Theorem 114 (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, 0], [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, 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, [0]], [%1, [1]], [%1, [0, 1]], [%1, [1, 0]], [%1, [0, 1, 1]], [%1, [1, 0, 1]], [%1, [0, 1, 1, 0]], [%1, [1, 0, 1, 1]], [%1, [0, 1, 1, 0, 1]], [%1, [1, 0, 1, 1, 0]], [%1, [0, 1, 1, 0, 1, 1]], [%1, [1, 0, 1, 1, 0, 1]], [%1, [0, 1, 1, 0, 1, 1, 0]]} %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] [-a - b + 2 a (1/4) , 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[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]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a (1/4) , a/2 + b/2 - 1/2 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[1] 1 - (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]], [2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a (1/4) , a + b - 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 0 = 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, 2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a/2 + b/2 - 1/2 a (1/4) , -3/2 a (1/4) + a/2 + b/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] (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]], [2, 1]] applying the rule, we see, by induction that the last two elements are m[1] [a + b - a (1/4) , -a - b] 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] 2 (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]], [1, 2, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [-3/2 a (1/4) + a/2 + b/2, -a - b + 2 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[1] 4 (1/4) + 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, 1], m[1]], [2, 1, 1]] applying the rule, we see, by induction that the last two elements are m[1] [-a - b, -a - b + 1/2 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[1] 6 + (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, 2, 1, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] 3 a 3 b [-a - b + 2 a (1/4) , 7/4 a (1/4) - --- - ---] 4 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] 11 - 7 (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]], [2, 1, 1, 2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [-a - b + 1/2 a (1/4) , 1/2 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 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, 2, 1, 1, 2]] applying the rule, we see, by induction that the last two elements are m[1] 3 a 3 b m[1] [7/4 a (1/4) - --- - ---, a/4 + b/4 - 1/4 a (1/4) ] 4 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] -3 (1/4) + 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, 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] [1/2 a (1/4) , a + b - 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 0 = 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, 2, 1, 1, 2, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a/4 + b/4 - 1/4 a (1/4) , -3/2 a (1/4) + a/2 + b/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] 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]], [2, 1, 1, 2, 1, 1]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [a + b - a (1/4) , -3/4 a (1/4) + a/2 + b/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] 5 (1/4) - 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]], [2, 1, 1, 2, 1, 1, 2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] [-3/4 a (1/4) + a/2 + b/2, - a/2 - b/2 + 1/4 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[1] (1/4) + 1 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)