Quadratic primes

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 \( 0 \leq n \leq 39 \). 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 \( 0 \leq n \leq 79 \). The product of the coefficients, \( −79 \) and \( 1601 \), is \( −126479 \).

Considering quadratics of the form:

\[ \begin{align} & n^2 + an + b,\ where\ |a|< 1000\ and\ |b|\ \leq 1000\\ \\ & where\ |n|\ is\ the\ modulus/absolute value\ of\ n\\ & e.g.\ |11|=11\ and\ |-4|=4 \end{align} \]

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 \( n=0 \).