# Task
 Consider the following operation:
 
 We take a positive integer `n` and replace it with the sum of its `prime factors` (if a prime number is presented multiple times in the factorization of `n`, then it's counted the same number of times in the sum). 
 
 This operation is applied sequentially first to the given number, then to the first result, then to the second result and so on.., until the result remains the same.

  Given number `n`, find the final result of the operation.

# Example

  For `n = 24`, the output should be `5`.
```
24 -> (2 + 2 + 2 + 3) = 9 -> (3 + 3) = 6 -> (2 + 3) = 5 -> 5.
So the answer for n = 24 is 5.```

# Input/Output


 - `[input]` integer `n`

  Constraints: `2 ≤ n ≤ 10000.`


 - `[output]` an integer

Difficulty: EASY
Tags: Ad-hoc

Starter code:
def factor_sum(n):
	

Solutions:
def factor_sum(n):
	while True:
		i = 2
		s = 0
		a = n
		while i <= n:
			if n % i == 0:
				s += i
				n /= i
				if n == 1:
					break
			else:
				i += 1
		if a != s:
			n = s
		else:
			return s


from functools import lru_cache

@lru_cache(maxsize=None)
def factor_sum(n):
	(i, res, save) = (2, 0, n)
	while i * i <= n:
		while not n % i:
			(n, res) = (n // i, res + i)
		i += 1 + i % 2
	if n > 2:
		res += n
	return res if res == save else factor_sum(res)


def prime_factors(n):
	factors = []
	for k in (2, 3):
		while n % k == 0:
			n //= k
			factors.append(k)
	k = 5
	step = 2
	while k * k <= n:
		if n % k:
			k += step
			step = 6 - step
		else:
			n //= k
			factors.append(k)
	if n > 1:
		factors.append(n)
	return factors

def factor_sum(n):
	while True:
		fs = sum(prime_factors(n))
		if fs == n:
			return fs
		n = fs


from collections import Counter

def factors(n):
	c = Counter()
	d = 2
	while d * d <= n:
		while n % d == 0:
			n //= d
			c[d] += 1
		d += 1 + (d > 2)
	if n > 1:
		c[n] += 1
	return c

def factor_sum(n):
	while True:
		c = factors(n)
		result = sum((key * value for (key, value) in c.items()))
		if result == n:
			return n
		n = result


r = lambda n: next((p + r(n // p) for p in range(2, int(n ** 0.5) + 1) if n % p < 1), n)
factor_sum = f = lambda n: (lambda s: s < n and f(s) or n)(r(n))


import math

def factor_sum(n):
	primes = [i for i in range(2, int(math.sqrt(n)) + 1)]
	while n not in primes:
		if n == sum(prime_factors(n, primes)):
			return n
		n = sum(prime_factors(n, primes))
	return n

def prime_factors(n, primes):
	for i in primes:
		for j in primes:
			if j > i and j % i == 0:
				primes.pop(primes.index(j))
	factors = []
	for p in primes:
		while n % p == 0:
			n //= p
			factors.append(p)
	if n != 1:
		factors.append(n)
	return factors


from math import ceil

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

def factor_sum(num):
	while not is_prime(num):
		n = num
		factors_sum = 0
		while num % 2 == 0:
			factors_sum += 2
			num /= 2
		i = 3
		max_factor = num ** 0.5
		while i <= max_factor:
			while num % i == 0:
				factors_sum += i
				num /= i
				max_factor = num ** 0.5
			i += 2
		if num > 1:
			factors_sum += num
		if n == factors_sum:
			return n
		num = factors_sum
	return num


def aux(x):
	s = 0
	i = 2
	while x > 1:
		while x % i == 0:
			s += i
			x /= i
		i += 1
	return s

def factor_sum(num):
	l = None
	while l != num:
		l = num
		num = aux(num)
	return l


import math

def factors(n):
	f = []
	while n > 1:
		for i in range(2, n + 1):
			if n % i == 0:
				f.append(i)
				n = n // i
				break
	return f

def factor_sum(n):
	f = factors(n)
	last = sum(f)
	while n != f[0]:
		n = sum(f)
		f = factors(n)
		if f == last:
			return n
		last = f
	return n


def factor_sum(n):
	while True:
		(t, s, d) = (n, 0, 2)
		while d * d <= t:
			while t % d == 0:
				s += d
				t //= d
			d += 1
		if t > 1:
			s += t
		if s == n:
			return s
		n = s