Euler’s Totient function, ?(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. 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 of n, 1 < n < 107, for which ?(n) is a permutation of n and the ratio n/?(n) produces a minimum.
// 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 – 1if 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.