I recently solved Project Euler Problem 14 in Python:
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.
But my code takes around 45 seconds to run. I'm looking for some basic ways to make my code more efficient, specifically in the function which calculates each sequence. I have heard in other answers that I should store operations that have already been completed, but how would I do this? I also feel like the method I use for sorting the sequence numbers is clunky and could be optimized.
import time start = time.time() def collatzSeq (n): chainNumber = 1 n1 = n while n1 != 1: if n1 % 2 == 0: n1 = n1/2 chainNumber += 1 else: n1 = (3*n1) + 1 chainNumber += 1 return [chainNumber, n] fullList =  for i in range(2, 1000000): fullList.append(collatzSeq(i)) sortedList = sorted(fullList, reverse=True) print(sortedList[:1]) print('Time:', 1000*(time.time() - start))