One of a number of prime number sieves, it is one of the most efficient ways to find all of the smaller primes. The multiples of a given prime are generated as a sequence of numbers beginning from that prime, with constant difference between them that is equal to that prime.

COMING SOON!

```
"""
Sieve of Eratosthones
The sieve of Eratosthenes is an algorithm used to find prime numbers, less than or equal to a given value.
Illustration: https://upload.wikimedia.org/wikipedia/commons/b/b9/Sieve_of_Eratosthenes_animation.gif
Reference: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
doctest provider: Bruno Simas Hadlich (https://github.com/brunohadlich)
Also thanks Dmitry (https://github.com/LizardWizzard) for finding the problem
"""
import math
def sieve(n):
"""
Returns a list with all prime numbers up to n.
>>> sieve(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
>>> sieve(25)
[2, 3, 5, 7, 11, 13, 17, 19, 23]
>>> sieve(10)
[2, 3, 5, 7]
>>> sieve(9)
[2, 3, 5, 7]
>>> sieve(2)
[2]
>>> sieve(1)
[]
"""
l = [True] * (n + 1) # noqa: E741
prime = []
start = 2
end = int(math.sqrt(n))
while start <= end:
# If start is a prime
if l[start] is True:
prime.append(start)
# Set multiples of start be False
for i in range(start * start, n + 1, start):
if l[i] is True:
l[i] = False
start += 1
for j in range(end + 1, n + 1):
if l[j] is True:
prime.append(j)
return prime
if __name__ == "__main__":
print(sieve(int(input("Enter n: ").strip())))
```