#### Problem

Euler’s Totient function, ?(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal tonwhich are relatively prime ton. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, ?(9)=6.

The number 1 is considered to be relatively prime to every positive number, so ?(1)=1.

Interestingly, ?(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value ofn, 1 <n< 10^{7}, for which ?(n) is a permutation ofnand the ration/?(n) produces a minimum.

#### Solution

// generate all prime numbers under <= this max let max = 10000I // initialise the list with 2 which is the only even number in the sequence let mutable primeNumbers = [2I] // only check the prime numbers which are <= the square root of the number n let hasDivisor n = primeNumbers |> Seq.takeWhile (fun n' -> n' <= bigint(sqrt(double(n)))) |> Seq.exists (fun n' -> n % n' = 0I) // only check odd numbers <= max let potentialPrimes = Seq.unfold (fun n -> if n > max then None else Some(n, n+2I)) 3I // populate the prime numbers list for n in potentialPrimes do if not(hasDivisor n) then primeNumbers <- primeNumbers @ [n] // use the same hasDivisor function instead of the prime numbers list as it offers // far greater coverage as the number n is square rooted so this function can // provide a valid test up to max*max let isPrime n = if n = 1I then false else not(hasDivisor(n)) // 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 = primeNumbers |> Seq.filter (fun x -> n % x = 0I) |> Seq.head getPrimeFactorsRec (denominators @ [denominator]) (n/denominator) getPrimeFactorsRec [] n // define Euler's totient function let totient n = if n = 1I then 1I else if isPrime n then n-1I else let primeFactors = getPrimeFactors n |> Seq.distinct n * (primeFactors |> Seq.map (fun n' -> n'-1I) |> Seq.reduce (*)) / (primeFactors |> Seq.reduce (*)) // define function to check if two numbers are permutations of each other let isPermutation a b = let aArray = a.ToString().ToCharArray() |> Array.sort let bArray = b.ToString().ToCharArray() |> Array.sort if Array.length aArray <> Array.length bArray then false else Array.forall2 (fun aChar bChar -> aChar = bChar) aArray bArray // check semi-primes less than 10 million let answer = primeNumbers |> Seq.collect (fun n -> primeNumbers |> Seq.filter (fun n' -> n' > n) |> Seq.map (fun n' -> n * n')) |> Seq.filter (fun n' -> n' > 8000000I && n' < 10000000I) |> Seq.map (fun n -> (n, totient n)) |> Seq.filter (fun (n, n') -> isPermutation n n') |> Seq.minBy (fun (n, n') -> double(n) / double(n'))

According to MathWorld the totient function of *n* equals *n – 1*if *n* is a prime, under normal circumstances *n* divided by *totient n* is minimal if *n* is a prime, but then they wouldn’t be permutations of each other as *n* cannot be the permutation of *n – 1*.

A semi-prime on the other hand, *n*, where *p* and *q* are primes and *n = pq*, its totient function equals *(p – 1)(q – 1)*, and therefore:

and is most likely the source of the answer.

The solution here generates all the semi-primes under 10 million and looks for the semi-prime n which returns the smallest* n/totient n*. The larger *n* is the smaller *n/totient n* is, hence to reduce the amount of computation required I only check for numbers greater than 8 million.

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