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?

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.