How about the following Haskell program? rec ((x:xs),p) = (filter (/= p+x) xs,p+...

sltkr · 2025-06-29T15:34:26 1751211266

That just enumerates the entire sequence; I think the challenge is to do it faster than that.

By the way, the use of `filter` makes your implementation unnecessarily slow. (The posted link also contains Haskell code, which uses `delete` from Data.List instead of `filter`, which is only slightly better.)

I'd solve it like this, which generates both sequences in O(n) time, and the mutual recursion is cute:

    a005228 = 1 : zipWith (+) a005228 a030124

    a030124 = go 1 a005228 where
        go x ys
            | x < head ys = x     : go (x + 1) ys
            | otherwise   = x + 1 : go (x + 2) (tail ys)