Suppose that $c$ is a nonnegative integer and $A_c = (a_n)$ and $B_c = (b_n)$ are strictly increasing complementary sequences satisfying

$$a_n = b_{2n} + b_{4n} + c,$$

where $b_0 = 1.$ Can someone prove that the sequence $A_1-A_0$ consists entirely of zeros and ones?

Notes:

$$ A_0 = (2, 10, 17, 23, 31, 38, 44, 52, 59, 65, 73, 80, 86, \ldots ) \\ A_1 = (3, 11, 17, 24, 31, 39, 45, 53, 59, 66, 74, 80, 87, \ldots ) $$

The sequence $A_0$ satisfies the linear recurrence $a_n = a_{n-1} + a_{n-3} - a_{n-4}$.

It may help to watch $A_1$ get started. Since $b_0=1$ we have $a_0=1+1+1=3$, and since $A_1$ and $B_1$ are complementary, we have $b_1=2$. Next, $a_1=b_2+b_2+1 \geq 4+6+1=11$, so that $b_2=4, b_3=5,\ldots,b_8=10$, and $a_1=11$. Then $a_2=b_4+b_8+1 \geq 17$, and so on.