def square_free(n):
pgen = primes()
p = next(pgen)
while p ** 2 <= n:
if n % p ** 2 == 0:
return False
p = next(pgen)
return True
while True:
while n != 1:
if n % plist[i] == 0:
while n % plist[i] == 0: # factor out current prime number
n /= plist[i]
distinct_count += 1
if distinct_count > 4:
return False
i += 1
return distinct_count == 4
start_time = time()
pgen = primes() # prime generator
p = next(pgen)
plist = []
while p < 1000000:
plist.append(p)
p = next(pgen)
# start with i=10 and check at each iteration if i, i+1, i+2, i+3 all have
# 4 distinct prime factors, when found, print i and exit loop
# h4dpf1 = has 4 distinct prime factor 1
# don't start at has_four_dpf(0), the function will get into a infinite loop
h4dpf1 = has_four_dpf(1)
h4dpf2 = has_four_dpf(2)
h4dpf3 = has_four_dpf(3)
h4dpf4 = has_four_dpf(4)
i = 4
n_str = str(n)
for i in range(1, len(n_str)):
if not is_prime(int(n_str[i:])):
return False
# test for right truncatable
for i in range(1, len(n_str)):
if not is_prime(int(n_str[:i])):
return False
return True
start_time = time()
pstream = primes()
for i in range(4): # skip first 4 primes
p = next(pstream)
l = []
while True:
p = next(pstream)
if truncatable(p):
l.append(p)
if len(l) == 11: # found all, we are done
break
print l
print sum(l)
# Circular primes
# 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 time import time
from UsefulFunctions import primes, is_prime, rotations
start_time = time()
p = primes()
candidate = next(p)
l = []
while candidate < 1000000:
still_prime = True
for e in rotations(candidate):
if not is_prime(e):
still_prime = False
break
if still_prime:
l.append(candidate)
candidate = next(p)
# A composite is a number containing at least two prime factors. For example,
# 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3.
# There are ten composites below thirty containing precisely two, not
# necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
# How many composite integers, n < 10^8, have precisely two, not necessarily
# distinct, prime factors?
from time import time
from UsefulFunctions import primes
start_time = time()
N = 100000000
pgen = primes()
pnumbers = [] # list of prime numbers < N/2
p = next(pgen)
while p <= N / 2:
pnumbers.append(p)
p = next(pgen)
print len(pnumbers)
count = 0
i = 0
while pnumbers[i] ** 2 < N:
j = i
while pnumbers[i] * pnumbers[j] <= N:
count += 1
j += 1
if j == len(pnumbers):
def ocgen():
n = 9
while True:
if not is_prime(n):
yield n
n += 2
start_time = time()
oc = ocgen() # odd composite stream
continue_ = True
while continue_:
c = next(oc) # get an odd composite candidate
pgen = primes() # start with 1st prime all over again
while True: # verification loop
p = next(pgen) # try a new prime number
i = 1
while p + 2 * (i ** 2) < c: # try to see if an i could make the current prime work
i += 1
if p + 2 * (i ** 2) == c: # if formula is verified, break
break
if p > c: # the formula fails
continue_ = False
break
print c
