import os
import math
import numpy as np


# Given the following sequence, find the longest chain that ends with 1 and starts with some number under a million

# n -> n/2    | n is even
# n -> 3n + 1 | n is odd

def next(n):
    if n % 2 == 0:
        return int(n/2)
    else:
        return (3*n + 1)

def sequence(n):
    res = [n]

    while res[-1] != 1:
        res.append(next(res[-1]))

    return res


def getLongestSequence(n):
    seqLengths = [0] * n
    seqLengths[0] = 1

    for x in range(1, n):
        i = x + 1
        l = 1

        while True:
            if i == 1:
                break
            elif i < n + 1 and seqLengths[i - 1] > 0:
                l = l + seqLengths[i - 1] - 1
                break
            else:
                l = l + 1
                i = next(i)

        seqLengths[x] = l
        print(x + 1, l)

    return seqLengths


def main():
    print("Hello, this is Patrick")

    t = getLongestSequence(1000000)

    m = t[0]
    j = 0
    for i in range(len(t)):
        if t[i] > m:
            m = t[i]
            j = i
    print("Max", j + 1, m)

if __name__ == "__main__":
    main()