This question is suggested by this one: prove: coefficients of $f(x)$ are rational numbers
What are the weakest sufficient conditions and strongest necessary conditions on a set of positive integers $S$ such that if $f$ is a polynomial with complex coefficients and $f(n)$ is an integer for all $n \in S$, then all the coefficients of $f$ are rational numbers?
That question shows that a sufficient condition is that $S$ contains all positive integers.
I conjecture that a necessary condition is that, for every prime $p$, $p \in S$.
There may be a similar sufficient condition something like "for all prime powers $p^k$ there is a $n \in S$ such that $p^k | n$."
Can there be such an $S$ with density zero? If not, how small can $\frac{\#\{k|(k \le n) \land (k \in S)\}}{n} $ be?
How about $\sum_{n \in S} \frac1{n}$ diverges?
An intuitive form of this might be:
How "thin" can $S$ be and still force $f(n) \in \mathbb{N}$ for all $n \in S$ implies that the coefficients of $f$ are rational.
How "thick" can $S$ be while a polynomial $f$ with at least one non-rational coefficient exists such that $f(n) \in \mathbb{N}$ for all $n \in S$.
