New order #4: World

JavaScript (ES6),  55 51  50 bytes

Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh

n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")

Try it online!

Jelly, 15 bytes

µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ

A full program accepting the integer, n (1-based), from STDIN which prints the result.

Try it online!

How?

µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ           µ¡  - repeat this monadic chain STDIN times (starting with x=0)
                -                   e.g. x = ...  0      [1,0]            [9,3,1,0]
 ×3             -   multiply by 3                 0      [3,0]            [27,9,3,0]
    H           -   halve                         0      [1.5,0]          [4.5,1.5,0.5,0]
   ż            -   zip together                  [0,0]  [[3,1.5],[0,0]]  [[27,4.5],[9,1.5],[3,0.5],[0,0]]
     Ḟ          -   floor                         [0,0]  [[3,1],[0,0]]    [[27,4],[9,1],[3,0],[0,0]]
      Ḣ         -   head                          0      [3,1]            [27,4]
       ḟ        -   filter discard if in x        []     [3]              [27,4]
        ȯ1      -   logical OR with 1             1      [3]              [27,4]
          Ṫ     -   tail                          1      3                4
           ;    -   concatenate with x            [1,0]  [3,1,0]          [4,9,3,1,0]
              Ḣ - head                            1      3                4
                - implicit print

Perl 6, 49 bytes

-2 bytes thanks to nwellnof

{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}

Try it online!

Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3

Explanation:

{                                             }  # Anonymous code block
 (                                   ...*)[$_]   # Index into the infinite sequence
  1,3                                            # That starts with 1,3
     ,{                             }            # And each element is
       (                 ).max(    )             # The first of
          @_[*-1]X                               # The previous element
        3*        div 6                          # Halved and floored
        3*        div  ,1                        # Or tripled
                               *∉@_             # That hasn't appeared in the sequence yet