Let's have matrix $(n-1) \times (n-1)$
$$ \begin{pmatrix} 3 & 1& 1& \cdots& 1 \\ 1 & 4& 1& \cdots& 1 \\ 1 & 1& 5& \cdots& 1 \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 1 & 1& 1& \cdots& n+1 \\ \end{pmatrix} $$
Let's mark it's determinant $h_n$. $h_2 = 3$ is trivial, but what about formula for the rest? My GF suggested, that formula for the determinant could be
$$ h_{n+1} = (n+1)h_n+n! $$
which seems to work just fine. But I have not idea how to go about proving it. Can someone push me in the right direction?
Thanks in advance.
Call your matrix $H_n$. For $n\ge3$, let $P$ be the bidiagonal matrix whose main diagonal entries are $1$ and whose superdiagonal entries are $-1$. Then $\det(P)=1$ and hence $\det(H_n)=\det(PH_n)$. Since $$ PH_n=\pmatrix{ 2&-3\\ &3&-4\\ &&\ddots&\ddots\\ &&&\ddots&-(n-1)\\ &&&&n-1&-n\\ 1&1&\cdots&1&1&n+1 }, $$ by Laplace expansion along the last row, we conclude that $h_n=n!\sum_{k=1}^{n}\frac1k$ for $n\ge3$. The same formula also holds for $n=2$. Now the rest is trivial.