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Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=100$,and such $$a_{n+2}=\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\rceil$$

Prove that the sequence $\{a_{n}\}$ is periodic.

I have use computer found the periodic is $T=874$,But how to prove

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    If I am not mistaken, then the period is in fact 2204, since $(a_{2205}, a_{2206}) = (1, 100)$ and no such pattern between 1 and 2205. – 2017-01-06
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    You are correct: $(a_1,\,a_2) = (1,\,100) = (a_{1+2204},\,a_{2+2204})$. Nevertheless, it seems that this case is not something exceptional. It is most likely that for every starting pair $(a_1,\,a_2)$, the resulting sequence is **periodic!** At least this is true for for all cases $\{-100 \le a_1,a_2 \le +100\}$. For example, corresponding to the question period, we have: $(a_1,\,a_2) = (1,\,99) = (a_{1+874},\,a_{2+874})$. – 2017-01-09

2 Answers 2

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This is not an answer, but rather an observation.

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

$\hspace{8em}$enter image description here

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 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.

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    Nice plot. Mind sharing the code you used ? – 2017-01-06
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    @LeGrandDODOM, Unfortunately I neither saved nor remember the code exactly. But it was a Mathematica code that looked like this: `l = NestList[{#[[2]], Ceiling[4/3 #[[2]] - #[[1]] + 1/2]} &, {1, 100}, 2204]; Show[Graphics[Point /@ l]]` – 2017-01-06
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Subtract the equation for $a_{n+1}$ from that for $a_{n+2}.$ This eliminates the constant, and you get $$ -a_{n-1}+\frac73 a_n -\frac73 a_{n+1}+a_{n+2} = 0. $$ Then go from there.

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    Kindly explain how to conclude above relation? $$ a_1=1,\, a_2=100,\, a_3=\left\lceil\frac{8\,a_2-6\,a_1+3}{6}\right\rceil=133,\, a_{n+2}=\frac{7\,a_{n+1}-7\,a_{n}+3\,a_{n-1}}{3} \\ \Rightarrow a_n=\left\{1,\,100,\,133,\,78,\,\color{red}{-\frac{85}{3}\,?}\right\} $$ – 2017-01-08