Evaluate the sum of all the amicable numbers under 10000.

let sumOfDivisors n = seq {for m in 2..int(sqrt(float n)) do if n % m = 0 then yield! [m;n/m]} |> Seq.sum |> (+) 1
let ans = {1..10000} |> Seq.filter (fun n -> sumOfDivisors (sumOfDivisors n) = n && sumOfDivisors n <> n) |> Seq.sum

The naive solution I came up for this returns a sequence of pairs. I’m not sure what the idiomatic and/or efficient solution is in F#, so I just yield!ed the pair as a list. Any F# experts have a better way?

I should add a disclaimer that while this sum of divisors function works for this problem, it’s actually wrong for perfect squares, just in case anyone stumbles on this code and tries to use it.

After seeing how explicitly calling DivRem sped up problem 16, I wondered if explicitly saving the sum would speed this up.

let sumOfDivisors n = seq {for m in 2..int(sqrt(float n)) do if n % m = 0 then yield! [m;n/m]} |> Seq.sum |> (+) 1
let amicable n =
let m = sumOfDivisors n
n = sumOfDivisors m && m <> n
let ans = {1..10000} |> Seq.filter amicable |> Seq.sum

But this gives the exact same performance; I guess the compiler does the optimization automatically.

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