AnthonyBloomer

import sys


# Take any positive integer n.
# If n is even, divide it by 2 to get n / 2.
# If n is odd, multiply it by 3 and add 1 to obtain 3n + 1.
# The conjecture is that no matter what number you start with, you will always eventually reach 1.

def collatz(n):
    if n % 2 == 0:

AnthonyBloomer

# convert a decimal to a given base
def d2b(dec, base):
    stack = []
    result = ''
    while dec > 0:
        rem = dec % base
        dec = dec / base
        stack.append(rem)
    while stack:
        result += str(stack.pop())