Theorem 117 (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 m[2] m[2] m[2] m[2] [2/3 %1 (-1) - 1/3 %2 (-1) + 4/3 %1 (1/2) + 4/3 %2 (1/2) , m[2] m[2] m[2] m[2] -2/3 %1 (-1) + 1/3 %2 (-1) + 2/3 %2 (1/2) + 2/3 %1 (1/2) ] m[1] m[1] m[1] m[1] %1 := -7/10 a (-1) - 7/10 b (-1) - 3/10 a (1/4) + 6/5 b (1/4) 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) 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] m[2] m[1] 15 + 20 (-1) (1/4) - 3 (1/2) (-1) - 2 (1/2) (1/4) m[2] m[1] m[2] m[1] - 15 (-1) (-1) (1/2) (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 m[2] m[2] m[2] m[2] [-2/3 %2 (-1) + 1/3 %1 (-1) + 2/3 %1 (1/2) + 2/3 %2 (1/2) , m[2] m[2] 2 %2 (1/2) + 2 %1 (1/2) ] m[1] m[1] m[1] m[1] %1 := 4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) m[1] m[1] m[1] m[1] %2 := -7/10 a (-1) - 7/10 b (-1) - 3/10 a (1/4) + 6/5 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[2] m[1] m[2] m[1] m[2] m[1] 15 (-1) (-1) (1/2) (1/4) - 11 (-1) (-1) m[2] m[1] m[2] m[1] m[2] m[1] - 4 (-1) (1/4) - 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[2] m[2] [2 %1 (1/2) + 2 %2 (1/2) , m[2] m[2] m[2] m[2] 2/3 %1 (-1) - 1/3 %2 (-1) - 8/3 %2 (1/2) - 8/3 %1 (1/2) ] m[1] m[1] m[1] m[1] %1 := -7/10 a (-1) - 7/10 b (-1) - 3/10 a (1/4) + 6/5 b (1/4) 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) 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] m[2] m[1] 15 (-1) (-1) (1/2) (1/4) + (1/2) (-1) m[2] m[1] m[2] m[1] m[2] m[1] + 19 (1/2) (1/4) + 11 (-1) (-1) - 16 (-1) (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 m[2] m[2] m[2] m[2] [2/3 %2 (-1) - 1/3 %1 (-1) - 8/3 %1 (1/2) - 8/3 %2 (1/2) , m[2] m[2] m[2] m[2] 7/3 %2 (1/2) + 7/3 %1 (1/2) - 1/3 %2 (-1) + 1/6 %1 (-1) ] m[1] m[1] m[1] m[1] %1 := 4/5 a (-1) + 4/5 b (-1) + 1/5 a (1/4) - 4/5 b (1/4) m[1] m[1] m[1] m[1] %2 := -7/10 a (-1) - 7/10 b (-1) - 3/10 a (1/4) + 6/5 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[2] m[1] m[2] m[1] m[2] m[1] 30 - 36 (1/2) (1/4) + 24 (-1) (1/4) + 11 (-1) (-1) m[2] m[1] m[2] m[1] m[2] m[1] + (1/2) (-1) - 15 (-1) (-1) (1/2) (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[2] m[2] m[2] m[2] [7/3 %1 (1/2) + 7/3 %2 (1/2) - 1/3 %1 (-1) + 1/6 %2 (-1) , m[2] m[2] m[2] m[2] 1/3 %1 (-1) - 1/6 %2 (-1) - 1/3 %2 (1/2) - 1/3 %1 (1/2) ] m[1] m[1] m[1] m[1] %1 := -7/10 a (-1) - 7/10 b (-1) - 3/10 a (1/4) + 6/5 b (1/4) 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) 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] m[2] m[1] 30 + 11 (1/2) (1/4) - 20 (-1) (1/4) - 6 (1/2) (-1) m[2] m[1] m[2] m[1] + 15 (-1) (-1) (1/2) (1/4) = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)