\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\coeff}{coeff} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf Quasi-Fibonacci Numbers of Order~11} \vskip 1cm \large Roman Witu{\l}a and Damian S{\l}ota \\ Institute of Mathematics \\ Silesian University of Technology \\ Kaszubska 23, \\ Gliwice 44-100 \\ POLAND \\ \href{mailto:r.witula@polsl.pl}{\tt r.witula@polsl.pl} \\ \href{mailto:d.slota@polsl.pl}{\tt d.slota@polsl.pl} \\ \end{center} \vskip .2in \begin{abstract} In this paper we introduce and investigate the so-called quasi-Fibonacci numbers of order~$11$. These numbers are defined by five conjugate recurrence equations of order five. We study some relations and identities concerning these numbers. We present some applications to the decomposition of some polynomials. Many of the identities presented here are the generalizations of the identities characteristic for general recurrence sequences of order three given by Rabinowitz. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newenvironment{sketchproof}{\textit{Sketch of the proof:} \ignorespaces}{\nopagebreak\null\hfill\hbox{$\Box$}} \newcommand{\cnd}{\hbox{\hspace*{-0.6ex}% \hbox{\raisebox{-0.6ex}{$\Box$}}}\vspace{0.5cm}\par} \newcommand{\cndd}{\mbox{\raisebox{-0.6ex}{$\Box$} \vspace{0.5cm}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section 1 %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction} Witu{\l}a, S{\l}ota and Warzy{\'n}ski \cite{wsw1} analyzed the relationships between the so-called quasi-Fibonacci numbers of order~$7$. In this paper we are focused on the generalization of the identities and some facts derived in~\cite{wsw1} to the quasi-Fibonacci numbers of order~$11$ $A_{n}(\delta)$, $B_{n}(\delta)$, $C_{n}(\delta)$, $D_{n}(\delta)$, and $E_{n}(\delta)$, $n\in \mathbb{N}$, $\delta\in \mathbb{C}$, defined by the relations (see also Section~\ref{roz3} in this paper for more details): \begin{multline*} \big( 1+\delta\, (\xi^m +\xi^{10m})\big)^{n}= A_n(\delta)+B_n(\delta)\,(\xi^m +\xi^{10m})+C_n(\delta)\,(\xi^{2m}+\xi^{9m}) +{}\\ {}+ D_n(\delta)\,(\xi^{3m}+\xi^{8m})+E_n(\delta)\,(\xi^{4m}+\xi^{7m}) \end{multline*} for $m,n\in \mathbb{N}$, $m\leq 5$, $\delta\in \mathbb{C}$, where $\xi\in \mathbb{C}$ is a~primitive root of unity of order~$11$. Additionally, the following important auxiliary sequence is considered: $$ {\cal A}_{n}(\delta):= 5A_n(\delta )-B_n(\delta )-C_n(\delta )-D_n(\delta )-E_n(\delta ). $$ Moreover, in order to make shorter expressions and formulas we introduce the following notation: $$ k_m := \cos \Big( \frac{m\, \pi}{11} \Big),\qquad s_m := \sin \Big( \frac{m\, \pi}{11} \Big),\qquad \xi := \exp \Big( \frac{i\, 2\, \pi}{11} \Big),\qquad $$ $$ \sigma_m:=\xi^{m} + \xi^{-m} \qquad \mbox{ and } \qquad \tau_m(\delta):=1+\delta\, \sigma_{m} $$ for $m\in \mathbb{Z}$ and $\delta\in \mathbb{C}$. We note that $$ \sigma_{m} = 2\, k_{2m} \qquad \mbox{ and } \qquad \tau_{m}(\delta) = 1 +2\, \delta\, k_{2m} $$ and \begin{align} &\sigma_m\, \sigma_n = \sigma_{m+n} + \sigma_{|m-n|},\label{a1}\\ &\sigma_{m\, n} = \sigma_{(11-m)n},\label{a2}\\ &\sigma_{m\, n} = \sigma_{|m|n},\label{a3}\\ &1+\sigma_1+\sigma_2 +\sigma_3+\sigma_4+\sigma_5=0.