Theorem 101 (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, -1, -1, 0]} 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 {[%1], [%1, [0]], [%1, [0, 1]], [%1, [1, 1, 1, 1, 1, 1, 1, 1, 1]], [%1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[0, 1, 1], [[1], m[1]]], [[0, 1, 1], [[1], m[1]], [0, 1, 1, 0, 1, 1, 0, 1, 1]], [[0, 1, 1], [[1], m[1]], [0, 1, 1, 0, 1, 1, 0, 1, 1, 0]]} %1 := [[0, 1, 1], m[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] m[1] [-a (-1) - b (-1) , 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] (-1) + 1 + (-1) (-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], m[1]], [2]] applying the rule, we see, by induction that the last two elements are m[1] m[1] m[1] m[1] [a (-1/2) , a (-1) + b (-1) - 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] ((-1/2) - 1) ((-1) - 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] m[1] [a (-1) + b (-1) - a (-1/2) , 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) - (-1) (-1/2) + (-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]], [1, 1, 1, 1, 1, 1, 1, 1, 1]] applying the rule, we see, by induction that the last two elements are 171 m[1] 85 m[1] 85 m[1] [--- a (-1/2) + --- a (-1) + --- b (-1) , 256 256 256 171 m[1] 171 m[1] 341 m[1] ---- a (-1) - --- b (-1) - --- a (-1/2) ] 512 512 512 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] -(-1) (-1/2) - 342 (-1/2) + (-1) + 512 = 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]], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]] applying the rule, we see, by induction that the last two elements are 171 m[1] 171 m[1] 341 m[1] [---- a (-1) - --- b (-1) - --- a (-1/2) , 512 512 512 683 m[1] 341 m[1] 341 m[1] ---- a (-1/2) + ---- a (-1) + ---- b (-1) ] 1024 1024 1024 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] (-1) + (-1) (-1/2) + 682 (-1/2) + 1024 = 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]]] applying the rule, we see, by induction that the last two elements are /3 a \ m[1] m[1] /3 a \ m[1] [2/3 |--- + b/2| (-1) + 1/6 a (-1) + 4/3 |--- + b/2| (1/2) \ 4 / \ 4 / m[1] /3 a \ m[1] m[1] - 2/3 a (1/2) , -2/3 |--- + b/2| (-1) - 1/6 a (-1) \ 4 / m[1] /3 a \ m[1] - 1/3 a (1/2) + 2/3 |--- + b/2| (1/2) ] \ 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] 3 (-1) (1/2) - 2 (-1) - 4 (1/2) + 6 = 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, 1, 1]] applying the rule, we see, by induction that the last two elements are /3 a \ m[1] m[1] m[1] [-2/3 |--- + b/2| (-1) - 1/6 a (-1) + 2/3 a (1/2) \ 4 / /3 a \ m[1] /3 a \ m[1] m[1] - 4/3 |--- + b/2| (1/2) , 1/12 |--- + b/2| (-1) + 1/48 a (-1) \ 4 / \ 4 / /3 a \ m[1] m[1] - 1/12 |--- + b/2| (1/2) + 1/24 a (1/2) ] \ 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] (-1) (1/2) + 10 (-1) + 6 (1/2) + 16 = 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, 1, 1, 2]] applying the rule, we see, by induction that the last two elements are /3 a \ m[1] m[1] /3 a \ m[1] [1/12 |--- + b/2| (-1) + 1/48 a (-1) - 1/12 |--- + b/2| (1/2) \ 4 / \ 4 / m[1] /3 a \ m[1] m[1] + 1/24 a (1/2) , 7/12 |--- + b/2| (-1) + 7/48 a (-1) \ 4 / 17 m[1] 17 /3 a \ m[1] - -- a (1/2) + -- |--- + b/2| (1/2) ] 24 12 \ 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] (-1) (1/2) - 6 (-1) - 11 (1/2) + 16 = 0 and this can never happen, unless it (possibly) coincides with the the above-found orbits (you prove it!)