Eigenvalues of offset multiplication tables Consider the $n$ x $n$  'multiplication table' constructed as (using Mathematica language)
$$
M_n^s = \text{M[n_,s_]:=Table[ k*m , {k,1+s, n+s}, {m, 1+s, n+s} ] }
$$
For example, 
$$M_4^0 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\3 & 6 & 9 & 12 \\4 & 7 & 12 & 16 \end{pmatrix} $$
and 
$$M_4^1 = \begin{pmatrix} 4 & 6 & 8 & 10 \\ 6 & 9 & 12 & 15 \\8 & 12 & 16 & 20 \\10 & 15 & 20 & 25 \end{pmatrix} $$
Amazingly, these matrices have all zero eigenvalues, except for a single value.  The non-zero eigenvalue exhibits an interesting pattern as a function of $n$ and $s.$  Starting with $n=2,$ the sequences read as follows 
$$ \text{eigv }M_n^0=\{1,5,14,30,55,91,140...\} = (n-1)(2n^2-n)/6 $$
$$ \text{eigv }M_n^1=\{4,13,29,54,90,139,203...\} = (n-1)(2n^2 + 5n + 6)/6 $$
$$ \text{eigv }M_n^2=\{9,25, 50, 86, 135, 199,280...\} = (n-1)(2n^2 + 11n + 24)/6 $$
$$ \text{eigv }M_n^3=\{25,61,110,174,255,355...\} = (n-1)(2n^2 + 17n + 54)/6 $$
I've worked out many of these, and it appears that, in considering the quadratic polynomial, and using $[n^1]$ to mean the coefficient of $n,$
$$ [n^2] = 2, \, [n^1]=6s-1, \, [n^0] = 6s^2 , \ s=1,2,3... $$
The question is: can these observations be proved?

All lines are linearly dependent (construction with entries $k*m$), so there is only one non zero eigenvalue.
This eigenvalue is the trace of the matrix (theorem that sum of eigenvalues is trace) so $\lambda=\sum_{j=1}^n (s+j)^2$
