Yan Cui

I help clients go faster for less using serverless technologies.

**This article is brought to you by**

I never fully recovered my workspace setup when I upgraded my laptop two years ago, and I still miss things today. If only I had known about Gitpod back then…

#### 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!

**Whenever you’re ready, here are 3 ways I can help you:**

**Production-Ready Serverless**: Join 20+ AWS Heroes & Community Builders and 1000+ other students in levelling up your serverless game. This is your one-stop shop for**quickly levelling up your serverless skills**.- I help clients
**launch product ideas**,**improve their development processes**and**upskill their teams**. If you’d like to work together, then let’s**get in touch**. **Join my community on Discord**, ask questions, and join the discussion on all things AWS and Serverless.

Keith Nicholasisn’t it better to do the filter then the takewhile?

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

AnonymousThe initial value passed to unfold should be (1, 2) instead of (0, 1), right? (I get that it doesn’t affect the final answer due to filtering for evens, but if someone is building up the solution like i was doing and printing out the intermediate results, the results of starting with (0, 1) could be confusing)