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. 



Difference of two integer sequences: all zeros and ones?
The same method as in [this answer to a previous your question](https://mathoverflow.net/a/342287/17581) 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 $(\*\*\_1)$.


