show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$ 
Let the sequence  $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and  $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$
Prove that the sequence $\{a_{n}\}$ is periodic.

I have used a computer and found the periodic is $T=874$, but how to prove it?
 A: This is not an answer, but rather an observation.

The following graph shows the set of points $P = \{(a_n, a_{n+1}) : n \geq 1\}$.
$\hspace{8em}$
Notice that they are confined in a very narrow region and are clustered near an ellipse. This ellipse is not hard to identify. Indeed, if a sequence $(b_n)$ satisfies
$$ b_{n+2} = \frac{4}{3}b_{n+1} - b_n, $$
then it follows that
$$ \det \begin{pmatrix} b_{n+1} & b_n \\ b_{n+2} & b_{n+1} \end{pmatrix} = b_{n+1}^2 - \frac{4}{3}b_{n+1}b_n + b_n^2 $$
is constant, since
$$ \begin{pmatrix} b_{n+2} & b_{n+1} \\ b_{n+3} & b_{n+2} \end{pmatrix}
= \begin{pmatrix} 0 & 1 \\ -1 & \frac{4}{3} \end{pmatrix}
\begin{pmatrix} b_{n+1} & b_n \\ b_{n+2} & b_{n+1} \end{pmatrix}
\quad \text{and} \quad
\det \begin{pmatrix} 0 & 1 \\ -1 & \frac{4}{3} \end{pmatrix} = 1. $$
Thus the points $(b_n, b_{n+1})$ stays forever on the ellipse
$$f(x, y) := x^2 - \frac{4}{3}xy + y^2 = \text{const}.$$
If we can somehow show that $f(a_n, a_{n+1})$ is also bounded by some perturbation argument, then since the region $f(x, y) \leq c$ is bounded and $P$ has only integer points, we can argue that $(a_n)$ is eventually periodic. But at this point, I am not sure if this observation will be really useful.
A: N.B. This is not complete answer, but just extrapolation of the clue provided by @SangchulLee.

First of all we can rewrite the recurrence as:
$$a_{n+2}=\frac{4a_{n+1}}{3}-a_{n}+v_{n+1}$$
where
$$v_{n+1}=\begin{cases}
  1 & a_{n+1} = 0 \bmod 3\\
  \frac{2}{3} & a_{n+1} = 1 \bmod 3 \\
  \frac{4}{3} & a_{n+1} = 2 \bmod 3
\end{cases}$$
Now we define
$$A_{n}  =    \begin{bmatrix}
  a_{n+1} & a_{n} \\
  a_{n+2} & a_{n+1}
  \end{bmatrix}$$
$$B  =  \begin{bmatrix}
    0&1  \\
    -1&\frac{4}{3}
    \end{bmatrix}$$
$$U_1 = \begin{bmatrix}
    0&0  \\
    0&1
    \end{bmatrix}$$
$$U_2 = \begin{bmatrix}
    0&0  \\
    1&0
    \end{bmatrix}$$
This means
$$A_{n+1}  =  BA_{n}+v_{n+1}U_1+v_{n+2}U_2$$
Then we can write
$$\det(A_{n+1}-v_{n+2}U_2) = \det(A_{n}+v_{n+1}B^{-1}U_1) $$
Simplifying this one obtains
$$a_{n+2}^2-\frac{4a_{n+2}a_{n+1}}{3}+a_{n+1}^2-v_{n+1}(a_{n+2}+a_{n+1})  =  a_{n+1}^2-\frac{4a_{n+1}a_n}{3}+a_n^2-v_{n+1}(a_{n+1}+a_n)$$
Notice that the equation $x^2-4xy/3+y^2-v(x+y)=C$ represents an ellipse for a given constant $C$ and $v\in\{1,2/3,4/3\}$
I was not able to proceed from here! Hope this helps although its a little late!
