// Solution itérative naive, utilise simplemment une boucle
let fib_it k =
    let mutable temp = 0
    let mutable na = 1;
    let mutable nb = 0
    for i = 1 to k+1 do
        temp <- nb
        nb <- na + nb
        na <- temp
    nb

// Solution récursive qui inverse le déroulement
let rec fib_rec = function
    | xs when xs < 2 -> 1
    | xs -> (fib_rec (xs - 2)) + (fib_rec (xs - 1))