I can satisfy conditions 1, 3 and almost satisfy condition 2. Letting $f_n(t) = {t+n-1 \choose n}$ we have the well-known generating function

$\displaystyle \sum_{n \ge 0} f_n(t) x^n = \frac{1}{(1 - x)^{-t}}$

which is rational in $x$ for any fixed *integer* value of $t$. I think condition 2 will end up being the hardest to satisfy because rational functions are quite rigid.

**Edit 1:** My current opinion is that the conditions are not satisfiable. Based on the analogous situation with linear homogeneous recurrences on the integers I am going to conjecture that any polynomial sequence which obeys a polynomial linear recurrence and is not essentially a geometric series has terms divisible by irreducible polynomials of arbitrarily high order.

**Edit 2:** A very strong result available in the integer case is Zsigmondy's theorem, but we don't need a result this strong. Here's a nice result in the integer case. Suppose an integer sequence $a_n$ satisfies a linear homogeneous recurrence with integer coefficients, and let $p$ be a prime not dividing those coefficients. Then the sequence $a_n \bmod p$ is periodic (not just eventually periodic) $\bmod p$ by Pigeonhole. If in addition there exists $n$ such that $a_n = 0$ and $a_n$ is unbounded, then it follows that there is a nonzero term of the sequence divisible by $p$. For example, this is true of the Fibonacci sequence; in fact we have the much stronger result that for $p > 5$, either $p | F_{p+1}$ or $p | F_{p-1}$.

My guess is that a result like this holds in the polynomial case with $p$ replaced by a monic irreducible polynomial (say, of degree $2$), although the argument above breaks down as written.