def pollard_p_minus_one(n, b=5, m=13, a=2):
if b < 1:
raise Exception('B must be bigger than 1')
if b >= m:
raise Exception('B must be less than sqrt(n)')
if m >= b**2:
raise Exception('M must be less than B^2')
if m >= math.sqrt(n):
raise Exception('M must be less than sqrt(n)')
if a < 1:
raise Exception('a must be bigger than 1')
mb_primes = utils.primes(2, b)
primes_mul = reduce(operator.mul, mb_primes, 1)
b = (a**primes_mul) % n
q = utils.gcd(b - 1, n)
if q > 1:
return q, n // q
m_primes = utils.primes(b, m)
for m in m_primes:
fm = (b**(m - 1)) % n
gm = utils.gcd(fm, n)
if gm > 1:
return gm, n // gm
return math.nan, math.nan
def main():
number = 600851475143
sqrt = int(math.sqrt(number))
for prime in reversed(primes(sqrt)):
if number % prime == 0:
return prime
def getFraction(self):
n = 1
d = 1
for (p, a) in zip(utils.primes(), self):
if a > 0: n *= p**a
elif a < 0: d *= p**(-a)
return (n, d)
def most_consecutive_primes(n):
# given n^2 + ax + b, assume the highest result we can get is 2n^2 + n
# generate all primes up to that
_primes = utils.primes(2*pow(n, 2) + n)
primes = {}
for p in _primes:
primes[p] = True
a, b = 0, 0
ps = 0
for _a in range(n):
for _b in range(n):
# consider all combos of positive/negative
for __a, __b in [(_a, _b), (-_a, _b), (-_a, -_b), (_a, -_b)]:
# create the function
def f(x):
return pow(x, 2) + __a*x + __b
# look for primes
x = 0
_ps = -1
while primes.get(f(x)):
_ps += 1
x += 1
if _ps > ps:
a, b = __a, __b
ps = _ps
return a*b
def main():
result = 0
while True:
result += 1
x = ways(take_upto(result, primes()), result)
if x > 5000:
break
return result
def getEpimoricZip(r):
"(6/5) -> [(3/2,1), (5/4,-1)]"
result = []
while r != Rational(1):
p, i = zip(utils.primes(), r)[-1]
e = Rational(p, p-1)
result.insert(0, (e, i))
r /= e ** i
return result
def testRandom3Digit():
nsamples = 10
for i in range(nsamples):
threeDigitPrimes = primes(1000)[2:]
random.shuffle(threeDigitPrimes)
p1, p2 = threeDigitPrimes[0], threeDigitPrimes[1]
n = p1 * p2
print("primes: " + str(p1) + ", " + str(p2))
print(quadSieve(n))
assert (quadSieve(n) == {p1, p2})
def longest_recurring_cycle(n):
for denom in utils.primes(n)[::-1]:
period_length = 1
# 1 % denom is always 1, so when it's seen again we know the period is over
while pow(10, period, denom) != 1:
period_length += 1
# reptend prime check
if denom - 1 == period_length:
return denom
def main():
trunc_count = 11
trunc_primes = []
for p in take_from(10, primes()):
if truncatable_prime(p):
trunc_count -= 1
trunc_primes.append(p)
if trunc_count == 0:
break
return sum(trunc_primes)
def compute():
PRIMES = primes(1000)
target = 10 # we are given that 10 only has 5 prime partitions
while True:
prime_partitions = [0] * (target + 1)
prime_partitions[0] = 1
for p in PRIMES:
for j in range(p, target + 1):
prime_partitions[j] += prime_partitions[j - p]
if prime_partitions[target] > 5000:
break
target += 1
return target
def compute():
# We are given that this pattern holds at least until 33
# We will cheat a bit here and assume that this pattern will not hold until 10,000
# This can be found with some experimentation
prime_list_full = primes(10000)
num = 35
while True:
prime_list = [p for p in prime_list_full if p < num]
if not (any(
sqrt(num / 2 - p / 2) == int(sqrt(num / 2 - p / 2))