\label{a4} \end{align} Moreover we have \begin{equation}\label{A} \tau_{m} \big( \tfrac{1}{2} \big) = 2 \, k_{m}^{2}, \qquad \tau_{m} \big( {-}\tfrac{1}{2} \big) = 2 \, s_{m}^{2}, \end{equation} and \begin{equation}\label{B} \tau_{m} \Big( \frac{T_{r}(k_{2s})}{2 k_{2m}} \Big) = \tau_{m} \Big( \frac{T_{2r}(k_{s})}{2 k_{2m}} \Big) = 1+k_{2rs}=2\, k_{rs}^{2}, \end{equation} where $T_r(x)$ denote the $r$-th Chebyshev polynomial of the first kind. For example, from the last formula we get \begin{align} \tau_{m} \big( 2\, k_{2m}^{2} - \tfrac{3}{2} \big) & = \tau_{m} \Big( \frac{T_{3}(k_{2m})}{2 k_{2m}} \Big) = 2\, k_{3m}^{2},\label{C}\\ \tau_{m} \big( 8\, k_{2m}^{4} - 10\, k_{2m}^{2} + \tfrac{5}{2} \big) & = \tau_{m} \Big( \frac{T_{5}(k_{2m})}{2 k_{2m}} \Big) = 2\, k_{5m}^{2},\label{D} \end{align} for every $m\in \mathbb{N}$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section 2 %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Minimal polynomials, linear independence over~$\mathbb{Q}$} Let $\Psi _n(x)$ be the minimal polynomial of $\cos (2\pi /n)$ for every $n\in\mathbb{N}$. W.~Watkins and J.~Zeitlin described in~\cite{qf2-lit11} the following identities $$ T_{s+1}(x)-T_s(x)=2^s\prod\limits_{d|n}\Psi _d(x) $$ if $n=2s+1$ and $$ T_{s+1}(x)-T_{s-1}(x)=2^s\prod\limits_{d|n}\Psi _d(x) $$ if $n=2s$. In the sequel if $n=2s+1$ is a prime number, then we get $$ T_{s+1}(x)-T_s(x)=2^s\Psi _1(x)\Psi _n(x), $$ which for example implies the formula \begin{align*} \Psi_{11}(x)&=\frac{1}{32(x-1)}(T_6(x)-T_5(x)) \\ &= \frac{1}{32(x-1)}(32x^6-16x^5-48x^4+20x^3+18x^2-5x-1) \\ &= x^5+\frac{1}{2}x^4-x^3-\frac{3}{8}x^2+\frac{3}{16}x+\frac{1}{32}. \end{align*} \begin{lemma}\label{qf2-lem1.4} Let $\xi =\exp (i2\pi /11)$. Then $$%\begin{align*} p_{11}(x) = \prod_{m=1}^{5} (x-\sigma_{m}) = x^5+x^4-4x^3-3x^2+3x+1 $$%\end{align*} is a minimal polynomial of the numbers $\sigma_m$, $m=1,2,\ldots,5$. Moreover, we have the identity $$ \Psi _{11}(x)=\frac{1}{32}p_{11}(2x). $$ \end{lemma} \begin{corollary}\label{qf2-cor1.4.2} The numbers $k_{2m}$, $m=0,1,\ldots,4$ are linearly independent over~$\mathbb{Q}$. \end{corollary} \begin{proof} If we suppose that $$ a+b\cos (2\pi /11)+c\cos (4\pi /11)+d\cos (6\pi /11)+e\cos (8\pi /11)=0 $$ for some $a,b,c,d,e\in\mathbb{Q}$, then by the formulas $$ \begin{array}{l} \cos 2\alpha =2\cos ^2\alpha -1, \\ \cos 3\alpha =4\cos ^3\alpha -3\cos\alpha , \\ \cos 4\alpha =8\cos ^4\alpha -8\cos ^2\alpha +1, \end{array} $$ the number $\cos (2\pi /11)$ would be a root of some polynomial $\in \mathbb{Q} [x]$ having degree $\leqslant 4$, which by