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≤n≤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≤n≤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|≤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.