Project Euler Problem 27

Problem 27

Euler discovered the remarkable quadratic formula:

n ² + 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 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.

# 考虑二次多项式：n ² + an + b, 满足|a| < 1000且|b| < 1000，这其中存在某个二次多项式能够对从0开始尽可能多的连续整数n都生成素数，求其系数a和b的乘积。

def isprime(n):
    if n <= 1:
        return False
    for i in range(2,int(n**0.5+1)):
        if n % i == 0:
            return False
    return True

num = 0
longest = 0
l = []
for a in range(-1000,1000):
    for b in range(-1000,1001):
        for n in range(100):
            f = n**2 + a*n + b
            if isprime(f):
                num = n
                continue
            else:
                break
        if longest < num:
            longest = num
            l = [a,b]
print(longest)
print(l[1]*l[0])

结果：-59231