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 $0<= n <= 40$. However, when
$n = 40, 40^{2} + 40 + 41 = 40 (40 + 1) + 41$ is divisable by 41, and certainly when $n = 41, 41^{2} + 41 + 41$ is
clearly divisable 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 $-126,479$.
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 $n = 0$.
You can find the algorithm to solve this challenge inside challenge.cpp.
Click here to see the result!
Result is: -59,231