An ode to 2019

We note that $1 + \omega^k + \omega^{2k} = 3$ when the integer $k$ is a multiple of $3$ , and is equal to $0$ otherwise. Thus $\begin{aligned} 3\sum_{k=0}^{673}\binom{2019}{3k} & = \; \sum_{k=0}^{2019}\binom{2019}{k}\big(1 + \omega^k+\omega^{2k}\big) \; = \; (1 + 1)^{2019} + (1 + \omega)^{2019} + (1 + \omega^2)^{2019} \\ & = \; 2^{2019} + (-\omega^2)^{2019} + (-\omega)^{2019} \; = \; 2^{2019} - 1 - 1 \; = \; 2^{2019} - 2 \end{aligned}$ so that $a=2$ , $b=2019$ , $c=2$ and $d=3$ , making the answer $2+2019 + 2 +3 = \boxed{2026}$ .