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Consider an $n \times n$ matrix $A_n$ with elements $a_{i,j}$ such that $a_{1,j},a_{2,j},a_{3,j},\ldots$ is the sequence of numbers not divisible by $j+1$ in increasing order starting from $1$ (e.g. $1,2,4,5,7,8,10,\ldots$ for $j = 2$).

For fun, I have computed the determinant for $n \le 8$ and then conjectured that:

$$\lvert A_n \rvert = (-1)^{n+1}$$

Is the conjecture true? How would you prove it?

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    $\begingroup$ To elaborate on the answer below, the sequence $$ a_{1,j} - 1, \quad a_{2,j} - 2, \quad a_{3,j} - 3, \quad \dots \quad a_{i,j} - i,\quad \dots $$ will consist of $j$ 0's, followed by $j$ 1's, followed by $j$ 2's, and so forth. The first non-zero entry of this sequence is a 1, and this 1 will always occur where $i = j$, corresponding to the diagonal entry of the column. $\endgroup$ Jul 18, 2023 at 19:24

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Your conjecture is true, and I'm surprised I've never seen this problem before; it's very charming.

The last column of the matrix is always $(1,2,3,\dots, n)^T$. Subtracting this from every other column, we obtain a matrix of the following form:

$$ \begin{bmatrix} 0,...,0 & 1 \\ L_n & *\end{bmatrix}$$

where $L_n$ is lower-triangular $(n-1)\times(n-1)$ with ones on the diagonal. (We can be more explicit if we like, but there is no need to.) Cofactor expansion along the first row shows that $\det(A_n) = (-1)^{n+1}\det(L_n)$, and $\det(L_n)=1$.

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  • $\begingroup$ What you describe as the bottom row is actually the bottom column; note that $a_{1,j}, a_{2,j}, \dots$ is a single column of the matrix. Of course, the answer is still basically correct because the transpose of a matrix has the same determinant $\endgroup$ Jul 18, 2023 at 19:24
  • $\begingroup$ Ah, thank you >.< Will fix that. $\endgroup$ Jul 18, 2023 at 19:26

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