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.

```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}";

```