Eigenvalues of some peculiar matrices While I was toying around with matrices, I chanced upon a family of tridiagonal matrices $M_n$ that take the following form: the superdiagonal entries are all $1$'s, the diagonal entries take the form $m_{j,j}=4 j (2 n + 3) - 8 j^2 - 6 n - 5$, and the subdiagonal entries take the form $m_{j+1,j}=4 j (2 j - 1) (2 n - 2 j + 1) (n - j)$.
For example, the $4\times 4$ member of this family looks like this:
$$M_4=\begin{pmatrix} 7 & 1 & 0 & 0 \\ 84 & 27 & 1 & 0 \\ 0 & 240 & 31 & 1 \\ 0 & 0 & 180 & 19\end{pmatrix}$$
I checked the eigenvalues of members of this family and I found that each member has the squares of the first few odd integers as eigenvalues. (For example, the eigenvalues of $M_4$ are $1,9,25,49$.) I couldn't find a way to prove this though.
I wish someone would help me! Thanks!
 A: Here is the eigendecomposition $M_4= VDV^{-1}$ confirming @Craig's observation:
$$\begin{pmatrix}
1 &1 &1 &1\\
42 &18 &2 &-6\\
840 &-120 &-120 &72\\
5040 &-3600 &2160 &-720
\end{pmatrix}
\begin{pmatrix}49\\&25\\&&9\\&&&1\end{pmatrix}\begin{pmatrix}
1 &1 &1 &1\\
42 &18 &2 &-6\\
840 &-120 &-120 &72\\
5040 &-3600 &2160 &-720
\end{pmatrix}^{-1}
$$
If you can provide a little hint on how you came up with this, then we can probably understand why $V$ has this special structure. Also, you can distribute the squares to the left and right matrices and obtain an even weirder situation.
A: If I'm not mistaken, the largest eigenvector is of the form 

$\begin{pmatrix} 1 \\ (2n-1)!/(2n-3)! \\ (2n-1)!/(2n-5)! \\ \vdots \\ (2n-1)!/1! \end{pmatrix}$.

This shouldn't be too difficult to prove, and I think the smallest eigenvector has a similarly compact form. My suspicion is that there is a simple rotation or pair of rotations you can do to see the eigenvalues explicitly.  I would look for something of the form $ULDL^{-1}U^{-1}$ or $LUDU^{-1}L^{-1}$, where $L$ is non-zero except on the diagonal and subdiagonal, and $U$ is zero except on the diagonal and superdiagonal.
