def extendSquareSpiral(sideLength, primeCountInDiagonales): edgeNumber = sideLength ** 2 for _ in range(4): edgeNumber += sideLength + 1 if isPrime(edgeNumber): primeCountInDiagonales += 1 return sideLength + 2, primeCountInDiagonales
def phi(n): result = n if isPrime(n): return n - 1 for p in getFactorization(n): result *= (1 - 1 / p[0]) return result
def consecutivePrimesFrom(primes, start, maxvalue):
result = []
for i in range(start + 1, len(primes)):
partsum = sum(primes[start : i + 1])
if isPrime(partsum) and partsum < maxvalue:
result.append(primes[start : i + 1])
if partsum > maxvalue:
return result
return result
def foursHasProperty(candidate, targetNumber, digitCount): for t in candidate[1]: result = [] for i in string.digits: pc = int(rep(candidate[0], t, i)) if isPrime(pc) and len(str(pc)) == digitCount: result.append(int(rep(candidate[0], t, i))) if(len(result) >= targetNumber): print(result) return True return False
'''
Created on Jun 2, 2015
n² + an + b = prime, where |a| < 1000 and |b| < 1000
b is prime
'''
from utilities.divisors import getPrimes, isPrime
if __name__ == '__main__':
maximum = (1, 1, 1)
for b in getPrimes(1000):
for a in range(-999, 1000, 2):
res = b
n = 0
while isPrime(res):
n += 1
res = n ** 2 + a * n + b
if(maximum[-1] < n):
print("scooby doo ", maximum[-1], n)
maximum = (a, b, n)
print(a, b, n)
print(maximum)
print(maximum[0] * maximum[1])
'''
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 utilities.divisors import getPrimes, isPrime
def allRotations(x):
return [str(x)[i:] + str(x)[:i] for i in range(len(str(x)))]
if __name__ == '__main__':
print(len([x for x in getPrimes(1000000) if all(isPrime(int(rot)) for rot in allRotations(x))]))
'''
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 itertools import permutations
from string import digits
from utilities.divisors import isPrime
if __name__ == '__main__':
for to in range(9, 1, -1):
candidates = ([int("".join(p)) for p in list(permutations([i for i in digits[1:to]]))
if isPrime(int("".join(p)))])
print(max(candidates) if candidates else "empty")
def foursHasProperty(x): s = str(x) return isPermutation(x + 3330, s) and isPermutation(x + 2 * 3330, s) \ and isPrime(x) and isPrime(x + 3330) and isPrime(x + 2 * 3330)
def allCombisArePrime(p1, p2): if(len(str(p1) + str(p2)) > 8): print(p1, p2) return False return isPrime(int(str(p1) + str(p2))) and isPrime(int(str(p2) + str(p1)))
def foursHasProperty(p):
return all(isPrime(i) for i in [int(p[i:]) for i in range(1, len(p))])\
and all(isPrime(i) for i in [int(p[:-i]) for i in range(1, len(p))])
def foursHasProperty(x):
for i in range(1, ceil(sqrt(x / 2))):
print(x, i, x - 2 * i ** 2)
if isPrime(x - 2 * i ** 2):
return True
return False
15 = 7 + 2×2**2
21 = 3 + 2×3**2
25 = 7 + 2×3**2
27 = 19 + 2×2**2
33 = 31 + 2×1**2
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum of a prime
and twice a square?
"""
from math import sqrt, ceil
from utilities.divisors import isPrime
def foursHasProperty(x):
for i in range(1, ceil(sqrt(x / 2))):
print(x, i, x - 2 * i ** 2)
if isPrime(x - 2 * i ** 2):
return True
return False
if __name__ == "__main__":
i = 1
while True:
i = i + 2
if not isPrime(i) and not foursHasProperty(i):
print(i)
break
def testIsPrime(self):
self.assertEqual(False, isPrime(4))
self.assertEqual(True, isPrime(2))
self.assertEqual(False, isPrime(1))
self.assertEqual(False, isPrime(0))