- Reciprocal cycles
Euler discovered the remarkable quadratic formula: n^2+n+41
It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when n=40,40^2+40+41=40(40+1)+41 is divisible by 41, and certainly when n=41,41^2+41+41 is clearly divisible by 41.
The incredible formula n^2−79n+1601 was discovered, which produces 80 primes for the consecutive values 0≤n≤79. The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
n^2+an+b, where |a|<1000 and |b|≤1000
where |n| is the modulus/absolute value of nhttps://projecteuler.net/problem=27
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.
問題は「2次式n^2 + an + bにおける，n = 0から始めて連続する整数で素数を生成したときに最長の長さとなる係数a, bの積を答求めよ」．
- 解答例 (Ruby)
require 'prime' def func(n, a, b) return n ** 2 + a * n + b end pNum = Prime.each(1000).to_a maxCount = 0 ansA = 0 ansB = 0 for i in 0..pNum.length - 1 do b = pNum[i] for a in -999..999 do n = 0 while Prime.prime?(func(n, a, b)) do n += 1 end if maxCount < n then ansA = a ansB = b maxCount = n end end end print(ansA, " * ", ansB, " = ", ansA * ansB, "\n")