Let $A$ be the $(n-1) \times (n-1)$ matrix whose entries $a_{ij}=d(\gcd(i+1,j+1))$. Here, $d(n)$ means the number of divisors of $n$. It seems that the determinant of it is the number of square-free positive integers less than or equal to $n$. How to prove it?
For $a_{ij}=d(\gcd(i,j))$, the determinant is $1$, because it is the product $CC^T$,where $c_{ij}=${if j|i then 1 else 0} But this method seems not to work on the $\gcd(i+1,j+1)$ case.
Let $X$ be the $n \times n$-matrix obtained as follows: in the first row, the $i$-th entry equals $\mu(i)$ where $\mu$ denotes the Möbius function, and in the $j$-th row ($j>1$), the $i$-th entry is $1$ if $i$ divides $j$ and $0$ otherwise.
For example, for $n=8$ we obtain $$ X=\left( \begin{array}{cccccccc} 1 & -1 & -1 & 0 & -1 & 1 & -1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right). $$
Now $XX^t$ contains the matrix of interest $A$ as a submatrix: $$ XX^t = Y = \left( \begin{array}{cccccccc} 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 2 & 1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 1 & 1 & 2 & 1 & 1 \\ 0 & 2 & 1 & 3 & 1 & 2 & 1 & 3 \\ 0 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 & 1 & 4 & 1 & 2 \\ 0 & 1 & 1 & 1 & 1 & 1 & 2 & 1 \\ 0 & 2 & 1 & 3 & 1 & 2 & 1 & 4 \\ \end{array} \right). $$ Let $k$ be the number of squarefree integers in $\{1,2,\ldots,n\}$. Then we claim that for general $n$ we have $\det X = k$ and that $XX^t$ is the block matrix $$ Y = \left( \begin{array}{cc} k & 0 \\ 0 & A \end{array} \right). $$ This would imply that $k \det(A) = \det(Y) = \det (XX^t) = \det(X)^2 = k^2$ or $\det(A)=k$, as desired.
Let us first show that $XX^t = Y$. We check the equality of the element in row $i$ and column $j$. For $i,j > 1$ this follows as in the $\gcd(i,j)$-case. For $i=j=1$ this follows from $k = \sum_{i=1}^n \mu(i)^2$. For the other cases this follows from the formula $$ \sum_{d \mid n} \mu(d) = 0 \qquad \mbox{for } n > 1. $$
It remains to show that $\det X = k$. For each $2 \leq i \leq n$, we add the $i$-th column $\mu(i)$ times to the first column. Then the determinant does not change and the first column will contain a $k$ in the first row and $0$ in other rows by the formula $\sum_{d \mid n} \mu(d) = 0$. In our $n=8$ example we find the following: $$ \left( \begin{array}{cccccccc} 6 & -1 & -1 & 0 & -1 & 1 & -1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right). $$ It is now clear that (in general) the determinant equals $k$ by noting that the term corresponding to the diagonal is the only nonzero term.
