/*
Description:
The Fibonacci numbers are the numbers in the following integer sequence (Fn):

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

such as

F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying

F(n) * F(n+1) = prod.

Your function productFib takes an integer (prod) and returns an array:

[F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)
depending on the language if F(n) * F(n+1) = prod.

If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prodyou will return

[F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)
F(m) being the smallest one such as F(m) * F(m+1) > prod.

Examples
productFib(714) # should return [21, 34, true], 
                # since F(8) = 21, F(9) = 34 and 714 = 21 * 34

productFib(800) # should return [34, 55, false], 
                # since F(8) = 21, F(9) = 34, F(10) = 55 and 21 * 34 < 800 < 34 * 55
Notes: Not useful here but we can tell how to choose the number n up to which to go: we can use the "golden ratio" phi which is (1 + sqrt(5))/2 knowing that F(n) is asymptotic to: phi^n / sqrt(5). That gives a possible upper bound to n.

You can see examples in "Example test".
*/

function productFib(prod) {
  let fib = [0, 1];
  for (let i = 0; fib[i + 1] < prod; i++) fib.push(fib[i] + fib[i + 1]);
  for (let j = 0; j < fib.length; j++) {
    if (fib[j] * fib[j + 1] === prod) return [fib[j], fib[j + 1], true];
    if (fib[j] * fib[j + 1] !== prod && fib[j] * fib[j + 1] > prod)
      return [fib[j], fib[j + 1], false];
  }
}
