integral transform of Fibonacci polynomials is integral The Fibonacci polynomials are defined recursively by $F_0(x)=0, F_1(x)=1$ and $F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq2$.
While computing certain integrals, I observe the following (numerically) which prompted me to ask:

Question. For $n, k\in\mathbb{N}$, are these always integers?
  $$\int_0^1F_n(k+nz)\,dz$$

To help clarify, here is a list of the first few polynomials:
$$F_2(x)=x, \qquad F_3(x)=x^2+1, \qquad F_4(x)=x^3+2x.$$
 A: The integral of each individual monomial will be integral. First we have the identity
$$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$
so my claim is that
$$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$
By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write
$$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$
it suffices to show that
$\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer.
However we can check that 
$$\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$$ 
and the claim follows.
A: A different view is given with direct use of the Lucas polynomials. Using the relation
$$\frac{d}{dx} \, L_{n}(x) = n \, F_{n}(x)$$
then the integral becomes
\begin{align}
\int_{0}^{1} F_{n}(k + n \, t) \, dt &= \frac{1}{n} \, \int_{k}^{k+n} F_{n}(u) \, du \\
&= \frac{1}{n^2} \, \int_{k}^{k+n} \frac{d}{du} \, L_{n}(u) \, du \\
&= \frac{L_{n}(n+k) - L_{n}(k)}{n^2}.
\end{align}
Since $L_{-n}(x)$ and $L_{n}(-x)$ have relations to $L_{n}(x)$ then the restrictions on $n$ and $k$ become $n \neq 0$. Since the Lucas polynomials evaluated at $x=\text{ integers}$ are integers then the integral evaluates to integers dependent upon $n$ and $k$. 
