An algorithem for finding the number of primes of the form 4k+3 under some n
Here is a simple algorithm in Python that count the primes of the form $4k+3$ that are less than a positive integer $n$. It uses the standard sieve of Eratosthenes.
import math

def number_of_primes_4kp3(n):
    is_prime = [True] * n

    for p in range(2, int(math.sqrt(n)) + 1):
        if is_prime[p]:
            for j in range(p * p, n, p):
                is_prime[j] = False

    count = 0
    for i in range(3, n, 4):
        if is_prime[i]:
            count += is_prime[i]

    return count

print(&quot;n    &quot;, &quot;number of primes 4k+3&quot;)
for i in range(1, 9):
    count = number_of_primes_4kp3(10 ** i)
    print(&quot;10**&quot; + str(i), count)

It takes about 20 seconds for the Python program above on an old Mac computer of mine to compute number_of_primes_4kp3(10 ** 8), returning 2880950.
When I use Java instead and only compute is_prime for odd numbers (which doubles the performance roughly), it takes less than half a second.