Fibonacci Recursion

# Own code based on the MIT 6.0001 Python Course Lecture 6 on Recursion and Dictionaries
# Function that generates the nth Fibonacci number and a for loop using a range to test it out
# This method is much slower than the alternative solution (using list iteration) that I submitted to solve the LeetCode #509 Fibonacci number problem faster than most
    # Using list iteration is faster than most LeetCode solutions, while using recursion is much slower to run through due to all the name spaces/scopes of recursion
    # List iteration solution to generate the Fibonacci numbers more quickly is available at "Leetcode #509 (E) Fibonacci Number" in "vixeyfox / LeetCode-Original-Solutions"
# 9/30/2022 Vixey Foxwish Douglas

def recursive_fibo(n):
    """ Recursive way to get nth Fibonacci number where F(0) = 0, F(1) = 1
    and F(n) = F(n-1) + F(n-2) and yields correct 0,1,1,2,3,5,8,13,... """
    if n == 0 or n == 1:
        return n # The first two values for the 0th and 1st Fibonacci number are 0 and 1, respectively
    else:
        next_fibo = recursive_fibo(n - 1) + recursive_fibo(n - 2) # this function's definition calls itself to list a Fibonacci number as sum of two preceding values
        return next_fibo
    
# Testing that the code works for the first 20 Fibonacci numbers which correctly come out as 1,1,2,3,5,8,13, ...
# Also includes code to approximate the Golden ratio as well as a more precise given value for this (1 + sqrt(5) )/2 irrational constant
for n in range(1, 21):
    if n == 1:
        print(recursive_fibo(n), 'is the', n, 'st Fibonacci number.') 
        # special case First -> 1st
    elif n == 2:
        print(recursive_fibo(n), 'is the', n, 'nd Fibonacci number.', end=' ')
        # special case Second -> 2nd
        golden_ratio_appx = recursive_fibo(n)/recursive_fibo(n-1)
        golden_rounded = round(golden_ratio_appx, 6)
        print('Golden ratio approximately', golden_rounded)
    elif n == 3:
        print(recursive_fibo(n), 'is the', n, 'rd Fibonacci number.', end=' ')
        # special case Third -> 3rd
        golden_ratio_appx = recursive_fibo(n)/recursive_fibo(n-1)
        golden_rounded = round(golden_ratio_appx, 6)
        print('Golden ratio approximately', golden_rounded)
    else:
        print(recursive_fibo(n), 'is the', n, 'th Fibonacci number.', end=' ')
        # 4th, 5th, etc. follow the regular "th" rule for naming the sequence
        golden_ratio_appx = recursive_fibo(n)/recursive_fibo(n-1)
        golden_rounded = round(golden_ratio_appx, 6)
        print('Golden ratio approximately', golden_rounded)
phi = 1.618033988749 # sourced from Wikipedia, not the true value, as it is irrational 
print('True golden ratio is actually closer to', phi)