Product of consecutive Fib numbers.js

// 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".

// References
// http://en.wikipedia.org/wiki/Fibonacci_number

// http://oeis.org/A000045

const productFib = prod => {
  let arr = [0, 1];
  let num1, num2, flag = false;
  for(let i = 2; i <= prod; i++){
      let muti = arr[i-2] * arr[i-1];
      if (muti < prod){
          arr[i] = arr[i-2] + arr[i-1]
      }else {
          if (muti === prod){
              flag = true;
          }
          num1 = arr[i-2];
          num2 = arr[i-1];
          return [num1, num2, flag];
      }
  }
}