Theorem 43 (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], [3, 1, 2, 1, -3, -1, -2, -1]} We have the following bounds for the general term, x[n] | x[n] | <= | x[-1] | + | x[0] | This is proved (rigorously!) by the existence of the following scheme [{[[1, 1], {[1, 1], [1, 2]}, {a, b}], [[1, 2], {[2, 1]}, {a, b, -a + b, a - 2 b}], [[2, 1], {[1, 1]}, {a, b, -a - b}]}, [1, 2]] 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, [[1], m[2]], [0, 1, 1, 0, 1]]} %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 m[1] m[1] m[1] m[1] [5/9 b (-1) + 5/9 a (-1) - 8/9 b (-1/4) + 4/9 a (-1/4) - 4/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 4/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] m[1] m[1] m[1] (-1/2) , 5/9 b (-1) + 5/9 a (-1) - 8/9 b (-1/4) m[1] + 4/9 a (-1/4) + 2/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 2/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-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] 9 - 4 (-1/2) (-1/4) + 9 (-1) (-1/2) (-1/4) m[2] m[1] m[1] m[1] + (-1/2) (-1) + 4 (-1/4) - 10 (-1) = 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 m[1] m[1] m[1] m[1] [5/9 b (-1) + 5/9 a (-1) - 8/9 b (-1/4) + 4/9 a (-1/4) + 2/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 2/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] (-1/2) , 2 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 2 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-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] 9 (-1) (-1/2) (-1/4) - 5 (-1) - 4 (-1/4) m[2] m[1] m[2] m[1] - 4 (-1/2) (-1) - 5 (-1/2) (-1/4) + 9 = 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 (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 2 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] m[1] m[1] m[1] (-1/2) , -5/9 b (-1) - 5/9 a (-1) + 8/9 b (-1/4) m[1] - 4/9 a (-1/4) - 8/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 8/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-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] 9 (-1) (-1/2) (-1/4) + (-1/2) (-1) m[2] m[1] m[1] m[1] + 11 (-1/2) (-1/4) + 5 (-1) - 8 (-1/4) + 9 = 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 m[1] m[1] m[1] m[1] [-5/9 b (-1) - 5/9 a (-1) + 8/9 b (-1/4) - 4/9 a (-1/4) - 8/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 8/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] (-1/2) , -1/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 1/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] m[1] m[1] m[1] (-1/2) - 5/18 b (-1) - 5/18 a (-1) + 4/9 b (-1/4) m[1] - 2/9 a (-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[2] m[1] m[1] m[2] m[1] 6 - 2 (-1/2) (-1/4) - 3 (-1) (-1/2) (-1/4) m[2] m[1] m[1] + 3 (-1/2) (-1) + 5 (-1) = 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] [-1/3 (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 1/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] m[1] m[1] m[1] (-1/2) - 5/18 b (-1) - 5/18 a (-1) + 4/9 b (-1/4) m[1] m[1] m[1] m[1] - 2/9 a (-1/4) , 5/18 b (-1) + 5/18 a (-1) - 4/9 b (-1/4) m[1] + 2/9 a (-1/4) + 7/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 7/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-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[2] m[1] 6 - 5 (-1/2) (-1/4) - 2 (-1/2) (-1) m[2] m[1] m[1] m[1] - 3 (-1/2) (-1) (-1/4) + 4 (-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, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [5/18 b (-1) + 5/18 a (-1) - 4/9 b (-1/4) + 2/9 a (-1/4) + 7/3 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) - 7/3 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-1/4) ) m[2] (-1/2) , -2 m[1] m[1] m[1] m[1] (2/3 b (-1) + 2/3 a (-1) - 2/3 b (-1/4) + 1/3 a (-1/4) ) m[2] (-1/2) + 2 m[1] m[1] m[1] m[1] (1/2 b (-1) + 1/2 a (-1) - b (-1/4) + 1/2 a (-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[1] m[1] -9 (-1/2) (-1) (-1/4) - 5 (-1) - 4 (-1/4) m[2] m[1] m[2] m[1] - (-1/2) (-1) + 19 (-1/2) (-1/4) + 18 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)