/
p27.go
77 lines (59 loc) · 1.97 KB
/
p27.go
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/*
Euler published 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, 41^2 + 41 + 41 is clearly divisible by 41.
Using computers, the incredible formula n^2 - 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:
n^2 + 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.
*/
package main
import (
"fmt"
"utils"
)
var primes map[int]struct{}
func init() {
prime := struct{}{}
primes = make(map[int]struct{})
for _, p := range utils.PrimesUpTo(10000) {
primes[int(p)] = prime
}
}
func main() {
maxConsecutive, ab := 0, 1
for a := 0; a < 1000; a++ {
if _, aPrime := primes[a]; aPrime {
for b := 0; b < 1000; b++ {
if _, bPrime := primes[b]; bPrime {
if consecutive, prod := consecutivePrimes(a, b), a*b; consecutive > maxConsecutive {
maxConsecutive, ab = consecutive, prod
}
if consecutive, prod := consecutivePrimes(-a, b), -a*b; consecutive > maxConsecutive {
maxConsecutive, ab = consecutive, prod
}
if consecutive, prod := consecutivePrimes(a, -b), a*-b; consecutive > maxConsecutive {
maxConsecutive, ab = consecutive, prod
}
if consecutive, prod := consecutivePrimes(-a, -b), -a*-b; consecutive > maxConsecutive {
maxConsecutive, ab = consecutive, prod
}
}
}
}
}
fmt.Println(maxConsecutive, ab)
}
func consecutivePrimes(a, b int) int {
n := 0
for {
fn := (n * n) + (a * n) + b
if _, prime := primes[fn]; !prime {
break
}
n++
}
return n
}