for p in prime_list)) and not is_prime(num):
return num
num += 2
def main():
prime_numbers = utils.primes(10000)
odd_composites = [n for n in range(2, 10000) if (utils.isprime(n) == False) & (n % 2 != 0)]
for odd_composite in odd_composites:
number_found = False
for prime in prime_numbers:
difference = odd_composite - prime
if difference > 0:
if np.sqrt(difference / 2) == int(np.sqrt(difference/2)):
number_found = True
if number_found == False:
return odd_composite
def compute():
LIMIT = 50_000_000
max_prime = int(sqrt(LIMIT)) # prime**2 must be less than LIMIT
PRIMES = primes(max_prime)
numbers = set()
for x in PRIMES:
for y in PRIMES:
for z in PRIMES:
value = x**2 + y**3 + z**4
if value < LIMIT:
numbers.add(value)
else:
break
return len(numbers)
def factor_map(max):
m = {}
primes = utils.primes(max)
sprimes = set(primes)
for n in xrange(1, max):
if n in sprimes:
m[n] = set()
continue
factors = []
for p in primes:
if p < n and n % p == 0:
factors.append(p)
m[n] = set(factors)
return m
def run(): to_proc = [] for prime in primes(1000,10000): to_proc.append(prime) to_proc.sort() permutations = [] for i in to_proc[:]: if len(permutations) == 2: break for j in xrange(1, (10000-i)/2): second = str(i+j) if is_prime(i+j) and is_permutation(str(i), second): third = i+j+j if is_prime(third) and is_permutation(second, str(third)): permutations.append((i,i+j,i+j+j)) print permutations
def main():
primes_less_1000 = list(takewhile(lambda x: x < 1000, primes()))
highest = [0, 0, 0]
for a in range(-999, 1001):
for b in primes_less_1000:
n = 0
while is_prime((n**2) + (a*n) + b):
n += 1
if n > highest[0]:
highest[0] = n
highest[1] = a
highest[2] = b
print(highest)
return highest[1] * highest[2]
def main():
total = 0
primes_less_limit = take_upto(MILL, primes())
for p in primes_less_limit:
n = p
circular = True
for _ in str(n):
if not is_prime(n):
circular = False
break
n = rotate_digits(n)
if circular:
total += 1
return total
def main():
length = 6
primes = [str(prime) for prime in utils.primes(10 ** length) if 10 ** (length - 1) < prime < 10 ** length]
counter = Counter()
for prime in primes:
digit_counter = Counter(prime)
for d, digit_count in digit_counter.iteritems():
if digit_count == 1:
continue
indices = [i for i, x in enumerate(prime) if x == d]
for length in xrange(2, digit_count + 1):
for i in itertools.combinations(indices, length):
counter.update([tuple([digit if digit_index not in i else -1 for digit_index, digit in enumerate(prime)])])
result = counter.most_common(1)[0][0]
for prime in primes:
if is_match(result, prime):
print prime
break
def factor(num, b):
"""
factor using Pollard's p-1 method
num: odd number to factor
b: smoothness cap
"""
primeList = utils.primes(b)
m = 1
for prime in primeList:
exp = math.floor(math.log(b, prime))
m *= pow(prime, exp)
a = 2
g = utils.xgcd(pow(a, m, num) - 1, num)[0]
if (1 < g and g < num):
return g
elif (g == 1):
# Increase B
return -1
elif (g == num):
# Decrease B
return -2
import utils
from itertools import combinations, takewhile, product, chain
primesSet = list(map(str, utils.primes(100000000)))
primes = list(takewhile(lambda x: int(x) < 10000, list(primesSet)))
primesSet = set(primesSet)
print len(primes), len(primesSet)
def checkAllPrimes(x):
if not len(set(x)) == len(x):
return False
for y, z in combinations(x, 2):
