Difference of two integer sequences: all zeros and ones? 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. 
 A: The same method as in this answer to a previous your question works as well.
As for $(A_0)$. Starting with a guess
$$
  7n+2\leq a_n\leq 7n+3,
$$
and trying to prove it inductionally, we arrive at $b_{6n+2}\geq 7n+4$ and $b_{6n}\leq 7n+1$, hence
$$
  t+\left\lfloor\frac{t+4}6\right\rfloor+1\leq b_t\leq t+\left\lceil\frac t6\right\rceil+1, \qquad(**_0)
$$
which agree for all $t=6k+2,\dots,6k+6$. So for all even $t$ we have $b_t=t+\lceil t/6\rceil+1$, which yields even an exact formula for $a_n$:
$$
  a_n=7n+\begin{cases}2,& n\mod 3=0; \\ 3,& n\mod 3\neq 0.\end{cases}
$$

As for $(A_1)$.
Starting with a guess
$$
  7n+3\leq a_n\leq 7n+4,
$$
and trying to prove it inductionally, we arrive at $b_{6n+3}\geq 7n+5$ and $b_{6n+1}\leq 7n+2$, hence
$$
  t+\left\lfloor\frac{t+3}6\right\rfloor+1\leq b_t\leq t+\left\lceil\frac{t-1}6\right\rceil+1,
  \qquad(**_1)
$$
which agree for all $t=6k+3,\dots,6k+7$. So we have $b_t=t+\lceil (t-1)/6\rceil+1$, except for $7k+3\leq b_{6k+2}\leq 7k+4$. This yields that 
$$
  a_n=7n+\begin{cases}3,& n\mod 3=0; \\ 3\text{  or  }4,& n\mod 3\neq 0.\end{cases}
$$
The required conclusion follows. 

In fact, that conclusion could be also derived directly from $(**_0)$ and $(**_1)$.
