# Quadratic primes

*Author: Shlomi Fish*

https://projecteuler.net/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.

Source code: prob027-shlomif.pl

use v6; sub is_prime(Int $n) returns Bool { if ($n <= 1) { return False; } for (2 .. $n.sqrt.floor) -> $i { if $n % $i == 0 { return False; } } return True; } my (Int $max_a, Int $max_b); my Int $max_iter = 0; for (0 .. 999) -> $b_coeff { for ((-$b_coeff+1) .. 999) -> $a_coeff { my $n = 0; while is_prime($b_coeff+$n*($n+$a_coeff)) { $n++; } $n--; if ($n > $max_iter) { ($max_a, $max_b, $max_iter) = ($a_coeff, $b_coeff, $n); } } } say "A sequence length of $max_iter, is generated by a=$max_a, b=$max_b, the product is {$max_a*$max_b}";