Quadratic primes

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本文探讨了Euler发现的奇特二次公式及其应用，通过编程实现找到系数a和b，使得二次多项式n² + an + b产生最多的连续质数，解决了Project Euler第27题。

https://projecteuler.net/problem=27

Quadratic primes

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.

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.

首先b一定大于零，其次本想第一个公式算出n，第二个公式只要去验证n+1即可，不需要从0开始计算，不过真正计算的时候，发现事实不是这样的。

看来还是每一个公式从0开始计算的实在，而且速度也不慢。

def QuadraticPrimes():
    coefficients = 0
    maxnum = 0
    for i in range(-999,1000):
        for j in range(1,1000):
            n = 0
            tmp = j
            #算出最大的连续n值
            while  tmp > 0 and isPrime(tmp):
                n += 1
                tmp = n * n + i * n + j
            if maxnum < n:
                maxnum = n
                coefficients = i * j
    return coefficients
#判断n是否是质数，记录一下字典，加快速度
primedict = {}
import math
def isPrime(n):
    if n in primedict:return primedict[n]
    temp = int(math.sqrt(n)) + 1
    for i in range(2,temp):
        if n % i == 0:
            primedict[n]=False
            return False
    primedict[n]=True
    return True

print(QuadraticPrimes())