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.
import math def IsPrime(x): if x<3: return False for i in range(2,int(math.sqrt(x))+1): if x%i==0: return False return True def func(a,b): k=0 while True: if IsPrime(k*k+a*k+b): k+=1 else: break return k-1 maxa,maxb=0,0 num=0 for j in range(-999,1000): if IsPrime(j): for i in range(-999,1000): temp=func(i,j) if temp>num: num=temp maxa,maxb=i,j print(maxa*maxb)
原文地址：http://blog.csdn.net/zhangzhengyi03539/article/details/46537323
