Generating double linear sequence in Haskell

Need some comments on making the below code more effective. I'm pretty new to Haskell.

The sequence in generateSeq is generated by the rule that if x is in the sequence then x2+1 and x3+1 are also in the sequence.

getNumberAt gets the n'th number in the (sorted) generated sequence.

The code works but somehow I feel that I'm overcomplicating this. Any pointers are appreciated!

generateSeq :: Int -> [Integer]
    generateSeq x
        | x == 0 = [1]
        | otherwise = sort $ prev++toAppend
                         where prev = generateSeq(x-1)
                               toAppend = [next*2+1,next*3+1]
                               next = head $ drop(x-1) prev

    getNumberAt :: Int -> Integer
    getNumberAt n = nub(generateSeq n)!!n

    -- Test this!
    main =
       print $ getNumberAt 321

Repeatedly sorting is asympotically slower than keeping a sorted data structure, nub takes quadratic time, head . drop (x-1) = (!!x), and try to use library-defined recursion combinators instead of explicitly using recursion.

import qualified Data.Set as S

generateSeq :: [Integer]
generateSeq = unfoldr (fmap foo . S.minView) $ singleton 1 where
  foo :: (Integer, S.Set Integer) -> (Integer, S.Set Integer)
  foo (x, s) = (x, S.insert (2*x+1) $ S.insert (3*x+1) s)

looks up comonad stuff on a hunch

import qualified Data.Set as S
import Control.Comonad

generateSeq :: [Integer]
generateSeq = unfoldr ((fmap . extend . uncurry) foo . S.minView) $ singleton 1 where
  foo :: Integer -> S.Set Integer -> S.Set Integer
  foo x = S.insert (2*x+1) . S.insert (3*x+1)

StackExchange Code Review Q#147413, answer score: 4