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 \(0 \le n \le 39\). 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 \(0 \le n \le 79\). The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
\(n^2 + an + b\),
where
\(|a|< 1000\)
and
\(|b|\le1000\)
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 \(n = 0\).