示例#1
0
def max_prime_factor(n):
	factors = [1]

	half = sqrt(int(n)) +1 #if we get here, we're too far ahead
	max_value = 1
	for i in itertools.count(3, 2): #no use in checking the pairs... I know they won't be primes
		if i >= half:
			break
		if n % i == 0 and i % 5 != 0 and util.is_prime(i): #let's trim a couple right now too...
			max_value = i
	return max_value
示例#2
0
def factors_of_a_number(n):
	prime_candidates = primes_util.erastot_sieve(int(round(n**0.5)))
	current = int(n)
	factors = {}
	while(current > 1):
		if primes_util.is_prime(current):
			prime = current
			if not current in factors:
				factors[current] =  1
			else:
				factors[current] = factors[current]+1
			break
		for prime in prime_candidates:
			if current % prime == 0: #if the remainder of the division is 0, it's a divisor
				if not prime in factors:
					factors[prime] =  1
				else:
					factors[prime] = factors[prime]+1
				current = current/prime
				break	
	return factors