######################################################################
# The following iterative sequence is defined for the set of positive integers:
#
# n → n/2 (n is even)
# n → 3n + 1 (n is odd)
#
# Using the rule above and starting with 13, we generate the following sequence:
#
# 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
# It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms.
# Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
#
# Which starting number, under one million, produces the longest chain?
#
# NOTE: Once the chain starts the terms are allowed to go above one million.
######################################################################

longestChain = []
currentChain = []

for i in range(2, 1000000):
    if i not in longestChain:
        print(i)
        currentChain.clear()
        n = i
        currentChain.append(n)

        while n != 1:
            if n % 2 == 0:  # even
                n /= 2
            else:
                n = 3*n + 1

            currentChain.append(n)

        if len(currentChain) > len(longestChain):
            longestChain = currentChain.copy()

print(longestChain[0])
# Solution: 837799