#### Problem

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

Find the sum of all the even-valued terms in the sequence which do not exceed four million.

#### Solution

Here’s my solution in F#:

let fibonacciSeq = Seq.unfold (fun (current, next) -> Some(current, (next, current + next))) (0, 1) let fibTotal = fibonacciSeq |> Seq.takeWhile (fun n -> n < 4000000) |> Seq.filter (fun n -> n % 2 = 0) |> Seq.sum

Here I’ve used a sequence, whilst a sequence is similar to a list or array in F# in that it holds a series of elements, there’s a crucial difference, each sequence element is computed only as required so it provides better performance than a list in situations which not all elements are used. If that sounds familiar to you, that’s because a sequence is basically an IEnumerable<T>!

In the first step of this code I’m building up the fibonacci sequence using the Seq.unfold function which given an initial value, generates a sequence by continuously applying some computation to work out each subsequent element in the sequence:

let fibonacciSeq = Seq.unfold (fun (current, next) -> Some(current, (next, current + next))) (0, 1)

This sequence, if iterated through, will contain all the numbers in the fibonacci sequence to infinity, which is why in the next line I’ve specified that we should take values from the sequence until the value exceeds 4 million:

fibonacciSeq |> Seq.takeWhile (fun n -> n < 4000000) [/code] The next two lines then identifies and sums all the even numbers in the sequence: [code lang="fsharp"] |> Seq.filter (fun n -> n % 2 = 0) // only interested in even numbers |> Seq.sum // add them up!

isn’t it better to do the filter then the takewhile?

yeah, you’re right, it’ll be more efficient to filter first