if not (y + z in primes and z + y in primes):
return False
return True
def checkAllPrimes2(x1, x2):
for y in x1:
if not (y + x2 in primes and x2 + y in primes):
return False
return True
# pairs = None
# print "Got primes"
# pairs = list(filter(checkAllPrimes,
from utils import primes
v = 600851475143
lim = v**.5
last = 1
for i in primes():
if v % i == 0:
last = i
if i > lim:
break
print(last)
from utils import primes
limit = 100
ps = primes(limit)
count = [[0 for _ in ps] for _ in range(limit+1)]
for n in range(1, limit+1):
for i, p in enumerate(ps):
if p == n: count[n][i] += 1
count[n][i] += count[n-p][i]
if i > 0:
count[n][i] += count[n][i-1]
for n, cs in enumerate(count):
if cs[-1] > 5000:
print n
break
r = pow(n, (q + 1) / 2, p)
while t != 1:
i = find_sq_num(t, p)
if i >= m: return
b = pow(c, 2 ** (m - i - 1), p)
m = i
c = b * b % p
t = t * b * b % p
r = r * b % p
yield r
yield p - r
limit = 5 * 10 ** 7
ps = list(primes(int(2 ** 0.5 * limit)))
els = array.array('B', '1' * (limit + 1))
for p in ps[1:]:
for r in calculate_r(p):
init = r if 2 * r * r - 1 != p else r + p
for k in xrange(init, limit+1, p):
els[k] = 0
count = 0
for i in xrange(2, limit + 1):
if els[i]:
count += 1
print count
from utils import primes
from math import log
def powers(l, limit):
res = l
while res <= limit:
yield res
res *= l
limit = 10**7
ps = list(primes(limit / 2))
res = 0
for lp in ps[::-1]:
for s in ps:
if lp * s > limit: break
if s >= lp: break
ans = 0
for l in powers(lp, limit):
if l * s > limit: break
ans = max(ans, l * list(powers(s, limit / l))[-1])
res += ans
print res
from utils import primes, modinv
limit = 1000000
ps = list(primes(limit + 100))[2:]
res = 0
for p1, p2 in zip(ps, ps[1:]):
if p1 > limit: break
power10 = 10 ** len(str(p1))
res += ((p2 - p1) * modinv(power10, p2) % p2) * power10 + p1
print res
from utils import primes
ty = 100
limit = 10**9
ps = list(primes(ty))
ns = [1]
res = set()
while ns:
n = ns.pop()
if n in res: continue
res.add(n)
for p in ps:
x = n * p
if x <= limit:
ns.append(x)
print len(res)
import utils
import itertools
import time
start = time.time()
BOUND = 1000000
current = (0,1)
primes = list(utils.primes(BOUND))
prime_arr = [False] * BOUND
for p in primes:
prime_arr[p] = True
print ("Primes generated")
max_length = 0
for i in range(0, len(primes)):
s = 0
curr_length = 0
for p in primes[i:]:
s += p
if s >= BOUND:
break
curr_length += 1
if prime_arr[s]:
if curr_length > max_length:
max_length = curr_length
print "New max: %s" % max_length
print sum(primes[i:i+curr_length])
print (time.time() - start)
import utils
from collections import Counter
from itertools import tee, izip
def window(iterable, size):
iters = tee(iterable, size)
for i in xrange(1, size):
for each in iters[i:]:
next(each, None)
return izip(*iters)
MAX = 1000000
primes = utils.primes(MAX)
def factor_consequtive(x):
factors = utils.factor(x, primes)
counts = Counter(factors)
for k in counts:
yield k ** counts[k]
def generate_consequtive():
i = 1
while True:
yield set(factor_consequtive(i))
i += 1
for a, b, c, d in window(generate_consequtive(), 4):
if len(a.union(b).union(c).union(d)) == 16:
print a
break
def coprimes(*numbers):
for p in utils.primes():
if all(n % p == 0 for n in numbers): return False
if all(n / p == 0 for n in numbers): return True # stop searching
prime.