Lemma~\ref{qf2-lem1.4} is impossible. \end{proof} \begin{corollary}\label{qf2-cor1.4.2a} Every five numbers which belong to the set $\{1,\sigma_{1},\sigma_{2},\sigma_{3}, \sigma_{4},\sigma_{5}\}$ are linearly independent over~$\mathbb{Q}$. \end{corollary} \begin{proof} It follows from the identity~(\ref{a4}). \end{proof} \medskip The following observation will play a~central role in our arguments. \begin{lemma}\label{qf11-lem2.20} Let $a_1, a_2,\ldots,a_n \in \mathbb{R}$ be linearly independent over~$\mathbb{Q}$ and let $f_k, g_k \in \mathbb{Q}[\delta]$, $k=1,2,\ldots,n$. If the identity holds \begin{equation}\label{new-g} \sum_{k=1}^{n} f_k (\delta) \, a_k = \sum_{k=1}^{n} g_k (\delta) \, a_k \end{equation} for every $\delta\in \mathbb{Q}$, then $f_k(\delta)=g_k(\delta)$ for every $k=1,2,\ldots,n$ and $\delta\in \mathbb{C}$. \end{lemma} \begin{proof} Since $f_k(\delta)\in \mathbb{Q}$ and $g_k(\delta)\in \mathbb{Q}$ for any $\delta\in \mathbb{Q}$ we get from~(\ref{new-g}) that $f_k(\delta)=g_k(\delta)$ for every $k=1,2,\ldots,n$ and $\delta\in \mathbb{Q}$. The last equality implies that all, respective coefficients of polynomials $f_k$ and $g_k$ are the same. So $f_k(\delta)=g_k(\delta)$ for all $\delta\in \mathbb{C}$ and $k=1,2,\ldots,n$. \end{proof} \medskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section 3 %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Quasi-Fibonacci numbers of order~11}\label{roz3} Let us start from the following basic result. \begin{lemma}\label{qf2-len3.1} Let $\delta\in\mathbb{C}$ and $n\in \mathbb{N}$. Then the following identities hold $$ \tau_{m}^{n}(\delta) = a_n(\delta ) +b_n(\delta )\, \sigma_{m} +c_n(\delta )\, \sigma_{2m} +d_n(\delta )\, \sigma_{3m} +e_n(\delta )\, \sigma_{4m} +f_n(\delta )\, \sigma_{5m} \hspace{40pt}{(3.2m-1)} $$ $$ \tau_{m}^{n}(\delta) = A_n(\delta ) +B_n(\delta )\, \sigma_{m} +C_n(\delta )\, \sigma_{2m} +D_n(\delta )\, \sigma_{3m} +E_n(\delta )\, \sigma_{4m} \hspace{110pt}{(3.2m)} $$ for every $m=1,2,\ldots,5$, i.e., \begin{align} \tau_{1}^{n}(\delta) =& a_n(\delta ) +b_n(\delta )\, \sigma_{1} +c_n(\delta )\, \sigma_{2} +d_n(\delta )\, \sigma_{3} +e_n(\delta )\, \sigma_{4} +f_n(\delta )\, \sigma_{5} \label{2.1} \\ =&A_n(\delta ) +B_n(\delta )\, \sigma_{1} +C_n(\delta )\, \sigma_{2} +D_n(\delta )\, \sigma_{3} +E_n(\delta )\, \sigma_{4}, \label{2.2} \end{align} \begin{align} \tau_{2}^{n}(\delta) =& a_n(\delta ) + b_n(\delta )\, \sigma_{2} + c_n(\delta )\, \sigma_{4} + d_n(\delta )\, \sigma_{5} + e_n(\delta )\, \sigma_{3} + f_n(\delta )\, \sigma_{1} \label{3.3a} \\ =&A_n(\delta ) +B_n(\delta )\, \sigma_{2} +C_n(\delta )\, \sigma_{4} +D_n(\delta )\, \sigma_{5} +E_n(\delta )\, \sigma_{3}, \label{3.4a} \end{align} \begin{align} \tau_{3}^{n}(\delta) =&a_n(\delta ) + b_n(\delta )\, \sigma_{3} + c_n(\delta )\, \sigma_{5} + d_n(\delta )\, \sigma_{2} + e_n(\delta )\, \sigma_{1} + f_n(\delta )\, \sigma_{4} \label{3.5a} \\ =&A_n(\delta ) +B_n(\delta )\, \sigma_{3} +C_n(\delta )\, \sigma_{5} +D_n(\delta )\, \sigma_{2} +E_n(\delta )\, \sigma_{1}, \label{3.6a} \end{align} \begin{align} \tau_{4}^{n}(\delta) =&a_n(\delta ) + b_n(\delta )\, \sigma_{4} + c_n(\delta )\, \sigma_{3} + d_n(\delta )\, \sigma_{1} + e_n(\delta )\, \sigma_{5} + f_n(\delta )\, \sigma_{2}\label{3.7a} \\ =&A_n(\delta ) +B_n(\delta )\, \sigma_{4} +C_n(\delta )\, \sigma_{3} +D_n(\delta )\, \sigma_{1} +E_n(\delta )\, \sigma_{5}, \label{3.8a} \end{align} \begin{align} \tau_{5}^{n}(\delta) =&a_n(\delta ) + b_n(\delta )\, \sigma_{5} + c_n(\delta )\, \sigma_{1} + d_n(\delta )\, \sigma_{4} + e_n(\delta )\, \sigma_{2} + f_n(\delta )\, \sigma_{3}\label{3.9a} \\ =&A_n(\delta ) +B_n(\delta )\, \sigma_{5} +C_n(\delta )\, \sigma_{1} +D_n(\delta )\, \sigma_{4} +E_n(\delta )\, \sigma_{2}, \label{3.10a} \end{align} where $$ a_0(\delta )=1,\quad\quad b_0(\delta )=\ldots=f_0(\delta )=0, $$ \begin{equation} \left\{\begin{array}{lll} a_{n+1}(\delta ) & = & a_n(\delta )+2\delta b_n(\delta ), \\ b_{n+1}(\delta ) & = & \delta a_n(\delta )+b_n(\delta )+\delta c_n(\delta ), \\ c_{n+1}(\delta ) & = & \delta b_n(\delta )+c_n(\delta )+\delta d_n(\delta ), \\ d_{n+1}(\delta ) & = & \delta c_n(\delta )+d_n(\delta )+\delta e_n(\delta ), \\ e_{n+1}(\delta ) & = & \delta d_n(\delta )+e_n(\delta )+\delta f_n(\delta ), \\ f_{n+1}(\delta ) & = & \delta e_n(\delta )+(1+\delta )f_n(\delta ) \end{array}\right. \label{2.3} \end{equation} and $$ A_n(\delta )=a_n(\delta )-f_n(\delta ),\quad B_n(\delta )=b_n(\delta )-f_n(\delta ), $$ $$ C_n(\delta )=c_n(\delta )-f_n(\delta ),\quad D_n(\delta )=d_n(\delta )-f_n(\delta ),\quad E_n(\delta )=e_n(\delta )-f_n(\delta ), $$ for every $n=0,1,2,\ldots$. Hence, the following identities are derived: \begin{align*} A_{n+1}(\delta ) &= a_n(\delta )+2\delta b_n(\delta )- \delta e_n(\delta )-(1+\delta )f_n(\delta ) \\ &= A_n(\delta )+2\delta B_n(\delta )-\delta E_n(\delta ),%\\ \end{align*} \begin{align*} B_{n+1}(\delta ) &= \delta a_n(\delta )+b_n(\delta )+ \delta c_n(\delta )-\delta e_n(\delta )-(1+\delta )f_n(\delta ) \\ &= \delta A_n(\delta )+B_n(\delta )+\delta C_n(\delta )-\delta E_n(\delta ),\\ %\end{align*} %\begin{align*} C_{n+1}(\delta ) &= \delta b_n(\delta )+c_n(\delta )+ \delta d_n(\delta )-\delta e_n(\delta )-(1+\delta )f_n(\delta ) \\ &= \delta B_n(\delta )+C_n(\delta )+\delta D_n(\delta )-\delta E_n(\delta ),\\ %\end{align*} %\begin{align*} D_{n+1}(\delta ) &= \delta c_n(\delta )+d_n(\delta ) -(1+\delta )f_n(\delta ) \\ &= \delta C_n(\delta )+D_n(\delta ),\\ %\end{align*} %\begin{align*} E_{n+1}(\delta ) &= \delta d_n(\delta )+e_n(\delta ) -\delta e_n(\delta )- f_n(\delta ) \\ &= \delta D_n(\delta )+(1-\delta )E_n(\delta ), \end{align*} i.e., the following system of linear equations holds \begin{equation} \left\{\begin{array}{l} A_{n+1}(\delta )=A_n(\delta )+2\delta