Question

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:

1. In the setting of the original question, is it true that some composition power of $P$ takes integers to integers?
2. The original question with rationals.

Answer 1

$P(x)= \frac{x(x+1)}{2} +1$.

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: