Theorem 108 (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, 0], [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, 2 b + a}], [[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[2] m[2] m[2] m[2] [-1/3 %2 (-1) + 2/3 %1 (-1) + 4/3 %2 (1/2) + 4/3 %1 (1/2) , m[2] m[2] m[2] m[2] 1/3 %2 (-1) - 2/3 %1 (-1) + 2/3 %2 (1/2) + 2/3 %1 (1/2) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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[2] m[1] m[2] 3 + 4 (-1) (1/4) - 3 (-1) (1/2) (1/4) - (1/2) = 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] [1/3 %2 (-1) - 2/3 %1 (-1) + 2/3 %2 (1/2) + 2/3 %1 (1/2) , m[2] m[2] 2 %2 (1/2) + 2 %1 (1/2) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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[2] m[1] m[2] m[2] m[1] 9 (-1) (1/2) (1/4) - 5 (-1) - 4 (-1) (1/4) m[2] m[2] m[1] - 4 (1/2) - 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[2] m[2] [2 %2 (1/2) + 2 %1 (1/2) , m[2] m[2] m[2] m[2] 1/3 %2 (-1) - 2/3 %1 (-1) + 8/3 %2 (1/2) + 8/3 %1 (1/2) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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[2] m[1] m[2] m[2] m[1] -9 (-1) (1/2) (1/4) - 7 (1/2) - 5 (1/2) (1/4) m[2] m[2] m[1] - 5 (-1) + 8 (-1) (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[2] m[2] m[2] m[2] [1/3 %2 (-1) - 2/3 %1 (-1) + 8/3 %2 (1/2) + 8/3 %1 (1/2) , m[2] m[2] m[2] m[2] -1/3 %2 (1/2) - 1/3 %1 (1/2) - 1/6 %2 (-1) + 1/3 %1 (-1) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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] 18 + 10 (1/2) (1/4) - 16 (-1) (1/4) m[2] m[2] m[1] m[2] m[2] + 9 (-1) (1/2) (1/4) - 7 (1/2) - 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[2] m[2] m[2] m[2] [-1/3 %2 (1/2) - 1/3 %1 (1/2) - 1/6 %2 (-1) + 1/3 %1 (-1) , m[2] m[2] m[2] m[2] 1/6 %2 (-1) - 1/3 %1 (-1) + 7/3 %2 (1/2) + 7/3 %1 (1/2) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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[2] m[1] m[2] m[1] 6 - 3 (1/2) (-1) (1/4) - 5 (1/2) (1/4) m[2] m[1] m[2] + 4 (-1) (1/4) - 2 (1/2) = 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[2] m[2] m[2] m[2] [1/6 %2 (-1) - 1/3 %1 (-1) + 7/3 %2 (1/2) + 7/3 %1 (1/2) , m[2] m[2] 2 %2 (1/2) + 2 %1 (1/2) ] m[1] m[1] %1 := - a/2 - b/2 - 1/2 a (1/4) + b (1/4) 2 a 2 b m[1] m[1] %2 := --- + --- + 1/3 a (1/4) - 2/3 b (1/4) 3 3 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[2] m[1] m[2] m[2] m[1] 9 (1/2) (-1) (1/4) - 5 (-1) - 4 (-1) (1/4) m[2] m[2] m[1] - 13 (1/2) - 5 (1/2) (1/4) + 18 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)