Ring of divisors of a natural number and the sum of divisors as an eigenvalue and the spectral norm of some matrix? Let $n$ be a natural number and $D_n$ be the set of divisors.
We can make this set to a ring by observing that each divisor $d$ has
$$0 \le v_p(d) \le v_p(n)$$
Hence we can add two divisors $d,e$ by setting:
$$d \oplus e := \prod_{p | n} p^{v_p(d)+v_p(e) \mod (v_p(n)+1)}$$
and similarily we can multiply them by setting:
$$d \otimes e := \prod_{p | n} p^{v_p(d) \cdot v_p(e) \mod (v_p(n)+1)}$$
Then, if $n = p_1^{a_1} \cdots p_r^{a_r}$, this ring will be isomorphic to the ring
$$\mathbb{Z}/(a_1+1) \times \cdots \times \mathbb{Z}/(a_r+1)$$
If $n$ is squarefree, than this reduces to :
$$d\oplus e = \frac{de}{\gcd(d,e)^2}$$
$$d\otimes e = \gcd(d,e)$$
and the ring of divisors is a boolean ring as defined here
https://mathoverflow.net/questions/369751/boolean-ring-of-unitary-divisors-structure-of-unitary-divisors and here
Does this characteristic polynomial factor into linear factors over the integers?
If we consider the addition table ($\oplus$) of this ring as a matrix, than it is clear that the sum of divisors $\sigma(n)$ is an eigenvalue to the eigenvector:
$$(1,\cdots,1)$$
Here is as an example the addition ($\oplus$) table for $n=12$:
$$\left(\begin{array}{rrrrrr}
1 & 2 & 3 & 4 & 6 & 12 \\
2 & 4 & 6 & 1 & 12 & 3 \\
3 & 6 & 1 & 12 & 2 & 4 \\
4 & 1 & 12 & 2 & 3 & 6 \\
6 & 12 & 2 & 3 & 4 & 1 \\
12 & 3 & 4 & 6 & 1 & 2
\end{array}\right)
$$
I have checked numerically ($n=1,\cdots,60$) that
$$\sigma(n) = |A_n^k|_2^{1/k}, \forall k \ge 1$$
where $A_n$ is the addition matrix of this ring.
Is there a proof for this?
Thanks for your help!
 A: You already note that $\sigma(n)$ is an eigenvalue of $A_n$, from which it follows that
$$|A_n^k|_2^{1/k}\geq\rho(A)\geq\sigma(n),$$
for every positive integer $k$. Here $\rho(A)$ denotes the spectral radius of $A$. Of course
$$|A_n^k|_2^{1/k}\leq|A_n|_2,$$
so it is now necessary and sufficient to show that $|A_n|_2\leq\sigma(n)$ for all $n$.
Note that for an appropriate ordering of $D_n$, the matrix $A_n$ is a block matrix: Writing $n=\prod_{p\mid n}p^{a_p}$ and setting $n_p:=\frac{n}{p^{a_p}}$ for each prime $p$ dividing $n$, we find that
$$A_n=\big[p^{i+j\pmod{a_p+1}}A_{n_p}\big]_{i,j=0\ldots a_p},$$
for every prime $p$ dividing $n$. Then setting
$$C(p^{a_p}):=\big[p^{i+j\pmod{a_p+1}}\big]_{i,j=0\ldots a_p},$$
yields $A_n=C(p^{a_p})\otimes A_{n_p}$, and hence by induction $A_n=\bigotimes_{p\mid n}C(p^{a_p})$, from which it follows that
$$|A_n|_2=\prod_{p\mid n}|C(p^{a_p})|_2.$$
For each prime power $p^{a_p}$ the entries of $C(p^{a_p})$ are positive and the sum over each row equals
$$\sum_{i=0}^{a_p}p^i=\frac{p^{a_p+1}-1}{p-1}=\sigma(p^{a_p}),$$
so by the Perron-Frobenius theorem (see point 11) we get $|C(p^{a_p})|_2=\sigma(p^{a_p})$, and hence
$$|A_n|_2=\prod_{p\mid n}|C_p|_2=\prod_{p\mid n}\sigma(p^{a_p})=\sigma(n).$$
We conclude that $|A_n^k|_2^{1/k}=\sigma(n)$ for every positive integer $k$.
