I recently read this paper in which the authors construct a matrix related to the Collatz conjecture such that
$$ m_{ij} = \begin{cases} 1 \text{ if } i = j\\ x \text{ if } c(i) = j\\ 0 \text{ otherwise} \end{cases} $$ where $$ \text{where } c(i) = \begin{cases} (3i+1)/2 \text{ if } i \text{ odd}\\ i/2 \text{ if } i \text{ even}\\ \end{cases} $$ is the Collatz function. They then show that for a matrix like this $M_k$ with $k$ rows and columns, if $det(M_k) = det(M_{k-1})$ then the Collatz conjecture follows.
But if $k$ is even, the only entry in the last column is a $1$ in the $(k,k)$ position. By Laplace expansion on the last column, $det(M_k)=det(M_{k-1})$ for $k$ even.
If $k$ is odd, the only entry in the last row is a $1$ in the $(k,k)$ position. So $M^T$ has only a single $1$ in the last column in the $(k,k)$ position. So by Laplace expansion on the last column, $det({M_k}^T) = det({M_{k-1}}^T)$. Since $det(M^T) = det(M)$ for any matrix $M$, $det(M_k)=det(M_{k-1})$ for $k$ odd as well.
Where is my error? Thanks.
When $k=8$, $m_{5,8}=x$ because $c(5)=8$, so it's not true that the only (nonzero) entry in the last column is a one in the $(8,8)$ position. More generally, this is the case when $k\equiv8\bmod{18}$.