Further Combinatorial Identities deriving from the $n$-th power of a $2\times 2$ matrix

Abstract

In this paper we use a formula for the $n$-th power of a $2\times 2$ matrix $A$ (in terms of the entries in $A$) to derive various combinatorial identities. Three examples of our results follow. 1) We show that if $m$ and $n$ are positive integers and $s\in \{0,1,2,\dots ,$ $\lfloor (mn-1)/2\rfloor \}$, then \begin{multline*} \sum_{i,j,k,t}2^{1+2t-mn+n} \frac{(-1)^{nk+i(n+1)}}{1+δ_{(m-1)/2,\,i+k}} \binom{m-1-i}{i} \binom{m-1-2i}{k}\times\\ \binom{n(m-1-2(i+k))}{2j}\binom{j}{t-n(i+k)} \binom{n-1-s+t}{s-t}\\ =\binom{mn-1-s}{s}. \end{multline*} 2) The generalized Fibonacci polynomial $f_m(x,s)$ can be expressed as \[ f_{m}(x,s)= \sum_{k=0}^{\lfloor (m-1)/2 \rfloor}\binom{m-k-1}{k}x^{m-2k-1}s^{k}. \] We prove that the following functional equation holds: \begin{equation*} f_{mn}(x,s)=f_{m}(x,s)\times f_{n}\left (\,f_{m+1}(x,s)+sf_{m-1}(x,s), \,-(-s)^{m}\right) . \end{equation*} 3) If an arithmetical function $f$ is multiplicative and for each prime $p$ there is a complex number $g(p)$ such that \begin{equation*} f(p^{n+1}) = f(p)f(p^{n})- g(p)f(p^{n-1}), \hspace{15pt} n \geq 1, \end{equation*} then $f$ is said to be \emph{specially multiplicative}. We give another derivation of the following formula for a specially multiplicative function $f$ evaluated at a prime power: \begin{equation*} f(p^{k})=\sum_{j=0}^{\lfloor k/2 \rfloor}(-1)^{j} \binom{k-j}{j}f(p)^{k-2j}g(p)^{j}. \end{equation*} We also prove various other combinatorial identities.

9 pages