RosettaCode- Count in factors created by tapicella - https://repl.it/Hthb/12

RosettaCode- Count in factors.py

'''
http://rosettacode.org/wiki/Count_in_factors

Write a program which counts up from   1,   displaying each number as the multiplication of its prime factors.

For the purpose of this task,   1   (unity)   may be shown as itself.


Example

      2   is prime,   so it would be shown as itself.
      6   is not prime;   it would be shown as   2 × 3 
      2144   is not prime;   it would be shown as   2 × 2 × 2 × 2 × 2 × 67

'''

#Note- I am going to try my own approach to get prime factors rather than looking something up 

prime_array = [0, 1, 2, 3] #This should allow me to optimize by testing division against primes once they are found
def getFactors(num):
  max_factor = num
  if num in prime_array: 
    return [num] 
  else:
    for i in range(2, len(prime_array)): #skip zero and one
      prime = prime_array[i]
      if num%prime == 0:
        return [prime]+getFactors(num//prime)
    #If dividing by a prime doesn't work, start iterating manually
    j = prime+1 
    #max_factor will be num//j.  Example: for 121, we wouldn't need to check 12 because 121//10 = 12, so 12 times anything greater will be more than 121 
    while j < max_factor: 
      if num%j == 0:
        return getFactors(j)+getFactors(num//j)
      else:
        max_factor = num//j 
        j = j+1 
    #if j hits max factor without being a factor, we know that num is prime 
    prime_array.append(num)
    prime_array.sort()
    return [num]
    
for x in range(50):
  str_factors = list(map(str, getFactors(x)))
  print("Factorized %s: %s" % (str(x),' x '.join(str_factors)))
print(getFactors(2144))