def p123_nice():
primes = prime_sieve(1000000)
for n in range(7038, len(primes), 2):
print([n], 2 * (n + 1) * primes[n])
if 2 * (n + 1) * primes[n] > pow(10, 10):
print(n + 1)
break
def p127(): # Answer 18407904, 17 min....Orz
sum_c = 0
L = 120000
primes = prime_sieve(L)
lrad = [0] * L
for a in range(1, L // 2):
if lrad[a] == 0:
lrad[a] = rad(a)
if a % 2 == 0:
inc = 2
else:
inc = 1
for b in range(a + 1, L - 1, inc):
if lrad[b] == 0:
lrad[b] = rad(b)
c = a + b
if c >= L:
break
if lrad[c] == 0:
lrad[c] = rad(c)
if (lrad[a] * lrad[b] * lrad[c]) >= c:
continue
if p127_check(a, b, c, primes):
print(a, b, c)
sum_c += c
print(sum_c)
return
def p132_nice():
primes = prime_sieve(10**6)
base = 10
exp = 10**9
result = sum(
islice((p for p in primes if modular_pow(base, exp, 9 * p) == 1), 0,
40))
print("The result is:", result)
def p133(): # Answer: 453647705
primes = prime_sieve(100000)
base = 10
for e in range(1, 20):
exp = 10**e
for p in primes:
if modular_pow(base, exp, 9 * p) == 1:
primes.remove(p)
print(sum(primes[0:100000]))
return
def p187():
cnt = 0
primes = prime_sieve(49999992)
lp = len(primes)
for i in range(0, lp - 1):
for j in range(i, lp):
if primes[i] * primes[j] > 100000000:
break
# print(primes[i]* primes[j], '\t', primes[i], primes[j])
cnt += 1
print(cnt)
return
def p132():
primes = prime_sieve(10**6)
base = 10
exp = 10**9
cnt = 0
s = 0
for p in primes:
if modular_pow(base, exp, 9 * p) == 1:
s += p
cnt += 1
if cnt == 40:
break
print(s)
def p123(): # Answer 21035
primes = prime_sieve(1000000)
for i in range(7038, len(primes), 2):
n = i + 1
p = primes[i]
s = (p - 1)**n + (p + 1)**n
p2 = p**2
r = s % p2
print([n], r)
if len(str(r)) > 10:
print(n)
break
return
def p131_slow(): # Answer: 173, 68.54s Mac pro 2016
cnt = 0
primes = prime_sieve(1000000)
for p in primes:
for i in count(1):
n = i**3
if is_perfect_cube(n + p):
if is_perfect_cube(n**2):
# print('[great] ', [p, n, i], n**2, n+p)
cnt += 1
break
if i > 600:
break
print(cnt)
def p134(): # 18613426663617118, 2sec < run time
total = 0
L = 1000000
primes = prime_sieve(L + 50)
for i in range(2, len(primes) - 1):
p1 = primes[i]
if p1 >= L:
break
p2 = primes[i + 1]
k = 1
while k < p1:
k *= 10
small_n = congruence(p2, p1, k)
total += (small_n * p2)
print(total)
def p200_slow():
res = []
primes = prime_sieve(400000)
for p1 in primes:
for p2 in primes:
if p1 == p2:
continue
sqube = p2q3(p1, p2)
if sqube > 7999728002312:
break
str_sqube = str(sqube)
if str_sqube.find('200') >= 0:
if prime_proof(str_sqube):
print(p1, p2, sqube)
res.append(sqube)
res = sorted(res)
print(res[199])
return
def p200(): # 229161792008, 6.12sec
res = []
primes = prime_sieve(200000)
for p1 in primes:
for p2 in primes:
if p1 == p2:
continue
sqube = p2q3(p1, p2)
if sqube > 7999728002312:
break
if sqube % 2 != 0 and sqube % 5 != 0:
continue
str_sqube = str(sqube)
if str_sqube.find('200') >= 0:
if prime_proof(str_sqube):
print(p1, p2, sqube)
res.append(sqube)
res = sorted(res)
print(res[199])
return
def solution():
sieve = prime_sieve(1000000)
return len([n for n in range(2, 1000000) if is_circular_prime(n, sieve)])
"""
Project Euler 35: Find the number of circular primes below one million
"""
from re import search
import util
L = 1000000
#return a list of rotations for string s
def rotate(s):
return [s[n:] + s[:n] for n in range(1, len(s))]
#make a set of primes < L and remove any that have {0,2,4,5,6,8} except 2 and 5
primes = set(['2','5']+[str(p) for p in util.prime_sieve(L) if not search('[024568]', str(p))])
for p in primes:
if all(pr in primes for pr in rotate(p)):
print p
n_circular_primes = sum(all(pr in primes for pr in rotate(p)) for p in primes)
print "Project Euler 35 Solution =", n_circular_primes
#!/usr/bin/python
import psyco
import util
primes = util.prime_sieve(3000)
def consecutive_primes(a, b):
num = 0
x = 0
while True:
y = x ** 2 + a * x + b
if y in primes:
num += 1
else:
break
x += 1
return num
if __name__ == "__main__":
max = 0
max_coeffs = None
for a in range(1000):
for b in range(1000):
if b not in primes:
continue
num = consecutive_primes( a, b)
if num > max:
max = num
max_coeffs = (a,b)
num = consecutive_primes(-a, b)
def solution():
sieve = prime_sieve(1000000)
return sum([n for n in range(10,1000000) if is_truncatable_prime(n, sieve)])
import util pi = 2000000 primes = util.prime_sieve(pi) psum = sum(primes) print(psum)
import psyco
import util
primes = util.prime_sieve(1000000)
sum = 0
for i in range(2, 1000001):
sum += util.totient(i, primes)
print sum
def solution():
sieve = prime_sieve(1000000)
return sum(
[n for n in range(10, 1000000) if is_truncatable_prime(n, sieve)])
"""Project Euler 7: Find the 10001st prime number By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10001st prime number?""" import util n = 10000 primes_list = util.prime_sieve(110000) print primes[n]
import util
import math
n = 10001
pi_block = 1000
# the PNT gives us a decent first guess to sieve through.
pi = int(n * math.log(n))
primes = util.prime_sieve(n)
while len(primes) < n:
pi += pi_block
primes = util.findprimes(pi, primes)
print(primes[n-1])
