Recursive numbers.

Consider the integer sequence $(w_n)$ defined by $w_0=0$ , $w_1=1$ and the recurrence relation $w_{n+1} \; = \; w_n + 2w_{n-1} \hspace{2cm} n \ge 1$ It is easy to solve this to show that $w_n \; = \; \tfrac13\big[2^n - (-1)^n\big] \hspace{2cm} n \ge 0$ Note also that $w_{n+1} - 2w_n \; = \; (-1)^n \hspace{2cm} n \ge 0$ Now define the integer sequence $(v_n)$ by setting $v_n \; = \; 2^{w_n} \hspace{2cm} n \ge 0$ so that $v_0=1$ , $v_1=2$ and we have the recurrence relation $v_{n+1} \; = \; v_n v_{n-1}^2 \hspace{2cm} n \ge 1$ We also observe that $\frac{v_{n+1}}{v_n^2} \; = \; 2^{(-1)^n} \hspace{2cm} n \ge 0$ so that $\frac{v_{n+1}}{v_n^2} \in \{2,\tfrac12\}$ for all $n \ge 0$ , and hence $\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}} \; = \; \tfrac52 \hspace{2cm} n \ge 0$ Finally, if we define $u_n \; = \; v_n + v_n^{-1}$ for $n \ge 0$ , then $u_0=2$ , $u_1=\tfrac52$ and $\begin{aligned} u_n(u_{n-1}^2-2)-u_1 & = \; \left(v_n + \frac{1}{v_n}\right)\left(v_{n-1}^2 + \frac{1}{v_{n-1}^2}\right) - \tfrac52 \\ & = \; \left(v_nv_{n-1}^2 + \frac{1}{v_nv_{n-1}^2}\right) + \left(\frac{v_n}{v_{n-1}^2} + \frac{v_{n-1}^2}{v_n} - \frac52\right) \\ & = \; v_{n+1} + \frac{1}{v_{n+1}} \; = \; u_{n+1} \end{aligned}$ so we have arrived at the sequence of this problem. In particular we note that $\lfloor u_n \rfloor = v_n$ for all $n \ge 1$ . When $N$ is odd we see that $\lfloor u_N \rfloor \; = \; v_N \; = \; 2^{w_N} \; = \; 2^{\frac13(2^N + 1)}$ In this case we have $N=2019$ , so that $a=2$ , $b=N=2019$ , $c=1$ and $d=3$ , so that $a+b+c+d=\boxed{2025}$ .