Another Lucas Number?

$\begin{aligned} \sqrt[16]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}}&=k \\ \\ 2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}&=k^{16} \\ \\ \implies k^{16}+\frac{1}{k^{16}}&=2207 \end{aligned}$ Using the identity, $\left(a^n+\dfrac{1}{a^n}\right)^2=a^{2n}+\dfrac{1}{a^{2n}}+2$ , we can scale down to $\begin{aligned} k^{8}+\frac{1}{k^{8}}&=\sqrt{2207+2}=47 \\ \\ k^{4}+\frac{1}{k^{4}}&=\sqrt{47+2}=7 \\ \\ k^{2}+\frac{1}{k^{2}}&=\sqrt{7+2}=3 \\ \\ k+\frac{1}{k}&=\sqrt{3+2}=\sqrt{5} \end{aligned}$ Solving the quadratic $k^2-\sqrt{5}k+1=0$ , we get $k=\dfrac{\sqrt{5}+1}{2}>1$ and $k=\dfrac{\sqrt{5}-1}{2}<1$ . So, we have, $\sqrt[16]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}}=\frac{1+\sqrt{5}}{2}$ Therefore, $a=1,\;b=1,\;c=5,\;d=2\implies a+b+c+d=9$ .