🏄Product of consecutive Fib numbers

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

Best Practices

Py First:

def productFib(prod):
  a, b = 0, 1
  while prod > a * b:
    a, b = b, a + b
  return [a, b, prod == a * b]

Py Second:

class Fib(object):

    """docstring for Fib"""

    def __init__(self):
        super(Fib, self).__init__()
        self._a = 0
        self._b = 1

    def __call__(self):
      self._a, self._b = self._b, self._a + self._b
      return self._a, self._b

def productFib(prod):
    # your code
    fib = Fib()
    a, b = fib()
    while (a * b) < prod:
      a, b = fib()

    return [a, b, (a * b) == prod]

Py Third:

def make_fib(a=1, b=1):
    while True:
        yield (a, b)
        a, b = b, a+b

def productFib(prod):
    for n1, n2 in make_fib():
        if n1 * n2 == prod:
            return [n1, n2, True]
        if n1 * n2 > prod:
            return[n1,n2, False]