Python Challenge - Semi-Prime Numbers

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def semi_prime_count(n):
count = 0
for i in range(4, n):
facts = numFactors(i)
for fact in facts:
if isPrime(fact[0]) and isPrime(fact[1]):
count += 1
return count

def numFactors(n):
newlst =
num = int(n**0.5)+1
for i in range(2, num):
if n % i == 0:
other = n // i
newlst.append((i, other))
return newlst

def isPrime(n):
if n == 1:
return False
if n == 2:
return True
num = int(n ** 0.5) + 1
for i in range(2, num):
if n % i == 0:
return False
return True

print(semi_prime_count(10))

1 Like
import numpy as np import math def semi_prime_count(n): list_of_primes = [n for n in range(2, n + 1) if is_prime(n)] list_of_semi_primes = [] for x in list_of_primes: for y in list_of_primes: if x * y < n and x * y not in list_of_semi_primes: list_of_semi_primes.append(x*y) return len(list_of_semi_primes) def is_prime(n): if n == 2: return True for i in range(2, math.ceil(np.sqrt(n)) + 1): if n % i == 0: return False return True semi_prime_count(10)
def semi_prime_count(n): def isprime(i): n= int(i/2) while n>=2: if i%n == 0 : return False n = n-1 return True def is_product_two(number): i = 2 count = [] while i<=number: if number%i == 0 and isprime(i): count.append(i) number = int(number/i) else: i += 1 if len(count)==2 or (len(count)==1 and count[0]*count[0]==number): print("True") print(count) return True print("False") print(count) return False count = 0 for i in range (1, n): if is_product_two(i): count +=1 return count print(semi_prime_count(10))