Find out the number under 1 million which produces the largest Collatz Sequence So, I have written a code which calculates the number under n( 1 million in this case) which produces the largest Collatz Sequence in that range().

The problem is given in detail below:



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.



My code:

def main(n):
    list = [1,2]
    for i in range(3,n):
        count = 0
        while i &gt; len(list):
            if i % 2 == 0:
                i = i/2
            else:
                i = 3*i+1
            count += 1
        list.append(count + list[int(i)-1])
    return list.index(max(list))+1

print(main(1000000))


So, what are the ways in which I can make my code more efficient since it takes quite a lot of time to execute?
I changed your main function a little bit to LargestCollatz which runs a bit faster on my machine. I'm not sure but I think the biggest improvement is in comparing two integers (c &lt; i) instead of the len function (i &gt; len(list)). Futhermore I used the impopular break statement and used // (which might be faster) instead of / so i could eliminate an int function. 

def LargestCollatz(n):

    # Collatz lengths, first entries are 'hand calculated'
    cl = [0, 1, 2]

    for i in range(3, n):
        c = i
        length = 0
        while True:
            if c % 2 == 0:
                c = c//2
            else:
                c = 3*c+1
            length += 1

            if c &lt; i:
                cl.append(length + cl[c])
                break

    return cl.index(max(cl))