Yan Cui

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#### 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*.

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