So today, I was pondering How difficult it’d be to implement the Sieve of Eratosthenes in Haskell.

Turns out, very easy after all.. and more elegant.

sieve [] = []
sieve (x:xs) = x : sieve [p | p <- xs, p `mod` x /= 0]

So With that figured out, A new challenge was to solve this for any n. O’course, my earlier much simpler solution would have sufficed –

foldl (lcm) 1 [1..20]

But that doesn’t quite use my own implementation of the Sieve of Eratosthenes algorithm, does it?

So that needed to be corrected 😉

So here’s my solution –