show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$

Question

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?

Answer 1

This is not an answer, but rather an observation.

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The following graph shows the set of points $P = \{(a_n, a_{n+1}) : n \geq 1\}$.

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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.

Answer 2

N.B. This is not complete answer, but just extrapolation of the clue provided by @SangchulLee.

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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!