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# Problem 27

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.

• 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.

n2 + n + 41

• n2 + an + b, 满足|a| < 1000且|b| < 1000

• 其中|n|指n的模或绝对值
例如|11| = 11以及|−4| = 4