Euler’s Totient function, ?(n) [sometimes called the phi function], is used to determine the number of numbers less than 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.
It can be seen that n=6 produces a maximum n/?(n) for n <= 10.
Find the value of n <= 1,000,000 for which n/?(n) is a maximum.
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 answer = [2I..1000000I] |> List.maxBy (fun n -> double(n) / double(totient n))
Having borrowed a number of functions I had used in previous solutions, the main function of interest here is the totient function. Going by the information on Euler’s totient function from wikipeidia and mathworld, the value of the totient function of n is be calculated using its prime factors:
In words, this says that the distinct prime factors of 36 are 2 and 3; half of the thirty-six integers from 1 to 36 are divisible by 2, leaving eighteen; a third of those are divisible by 3, leaving twelve coprime to 36. And indeed there are twelve: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, and 35.
Which in essence is what the totient function in my solution is doing, except to avoid any rounding errors it actually multiplies the number n by the product of the numerators before dividing by the product of denominators, i.e. 36 * (1 * 2) / (2 * 3) instead of 36 * 0.5 * 0.66666666667 which introduces a rounding error which may or may not has an influence in the result.
This solution takes a long time and doesn’t comply with the under one minute rule, by using a prime sieve like this one I was able to get the execution time down to under 50s on my machine. Here is the revised code using the aforementioned sieve implementation:
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