def testPrimesLowerThan100(self): soa = SieveOfAtkin(100) primesLowerThan100 = [x for x in self.primes if x<100] primes = soa.getPrimes() primes.sort() primesLowerThan100.sort() self.assertEqual(primes, primesLowerThan100)
def testPrimesLowerThan2000(self):
soa = SieveOfAtkin(2000)
primesLowerThan2000 = [x for x in self.primes if x < 2000]
primes = soa.getPrimes()
primes.sort()
primesLowerThan2000.sort()
self.assertEqual(primes, primesLowerThan2000)
from sieve_of_atkin import SieveOfAtkin
soa = SieveOfAtkin(1000000)
primes = set(soa.getPrimes())
s = 0
def rotate(n):
return int(str(n)[-1] + str(n)[:-1])
def isCircularPrime(n):
v = rotate(n)
while n != v:
if v not in primes:
return False
v = rotate(v)
return True
for i in primes:
if isCircularPrime(i):
s += 1
print s
# The intended value has to be in the primes list, meaning < limit. # The biggest possible sequence will involve the smallest primes that add up # to limit. def get_max_sequence(): seq = 0 s = 0 while(s < limit): s += primes_list[seq] seq +=1 return seq limit = 1000000 # Generate list of primes. soa = SieveOfAtkin(limit) primes = set(soa.getPrimes()) # Because result is given as a set, turn into a list and sort it. primes_list = list(primes) primes_list.sort() # We know that the sequence will have a maximum: # - only consider sequences of that size or smaller for l in xrange(get_max_sequence(), 1, -1): i = 0 value = sum(primes_list[i:i+l]) # Because the value has to be in primes_list, no need to check values bigger # than the limit. while value < limit: if value in primes: print l, value exit(0) i +=1
from sieve_of_atkin import SieveOfAtkin
from math import sqrt, pow
soa = SieveOfAtkin(1000)
primes = soa.getPrimes()
nmax = 0
# Taken from here: http://blog.dreamshire.com/2009/03/26/94/
def is_prime(n):
if n == 2 or n == 3: return True
if n < 2 or n % 2 == 0: return False
if n < 9: return True
if n % 3 == 0: return False
r = int(sqrt(n))
f = 5
while f <= r:
if n % f == 0: return False
if n % (f + 2) == 0: return False
f += 6
return True
# b has to be prime, so that when n = 0 the result is prime
for a, b in [(a, b) for a in range(-1000, 1000) for b in primes]:
n = 1
while is_prime(pow(n, 2) + a * n + b):
n += 1
if n > nmax:
nmax, product = n, a * b
return nod * 2
exponent = 1
while remain % prime_list[i] == 0:
exponent += 1
remain /= prime_list[i]
nod *= exponent
# If there is no remainder, return the count
if remain == 1:
return nod
return nod
number = 0
i = 1
soa = SieveOfAtkin(75000)
prime_list = soa.getPrimes()
"""
while NumberOfDivisors(number) < 500:
number += i
i +=1
"""
while PrimeFactorisationNoD(number, prime_list) < 500:
number += i
i +=1
print number
""" - Fermat's little theorem: 1/d has a cicle of n digits if 10^n - 1 mod n = 0 - If d is prime, than it can have a up to d - 1 repeating decimal digits - So, find the last prime p, under 1000, that has p - 1 digits """ from sieve_of_atkin import SieveOfAtkin soa = SieveOfAtkin(1000) primes = soa.getPrimes() primes.sort() # just to be sure primes.reverse() for prime in primes: cycle = 1 while (pow(10, cycle) - 1 ) % prime != 0: cycle += 1 if prime - cycle == 1: break print prime
