The first one repeats $ a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,c,-a,-b,a-b,\ldots $
This does follow, eventually, from the description $ H(n) = A \, r_1^n + B \, r_2^n $ for some complex constants $A,B.$
Indeed, $ \left( \begin{array}{r} A \\ B \end{array} \right) = \left( \begin{array}{rr} \frac{1}{2} - \frac{i}{2 \sqrt 3} & -\frac{1}{2} - \frac{i}{2 \sqrt 3} \\ \frac{1}{2} + \frac{i}{2 \sqrt 3} & -\frac{1}{2} + \frac{i}{2 \sqrt 3} \end{array} \right) \cdot \left( \begin{array}{r} a \\ b \end{array} \right) $
So, for the sequence $ 1,-1,-2,-1,1,2, 1,-1,-2,-1,1,2,\ldots $ we have $a=1,b=-1,$ then $A=1,B=1$ and $ H(n) = r_1^n + r_2^n. $
For the sequence $ 1,1,0,-1,-1,0, 1,1,0,-1,-1,0,\ldots $ we have $a=1,b=1,$ then $A=- \frac{i}{ \sqrt 3} ,B= \frac{i}{ \sqrt 3}$ and $ H(n) = - \frac{i}{ \sqrt 3} \left( r_1^n - r_2^n \right). $
The repetition of length 6 and the half-repetition with negation of length 3 can be seen from both roots satisfying $r_j^6 = 1$ and $r_j^3 = -1.$ Or, you can just start a sequence with symbols $a,b$ and confirm the pattern I gave first.