers into the integers....
Disregard the following part, as it was pointed in the comments, it only works if  for each $k$ we can find an $l$ and $n_1,..., n_k$ so that $f^{(n_i)}(0)$ and $f^{(n_i+l)}(0)$ are integers(or rational for the second question). 
EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):
Let $P$ be such a polynomial, and let $m$ be the degree of $P$. Then using the Lagrange interpolation formula, you can reconstruct $P(x)$ from $m+1$ distinct integer values of the type $P^{(k)}(0)$, and since all of these are rational, all the coefficients are rational. Actually this way one can prove the following Lemma:
A: A slight variant that turns out to be non-elemenatry is to replace the polynomial with a rational function. This leads to:
Theorem: Let $R(x)\in\mathbf{Q}(x)$ be a rational function of degree at least 2, let $\alpha\in\mathbf{Q}$ be an initial value, and suppose that the orbit $O_R(\alpha)=\{R^{(n)}(\alpha) : n\ge0\}$ contains infinitely many integers. Then the second iterate $R^{(2)}(x)$ of $R$ is a polynomial.
For specific $R$ there are often easy proofs, but in general the proof seems to require some non-trivial result on Diophantine approximation such as Thue's theorem. There's an exposition of the proof in The Arithmetic of Dynamical Systems (Springer 2007), Section 3.7. See Section 3.8 for a stronger result saying roughly that as $n$ gets large, then the numerator and denominator of $R^{(n)}(\alpha)$ have about the same number of digits. (There's one extra technical condition for this last result.)
