Theorem 10 (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, 0], [5, -1, -2, 3, -5, 1, 2, -3]} 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] | + | x[0] | If x[n-1] mod, 2, equals , 2, and x[n] mod, 2, equals, 1, then | x[n] | <= | 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 {[%1], [%1, [0]], [%1, [1]], [%1, [0, 1]], [%1, [0, 1, 1]], [%1, [0, 1, 1, 0]], [%1, [0, 1, 1, 0, 1]]} %1 := [[0, 1, 1, 1], m[1]] For the regular expression, [[[2, 1, 1, 1], m[1]]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [-6/5 a (1/4) - 4/5 b (1/4) + 1/5 a (-1) - 1/5 b (-1) , m[1] m[1] m[1] m[1] 2/5 a (-1) - 2/5 b (-1) + 3/5 a (1/4) + 2/5 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[1] m[1] m[1] 5 (1/4) (-1) + 4 (1/4) + (-1) + 5 = 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]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [2/5 a (-1) - 2/5 b (-1) + 3/5 a (1/4) + 2/5 b (1/4) , m[1] m[1] m[1] m[1] 3/10 a (1/4) + 1/5 b (1/4) - 3/10 a (-1) + 3/10 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] 5 (1/4) (-1) - 7 (-1) - 8 (1/4) + 10 = 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]], [2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [2/5 a (-1) - 2/5 b (-1) + 3/5 a (1/4) + 2/5 b (1/4) , m[1] m[1] m[1] m[1] 3/5 a (1/4) + 2/5 b (1/4) - 3/5 a (-1) + 3/5 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] ((-1) - 1) ((1/4) - 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]], [2, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [3/5 a (1/4) + 2/5 b (1/4) - 3/5 a (-1) + 3/5 b (-1) , 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] -5 (1/4) (-1) - 3 (1/4) + 8 (-1) + 5 = 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]], [2, 1, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a (-1) - b (-1) , m[1] m[1] m[1] m[1] -3/10 a (1/4) - 1/5 b (1/4) - 1/5 a (-1) + 1/5 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] -5 (1/4) (-1) - 12 (-1) + 2 (1/4) + 10 = 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]], [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] [-3/10 a (1/4) - 1/5 b (1/4) - 1/5 a (-1) + 1/5 b (-1) , m[1] m[1] m[1] m[1] -4/5 a (-1) + 4/5 b (-1) + 3/10 a (1/4) + 1/5 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[1] m[1] m[1] -5 (1/4) (-1) + (1/4) - 6 (-1) + 10 = 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]], [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] [-4/5 a (-1) + 4/5 b (-1) + 3/10 a (1/4) + 1/5 b (1/4) , m[1] m[1] m[1] m[1] -3/5 a (1/4) - 2/5 b (1/4) + 3/5 a (-1) - 3/5 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] 5 (1/4) (-1) + 14 (-1) + (1/4) + 10 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)