def main(top=1000000):
prime_found = None
# Generate a list with cumalitive sum
cum_sum_primes = [0]
cum_sum = 0
for p in primes_erat():
cum_sum += p
cum_sum_primes.append(cum_sum)
if cum_sum > top:
break
num_primes = len(cum_sum_primes)
num_terms = num_primes
while True:
for offset in range(0, 10):
if offset + num_terms >= num_primes:
break
sum_of_primes = cum_sum_primes[offset +
num_terms] - cum_sum_primes[offset]
if sum_of_primes < top and is_prime(sum_of_primes):
prime_found = sum_of_primes
break
if prime_found:
break
num_terms -= 1
return prime_found
def find_count(n):
sump = 0
i = n
prime = 0
while sump + primes[i + 1] < 10**6:
sump += primes[i]
i += 1
if is_prime(sump): prime = sump
return prime
def f(n):
"""
#11914460
>>> sum(f(15 * (10 ** 7)))
676333270
>>> sum(f(2000000))
1242490
>>> sum(f(100))
10
"""
ps = list(primes(5000))
steps = list(get_steps(ps, n))
for i in steps:
if i >= n:
break
sq_i = i * i
if (
i < n
and all(map(lambda x: is_prime(sq_i + x), ADD_List))
and all(map(lambda x: not is_prime(sq_i + x), [11, 17, 19, 21, 23]))
):
yield i
def try_formula(a, b):
"""
>>> try_formula(1, 41)
40
>>> try_formula(-79, 1601)
80
"""
n = 0
while True:
x = (n**2) + (a * n) + b
if not is_prime(x):
break
n += 1
return n
def prime_proof(x):
sx = str(x)
n = len(sx)
for i in range(n):
if i == 0:
Q = (1, 2, 3, 4, 5, 6, 7, 8, 9)
else:
Q = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
for q in Q:
sy = list(sx)
sy[i] = str(q)
y = int(''.join(sy))
if eulertools.is_prime(y):
return False
return True
def main(max_prime=1000000):
prime_list = set(
p for p in itertools.takewhile(lambda x: x < max_prime, primes_erat()))
count = 0
for n in prime_list:
is_circular = True
for x in rotations(str(n)):
x = int(x)
if x == n:
continue
if x < max_prime and not x in prime_list:
is_circular = False
break
elif x >= max_prime and not is_prime(x):
is_circular = False
break
if is_circular:
count += 1
return count
#_______Description_____
#https://projecteuler.net/problem=51
#By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
#By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.
#Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.
##_______end_description_____:
from eulertools import Sieve, is_prime
primes_set = set(Sieve(999999))
primes_reduce = set()
for i in primes_set:
if i > 100000:
digits = [n for n in str(i)]
for n in digits:
if digits.count(n) == 3 and n != digits[5] and any(
(n == '0', n == '1', n == '2')):
primes_reduce.add(i)
for i in primes_reduce:
digits = str(i)
count = 0
for n in digits:
if digits.count(n) == 3:
repeating = n
break
for d in range(0, 10):
new_prime = int(digits.replace(repeating, str(d), 3))
if is_prime(new_prime) and new_prime > 100000: count += 1
if count == 8:
print(i)
break
#_______Description_____
#https://projecteuler.net/problem=41
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
#What is the largest n-digit pandigital prime that exists?
##_______end_description_____:
#only 4 and 7 digit pandigital numbers can be prime because otherwise sum of all digit is divisble by 3 meaning whole number
#is divisible by 3. That's why limit is largest pandigital 7-digit number.
from eulertools import is_prime, is_pandigital
for n in range(7654321, 1, -2): # looping down, decrementing by 2
if is_pandigital(n):
if is_prime(n):
print(n)
break