Lucas Sequences
October 1, 2013
We begin with a function that computes the first n elements of a Lucas sequence, given the first two elements of the sequence:
(define (lucas l2 l1 n)
(let loop ((n n) (l2 l2) (l1 l1) (ls (list l1 l2)))
(if (zero? n) (reverse ls)
(let ((l (+ l1 l2)))
(loop (- n 1) l1 l (cons l ls))))))
This can be used to calculate the Fibonacci and Lucas numbers:
> (lucas 1 1 20) ; fibonacci numbers
(1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584
4181 6765 10946 17711)
> (lucas 1 3 20) ; lucas numbers
(1 3 4 7 11 18 29 47 76 123 199 322 521 843 1364 2207 3571
5778 9349 15127 24476 39603)
Another approach uses P, Q and the first two elements of the sequence:
(define (lucas p q l2 l1 n)
(let loop ((n n) (l2 l2) (l1 l1) (ls (list l1 l2)))
(if (zero? n) (reverse ls)
(let ((l (- (* p l1) (* q l2))))
(loop (- n 1) l1 l (cons l ls))))))
And here are the Fibonacci and Lucas numbers again:
> (lucas 1 -1 1 1 20)
(1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584
4181 6765 10946 17711)
> (lucas 1 -1 1 3 20)
(1 3 4 7 11 18 29 47 76 123 199 322 521 843 1364 2207 3571
5778 9349 15127 24476 39603)
Our last functions are mutually-recursive functions that compute the nth element of a U or V sequence in time O(log n):
(define (u n p q)
(if (< n 1) 1 (if (= n 1) p
(let ((k (quotient n 2)))
(if (odd? n)
(- (* (u (+ k 1) p q) (v k p q)) (expt q k))
(* (u k p q) (v k p q)))))))
(define (v n p q)
(if (< n 1) 2 (if (= n 1) p
(let ((k (quotient n 2)))
(if (odd? n)
(- (* (v (+ k 1) p q) (v k p q)) (* p (expt q k)))
(- (expt (v k p q) 2) (* 2 (expt q k))))))))
Then we can compute particular elements of the Fibonacci or Lucas sequences:
> (u 22 1 -1) ; fibonacci
17711
> (v 22 1 -1) ; lucas
39603
You can run the program at http://programmingpraxis.codepad.org/90DQYRBD.
Here’s a Haskell version that gives two implementations: a linear update a la
the usual iterative Fibonacci procedure, and a (mutually) recursive one like
the last solution.
I’m away from my computer at the moment so I’m not sure if this works, but I like the following definition better (if it works):
[sourecode lang=”css”]
lucas x0 x1 p q = x0 : scanl (\x y -> p * x – q * y) x1 $ lucas x0 x1 p q
[/sourcecode]
Apologies for the phone keyboard typo.
In Python.
Oops, forgot the recursive formula for Lucas V.
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