def quad_primes(limit):
res = []
for a in xrange(-limit+1,limit):
for b in [1] + primes.generate(limit):
res.append((seq(a,b), (a, b), a*b))
res.append((seq(a,-b), (a, -b), a*b))
return max(res)
def solve():
# Prime Number Theorem:
# Roughly speaking, the prime number theorem states that if you randomly
# select a number nearby some large number N, the chance of it being prime
# is about 1 / ln(N), where ln(N) denotes the natural logarithm of N. For
# example, near N = 10,000, about one in nine numbers is prime, whereas
# near N = 1,000,000,000, only one in every 21 numbers is prime.
return primes.generate(10000)[1001]
def factor_count(n):
"""
returns a dictionary of {factor: times_occuring} for the number n
"""
ret = collections.defaultdict(int)
for p in itertools.takewhile(lambda p: p<=n, primes.generate()):
while not n % p:
ret[p] += 1
n = n // p
return ret
def run():
print(sum(itertools.takewhile(lambda n:n<2e6, primes.generate())))
def iter_factors(n):
for p in itertools.takewhile(lambda p: p<=n, primes.generate()):
while not n % p:
yield p
n = n // p
#!/usr/bin/python
# https://projecteuler.net/problem=10
# Find the sum of all the primes below two million.
from primes import generate
sum = 0L
n = 2000000
for prime in generate():
if prime < n:
sum += prime
else:
print sum
break
import primes as prim
print prim.generate(10)
limit=100
count=0
yaks=''
for p in prim.P:
yaks=yaks+ '''
#######################################
'''
yaks=yaks+ str(p)
yaks=yaks+ '''
#######################################
'''
yaks=yaks+ str(prim.YaksicPrimes(p))
count+=1
if count>limit:
f = open('YaksicPrimes.txt', 'w')
f.write(yaks)
break
def test_primes(self):
for x in imap(eq, self.primes, generate()):
self.assertTrue(x)
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
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 collections import deque
from primes import generate, check as isprime
def is_circular_prime(p):
rotations = len(str(p))
prime = deque(str(p))
for x in range(rotations):
prime_int = int("".join(prime))
if not isprime(prime_int):
return False
prime.rotate(1)
return True
primes = generate(1000000)
count = 0
for i, prime in enumerate(primes):
if prime and is_circular_prime(i):
count += 1
print(count)
#!/usr/bin/python
# https://projecteuler.net/problem=7
# What is the 10,001st prime number?
from primes import generate
for n, prime in enumerate(generate()):
if n == 10000:
print prime
break
def nth_prime(n):
for idx, p in enumerate(primes.generate(), start=1):
if idx==n:
return p