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"""

Project Euler Problem 27

https://projecteuler.net/problem=27

Problem Statement:

Euler discovered the remarkable quadratic formula:

n2 + n + 41

It turns out that the formula will produce 40 primes for the consecutive values

n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible

by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41.

The incredible formula n2 − 79n + 1601 was discovered, which produces 80 primes

for the consecutive values n = 0 to 79. The product of the coefficients, −79 and

1601, is −126479.

Considering quadratics of the form:

n² + an + b, where |a| < 1000 and |b| < 1000

where |n| is the modulus/absolute value of ne.g. |11| = 11 and |−4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that

produces the maximum number of primes for consecutive values of n, starting with

n = 0.

"""

import math

def is_prime(k: int) -> bool:

"""

Determine if a number is prime

>>> is_prime(10)

False

>>> is_prime(11)

True

"""

if k < 2 or k % 2 == 0:

return False

elif k == 2:

return True

else:

for x in range(3, int(math.sqrt(k) + 1), 2):

if k % x == 0:

return False

return True

def solution(a_limit: int = 1000, b_limit: int = 1000) -> int:

"""

>>> solution(1000, 1000)

-59231

>>> solution(200, 1000)

-59231

>>> solution(200, 200)

-4925

>>> solution(-1000, 1000)

>>> solution(-1000, -1000)

"""

longest = [0, 0, 0] # length, a, b

for a in range((a_limit * -1) + 1, a_limit):

for b in range(2, b_limit):

if is_prime(b):

count = 0

n = 0

while is_prime((n ** 2) + (a * n) + b):

count += 1

n += 1

if count > longest[0]:

longest = [count, a, b]

ans = longest[1] * longest[2]

return ans

if __name__ == "__main__":

print(solution(1000, 1000))

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