#### Problem

Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d <= 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d <= 1,000,000?

#### Solution

let hasDivisor(n) = let upperBound = bigint(sqrt(double(n))) [2I..upperBound] |> Seq.exists (fun x -> n % x = 0I) let isPrime(n) = if n = 1I then false else not(hasDivisor(n)) // define the sequence of prime numbers let primeSeq = Seq.unfold (fun state -> if state >= 3I then Some(state, state+2I) else Some(state, state+1I)) 1I |> Seq.filter isPrime |> Seq.cache // define function to find the prime denominators for a number n let getPrimeFactors n = let rec getPrimeFactorsRec denominators n = if n = 1I then denominators else let denominator = primeSeq |> Seq.filter (fun x -> n % x = 0I) |> Seq.head getPrimeFactorsRec (denominators @ [denominator]) (n/denominator) getPrimeFactorsRec [] n let totient n = let primeFactors = getPrimeFactors n |> Seq.distinct n * (primeFactors |> Seq.map (fun n' -> n'-1I) |> Seq.reduce (*)) / (primeFactors |> Seq.reduce (*)) let f n = [2I..n] |> List.map totient |> List.sum let answer = f 1000000I

This is almost identical to the solution to problem 69, with the exception of the last two steps. According to the information on Farey sequence on wikipedia:

The Farey sequence of order n contains all of the members of the Farey sequences of lower orders. In particular F_{n}contains all of the members of F_{n?1}, and also contains an additional fraction for each number that is less than n andcoprimeto n. Thus F_{6}consists of F_{5}together with the fractions^{1}?_{6}and^{5}?_{6}. The middle term of a Farey sequence F_{n}is always^{1}?_{2}, for n > 1.

From this, we can relate the lengths of F_{n}and F_{n?1}usingEuler’s totient function?(n) :

Given that *F1 = 0* in our case, *F2 = 0 + totient 2*, *F3 = totient 2 + totient 3*, and so on, and therefore *Fn = totient 2 + totient 3 + … totient n*.

Hi, I’m **Yan**. I’m an **AWS Serverless Hero** and the author of **Production-Ready Serverless**.

I specialise in rapidly transitioning teams to serverless and building production-ready services on AWS.

Are you struggling with serverless or need guidance on best practices? Do you want someone to review your architecture and help you avoid costly mistakes down the line? Whatever the case, I’m here to help.

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Further reading

Here is a complete list of all my posts on serverless and AWS Lambda. In the meantime, here are a few of my most popular blog posts.

- Lambda optimization tip – enable HTTP keep-alive
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- Many faced threats to Serverless security
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- Yubl’s road to Serverless
- AWS Lambda – should you have few monolithic functions or many single-purposed functions?
- AWS Lambda – compare coldstart time with different languages, memory and code sizes
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