import os
import math
import numpy as np


# Find the sum of all prime numbers below two million

def isDivisible(n, ms):
    res = False
    for m in ms:
        if n % m == 0:
            res = True
            break
    return res


def getNextPrime(ps):
    if len(ps) == 0:
        return [2]
    else:
        p = ps[-1] + 1
        while isDivisible(p, [q for q in ps if q <= np.sqrt(p)]):
            p = p + 1
        ps.append(p)
        return ps

def getFirstNPrimesBad(n):
    assert n >= 0

    l = 0
    ps = []

    while l < n:
        ps = getNextPrime(ps)
        l = l + 1

    return ps

def sieve(n):
    assert n > 1

    ns = [True] * n

    for i in range(2, math.ceil(np.sqrt(n))):
        if ns[i]:
            j = pow(i, 2)

            while j < n:
                ns[j] = False
                j = j + i

    return [i for i,val in enumerate(ns) if val][2:]


def product(xs):
    res = 1

    for x in xs:
        res = res * x

    return res


# Found a formula for this exact question:
# https://oeis.org/A007504
# a(n) = (n^2/2)*(log n + log log n - 3/2 + (log log n - 3)/log n + (2 (log log n)^2 - 14 log log n + 27)/(4 log^2 n) + O((log log n/log n)^3))

# def getSumOfPrimes(n):
#     return 0.5 * pow(n, 2) * (np.log(n) + np.log(np.log(n)) - 1.5 + (np.log(np.log(n)) - 3) / np.log(n) + 
#         (2 * pow(np.log(np.log(n)), 2) - 14 * np.log(np.log(n)) + 27) / (4 * np.log2(n)) + some O term)

# Unusable because of the o term, unfortunately. Also, I don't know exactly for which n prime numbers I want to calculate this sum.
# We will have to do it the hard way


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

    ps = sieve(2000000)
    print(sum(ps))


if __name__ == "__main__":
    main()