Theorem 58 (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], [5, 3, 4, -1, -5, -3, -4, 1]} 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] | + 2 | x[0] | If x[n-1] mod, 2, equals , 1, and x[n] mod, 2, equals, 2, then | x[n] | <= | x[-1] | + 2 | x[0] | If x[n-1] mod, 2, equals , 2, and x[n] mod, 2, equals, 1, then | x[n] | <= | x[-1] | + | 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 {[%1, [[1], m[2]]], [%1, [[1], m[2]], [0]], [%1, [[1], m[2]], [0, 1]], [%1, [[1], m[2]], [0, 1, 1]], [%1, [[1], m[2]], [0, 1, 1, 0]]} %1 := [[0, 1, 1, 1], m[1]] For the regular expression, [[[2, 1, 1, 1], m[1]], [[1], m[2]]] applying the rule, we see, by induction that the last two elements are 11 m[1] 11 m[1] m[1] 16 m[1] [-- a (-1) + -- b (-1) + 4/15 a (1/4) - -- b (1/4) - 4/3 15 15 15 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 4/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] 11 m[1] 11 m[1] m[1] (-1/2) , -- a (-1) + -- b (-1) + 4/15 a (1/4) 15 15 16 m[1] - -- b (1/4) + 2/3 15 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) - 2/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-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] m[2] m[1] m[2] m[1] -15 + 15 (-1) (-1/2) (1/4) - 2 (-1/2) (1/4) m[2] m[1] m[1] - 3 (-1/2) (-1) + 20 (1/4) = 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, 1, 1], m[1]], [[1], m[2]], [2]] applying the rule, we see, by induction that the last two elements are 11 m[1] 11 m[1] m[1] 16 m[1] [-- a (-1) + -- b (-1) + 4/15 a (1/4) - -- b (1/4) + 2/3 15 15 15 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) - 2/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-1/2) , -2 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 2 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-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] m[2] m[1] m[1] m[1] -15 (-1) (-1/2) (1/4) + 11 (-1) + 4 (1/4) m[2] m[1] m[2] m[1] + 4 (-1/2) (-1) + 11 (-1/2) (1/4) - 15 = 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, 1, 1], m[1]], [[1], m[2]], [2, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [-2 (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 2 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] 11 m[1] 11 m[1] m[1] (-1/2) , --- a (-1) - -- b (-1) - 4/15 a (1/4) 15 15 16 m[1] + -- b (1/4) - 8/3 15 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 8/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-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] m[2] m[1] m[2] m[1] -15 (-1) (-1/2) (1/4) + (-1/2) (-1) m[2] m[1] m[1] m[1] + 19 (-1/2) (1/4) + 11 (-1) - 16 (1/4) - 15 = 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, 1, 1], m[1]], [[1], m[2]], [2, 1, 1]] applying the rule, we see, by induction that the last two elements are 11 m[1] 11 m[1] m[1] 16 m[1] [--- a (-1) - -- b (-1) - 4/15 a (1/4) + -- b (1/4) - 8/3 15 15 15 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 8/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-1/2) , -7/3 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 7/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] 11 m[1] 11 m[1] m[1] (-1/2) - -- a (-1) - -- b (-1) - 2/15 a (1/4) 30 30 m[1] + 8/15 b (1/4) ] 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] m[2] m[1] m[2] m[1] -30 + 15 (-1) (-1/2) (1/4) + 36 (-1/2) (1/4) m[2] m[1] m[1] m[1] - (-1/2) (-1) - 11 (-1) - 24 (1/4) = 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, 1, 1], m[1]], [[1], m[2]], [2, 1, 1, 2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [-7/3 (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 7/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] 11 m[1] 11 m[1] m[1] (-1/2) - -- a (-1) - -- b (-1) - 2/15 a (1/4) 30 30 m[1] 11 m[1] 11 m[1] m[1] + 8/15 b (1/4) , --- a (-1) - -- b (-1) - 2/15 a (1/4) 30 30 m[1] + 8/15 b (1/4) - 1/3 m[1] m[1] m[1] m[1] (4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) ) m[2] (-1/2) + 1/3 m[1] m[1] m[1] m[1] (7/10 a (-1) + 7/10 b (-1) + 3/10 a (1/4) - 6/5 b (1/4) ) m[2] (-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[2] m[1] m[1] m[2] m[1] -30 - 15 (-1/2) (-1) (1/4) + 11 (-1/2) (1/4) m[2] m[1] m[1] - 6 (-1/2) (-1) - 20 (1/4) = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)