Project Euler#14 Longest Collatz sequence

#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.

def isEven(n):
    if n%2 == 0:
        return True
    else:
        return False

def isOdd(n):
    if n%2 == 0:
        return False
    else:
        return True


maxChain = 0
maxStart = 0

for i in range(2,1000000):
    n = i
    count = 1


    while n != 1:
        if isEven(n):
            n = n/2
            count += 1

        else:
            n = (3*n) + 1
            count += 1


    if count > maxChain:
        maxChain = count
        maxStart = i

print maxChain,maxStart

#Improvement (Not implemented): only check odd n. it was proved empirically that the chain length of odd starting numbers was always equal to or larger than the previous even starting number.