Prime numbers sum - for loop and big numbers

Question

I'm running the following code to find the sum of the first 10,000,000 prime numbers. How can I optimize it such that it doesn't take forever to obtain the result(the sum of prime numbers)?

sum=0
num=2
iterator=0

while iterator<10000000:
    prime = True

    for i in range(2,num):
        if (num%i==0):
            prime = False

    if prime:
        sum=sum+num
         # print (num, sum, iterator)
        iterator=iterator+1
    num=num+1

print(sum)

Answer 1

the 10,000,000 th prime is approximately n * ln(n) + n * ln( ln(n) ) or ~188980383 ... then you can use a sieve to find all primes under that value (discard any extras ... (ie you will get about 50k extra prime numbers when using 10million, note this took approximately 8 seconds for me))

see also : Finding first n primes?

see also : Fastest way to list all primes below N

Answer 2

You can use the Sieve of Eratosthenes . It's a much faster method to find the first n prime numbers.

Answer 3

It turns out the Sieve of Eratosthenes is fairly simple to implement in Python by using correct slicing. So you can use it to recover the n first primes and sum them.

It was pointed out in Joran Beasley's answer that an upper bound for the n -th prime is n * ln(n) + n * ln( ln(n) ) , which we can use and then discrard the extra primes. Note that this bound does not work for n smaller than 6.

from math import log

def sum_n_primes(n):
    # Calculate the upper bound
    upper = int(n * ( log(n) + log(log(n))))

    # Prepare our Sieve, for readability we make index match the number by adding 0 and 1
    primes = [False] * 2 + [True] * (upper - 1)

    # Remove non-primes
    for x in range(2, int(upper**(1/2) + 1)):
        if primes[x]:
            primes[2*x::x] = [False] * (upper // x - 1) # Replace // by / in Python2

    # Sum the n first primes
    return sum([x for x, is_prime in enumerate(primes) if is_prime][:n])

It takes a few seconds, but outputs that the sum of the 10,000,000 first primes is 870530414842019 .

If you need n to be any higher, one solution would be to improve your upper bound .