Comment by mlochbaum

It's possible to get O(N^(3/2)/log N) in an ordinary interpreter with some changes to the code, assuming linear-time ~ with hashing. The idea is to leave the primes in ⍵ and stop when it stops changing, which will happen at the first prime over √N. To get the complexity multiply by O(N) for each step. It's also a small change to get the multiples to remove to start at n×n instead of n.

    i←¯1⋄{⍵~n×n↓⍳1+⌊N÷n←i⊃⍵⊣i+←1}⍣≡1↓1+⍳N

I think this is about as good as can be done with ~ instead of marking out bits. And I wouldn't say it's as easy as the imperative version!