Theorem 39 (by Shalosh B. Ekhad)

                       Consider the following recurrence

if x[n-1], x[n] mod 2 equal, resp. , 1, 1, then,

    x[n + 1] = 1/2 x[n - 1] + 1/2 x[n]

  if x[n-1], x[n] mod 2 equal, resp. , 1, 0, then, x[n + 1] = -x[n - 1] - x[n]

  if x[n-1], x[n] mod 2 equal, resp. , 0, 1, then, x[n + 1] = -x[n - 1] - x[n]

if x[n-1], x[n] mod 2 equal, resp. , 0, 0, then,

    x[n + 1] = 1/2 x[n - 1] - 1/2 x[n]

          based on empirical observation, we are lead to the following

                   Conjecture: Every trajectory of the rule F

                       with gcod(x[-1],x[0])=1 eventually

                      ends in one of the following orbits

                     {[0, 0], [1, -1], [1, 1], [1, -1, 0]}

            We have the following bounds for the general term, x[n]

                       If x[n-1] mod, 2, equals , 1, and

                         x[n] mod, 2,  equals, 1, then

                        | x[n] | <= | x[-1] | + | x[0] |

                       If x[n-1] mod, 2, equals , 1, and

                         x[n] mod, 2,  equals, 2, then

                        | x[n] | <= | x[-1] | + | x[0] |

                       If x[n-1] mod, 2, equals , 2, and

                         x[n] mod, 2,  equals, 1, then

                      | x[n] | <= 2 | x[-1] | + 2 | x[0] |

        This can presumbly be proved rigorously by an appropriate scheme

                         but it is not implemented yet

    Without loss of generality we can have the first two elements as, [a, b]

           Because of the above inequalities, it is possible to prove

            (rigorously!) that the only words in the alphabet, {0,1}

              that a trajectory of the rule modulo 2, can have are

{[[0], [[1], m[1]]], [[0], [[1], m[1]], [0]], [[0], [[1], m[1]], [0, 1]],

    [[0], [[1], m[1]], [0, 1, 1]]}

                 For the regular expression, [[2], [[1], m[1]]]

                    applying the rule, we see, by induction

                         that the last two elements are

         2 a               m[1]                      m[1]
[- b/3 - --- - 4/3 b (-1/2)     + 4/3 (-a - b) (-1/2)    ,
          3

            2 a               m[1]                      m[1]
    - b/3 - --- + 2/3 b (-1/2)     - 2/3 (-a - b) (-1/2)    ]
             3

         By equating the first two elements with the last two elements

        we see that a and b must be 0 (or one of the above-found orbits)

                      unless the following condition holds

                                          m[1]
                              2 - 2 (-1/2)     = 0

       and this can never happen, unless it (possibly) coincides with the

                     the above-found orbits (you prove it!)

              For the regular expression, [[2], [[1], m[1]], [2]]

                    applying the rule, we see, by induction

                         that the last two elements are

         2 a               m[1]                      m[1]
[- b/3 - --- + 2/3 b (-1/2)     - 2/3 (-a - b) (-1/2)    ,
          3

    2 b   4 a               m[1]                      m[1]
    --- + --- + 2/3 b (-1/2)     - 2/3 (-a - b) (-1/2)    ]
     3     3

         By equating the first two elements with the last two elements

        we see that a and b must be 0 (or one of the above-found orbits)

                      unless the following condition holds

                                          m[1]
                              1 - 4 (-1/2)     = 0

       and this can never happen, unless it (possibly) coincides with the

                     the above-found orbits (you prove it!)

             For the regular expression, [[2], [[1], m[1]], [2, 1]]

                    applying the rule, we see, by induction

                         that the last two elements are

 2 b   4 a               m[1]                      m[1]
[--- + --- + 2/3 b (-1/2)     - 2/3 (-a - b) (-1/2)    ,
  3     3

            2 a               m[1]                      m[1]
    - b/3 - --- - 4/3 b (-1/2)     + 4/3 (-a - b) (-1/2)    ]
             3

         By equating the first two elements with the last two elements

        we see that a and b must be 0 (or one of the above-found orbits)

                      unless the following condition holds

                                     0 = 0

       and this can never happen, unless it (possibly) coincides with the

                     the above-found orbits (you prove it!)

           For the regular expression, [[2], [[1], m[1]], [2, 1, 1]]

                    applying the rule, we see, by induction

                         that the last two elements are

         2 a               m[1]                      m[1]
[- b/3 - --- - 4/3 b (-1/2)     + 4/3 (-a - b) (-1/2)    ,
          3

                            m[1]                      m[1]
    b/6 + a/3 - 1/3 b (-1/2)     + 1/3 (-a - b) (-1/2)    ]

         By equating the first two elements with the last two elements

        we see that a and b must be 0 (or one of the above-found orbits)

                      unless the following condition holds

                                          m[1]
                              3 + 6 (-1/2)     = 0

       and this can never happen, unless it (possibly) coincides with the

                     the above-found orbits (you prove it!)