Hamming numbers are numbers of the form  

    H = 2i × 3j × 5k
           where 
     i,  j,  k  ≥  0 

Hamming numbers   are also known as   ugly numbers   and also   5-smooth numbers   (numbers whose prime divisors are less or equal to 5).



Task

Generate the sequence of Hamming numbers, in increasing order.   In particular:

 Show the   first twenty   Hamming numbers.
 Show the   1691st   Hamming number (the last one below   231).
 Show the   one millionth   Hamming number (if the language – or a convenient library – supports arbitrary-precision integers).


Related tasks

 Humble numbers
 N-smooth numbers


References

 Wikipedia entry:   Hamming numbers     (this link is re-directed to   Regular number).
 Wikipedia entry:   Smooth number
 OEIS entry:   A051037   5-smooth   or   Hamming numbers
 Hamming problem from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread here and here).


type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>> // ': fix colouring
 
let rec hammings() =
  let rec (-|-) (Cons(x, nxf) as xs) (Cons(y, nyf) as ys) =
    if x < y then Cons(x, lazy(nxf.Value -|- ys))
    elif x > y then Cons(y, lazy(xs -|- nyf.Value))
    else Cons(x, lazy(nxf.Value -|- nyf.Value))
  let rec inf_map f (Cons(x, nxf)) =
    Cons(f x, lazy(inf_map f nxf.Value))
  Cons(1I, lazy(let x = inf_map ((*) 2I) hamming
                let y = inf_map ((*) 3I) hamming
                let z = inf_map ((*) 5I) hamming
                x -|- y -|- z))
 
// testing...    
[<EntryPoint>]
let main args =
  let rec iterLazyListFor f n (Cons(v, rf)) = 
    if n > 0 then f v; iterLazyListFor f (n - 1) rf.Value
  let rec nthLazyList n ((Cons(v, rf)) as ll) =
    if n <= 1 then v else nthLazyList (n - 1) rf.Value
  printf "( "; iterLazyListFor (printf "%A ") 20 (hammings()); printfn ")"
  printfn "%A" (hammings() |> nthLazyList 1691)
  printfn "%A" (hammings() |> nthLazyList 1000000)
  0