Theorem 112 (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], [3, 2, 1, -3, -2, -1], [1, -1, -1, 0, -1, 1, 1, 0], [1, 1, 0, 1, -1, -1, 0, -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}], [[2, 1], {[1, 1]}, {a, b, -a - b}]}, [1, 1]] 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, 1], m[1]]], [[[0, 1, 1], m[1]], [0]], [[[0, 1, 1], m[1]], [0, 1]]} For the regular expression, [[[2, 1, 1], m[1]]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [b (-1) + a (-1) , m[1] m[1] m[1] m[1] 2/3 a (-1) + 2/3 b (-1) - 2/3 b (1/2) + 1/3 a (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[1] m[1] m[1] -3 (-1) (1/2) - 5 (-1) + 2 (1/2) + 3 = 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], m[1]], [2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [2/3 a (-1) + 2/3 b (-1) - 2/3 b (1/2) + 1/3 a (1/2) , m[1] m[1] m[1] m[1] 1/3 b (-1) + 1/3 a (-1) + 2/3 b (1/2) - 1/3 a (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[1] ((-1) - 1) ((1/2) - 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], m[1]], [2, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [1/3 b (-1) + 1/3 a (-1) + 2/3 b (1/2) - 1/3 a (1/2) , m[1] m[1] -a (-1) - b (-1) ] 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[1] m[1] m[1] 2 (-1) + 3 (-1) (1/2) + (1/2) + 3 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)