Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:

$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.

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

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian).

Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.

Some discussion regarding this question can be found at Mathlinks, and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.

Any help would be appreciated, thanks.




I would like to clarify some points about this problem:

1. There is a ceiling function ($\lceil x \rceil$) at the recursive sequence which makes it considerably harder. The period is not 5 as claimed by some answers.

2. Miklos Schweitzer is not a conventional competition. This competition for undergraduate students is unique. The contest lasts 10 days with 10-12 problems, which are quite challenging and basically of research level. Moreover any literature can be used.

3. An example of Miklos Schweitzer problem can be found here at MO. It was indeed a very nice question with an even nicer solution. I'm not sure Art of Problem Solving would be better.

I am sorry for any inconvenience caused and I hope this question do not get closed.

Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:

$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.

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

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian).

Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.

Some discussion regarding this question can be found at Mathlinks, and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.

Any help would be appreciated, thanks.




I would like to clarify some points about this problem:

1. There is a ceiling function ($\lceil x \rceil$) at the recursive sequence which makes it considerably harder. The period is not 5 as claimed by some answers.

2. Miklos Schweitzer is not a conventional competition. This competition for undergraduate students is unique. The contest lasts 10 days with 10-12 problems, which are quite challenging and basically of research level. Moreover any literature can be used.

3. An example of Miklos Schweitzer problem can be found here at MO. It was indeed a very nice question with an even nicer solution. I'm not sure Art of Problem Solving would be better.

I am sorry for any inconvenience caused and I hope this question do not get closed.

Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:

$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.

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

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian).

Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.

Some discussion regarding this question can be found at Mathlinks, and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.

Any help would be appreciated, thanks.

Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:

$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.

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

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian).

Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.

Some discussion regarding this question can be found at Mathlinks, and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.

Any help would be appreciated, thanks.




I would like to clarify some points about this problem:

1. There is a ceiling function ($\lceil x \rceil$) at the recursive sequence which makes it considerably harder. The period is not 5 as claimed by some answers.

2. Miklos Schweitzer is not a conventional competition. This competition for undergraduate students is unique. The contest lasts 10 days with 10-12 problems, which are quite challenging and basically of research level. Moreover any literature can be used.

3. An example of Miklos Schweitzer problem can be found here at MO. It was indeed a very nice question with an even nicer solution. I'm not sure Art of Problem Solving would be better.

I am sorry for any inconvenience caused and I hope this question do not get closed.