Answer:
"""
from __future__ import print_function
from utils import timer, primes, take_n, is_prime
import itertools
ANSWER = None
def test_answer():
if ANSWER is None:
assert 0, 'Not Completed'
else:
assert ANSWER == main()
PRIMES = list(take_n(10000, primes()))
PRIMES_SET = frozenset(PRIMES)
@timer
def main():
pass
if __name__ == '__main__':
print(main())
from __future__ import print_function
from utils import timer, primes
import itertools
ANSWER = 65537
def test_answer():
if ANSWER is None:
assert 0, 'Not Completed'
else:
assert ANSWER == main()
primes_l = set(itertools.islice(primes(), 10000))
square_sums = [2*(x**2) for x in range(40)]
@timer
def main():
for n in itertools.count(3, 2):
for p in primes_l:
if p > n:
break
for s in square_sums:
t = p + s
if t == n:
break
else:
return n
from utils import primes, memoize
primes = primes(10000000)
primeSet = set(primes)
@memoize
def factors(x):
if x in primeSet:
return {x}
t = x
for i in primes:
if t % i == 0:
t = t / i
break
return {t, i} | factors(t)
###Calculates the number of relatively prime numbers up to x
@memoize
def phi(x):
if x in primeSet:
return x - 1.
for p in primes:
c = 0
s = x
if x % p == 0:
while s % p == 0:
c += 1
s = s / p
if c == 1:
return phi(p) * phi(x / p)
else:
return phi(s * p) * (p ** (c - 1) * (p - 1))
from utils import is_prime, primes consecutive_primes = [] total = 0 for prime in primes(): total += prime if is_prime(total): consecutive_primes.append(prime) else: break print consecutive_primes
from utils import primes
from bisect import bisect_right
limit = 10**8
ps = list(primes(limit))
res = 0
for i, p in enumerate(ps):
if p * p > limit: break
x = limit / p
if limit % p:
j = bisect_right(ps, x)
else:
j = bisect_right(ps, x-1)
res += j - i
print res
import itertools
import utils
primes = utils.primes(17)
def f(n):
sn = str(n)
for i in range(1, 8):
if int(sn[i:i+3]) % primes[i - 1] != 0:
return False
return True
matches = []
for i in itertools.permutations('0123456789', 10):
n = int(''.join(i))
if f(n):
matches.append(n)
print matches
print sum(matches)
from itertools import islice
from utils import primes
# https://docs.python.org/2/library/itertools.html#recipes
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)
n = nth(primes(), 10000)
print(n)
def __float__(self):
f = 1.0
for (p, a) in zip(utils.primes(), self):
f *= p**a
return f
import utils
primes = utils.primes(7654321)
def p(n):
sn = str(n)
lsn = len(sn)
if lsn > 10:
return False
return set([ int(d) for d in sn ]) == set(range(1, len(sn) + 1))
primes.reverse()
for prime in primes:
if p(prime):
print prime
break
def main():
s = 0
primes_larger_than_7 = dropwhile(lambda i: i<10, primes())
print sum(take_11(truncatable_primes(primes_larger_than_7)))
from utils import prime_cache, primes
LIMIT = 1_000_000
CACHE = prime_cache(LIMIT)
PRIMES = primes(LIMIT)
def compute():
max_consec_count = 0
max_consec_count_num = 0
len_primes = len(PRIMES)
for i in range(len_primes):
consecutive_count = 1
consecutive_sum = PRIMES[i]
for j in range(i + 1, len_primes):
consecutive_sum += PRIMES[j]
consecutive_count += 1
if consecutive_sum >= LIMIT:
break
elif consecutive_count > max_consec_count and CACHE[
consecutive_sum]:
max_consec_count = consecutive_count
max_consec_count_num = consecutive_sum
return max_consec_count_num
if __name__ == "__main__":
print(compute())
