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Let $B_n$ denote the Boolean algebra of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$.

Let $M_n:=C_n+C_n^T$ and $f_n$ the characteristic polynomial of $M_n$ (this is the Cartan matrix of a Frobenius algebra associated to the incidence algebra of $B_n$, thus is appears quite naturally).

Question: Is it true that $f_n$ factors over $\mathbb{Q}$ in exactly $n+1$ different irreducible factors? What are the numbers appearing in the multiplicity of the irreducible factors?

This is true for $n \leq 14$.

For example for $n=6$ the 7 irreducible factors are:

(x - 2)^5 * (x + 1)^10 * (x - 1)^15 * (x^2 + 7x + 2) * (x^2 - 11x + 2)^5 * (x^2 - 5x + 2)^9 * (x^4 - 20x^3 - 161x^2 - 40x + 4)

For $n=8$ the 9 irreducible factors are:

(x - 2)^14 * (x + 1)^42 * (x - 1)^56 * (x^2 + 25x + 2) * (x^2 + 7x + 2)^7 * (x^2 - 11x + 2)^20 * (x^2 - 5x + 2)^28 * (x^4 - 44x^3 - 1313x^2 - 88x + 4) * (x^4 - 20x^3 - 161x^2 - 40x + 4)^7

The next two questions indicate that something interesting might be going on for the powers of the irreducible factors:

Question 2: The maximal power $p_n$ such that $(x-2)^{p_n}$ is a factor of $f_n$ starts for even $n \geq 2$ with $1,2,5,14,42,132$. Is it the Catalan sequence?

Question 3: The maximal power $q_n$ such that $(x-3)^{q_n}$ is a factor of $f_n$ starts for odd $n \geq 1$ with $1,2,5,14,42,132$ . Is it the Catalan sequence?

Sadly I can only test those things for $ n \leq 14$ at the moment. One might find many more sequences appearing, for example for even $n$ the maximal power of (x^2 - 5*x + 2) appearing starts with 1, 3, 9, 28, 90 and might be given by https://oeis.org/A000245 .

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    $\begingroup$ The matrix $M_n$ commutes with the action of the symmetric group $S_n$ on $B_n$. In this action, the irrep indexed by the partition $(n-i,i)$, $0\leq i\leq \lfloor n/2\rfloor$, has multiplicity $n+1-2i$ (and no other irreps appear). One should try to compute the characteristic polynomial of the action of $M_n$ on each isotypic component. The multiplicity of each irreducible factor of $f_n$ must be a sum of dimensions of the irreps indexed by $(n-i,i)$. For instance, when $n=6$ the multiplicity 9 of $x^2-5x+2$ can only correspond to $\dim( 4,2)$, but in general there will be ambiguity. $\endgroup$
    – Richard Stanley
    Commented Aug 19, 2020 at 2:50
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    $\begingroup$ Continuing my comment, the matrix $M_n$ actually commutes with the action of the larger group $G=S_n\times(\mathbb{Z}/2\mathbb{Z})$, where the generator of $\mathbb{Z}/2\mathbb{Z}$ is complementation. Exactly $n+1$ distinct irreps of $G$ appear in this action. This suggests that the characteristic polynomial of the action of $G$ on each isotypic component will have a "dominant" factor. All other factors are dominant factors of other isotypic components. $\endgroup$
    – Richard Stanley
    Commented Aug 19, 2020 at 13:08
  • $\begingroup$ There is also a galois action of $Gal(\bar{\mathbb{Q}} /\mathbb{Q} ) $ on the eigenvalues of each isotypic component. Grouping eigenvalies in orbits produces exactly an irreducible factor over $\mathbb{Q}$. Since @RichardStanley pointed out there are exactly $(n+1) $ components, the thesis is equivalent to such galois action to be transitive on each (eigenvalues set of) isotypic component. $\endgroup$
    – Andrea Marino
    Commented Jul 4, 2021 at 23:44
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    $\begingroup$ Maybe oeis.org/A000344 for $(x^2 -11 x+ 2)$ ? Part of oeis.org/A009766 $\endgroup$
    – F. C.
    Commented Jul 8, 2021 at 19:36
  • $\begingroup$ For n=10, one gets $(x - 2)^{42} \cdot (x + 1)^{170} \cdot (x - 1)^{212} \cdot (x^{2} + 79 x + 2) \cdot (x^{2} + 25 x + 2)^{9} \cdot (x^{2} + 7 x + 2)^{35} \cdot (x^{2} - 11 x + 2)^{75} \cdot (x^{2} - 5 x + 2)^{90} \cdot (x^{4} - 100 x^{3} - 9601 x^{2} - 200 x + 4) \cdot (x^{4} - 44 x^{3} - 1313 x^{2} - 88 x + 4)^{9} \cdot (x^{4} - 20 x^{3} - 161 x^{2} - 40 x + 4)^{35}$. $\endgroup$
    – F. C.
    Commented Jul 9, 2021 at 5:55

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