def factorGenerator(n):
p = prime_factors(n)
factors = {}
for p1 in p:
try:
factors[p1] += 1
except KeyError:
factors[p1] = 1
return factors
def smallest_multiple(values):
decompositions = []
for v in values:
factors, powers = prime_factors(v)
decompositions.append(pd.Series(powers, index=factors))
decompositions = pd.concat(decompositions, axis=1, keys=values)
smallest_multiple_decomposition = decompositions.max(axis=1)
factors = pd.Series(smallest_multiple_decomposition.index)
powers = smallest_multiple_decomposition.reset_index(drop=True)
return int((factors**powers).prod())
def is_circular_prime(n):
ret = True
s = str(n)
s_cycle = s + s # double string to generate circular number
if len(s) > 1:
for i in xrange(0, len(s)):
tmp = s_cycle[i:i+len(s)]
if len(prime_factors(int(tmp))) > 1:
ret = False
break
return ret
def n_divisors(number):
'''
Form of the sigma function for counting the number of divisors
of a given number that relies on the prime factorization of the
given number
Source:
https://en.wikipedia.org/wiki/Divisor_function#Formulas_at_prime_powers
'''
pf = prime_factors(number)
return product((x + 1) for x in pf.values())
def main():
'''
2520 is the smallest number that can be divided by each of the numbers from 1 to
10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the
numbers from 1 to 20?
'''
# Figure out the max power of each prime factor in 2..20
factors = defaultdict(int)
for pf in [prime_factors(i) for i in range(2, 21)]:
for k, v in pf.items():
factors[k] = max(factors[k], v)
return product(k**v for k, v in factors.items())
def is_prime(n):
f, p = prime_factors(n)
return len(f) == 1 and p[0] == 1
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
'''
from problem3 import prime_factors
def is_circular_prime(n):
ret = True
s = str(n)
s_cycle = s + s # double string to generate circular number
if len(s) > 1:
for i in xrange(0, len(s)):
tmp = s_cycle[i:i+len(s)]
if len(prime_factors(int(tmp))) > 1:
ret = False
break
return ret
N = 1000000
count = 0
for i in xrange(2, N+1):
if len(prime_factors(i)) == 1:
if is_circular_prime(i):
count += 1
print i, count
print '==result=='
print count
def test_prime_factors(self):
self.assertEqual(problem3.prime_factors(2), [2])
self.assertEqual(problem3.prime_factors(4), [2,2])
self.assertEqual(problem3.prime_factors(6), [2,3])
self.assertEqual(problem3.prime_factors(8), [2,2,2])
self.assertEqual(problem3.prime_factors(13195), [5, 7, 13, 29])
