Fibonacci Nested Radical Product

Note that $x \; = \; \frac{\Delta_n}{F_n} \; = \; \sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \cdots}}}}$ is such that $x > \sqrt{2}$ and $\begin{aligned} x^2 & = \; 2 + \sqrt{2 - x} \\ (x^2 - 2)^2 + x - 2 & = \; 0 \\ x^4 - 4x^2 + x + 2 & = \; 0 \\ (x - 1)(x + 2)(x^2 - x - 1) & = \; 0 \end{aligned}$ and hence $x = \tfrac12(\sqrt{5}+1) = \phi$ . Now $F_n \; = \; \tfrac{1}{\sqrt{5}} \big[\phi^n - (-\phi^{-1})^n\big] \hspace{2cm} \Delta_n \; =\; \phi F_n \; = \; \tfrac{1}{\sqrt{5}}\big[ \phi^{n+1} + (-\phi^{-1})^{n-1} \big]$ so that $F_{n-1} + \Delta_n \; = \; \tfrac{1}{\sqrt{5}}\big[\phi^{n-1} + \phi^{n+1}\big] \; = \; \phi^n$ Thus $\prod_{n=1}^N\big(F_{n-1} + \Delta_n) \; = \; \prod_{n=1}^N \phi^n \; =\; \phi^{\frac12N(N+1)} \; = \; F_{a-1} + \Delta_a$ where $a =\tfrac12N(N+1)$ . In this case $N=2017$ , and so $a = \boxed{2035153}$ .