def sieve_of_eratosthenes(limit):
    primes = []
    is_prime = [True] * (limit + 1)
    p = 2
    while p * p <= limit:
        if is_prime[p]:
            for i in range(p * p, limit + 1, p):
                is_prime[i] = False
        p += 1
    for p in range(2, limit + 1):
        if is_prime[p]:
            primes.append(p)
    return primes

# Estimate an upper bound for the 1000th prime number
# Using the approximation n * log(n) + n * log(log(n)) for the nth prime
import math
n = 1000
upper_bound = int(n * math.log(n) + n * math.log(math.log(n)))

# Calculate the first 1000 prime numbers
first_1000_primes = sieve_of_eratosthenes(upper_bound)[:1000]

# Print the first 1000 prime numbers
print(first_1000_primes)