1
$\begingroup$

$\mathbf{SETUP}$

For my theoretical physics PhD I have been studying a model that requires the inversion of an $n \times n$ matrix of this form:

$$ \mathbf{A}_n= \begin{pmatrix} 1 & -a_{12} & -a_{13} & -a_{14} & ... & -a_{1n} \\ -a_{21} & 1 & -a_{23} & -a_{24} & ... & -a_{2n} \\ -a_{31} & -a_{32} & 1 & -a_{34} & ... & -a_{3n} \\ -a_{41} & -a_{42} & -a_{43} & 1 & ... & -a_{4n} \\ ... & ... & ... & ... & ... & ... \\ -a_{n1} & -a_{n2} & -a_{n3} & -a_{n4} & ... & 1 \\ \end{pmatrix} $$

(The minus signs are important to the model but not relevant for this question.)

In the determinant of this matrix, the number of terms of different algebraic order in the matrix entries (e.g. $a_{12}a_{34}a_{43}$ is 3rd order, etc.) is exactly the partial derangement numbers, a.k.a. the Rencontres numbers:

https://oeis.org/wiki/Rencontres_numbers

https://oeis.org/A008290

This of course means the number of terms of $n^\text{th}$ order are the (full) derangement numbers, a.k.a. the subfactorial $!n$

https://oeis.org/wiki/Number_of_derangements

https://oeis.org/A000166

$\mathbf{QUESTION}$

Is the above connection of this fairly general form of matrix to the these famous number sequences known?

I haven't found it in any wiki or comments of the sequence, and numerous textbooks and university courses I found go through the well-trodden approach of finding just the subfactorial via an inclusion-exclusion argument, the same shown on its Wikipedia page (I haven't found a derivation of the Rencontres numbers specifically).

I am not a mathematician so I am probably missing somewhere obvious this is talked about where the notation is not familiar to me.

The one thing I did find is a connection to MacMahon's Master Theorem:

https://en.wikipedia.org/wiki/MacMahon%27s_master_theorem

P.A. MacMahon, Combinatory analysis, vols 1 and 2, Cambridge University Press, 1915–16

where he derives (on page 114) the subfactorial $!6 = 265$ using his theorem (that counts terms of different algebraic orders in a similarly general matrix), and some version of what seems to often be called Cauchy's formula for number of cycles of different type.

But, even here, he does not report the full relation to the partial derangement Rencontres numbers.

$\mathbf{EXPLANATION \ OF \ RESULT}$

The reason for the connection is fairly intuitive once using a definition for the determinant via all permutations $\sigma$ of a given symmetric group $S_n$:

$$ \det(\mathbf{A}_n) = \sum_{\sigma \in S_n} \text{sgn}(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)} $$

noting that fixed points (1-cycles) in the cyclic structure of these permutations correspond to the diagonal entries $a_{ii}$, i.e. where $\sigma(i)=i$.

For my matrix $\mathbf{A}_n$, these diagonal entries are all $a_{ii}=1$, and hence 'disappear' from the algebraic expression, leaving behind all the fully and partially deranged permutations.

Thus the terms of $n^\text{th}$ order come from permutations with no fixed points: meaning the number of these terms is the (full) derangement numbers.

Then, the terms of $(n-1)^\text{th}$ order come from permutations with 1 fixed point; terms of $(n-2)^\text{th}$ order from permutations with 2 fixed points; etc.; meaning the number of these sub-$n$ order terms is the partial derangement numbers.

P.S. Even if this is known and not new, I would like to add it to the OEIS comments if possible - I am in the process of requesting an account.

$\endgroup$
1
  • $\begingroup$ This question gets at the same pattern based on the permanent of a matrix all 0s on the diagonal, all 1s elsewhere. $\endgroup$ Commented May 13 at 12:03

0

You must log in to answer this question.