Prime factorization program Compute the prime factors of a given natural number.
A prime number is only evenly divisible by itself and 1.
Note that 1 is not a prime number.

My solution will catch every prime up to and including 896803. This is the fastest solution I could come up with. I know that hard code can only go so far.

def prime_factors(n):
    factors = []
    found_last_factor = False
    if n == 1: return factors
    if is_prime(n): return [n]
    while found_last_factor == False: 
        for f in range(2, int(n/2+1)):
            if n % f == 0 and is_prime(f):
                n /= f
                factors.append(f)
                if is_prime(n): 
                    factors.append(n)
                    found_last_factor = True
                    break
    return sorted(factors)

def is_prime(f: int) -&gt; bool:
    if f is 1: return False
    for n in (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 
            61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 
            137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 
            199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 
            277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 
            359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 
            439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 
            521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 
            607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 
            683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 
            773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 
            863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947):
        if f == n: return True
        elif f % n == 0: return False
        elif f / n &lt; n: return True
while found_last_factor == False:



Style is subjective, but I know I'm not the only develop who considers comparisons to Boolean literals to be bad style. while not found_last_factor: is an alternative which I consider preferable.



There are various inefficiencies in this code:


    while found_last_factor == False:
        for f in range(2, int(n/2+1)):
            if n % f == 0 and is_prime(f):
                n /= f
                factors.append(f)
                if is_prime(n): 
                    factors.append(n)
                    found_last_factor = True
                    break
    return sorted(factors)




If it were modified to handle repeated prime factors in an inner loop rather than in the outmost loop then


The test is_prime(f) would be unnecessary, because n % f == 0 would already imply the primality of f.
Primes would be discovered in order, removing the need to call sorted.
A lot of unnecessary iterations of f would be saved. You rebutted AJNeufeld's observation on this subject, so consider an alternative perspective: add a line print(f) at the top of the for f loop, and then call prime_factors(1024).

The call to is_prime(n) tests primality by trial division, and is called from a loop which is factoring n by trial division. That's duplication of both code and run time.




/= does floating point division. Use //= for integer division. Similarly int(n/2+1) could be n//2 + 1.