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For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant integer choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. $$ *$$

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$) $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the example in Barry Cipra's comment.

For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.

Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.

For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant integer choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. $$ *$$

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$) $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $ in the spectral decomposition, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the example in Barry Cipra's comment.

For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.

Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.

For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant initial choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. $$ *$$

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated with multiplicity ) $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $L^k v$ is unbounded. Note that $v$ is certainly not a constant vector, as required. (Note that for even $n$, $\lambda_{n/2}=-3$ corresponds to Barry Cipra's example).

For what $n$ it is possible to find an initial non-constant integer labeling $v$ of the vertices for which $A$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.

Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$, that is $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.

For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant integer choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. $$ *$$

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$) $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the example in Barry Cipra's comment.

For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.

Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.

I assume $n$ is the number of vertices of the polygon. Then, for any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant initial choice of labels for which the set of numbers $A$ is bounded. A partial answer is then: yes is $n$ is a multiple of $4$ (I'm not sure if this is what you wanted, though).

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ simple eigenvalues, $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $L^k v$ is unbounded. Note that $v$ is certainly not a constant vector, as required. (Note that for even $n$, $\lambda_{n/2}=-3$ corresponds to Barry Cipra's example).

You may ask, for what $n$ it is possible to find a non-constant integer initial labeling $v$ of the vertices for which $A$ is bounded. This is certainly the case if $n$ is a multiple of $4$: there is the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$.

A more subtle question: is the above possible, for a given number $n$ (possibly not a multiple of $4$) ? This is equivalent to the question: are there non-constant integer vectors in the eigenspace of $L$ corresponding to eigenvalues not larger than $1$ in absolute value? The eigenvectors of $L$ are: $$v_k:=(1,\omega^k, \omega^{2k},\dots,\omega^{(n-1)k})^T,\qquad k=1,\dots, n\ ,$$

where $\omega:= \exp(2\pi i /n)$, so the question boils down to asking if a linear combination of the $v_k$ with $|k|\le n/4 $ can be a non-constant integer vector.

For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded.

The question is more subtle if you ask for which $n$ it is possible to find a non-constant initial choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. $$ *$$

The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated with multiplicity ) $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$

In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $L^k v$ is unbounded. Note that $v$ is certainly not a constant vector, as required. (Note that for even $n$, $\lambda_{n/2}=-3$ corresponds to Barry Cipra's example).

For what $n$ it is possible to find an initial non-constant integer labeling $v$ of the vertices for which $A$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.

Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$, that is $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.