def main():
prime = Prime()
prime_list = []
prime_under_one_million = []
for p in prime.get_prime():
if p < 1000:
prime_list.append(p)
prime_under_one_million.append(p)
elif p < 1000000:
prime_under_one_million.append(p)
else:
break
negative_prime_list = [0 - v for v in sorted(prime_list, reverse=True)]
negative_prime_list.extend(prime_list)
# b is try to be big, then n&b co efficient will be big
# a is try to be small, then
current_max_combination = 0
for a in negative_prime_list:
for b in negative_prime_list:
combi_cnt = 0
for n in count():
if n * n + a * n + b not in prime_under_one_million:
break
else:
combi_cnt += 1
if combi_cnt > current_max_combination:
print 'find an greater combination: a={}, b={} has {} combinations'.format(
a, b, combi_cnt)
current_max_combination = max(current_max_combination,
combi_cnt)
def test_get_prime(self):
self.assertEqual(Prime.get_prime(0), 2)
self.assertEqual(Prime.get_prime(1), 3)
self.assertEqual(Prime.get_prime(2), 5)
self.assertEqual(Prime.get_prime(3), 7)
self.assertEqual(Prime.get_prime(4), 11)
self.assertEqual(Prime.get_prime(5), 13)
self.assertEqual(Prime.get_prime(6), 17)
self.assertEqual(Prime.get_prime(7), 19)
def main():
prime_ins = Prime()
prime_hash = dict()
circular_prime_list = list()
for i in prime_ins.get_prime():
if i < TARGET:
prime_hash[i] = True
else:
break
for prime in prime_hash:
circular_list = []
for i in range(1, len(str(prime))):
circular_list.append(int(''.join(str(prime)[i:] + str(prime)[0:i])))
for num in circular_list:
if not num in prime_hash:
break
else:
print "circular prime found: {}".format(prime)
circular_prime_list.append(prime)
print "count of all circular prime is {}".format(len(circular_prime_list))
from prime import Prime
"""
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
prime_list = Prime()
prime_sum = 0
for i in prime_list.get_prime():
if i < 2000000:
prime_sum += i
else:
print prime_sum
break
Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
"""
prime_list = list()
prime_ins = Prime()
for prime in prime_ins.get_prime():
if prime > 100:
break
else:
prime_list.append(prime)
def is_prime(n=None):
for i in range(2, n):
if n % i == 0:
return False
else:
return True
def find_ingredient(n=None):
