027.py

#!/usr/bin/python
# -*- coding: utf-8 -*-

#Euler published 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.

#Using computers, 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 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.

#Answer:
	#-59231

from time import time; t=time()
from mathplus import sieve, product

M = 15000
primes = sieve(M)
# one of (1+a+b, b*b+a*b+b) is not a prime, so n*n+a*n+b has less than range(b) primes
ab = [[a, b, b] for a, b in product(
    range(-1000+1, 1000, 2),
    [i for i in range(83, 1000, 2) if primes[i]]
    )]

n = 0
lastone = None
while len(ab) > 1:
    lastone = ab[0]
    x = 2*n+1
    for v in ab:
        v[2] += x+v[0]
    ab = [[a, b, x] for a, b, x in ab if x > 0 and (x >= M or primes[x])]
    n += 1

lastone = ab[0]
print(lastone[0]*lastone[1])#, time()-t