The (pseudo-)linear time algorithm for the Fibonacci numbers can easily be extended to the Lucas numbers:
(define (lucas n)
(let loop ((a 2) (b 1) (n n))
(if (= n 0)
a
(loop b (+ a b) (- n 1)))))
This can be mapped over the integers to get the desired result:
> (map lucas '(0 1 2 3 4 5))
(2 1 3 4 7 11)
> (reverse (map lucas '(0 1 2 3 4 5)))
(11 7 4 3 1 2)
But there's actually a faster way: if you realize that the above algorithm computes Lᵢ₋₁ and Lᵢ₋₂ before it computes Lᵢ, then saving those in a list should save a lot of time. That gives:
(define (lucas n)
(let loop ((a 2) (b 1) (n n) (l '()))
(if (= n 0)
l
(loop b (+ a b) (- n 1) (cons a l)))))
In action:
> (lucas 20)
(9349 5778 3571 2207 1364 843 521 322 199 123 76 47 29 18 11 7 4 3 1 2)