Too long for a comment.
If you are restricting to $n$ being an integer, unless I made a mistake, the problem can be rephrased as Let $$a_{n+1}=4a_n+a_{n-1} \\ a_1=1 \\ a_0=\frac{1}{2}$$$$a_{n+1}=4a_n-a_{n-1} \\ a_1=1 \\ a_0=\frac{1}{2}$$
Then, show that $\lfloor a_n \rfloor$ is prime implies that $n$ is prime.
Here are some notes, I didn't check the details so there may be many mistakes.
Note 1: I think that $a_{2n} \in \mathbb Z +\frac{1}{2}$ and $a_{2n+1} \in \mathbb Z$. This implies that $$ \lfloor a_{2n} \rfloor = a_{2n}-\frac{1}{2} \\ \lfloor a_{2n+1} \rfloor = a_{2n+1} $$
This suggests splitting the problem into odd and even $n$.
Note 2: The odd $n's$. Let $$ b_n=a_{2n+1}=\frac{(2-\sqrt{3})^{2n+1}+(2+\sqrt{3})^{2n+1}}{4}=\frac{(2-\sqrt{3})(7-4\sqrt{3})^{n}+(2+\sqrt{3})(7+4\sqrt{3})^{2n+1}}{4} $$ I think that this is the solution to the recurrence $$ b_{n+1}=14b_n+b_{n-1} \\ b_{1}= \mbox{something} \\ b_0=\mbox{something} $$$$ b_{n+1}=14b_n-b_{n-1} \\ b_{1}= \mbox{something} \\ b_0=\mbox{something} $$
The problem then becomes
Question A Show that $b_{n}$ prime implies that $2n+1$ is prime.
Similarly, $$c_{n}=a_{2n}$$ is the solution to the same recurrence, with different innitial condition.
Setting $$ d_{n}=c_n-\frac{1}{2} $$ we get $$ c_{n}=d_n+\frac{1}{2} $$ and hence the recurrence becomes $$ c_{n+1}=14c_n+c_{n-1} \Rightarrow d_{n+1}=14d_n+d_{n-1}+7 $$$$ c_{n+1}=14c_n-c_{n-1} \Rightarrow d_{n+1}=14d_n-d_{n-1}+6 $$ and the question becomes:
Question B: show that $d_n$ is not prime for $n \geq 2$.
There are some techniques of solving such problems for recurrences, so these comments could help or could be totally useless.
ANd keep in mind that there may be mistakes above.