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?

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!