#
# It turns out that the formula will produce 40 primes for the consecutive values n=0 to 39
#
# howver, when n = 40, the result is not prime
#
# n**2 - 79n + 1601 produces 80 consecutive primes for the consecutive values n=0 to 79. the produce of the coefficients -79 and 1601 is -126479
#
# considering quadratics of the form n**2 + an + b where abs(a) < 1000 and abs(b) < 1000, find the product of the coefficients a and b for the quadratic expression which produces the maximum number of primes for consective values of n starting with 0
import utils
def f(b, c, n):
return n**2 + b*n + c
primes = utils.primes(10000)
maximum = {'prime': 0}
for b in range(-1000, 1001):
for c in range(-1000, 1001):
#print b,c
for n in xrange(0, 10000):
res = f(b, c, n)
if res not in primes:
if n > maximum['prime']:
maximum['prime'] = n - 1
maximum['b'] = b
maximum['c'] = c
break
print maximum
from utils import primes
from utils import is_prime
from itertools import combinations
PRIMES = primes(10000)
def concat_prime(tuple):
if len(tuple) != 2:
raise AssertionError("concat_prime requires a tuple with 2 numbers")
else:
if is_prime(int(str(tuple[0]) + str(tuple[1]))) and is_prime(
int(str(tuple[1]) + str(tuple[0]))):
return True
return False
def compute():
prime_combos = combinations(PRIMES[1:], 2) # 2 cannot be included
return [v for v in prime_combos if concat_prime(v)][:10]
if __name__ == "__main__":
print(compute())
from utils import primes, is_prime
c = 0
s = 0
for p in primes():
if p < 10:
continue
if c >= 11:
break
truncations = set([int(str(p)[:i]) for i in range(1, len(str(p)))] +
[int(str(p)[i:]) for i in range(1, len(str(p)))])
if all(map(is_prime, truncations)):
c += 1
s += p
print(s)
from utils import primes
ps = set(primes(10**6))
count = 0
for a in range(1, 10000):
b = a + 1
p = b**3 - a**3
if p > 10**6: break
if p in ps:
count += 1
print count
def calc_mods(p):
res = []
for i in range(p):
j = i * i % p
for off in (1, 3, 7, 9, 13, 27):
if (j + off) % p == 0:
break
else:
res.append(i)
return res
limit = 15 * 10**7
res = 0
mods = {p: calc_mods(p) for p in primes(500)}
for n in xrange(0, limit + 1, 210):
for off in (10, 80, 130, 200):
k = n + off
possib = True
for p in mods:
if k * k > p and k % p not in mods[p]:
possib = False
break
if not possib: continue
k = (n + off)**2
ps = []
for j in range(k + 1, k + 28, 2):
if isprime(j):
ps.append(j)
def main():
for p in itertools.islice(utils.primes(), 10000, 10001):
print p
import utils
primes = set(utils.primes(10**6))
def rotations(n):
sn = str(n)
return set([ int(sn[i:] + sn[:i]) for i in range(len(sn)) ])
print len(filter(lambda n: rotations(n).issubset(primes), primes.copy()))
return inner @memWays def ways(n, nums): #print n, nums if not nums: return 0 m = nums[0] if(m == n): return 1 + ways(n, nums[1:]) elif(m < n): return ways(n - nums[0], nums) + ways(n, nums[1:]) else: while True: nums = nums[1:n] if not nums: return 0 elif(nums[0] >= n): return ways(n, nums) from utils import primes p = primes(100) from time import time start = time() for x in range(1, len(p)): if ways(x, p) > 5000: print x break print time() - start
def main():
guard = 2000000
print sum(takewhile(lambda x: x<guard, primes()))
from __future__ import print_function
from utils import timer, primes, is_permutation, take_between
from itertools import dropwhile, takewhile
ANSWER = 296962999629
