Recursive sequence and floor function in 2019!

Let us assume that there is a sequence $b_n,$ such that $b_n \neq 0$ and $a_n=\frac{b_n}{b_{n-1}}$ for all $n.$ By substituting $a_n$ by $\frac{b_n}{b_{n-1}}$ into the given recursive relation, we obtain that $b_n$ must satisfy $b_{n+1}= 2 \cos \frac{\pi}{\sqrt 2 }b_n- b_{n-1}.$ Let us also assume that $b_0=b_1=1,$ such that $a_1=\frac{b_1}{b_{0}}=1.$ Then the sequence $b_n$ is a sequence defined by the given linear recursion. The characteristic polynomial of the linear recurrence is $r^2-2 \cos \frac{\pi}{\sqrt 2 }+1,$ with complex zeros $\cos \frac{\pi}{\sqrt 2 }\pm i \sin \frac{\pi}{\sqrt 2 }.$ Therefore, $b_n= c_1 \cos {(n \frac{\pi}{\sqrt 2 })}+ c_2 \sin {(n \frac{\pi}{\sqrt 2 })}.$ Using the given values of $b_0$ and $b_1,$ to find the values of $c_1$ and $c_2,$ and transforming the expression a bit, we obtain that $b_n=\sec{\frac{\pi}{2\sqrt{2}}} \cos {(\frac{\pi}{2\sqrt{2}}(1-2n))}.$ Now it is not difficult to see that $b_n\neq 0$ for any value of $n,$ because the argument of the cosine function wont be an integer multiple of $\pi$ ever. So, $a_n= \frac{b_n}{b_{n-1}}= \frac {\cos {(\frac{\pi}{2\sqrt{2}}(1-2n)})}{\cos {(\frac{\pi}{2\sqrt{2}}(3-2n)})}.$ Using this formula, we obtain that $a_{2019} =2.20106,$ and, therefore, $\lfloor a_{2019} \rfloor=\boxed{2}.$

By direct evaluation $\approx 2.20106122839546$ .