B_n(\delta )-\delta E_n(\delta ), \\ B_{n+1}(\delta )=\delta A_n(\delta )+B_n(\delta )+\delta C_n(\delta )-\delta E_n(\delta ), \\ C_{n+1}(\delta )=\delta B_n(\delta )+C_n(\delta )+\delta D_n(\delta )-\delta E_n(\delta ), \\ D_{n+1}(\delta )=\delta C_n(\delta )+D_n(\delta ), \\ E_{n+1}(\delta )=\delta D_n(\delta )+(1-\delta )E_n(\delta ), \end{array}\right. \label{new3.1} \end{equation} where $A_0(\delta )=1$, $B_0(\delta )=C_0(\delta )=D_0(\delta )=E_0(\delta )=0$. \end{lemma} \begin{proof} Proofs of the identities ($3.2m-1$) for every $m=1,2,\ldots,5$ by induction follow. For $n=1$ we have $$ \tau_{m}(\delta) = a_1(\delta ) +b_1(\delta )\, \sigma_{m} +c_1(\delta )\, \sigma_{2m} +d_1(\delta )\, \sigma_{3m} +e_1(\delta )\, \sigma_{4m} +f_1(\delta )\, \sigma_{5m}. $$ Suppose that for some $n\in \mathbb{N}$ the equalities ($3.2m-1$), $m=1,2,\ldots,5$, holds. Then by~(\ref{a1}), (\ref{a2}) and~(\ref{2.3}) we get \begin{multline*} \tau_{m}^{n+1}(\delta) = \tau_{m}(\delta)\, \tau_{m}^{n}(\delta) = {}\\ {}= \big(1+\delta\, \sigma_m \big) \big( a_n(\delta ) +b_n(\delta )\, \sigma_{m} +c_n(\delta )\, \sigma_{2m} +d_n(\delta )\, \sigma_{3m} +e_n(\delta )\, \sigma_{4m} +f_n(\delta )\, \sigma_{5m} \big) ={}\\ {}= \big(1+\delta\, \sigma_m \big)\, a_n(\delta ) + \big( \sigma_{m} + \delta\, (\sigma_{2m} +2 )\big) \, b_{n}(\delta) + \big( \sigma_{2m} + \delta\, (\sigma_{3m} +\sigma_{m} )\big) \, c_{n}(\delta) + {}\\ {}+ \big( \sigma_{3m} + \delta\, (\sigma_{4m} +\sigma_{2m} )\big) \, d_{n}(\delta) + \big( \sigma_{4m} + \delta\, (\sigma_{5m} +\sigma_{3m} )\big) \, e_{n}(\delta) + \big( \sigma_{5m} + \delta\, (\sigma_{5m} +\sigma_{4m} )\big) \, f_{n}(\delta) ={}\\ {}= \big( a_n(\delta) + 2\, \delta\, b_n(\delta) \big) + \big( \delta\, a_n(\delta) + b_n(\delta) +\delta\, c_{n}(\delta) \big) \, \sigma_{m} + \big( \delta\, b_n(\delta) + c_n(\delta) +\delta\, d_{n}(\delta) \big) \, \sigma_{2m} + {}\\ {}+ \big( \delta\, c_n(\delta) + d_n(\delta) +\delta\, e_{n}(\delta) \big) \, \sigma_{3m} + \big( \delta\, d_n(\delta) + e_n(\delta) +\delta\, f_{n}(\delta) \big) \, \sigma_{4m} + \big( \delta\, e_n(\delta) + ( 1 + \delta ) \, f_{n}(\delta) \big) \, \sigma_{5m} ={}\\ {}= a_{n+1}(\delta ) +b_{n+1}(\delta )\, \sigma_{m} +c_{n+1}(\delta )\, \sigma_{2m} +d_{n+1}(\delta )\, \sigma_{3m} +e_{n+1}(\delta )\, \sigma_{4m} +f_{n+1}(\delta )\, \sigma_{5m}, \end{multline*} which by the principle of mathematical induction means that~($3.2m-1$) hold for every $n\in \mathbb{N}$ and $m=1,2,\ldots,5$. Formula ($3.2m$) from~($3.2m-1$) and~(\ref{a4}) follows for every $m=1,2,\ldots,5$. \end{proof} \begin{definition}\label{qf2-def3.1} We call all elements of the sequences $\{ A_n(\delta )\}$, $\{ B_n(\delta )\}$, $\ldots$, $\{ E_n(\delta )\}$ the \textit{quasi-Fibonacci numbers of order $(11,\delta )$} (see Table~\ref{qf2-tab-d2} at the end of the paper). To simplify notation we shall write $\{A_n\}$, $\{B_n\}$, $\ldots$, $\{E_n\}$ instead of $\{ A_n(1 )\}$, $\{ B_n(1 )\}$, $\ldots$, $\{ E_n(1 )\}$, respectively. We call the elements of the sequences $\{ A_n\}$, $\{ B_n\}$, $\ldots$, $\{ E_n\}$ the \textit{quasi-Fibonacci numbers of order~$11$} (see Table~\ref{qf2-tab3} at the end of the paper). \end{definition} \begin{corollary} In the sequel, for $\delta =1$ we obtain the following recurrence relations: \begin{equation} \left\{\begin{array}{l} A_{n+1}=A_n+2B_n-E_n, \\ B_{n+1}=A_n+B_n+C_n-E_n, \\ C_{n+1}=B_n+C_n+D_n-E_n, \\ D_{n+1}=C_n+D_n, \\ E_{n+1}=D_n. \end{array}\right. \label{28} \end{equation} \end{corollary} \begin{corollary}\label{qf2-cor3.1.1} Adding identities ($3.k$) for odd and even $k=1,\ldots ,10$ respectively we obtain the identity: \begin{align} \tau_{1}^{n}(\delta) + \tau_{2}^{n}(\delta) &+ \tau_{3}^{n}(\delta) + \tau_{4}^{n}(\delta) + \tau_{5}^{n}(\delta) = \nonumber \\ &= 5a_n(\delta )-b_n(\delta )-c_n(\delta )-d_n(\delta )-e_n(\delta )-f_n(\delta )\nonumber \\ &= 5A_n(\delta )-B_n(\delta )-C_n(\delta )-D_n(\delta )-E_n(\delta )\nonumber\\ &:= {\cal A}_n(\delta ).\label{qf11-brak} \end{align} Hence, by~(\ref{A}) the following identities holds: \begin{equation}\label{3.15} 2^{-n} {\cal A}_n\big(\tfrac{1}{2}\big) = \sum_{m=1}^{5} k_{m}^{2n} \end{equation} and \begin{equation} 2^{-n} {\cal A}_n\big({-}\tfrac{1}{2}\big) = \sum_{m=1}^{5} s_{m}^{2n}. \end{equation} Moreover, from~(\ref{qf11-brak}) we obtain \begin{multline*} \sum_{m=1}^{5} \coeff \big( \tau_{m}^{n}(\delta); \delta^n \big) = \sum_{m=1}^{5} \sigma_{m}^{n} = 2^n\, \sum_{m=1}^{5} k_{2m}^{n} = {}\\ {}= 5\, \coeff \big( A_{n}(\delta); \delta^n\big) - \coeff \big( B_{n}(\delta); \delta^n\big) - \ldots -\coeff \big( E_{n}(\delta); \delta^n\big) = \coeff \big( {\cal A}_{n}(\delta); \delta^n\big). \end{multline*} Hence, by~(\ref{3.15}) we get $$ \coeff \big( {\cal A}_{2n}(\delta); \delta^{2n}\big) = 2^n\, {\cal A}_{n} \big(\tfrac{1}{2}\big) $$ and $$ \coeff \big( {\cal A}_{2n-1}(\delta); \delta^{2n-1}\big) = 2^{2n-1}\, \sum_{m=1}^{5} (-1)^m\, k_{m}^{n}. $$ In Table~\ref{qf2-tab3b} twelve initial values of the sequences $\{{\cal A}(1)\}$ (see also \seqnum{A062883} in~\cite{sloan}), $\{{\cal A}(1/2)\}$ and $\{{\cal A}(-1/2)\}$ are presented. \end{corollary} \begin{corollary} By~(\ref{C}) and ($3.2m$) we have the identity \begin{multline*} 2^n\, k_{3m}^{2n} = A_{n}(\delta) + B_{n}(\delta)\, \sigma_{m} + C_{n}(\delta)\, \sigma_{2m} + D_{n}(\delta)\, \sigma_{3m} + E_{n}(\delta)\, \sigma_{4m} ={}\\ {}= A_{n}(\delta) + 2\, \Big( B_{n}(\delta)\, k_{2m} + C_{n}(\delta)\, k_{4m} + D_{n}(\delta)\, k_{5m} + E_{n}(\delta)\, k_{3m} \Big) \end{multline*} for $\delta:=2\, k_{2m}^{2} - \frac{3}{2}$ and for every $m=1,2,\ldots,5$. Furthermore, by~(\ref{B}) and ($3.2m$) the following general formula hold \begin{multline*} 