def test_answer():
if ANSWER is None:
assert 0, 'Not Completed'
else:
assert ANSWER == main()
PRIMES_WITH_4_DIGITS = list(take_between(1000, 9999, primes()))
@timer
def main():
for p1 in PRIMES_WITH_4_DIGITS:
for p2 in PRIMES_WITH_4_DIGITS:
if p2 <= p1:
continue
p3 = p2 + (p2 - p1)
if p3 in PRIMES_WITH_4_DIGITS:
if is_permutation(p1, p2) and is_permutation(p2, p3):
if p1 == 1487:
continue
return int(str(p1) + str(p2) + str(p3))
from utils import primes
res = 0
for p in primes(10**8):
if p < 5: continue
res += (p - 3) * pow(8, p - 2, p) % p
print res
def main():
cnt = 0
for p in itertools.takewhile(less_than_one_million, utils.primes()):
if all(utils.is_prime(i) for i in rotated(p)):
cnt += 1
print cnt
def main():
upper_limit = 1000000
prime_numbers = utils.primes(upper_limit)
return sum(
[circular_prime(prime, prime_numbers) for prime in prime_numbers])
def solve_0(N):
return utils.primes(N)[-1]
from utils import primes
def f(n, k):
res = 0
while n:
res += n // k
n //= k
return res * k
n = 20000000
k = 15000000
res = 0
for p in primes(n):
res += f(n, p) - f(k, p) - f(n - k, p)
print(res)
def product(seq):
res = 1
for s in seq:
res *= s
return res
def muls(ps):
res = []
for i in range(len(ps) + 1):
for comb in combinations(ps, i):
res.append(product(comb))
return sorted(res)
ps = list(primes(190))
n = len(ps)
left = muls(ps[:n // 2])
right = muls(ps[n // 2:])
ans = 0
limit = product(ps)**0.5
from bisect import bisect_left
for val in left:
i = bisect_left(right, limit / val)
if i:
ans = max(ans, val * right[i - 1])
print(ans % 10**16)
import utils
def is_perm(a, b):
return sorted(str(a)) == sorted(str(b))
best = (10000, 1)
primes = [ i for i in utils.primes(4000) if i > 2000 ]
for i in primes:
for j in primes:
n = i * j
if n > 10**7:
break
phi = (i - 1) * (j - 1)
ratio = (n * 1.0) / phi
curr = (ratio, n)
if is_perm(n, phi) and curr < best:
best = curr
print best[1]
def main(n=MILL*2):
return sum(takewhile(lambda x: x < n, primes()))
from utils import primes, is_prime as isprime
from collections import defaultdict
ps = primes(10**4)[1:]
concatenable = {p: set() for p in ps}
n = len(ps)
for p in ps:
for q in ps:
if isprime(int(str(p) + str(q))):
concatenable[p].add(q)
for p in ps:
for q in concatenable[p]:
if q <= p: continue
if p not in concatenable[q]: continue
for r in concatenable[p] & concatenable[q]:
if r <= q: continue
if p not in concatenable[r] or q not in concatenable[r]: continue
for s in concatenable[p] & concatenable[q] & concatenable[r]:
if s <= r: continue
if p not in concatenable[s] or q not in concatenable[
s] or r not in concatenable[s]:
continue
for t in concatenable[p] & concatenable[q] & concatenable[
r] & concatenable[s]:
if t <= s: continue
if set([p, q, r, s]) <= concatenable[t]:
print p, q, r, s, t, p + q + r + s + t
"""
from __future__ import print_function
from utils import timer, MILL, take_upto, primes
ANSWER = 997651
def test_answer():
if ANSWER is None:
assert 0, 'Not Completed'
else:
assert ANSWER == main()
PRIMES = list(take_upto(MILL, primes()))
PRIME_SET = set(PRIMES)
PRIME_SUMS = [PRIMES[0]]
# Calculate the cumaltive sum of primes for each step.
for psi, p in enumerate(PRIMES[1:]):
PRIME_SUMS.append(PRIME_SUMS[psi] + p)
@timer
def main():
# We use a sliding window to calculate the longest run and its associated
# prime.
psl = len(PRIME_SUMS)
high_run = 0
high_prime = 0