The question asks to find an explicit formula for a sequence that satisfies $a_0=1$ and $a_n=3a_{n-1}+\lfloor a_{n-1}\sqrt{5}\rfloor$. The solution can be found here. However, I tried doing an approach that did not follow any of these solutions. I need help continuing the solution as I have hit a roadblock where I believe it is computationally difficult to continue on.
What I started with is the fact that $a_n=\lfloor (3+\sqrt{5})a_{n-1}\rfloor$. Now, I let $\alpha=3+\sqrt{5}$ and $\beta=3-\sqrt{5}$. I then computed that $$\alpha^2=14+6\sqrt{5}$$ $$\alpha^3=72+52\sqrt{5}$$ I then tried finding some patterns. First we have $$\boxed{a_1=\alpha+\beta-1}$$ Then we have that $$a_2=\lfloor \alpha(\alpha+\beta-1)\rfloor$$ $$a_2=\lfloor \alpha^2-\alpha+\alpha\beta\rfloor$$ Note that $-1<\beta^2-\beta<0$. Since $\alpha^2-\alpha+\alpha\beta+\beta^2-\beta$ is an integer, we must have that $$\boxed{a_2=\alpha^2-\alpha+\beta^2-\beta+\alpha\beta}$$ Similarly, we have that $$a_3=\lfloor \alpha(\alpha^2-\alpha+\beta^2-\beta+\alpha\beta)\rfloor$$ $$a_3=\lfloor \alpha^3-\alpha^2+\alpha\beta(\alpha+\beta-1)\rfloor$$ Again, note that $-1<\beta^3-\beta^2<0$. Since $\alpha^3-\alpha^2+\alpha\beta(\alpha+\beta-1)+\beta^3-\beta^2$ is an integer, we have that $$\boxed{a_3=\alpha^3-\alpha^2+\beta^3-\beta^2+\alpha\beta(\alpha+\beta-1)}$$ We then claim that $$a_n=\alpha^n-\alpha^{n-1}+\beta^n-\beta^{n-1}+\alpha\beta(a_{n-2})$$ This is probably not hard to verify (I haven't yet), but I computationally verified this formula for some sequential terms of $a_n$ Note that $\alpha+\beta=6$ and $\alpha\beta=4$. Hence $\alpha$ and $\beta$ are roots of the equation $r^2-6r+4$. We can then get the following $3$ equations $$\begin{cases}a_n=\alpha^n-\alpha^{n-1}+\beta^n-\beta^{n-1}+4a_{n-2}\\ -6a_{n-1}=-6\alpha^{n-1}+6\alpha^{n-2}-6\beta^{n-1}+6\beta^{n-2}-24a_{n-3}\\ 4a_{n-2}=4\alpha^{n-2}-4\alpha^{n-3}+4\beta^{n-2}-4\beta^{n-3}+16a_{n-4}\\ \end{cases}$$ Adding all these 3 equations gives us $$a_n-6a_{n-1}+4a_{n-2}-4(a_{n-2}-6a_{n-3}+4a_{n-4})=0$$ Note that the equation $r^4-6r^3+4r^2-4(r^2-6r+4)=0$ has roots $-2,2,\alpha,\beta$. Hence, using the properties of linear recurrences, we have that $a_n$ must be a linear combination of $(-2)^n,2^n,\alpha^n,\beta^n$
This is where I am stuck. I did try saying that $a_n=A(-2)^n+B(2)^n+C(\alpha)^n+D(\beta)^n$ and plugging in $n=0,1,2,3$. However, that looks pretty hard to solve for $A,B,C,D$. I'm not sure if there is any way to go about this, so I would like some input on if there is anywhere to continue from where I am. If that is not feasible, are there any deviations (aside from those that lead to the official solutions posted) from my solution that I could make to still solve the problem