Euler discovered the remarkable quadratic formula:
n2+n+41
It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤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.
The incredible formula n2−79n+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:
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.
Well, first I need to find a fast way to check for primes. I should probably use one of the prime sieves I built for other problems.
...
Huh. The first sieve I chose to use worked fine, even though it uses
the inefficient Vec<bool>.