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.