Quadratic primes
Euler published 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.

Using computers, 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.