Some handcrafted, artisanal n'th number Fibonacci finder functions

fibonacci_examples.py

def fib1(n):
    # Using array operations
    fibArr = [0,1,1]
    for i in range(0, n):
        fibArr.pop(0)
        fibArr.append(fibArr[0] + fibArr[1])
    return fibArr[0]

def fib2(n):
    # Using a recursive inner function
    fibArr = [0, 1, 1]  # [last, current, iteration]
    def fibme(myFibArr):
        if n >= myFibArr[2]: 
            return fibme([
                      myFibArr[1], 
                      myFibArr[0] + myFibArr[1], 
                      myFibArr[2] + 1]
                      )
        else:
            return myFibArr[0]
    return fibme(fibArr)

def fib3(n):
    # Using Binet's formula for the nth Fibonacci number (found here: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html)
    from math import sqrt
    phi = (1 + 5 ** 0.5) / 2
    return int((phi**n - (phi*-1)**(n*-1))/sqrt(5))  # Binet's equation as implemented on python gives results that need deburring, e.g. 6765.000000000005

format_spec = "{:>6}" * 4
print(format_spec.format("n", "fib1", "fib2", "fib3"))
for i in range(0, 21):
    print(format_spec.format(i, fib1(i), fib2(i), fib3(i)))

# Output:
#     n  fib1  fib2  fib3
#      0     0     0     0
#      1     1     1     1
#      2     1     1     1
#      3     2     2     2
#      4     3     3     3
#      5     5     5     5
#      6     8     8     8
#      7    13    13    13
#      8    21    21    21
#      9    34    34    34
#     10    55    55    55
#     11    89    89    89
#     12   144   144   144
#     13   233   233   233
#     14   377   377   377
#     15   610   610   610
#     16   987   987   987
#     17  1597  1597  1597
#     18  2584  2584  2584
#     19  4181  4181  4181
#     20  6765  6765  6765