Solution to problem 27

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#!/usr/bin/env python3
"""
Created on 15 Sep 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 27 of Project Euler
https://projecteuler.net/problem=27
"""
from utils import timeit, is_prime
@timeit("Problem 27")
def compute():
"""
Euler discovered the remarkable quadratic formula:
n^2 + n + 41
It turns out that the formula will produce 40 primes for the consecutive
integer values 0n39. However, when n=40, 40^2+40+41=40(40+1)+41 is
divisible by 41, and certainly when n=41,41^2+41+41 is clearly divisible
by 41.
The incredible formula n^279n+1601 was discovered, which produces 80
primes for the consecutive values 0n79. The product of the coefficients,
79 and 1601, is 126479.
Considering quadratics of the form:
n^2 + an + b
where |a|<1000, |b|1000 and |n| is the modulus/absolute value of n
e.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.
"""
LIMIT = 1000
consecutive_values = 0
for a in range(-999, LIMIT):
for b in range(LIMIT + 1):
n = 0
while is_prime(abs((n ** 2) + (a * n) + b)):
n += 1
if n > consecutive_values:
consecutive_values = n
c = a * b
return c
if __name__ == "__main__":
print(f"Result for Problem 27: {compute()}")