next up previous
Next: Bibliography

A few (conjectural) modular properties of determinants associated to Pascal's triangle

Roland Bacher

started in september 04, last update: september 04

Under construction.

The aim of these notes is to present a few problems and conjectures related to the my joint paper with R. Chapman Symmetric Pascal matrices modulo $p$ (European J. of Combinatorics 25 (2004) 459-473).

Denote by $P$ the infinite symmetric matrix with coefficients

\begin{displaymath}p_{i,j}={i+j\choose i}=\frac{(i+j)!}{i! j!}, 0\leq i,j\end{displaymath}

given by the binomial coefficients arising as entries in Pascal's triangle


We denote by $P(n)$ the symmetric $n\times n$ submatrix of $P$ defined by the $n$ first rows and columns. The matrix $P(n)$ has thus coefficients ${i+j\choose i}$ with $0\leq i,j<n$. Denote by $\chi_n(t)=\det(t I(n)-P(n))$ the characteristic polynomial of $P(n)$ (where $I(n)$ denotes of course the identity matrix of order $n\times n$).

The main result of [1] (Theorem 1.3 in [1]) reads now:

Theorem 0.1   When $q=p^l$ is a power of a prime $p$ and $0\leq k\leq q/2$ then

\begin{displaymath}\chi_{q-k}(t)\equiv(t^2+t+1)^{(q-\epsilon(q))/3-k}(t-1)^{(q+2\epsilon(q))/3-k}\det (t^2I(k)+P(k))\pmod p\end{displaymath}

where $\epsilon(q)\in\{-1,0,1\}$ satisfies $\epsilon(q)\equiv q\pmod 3$.

This result determines the reduction modulo $2$ of the characteristic polynomial $\chi_n$ completely.

Theorem 0.1 seems to be the tip of an iceberg. Many similar formulae seem to exist. The easiest one (Conjecture 1.6 of [1]) can be stated as follows:

Conjecture 0.2   For each integer $k\geq 0$ there exists a monic polynomial $c_k\in{\mathbf Z}[t]$ of degree $4k$ such that $c_k(t)=t^{4t}c_k(\frac{1}{t})$ with the following property: if $q$ is a power of a prime $p$ and $0\leq k\leq q/2$ then

\begin{displaymath}\chi_{q+k}(t)\equiv (t^2+t+1)^{(q-\epsilon(q))/3-k}(t-1)^{(q+2\epsilon(q))/3-k}c_k(t)\pmod p\end{displaymath}

where $\epsilon(q)\in\{-1,0,1\}$ satisfies $\epsilon(q)\equiv q\pmod 3$.

The equality $c_k(t)=t^{4t}c_k(\frac{1}{t})$ states of course that the polynomials $c_k=\sum_{i=0}^{4k}\gamma_{i,k}t^i$ are palindromic: $\gamma_{i,k}=\gamma_{4k-i,k}$ and their roots are symmectric with respect to the unit circle: $\rho\in{\mathbf C}$ is a root of $c_k$ if and only if $\frac{1}{\rho}$ is a root of $c_k$.

The first few of these conjectural polynomials $c_k(t)$ are seemingly

\qquad -2972t^9+17424t^8-2972t^7+\dots -196t+1)\end{array}\end{displaymath}

If this conjecture holds then

\begin{displaymath}c_k(t)\equiv\Big(\det(tI(k)+P(k))\Big)^4\pmod 2\end{displaymath}

by Theorem 0.1 (and some easy corollaries of it).

Computations with $P(n)\pmod 3$ for $n\leq 400$ suggest (cf. Conjecture 1.7 in [1]):

Conjecture 0.3   We have

\begin{displaymath}c_k(t)\equiv(t+1)^{3k}\det(tI(k)+P(k))\pmod 3 .\end{displaymath}

Theorem 0.1 and Conjectures 0.2 and 0.3 define now $\chi_n(t)$ completely modulo $3$ from the trivial initial values $\chi_0=1$ and $\chi_1=t-1$.

Proofs of these conjectures would thus be very interesting.

next up previous
Next: Bibliography
Roland Bacher 2004-09-13