def composite(limit=10 ** 3):
numbers = set([_ for _ in range(1, limit) if _ % 2 == 1])
primes = set(primes_sieve(limit))
for _ in range(0, limit):
for p in primes:
n = int(p + 2 * _ ** 2)
if n in numbers:
numbers.remove(n)
return numbers
def composite(limit=10**3):
numbers = set([_ for _ in range(1, limit) if _ % 2 == 1])
primes = set(primes_sieve(limit))
for _ in range(0, limit):
for p in primes:
n = int(p + 2 * _**2)
if n in numbers:
numbers.remove(n)
return numbers
def consecutive_primes(count=2):
memo = list()
primes = set(primes_sieve())
uniq_primes = list()
for n in range(0, 10**6):
if len(memo) == count:
return memo
factors = Counter(prime_factors(n))
if len(set(factors)) == count and factors not in uniq_primes:
memo.append(n)
uniq_primes.append(factors)
else:
memo = list()
uniq_primes = [factors]
def truncate_primes(limit=10**6):
primes = set(primes_sieve(limit))
memo = list()
for prime in primes:
prime_string = str(prime)
front = prime_string
back = prime_string
if len(prime_string) < 2:
continue
for _ in range(len(prime_string) - 1):
back = back[:-1]
front = front[1:]
if int(back) not in primes or int(front) not in primes:
break
else:
memo.append(prime)
return memo
def truncate_primes(limit=10**6):
primes = set(primes_sieve(limit))
memo = list()
for prime in primes:
prime_string = str(prime)
front = prime_string
back = prime_string
if len(prime_string) < 2:
continue
for _ in range(len(prime_string)-1):
back = back[:-1]
front = front[1:]
if int(back) not in primes or int(front) not in primes:
break
else:
memo.append(prime)
return memo
'''
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?
'''
from utils.divisors import primes_sieve, is_prime
from collections import Counter
def pandigital_primes(primes):
memo = list()
print('starting')
for prime in primes:
counter = Counter(str(prime))
v = Counter([str(_) for _ in range(1, len(str(prime))+1)])
if counter == v:
memo.append(prime)
return memo
if __name__ == "__main__":
answer = pandigital_primes(primes_sieve(10**7))
print(answer[-1])
What 12-digit number do you form by concatenating the three terms in this sequence?
'''
from utils.divisors import primes_sieve, is_prime
from collections import defaultdict, Counter
def same_permutation(a, b):
d = defaultdict(int)
for x in a:
d[x] += 1
for x in b:
d[x] -= 1
return not any(d.values())
if __name__ == "__main__":
primes = [ p for p in primes_sieve(10000) if len(str(p)) == 4]
memo = dict()
for p in primes:
for item, value in memo.items():
if same_permutation(str(p), str(item)):
memo[item] = memo[item] + [p]
else:
memo[p] = [p]
primes = [sorted(vs, reverse=True) for vs in memo.values()]
c = primes
for p in primes:
memo = defaultdict(int)
'''
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
'''
from utils.divisors import primes_sieve, is_prime
from collections import defaultdict, Counter
if __name__ == "__main__":
primes = primes_sieve()
prime_set = set(primes)
largest_prime_possible = 0
prime_sums = [0]
for p in primes:
prime_sums.append(prime_sums[-1] + p)
if prime_sums[-1] > 10**6:
break
count = 0
for i in range(len(prime_sums)):
for i_2 in range(len(prime_sums) - 1, i + count, -1):
n = prime_sums[i_2] - prime_sums[i]
'''
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?
'''
from utils.divisors import primes_sieve, is_prime
from collections import Counter
def pandigital_primes(primes):
memo = list()
print('starting')
for prime in primes:
counter = Counter(str(prime))
v = Counter([str(_) for _ in range(1, len(str(prime)) + 1)])
if counter == v:
memo.append(prime)
return memo
if __name__ == "__main__":
answer = pandigital_primes(primes_sieve(10**7))
print(answer[-1])
