TITLE: Do there exist integers $a_1,a_2,\ldots,a_{n-1},a_n$ such that $a_i+a_{i+1}+a_{i+3}=1$?
QUESTION [6 upvotes]: I am curious about the existence of a sequence of integers $a_1,a_2,\ldots,a_{n-1},a_n$ for some $n\ge1$ with the property that for all positive integers $i$,
$$a_i+a_{i+1}+a_{i+3}=1$$
where we compute indices modulo $n$ as necessary.
My interest lies in the more general case of any finite weighted sum of indices relative to $i$; the case above is simply the first which seems nontrivial. (The motivation for this comes in thinking about "tiling" $\mathbb{Z}$ with finite weighted tiles, where we want to lay down an integer number of copies in each position: the existence of such a series of $a_i$ amounts to the existence of a periodic tiling.)
It is easy to see that $n$ must be a multiple of three, by adding all congruences together; I can also rule out individual cases of $n=3$ or $n=6$, but I don't see how to get all $n$ in general. It would suffice to show that the vectors $(1,1,0,1,0,\ldots,0)$ and its cyclic rotations are linearly independent over $\mathbb{R}$, but I'm not sure how to do this either.
As a remark, this is not true if we replace $(i,i+1,i+3)$ with $(i,i+1,i+5)$, so the argument must hinge on the specific offsets somehow.

REPLY [4 votes]: Consider the recurrence relation $a_{k+3}=1-a_{k+1}-a_k$. From the theory of linear recurrence relations we have $a_k = \frac{1}{3}+c_1\alpha^k+c_2\beta^k+c_3\gamma^k$ where $\alpha$, $\beta$ and $\gamma$ are the roots of $x^3+x+1=0$, for some constants $c_1$, $c_2$ and $c_3$.
We want the sequence to also satisfy for some $n$ the equations $a_{n-2}+a_{n-1} +a_1=1$, $a_{n-1}+a_n +a_2=1$ and $a_n+a_1 +a_3=1$. This translate into the system
$$
\begin{array}{l}
c_1(\alpha^{n-2}+\alpha^{n-1}+\alpha)+c_2(\beta^{n-2}+\beta^{n-1}+\beta)+c_3(\gamma^{n-2}+\gamma^{n-1}+\gamma)=0\\
c_1(\alpha^{n-1}+\alpha^n+\alpha^2)+c_2(\beta^{n-1}+\beta^n+\beta^2)+c_3(\gamma^{n-1}+\gamma^n+\gamma^2)=0\\
c_1(\alpha^n+\alpha+\alpha^3)+c_2(\beta^n+\beta+\beta^3)+c_3(\gamma^n+\gamma+\gamma^3)=0\\
\end{array}
$$
After some calculations, the determinant of the system is
$$
\frac{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha^n-1)(\beta^n-1)(\gamma^n-1)}{\alpha^2\beta^2\gamma^2}
$$
None of the factors can be $0$, so the determinant of the system is not $0$ and the only solution for the system is $c_1=c_2=c_3=0$ which means that the only sequence satisfying the requested conditions is constant sequence $a_k = \frac{1}{3}$.