def factors(n):
p = list(sieve(int(n / 2)))
primes = []
for ii in range(len(p)):
if n % p[ii] == 0:
primes.append(p[ii])
d = n
e = []
for a in range(len(primes)):
while d % primes[a] == 0:
d = d / primes[a]
e.append(primes[a])
if np.prod(e) != n:
print('oops!, n may be prime')
n_occurence = []
for c in range(len(primes)):
n_occurence.append(e.count(primes[c]))
n_factors = np.prod(np.array(n_occurence) + 1)
return n_factors, primes, e
def main():
MAX = 10 ** 6
primes = sieve(MAX + 1)
sprimes = set(primes)
longest = 0
sums = {}
for n in xrange(1, MAX + 1):
s, nums, snums = n, [], set()
while s <= MAX:
if s in sums:
break
nums.append(s)
snums.add(s)
t = spd(s, primes, sprimes)
sums[s] = t
s = t
if s in snums:
length = len(nums) - nums.index(s)
if length > longest:
longest = length
ans = nums[nums.index(s)]
print ans
break
print ans
def main():
MAX = 10**6
primes = sieve(MAX + 1)
sprimes = set(primes)
longest = 0
sums = {}
for n in xrange(1, MAX + 1):
s, nums, snums = n, [], set()
while s <= MAX:
if s in sums:
break
nums.append(s)
snums.add(s)
t = spd(s, primes, sprimes)
sums[s] = t
s = t
if s in snums:
length = len(nums) - nums.index(s)
if length > longest:
longest = length
ans = nums[nums.index(s)]
print ans
break
print ans
def main(): MAX = 5 * 10 ** 7 d = sieve(int(MAX ** 0.5) + 1) t = sieve(int(MAX ** (1.0/3)) + 1) q = sieve(int(MAX ** (0.25)) + 1) numbers = set() for qi in q: for ti in t: for di in d: n = qi * qi * qi * qi + ti * ti * ti + di * di if n < MAX: numbers.add(n) else: break print(len(numbers))
def main(): MAX = 100 primes = sieve(MAX) ways = [ 1 ] + [ 0 ] * MAX for p in primes: for i in xrange(p, MAX + 1): ways[i] += ways[i - p] for i in xrange(len(ways)): if ways[i] > 5000: print i return
def main():
MAX = 100
primes = sieve(MAX)
ways = [1] + [0] * MAX
for p in primes:
for i in xrange(p, MAX + 1):
ways[i] += ways[i - p]
for i in xrange(len(ways)):
if ways[i] > 5000:
print i
return
def n_cons_primes(limit, sieve_limit):
primes = sieve(sieve_limit)
n_max = 0
ab = 0
bb = 0
for a in range(-limit, 1, limit):
for b in range(-limit, 1, limit):
n = 0
break_ = 0
value = (n**2) + (a * n) + b
while break_ == 0:
if value in primes:
n += 1
else:
break_ = 1
n_temp = n
if n_temp > n_max:
n_max = n
ab = a
bb = b
return n_max - 1, a, b
def main():
MAX = 10**7
lb = int(MAX**0.5) - 1000
ub = int(MAX**0.5) + 1000
primes = [p for p in sieve(ub) if p > lb]
min_ratio = MAX
dat_n = None
for i in xrange(len(primes)):
for j in xrange(i + 1, len(primes)):
p1, p2 = primes[i], primes[j]
n = p1 * p2
if n > MAX:
break
phi = (p1 - 1) * (p2 - 1)
ratio = n / phi
if ratio < min_ratio and is_perm(n, phi):
min_ratio = ratio
dat_n = n
print dat_n
def main(): MAX = 10 ** 7 lb = int(MAX ** 0.5) - 1000 ub = int(MAX ** 0.5) + 1000 primes = [ p for p in sieve(ub) if p > lb ] min_ratio = MAX dat_n = None for i in xrange(len(primes)): for j in xrange(i + 1, len(primes)): p1, p2 = primes[i], primes[j] n = p1 * p2 if n > MAX: break phi = (p1 - 1) * (p2 - 1) ratio = n / phi if ratio < min_ratio and is_perm(n, phi): min_ratio = ratio dat_n = n print dat_n
from common import sieve
N = 20
lst = range(1, N + 1)
primes = sieve(N)
lcm = 1
for prime in primes:
while (True):
found_a_multiple = False
for i, n in enumerate(lst):
if n % prime == 0:
lst[i] /= prime
found_a_multiple = True
if found_a_multiple:
lcm *= prime
else:
break
print lcm
import numpy as np
from common import sieve
def multi(b):
out = 1
for i in range(len(b)):
out = out * b[i]
return out
n = 451000
p = list(reversed(sieve(int(n / 2))))
# print(p)
primes = []
for ii in range(len(p)):
if n % p[ii] == 0:
primes.append(p[ii])
amax = int(np.ceil(np.log(n) / np.log(primes[0])))
print(amax)
e = np.zeros(len(primes))
print(e)
e_next = np.zeros(len(primes))
e_next[0] = amax
print(e - e_next)
i_last = np.nonzero(e - e_next)[0][0]
print(len(primes))
from common import sieve N = 20 lst = range(1, N+1) primes = sieve(N) lcm = 1 for prime in primes: while(True): found_a_multiple = False for i, n in enumerate(lst): if n%prime == 0: lst[i] /= prime found_a_multiple = True if found_a_multiple: lcm *= prime else: break print lcm
from common import sieve n = 2000000 print sum(sieve(n))
# What is the value of the first triangle number
# to have over five hundred divisors?
import numpy as np
from common import sieve
def gen_triangle(x):
tri = 0
for i in range(1, x + 1):
tri = tri + i
return tri
p = list(sieve(70000))
def factors(n, primes):
if n in primes:
n_factors = 2
primes = []
e = []
else:
primes = []
for ii in range(len(p)):
if n % p[ii] == 0:
primes.append(p[ii])
d = n
e = []
for a in range(len(primes)):
