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Moritz Firsching
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Too long for a comment.

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ $A^{182}$ is the identity.

Unfortunately, not even the reasoning $\chi(A^k)(1)\neq0$ implies there is no $w$, such that $Aw=w$ seems to apply in the $\mathbb{Z}/6\mathbb{Z}$ situation; for example we have $$A^7\cdot (3,3,3,3,3,3)^t = (3,3,3,3,3,3)^t$$

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

Too long for a comment.

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ $A^{182}$ is the identity.

Unfortunately, not even the reasoning $\chi(A^k)(1)\neq0$ implies there is no $w$, such that $Aw=w$ seems to apply in the $\mathbb{Z}/6\mathbb{Z}$ situation; for example we have $$A^7\cdot (3,3,3,3,3,3)^t = (3,3,3,3,3,3)^t$$

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Joseph O'Rourke
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Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

added 228 characters in body
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Moritz Firsching
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Let $A$$$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 182nd91st entry in the list.

$$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

Let $A$ be the adjacency matrix of the path graph. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 182nd entry in the list.

$$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$

Let $$A=\left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$ be the adjacency matrix of the path graph. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$): $$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$ A^{91} = \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

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Moritz Firsching
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