Project Euler – Problem 27 Solution

Problem

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.

Solution

let hasDivisor(n) =
    let upperBound = int64(sqrt(double(n)))
    [2L..upperBound] |> Seq.exists (fun x -> n % x = 0L)

// need to consider negative values
let isPrime(n) = if n <= 1L then false else not(hasDivisor(n))

// the quadratic expression
let F (n:int64) (a:int64) (b:int64) = (n*n) + (a*n) + b

// function to return the number of consecutive primes the coefficients generate
let primeCount a b =
    Seq.unfold (fun state -> Some(state, state + 1L)) 0L
    |> Seq.takeWhile (fun n -> isPrime (F n a b))
    |> Seq.length

let aList, bList = [-999L..999L], [2L..999L] |> List.filter isPrime

let answer =
    let (a, b, _) =
        aList
        |> List.collect (fun a ->
            bList
            |> List.filter (fun b -> a + b >= 1L)
            |> List.map (fun b -> (a, b, primeCount a b)))
        |> List.maxBy (fun (_, _, count) -> count)
    a * b

Whilst I started off using a brute force approach (which took quite a while to run) I realised that most of the combinations of a and b don’t generate any primes at all. In order to reduce the number of combinations you need to check, consider the cases when n = 0 and n = 1:

n = 0: expression evaluates to b, in order for this to generate a prime b must be >= 2.

n = 1: expression evaluates to 1 + a + b, in order for this to generate a prime 1 + a + b must be >= 2, therefore a + b >= 1.