2^n\, k_{r\, m}^{2n} = A_{n}(\delta) + B_{n}(\delta)\, \sigma_{m} + C_{n}(\delta)\, \sigma_{2m} + D_{n}(\delta)\, \sigma_{3m} + E_{n}(\delta)\, \sigma_{4m} ={}\\ {}= A_{n}(\delta) + 2\, \Big( B_{n}(\delta)\, k_{2m} + C_{n}(\delta)\, k_{4m} + D_{n}(\delta)\, k_{5m} + E_{n}(\delta)\, k_{3m} \Big) \end{multline*} for $\delta:= T_r (k_{2m})/ (2\, k_{2m})$, $m,r\in \mathbb{N}$. We note that $x$ divides $T_r(x)$ iff $r$ is an odd positive integer. \end{corollary} \begin{corollary} Since $\deg_{\delta} \big( \tau_{m}^{n} \big) =n$ for every $m=1,2,\ldots,5$, by~($3.2m$) for $m=1,2,\ldots,5$ it can be easily deduced the following formula $$ \max \big\{ \deg \big( A_{n}(\delta)\big), \deg \big( B_{n}(\delta)\big), \deg \big( C_{n}(\delta)\big), \deg \big( D_{n}(\delta)\big), \deg \big( E_{n}(\delta)\big) \big\} =n. $$ Hence, by~(\ref{new3.1}) and Table~\ref{qf2-tab-d2}, if $\deg (D_{n}(\delta))=n$ for $n=5,6,\ldots$ then also $\deg (C_{n}(\delta))=n$ for $n=5,6,\ldots$, and $\deg (E_{n}(\delta))=n$ for infinite many $n\in \mathbb{N}$ (see also the Section~\ref{roz6} in this paper). \end{corollary} \begin{corollary} Taking into account the equations (from the last to the first one, respectively) of the recurrence system~(\ref{new3.1}), we obtain: \begin{equation}\left\{ \begin{array}{rl} \delta D_n(\delta )\!\!\! &= E_{n+1}(\delta )-(1-\delta )E_n(\delta ), \\ \delta ^2C_n(\delta )\!\!\! &= \delta D_{n+1}(\delta )-\delta D_n(\delta ) \\ &= E_{n+2}(\delta )+(\delta -2)E_{n+1}(\delta )+(1-\delta )E_n(\delta ), \\ \delta ^3B_n(\delta )\!\!\! &= \delta ^2C_{n+1}(\delta )-\delta ^2C_n(\delta )- \delta ^3D_n(\delta )+\delta ^3E_n(\delta ) \\ &= E_{n+3}(\delta )+(\delta -3)E_{n+2}(\delta )+(3-2\delta -\delta ^2)E_{n+1}(\delta ) \\ & +(\delta ^2+\delta -1)E_n(\delta ), \\ \delta ^4A_n(\delta )\!\!\!&= \delta ^3B_{n+1}(\delta )-\delta ^3B_n(\delta )- \delta ^4C_n(\delta )+\delta ^4E_n(\delta ) \\ &= E_{n+4}(\delta )+(\delta -4)E_{n+3}(\delta )+(-2\delta ^2-3\delta +6)E_{n+2}(\delta ) \\ & +(-\delta ^3+4\delta ^2+3\delta -4)E_{n+1}(\delta ) +(\delta ^4+\delta ^3-2\delta ^2-\delta +1)E_n(\delta ) \end{array}\right. \label{adam1} \end{equation} and, finally: \begin{equation}\label{wz314p} \delta ^4A_{n+1}(\delta )-\delta ^4A_n(\delta )-2\delta ^5B_n(\delta )+ \delta ^5E_n(\delta )=0, \end{equation} i.e. \begin{multline} E_{n+5}(\delta ) + (\delta -5)E_{n+4}(\delta )+(-4\delta ^2-4\delta +10)E_{n+3}(\delta ) +\\ + (-3\delta ^3+12\delta ^2+6\delta -10)E_{n+2}(\delta ) + (3\delta ^4+6\delta ^3-12\delta ^2-4\delta +5)E_{n+1}(\delta )+ \\ + (\delta ^5-3\delta ^4-3\delta ^3+4\delta ^2+\delta -1)E_n(\delta )=0. \label{adam} \end{multline} \end{corollary} Immediately from equations~(\ref{28}) or~(\ref{adam1}) and~(\ref{adam}) for $\delta=1$ we obtain the following formulas: \begin{equation}\label{3.20} \left\{ \begin{array}{l} E_{n+1}=D_n, \\ C_n=D_{n+1}-D_n, \\ B_n=D_{n+2}-2D_{n+1}+D_{n-1}, \\ A_n=D_{n+3}-3D_{n+2}+D_{n+1}+2D_n, \end{array} \right. \end{equation} and \begin{equation} D_{n+4}-4D_{n+3}+2D_{n+2}+5D_{n+1}-2D_n-D_{n-1}=0. \label{2.4} \end{equation} The characteristic polynomial $p_{11}({\mathbb X};\delta )$ of the recurrence equation~(\ref{adam}) has the following decomposition (see general formula~(\ref{w3.40}) below): \begin{multline}\label{w3.22} p_{11}({\mathbb X};\delta ) := \mathbb{X}^5+(\delta -5)\,\mathbb{X}^4 +(-4\delta ^2-4\delta +10)\,\mathbb{X}^3 +(-3\delta ^3+12\delta ^2+6\delta -10)\,\mathbb{X}^2 +{}\\ {} +(3\delta ^4+6\delta ^3-12\delta ^2-4\delta +5)\,\mathbb{X} +(\delta ^5-3\delta ^4-3\delta ^3+4\delta ^2+\delta -1)={}\\ {}= \prod_{m=1}^{5} \big(\mathbb{X} - \tau_{m} (\delta) \big). \end{multline} %where $\xi =\exp (i2\pi /11)$. \begin{sketchproof} We have $$ \sum_{m=1}^{5} \tau_{m}(\delta) = \sum_{m=1}^{5} \big( 1+\delta\, \sigma_m \big) = 5+\delta\, \sum_{m=1}^{5} \sigma_m \stackrel{(\ref{a4})}{=} 5-\delta, $$ \begin{multline*} \sum_{1\leq m < n \leq 5} \tau_{m}(\delta)\, \tau_{n}(\delta) = \frac{1}{2} \, \bigg( \Big(\sum_{m=1}^{5} \tau_{m}(\delta) \Big)^2 - \sum_{m=1}^{5} \tau_{m}^{2}(\delta) \bigg) = \frac{1}{2} \, \Big( \big(5-\delta\big)^2 - \sum_{m=1}^{5} \big( 1+\delta\, \sigma_m \big)^2 \Big) ={}\\ {}= \frac{1}{2} \, \Big( \big(5-\delta\big)^2 - 5 - 2\, \delta\, \sum_{m=1}^{5} \sigma_m - \delta^2\, \sum_{m=1}^{5} \sigma_{m}^{2} \Big) \stackrel{(\ref{a4}),(\ref{a1})}{=} {}\\ {}= \frac{1}{2} \, \Big( 25-10\, \delta +\delta^2 -5+2\, \delta - \delta^2\, \sum_{m=1}^{5} \big(\sigma_{2m} +2\big) \Big) \stackrel{(\ref{a4})}{=} 10 - \, \delta -4\, \delta^2, \end{multline*} etc. \end{sketchproof} It follows from decomposition~(\ref{w3.22}) that there exist numbers $\alpha, \beta, \gamma,$ $\varepsilon, \varphi\in\mathbb{R}$ such that: $$%\begin{align*} E_n(\delta ) = \alpha\, \tau_{1}^{n}(\delta) + \beta\, \tau_{2}^{n}(\delta) + \gamma\, \tau_{3}^{n}(\delta) + \varepsilon\, \tau_{4}^{n}(\delta) + \varphi\, \tau_{5}^{n}(\delta) $$%\end{align*} for every $n\in\mathbb{N}$. Solving the respective system of linear equations: $$ \left\{\begin{array}{lcl} E_n(\delta ) &=& 0, \qquad\qquad n=1,2,3,\\ E_4(\delta ) &=& \delta^4, \\ E_5(\delta ) &=& 5\delta^4-\delta^5, \\ \end{array}\right. $$ we obtain: \begin{align} 11\, E_n(\delta ) &= \sum_{m=1}^{5} \big( \sigma_{4m} - \sigma_{5m} \big) \, \tau_{m}^{n}(\delta)\label{110}\\ &= \big( \sigma_{4} - \sigma_{5} \big) \, \tau_{1}^{n}(\delta) + \big( \sigma_{3} - \sigma_{1} \big) \, \tau_{2}^{n}(\delta) + \big( \sigma_{1} - \sigma_{4} \big) \, \tau_{3}^{n}(\delta) {} \nonumber\\ &\phantom{=} + \big( \sigma_{5} - \sigma_{2} \big) \, \tau_{4}^{n}(\delta) + \big( \sigma_{2} - \sigma_{3} \big) \, \tau_{5}^{n}(\delta) \nonumber \\ &= 2 \big( k_1 - k_3 \big) \big( 1 + 2\, \delta\, k_2 \big)^{n} + 2 \big( k_6 - k_2 \big) \big( 1 + 2\, \delta\, k_4 \big)^{n} + 2 \big( k_2 + k_3 \big) \big( 1 + 2\, \delta\, k_6 \big)^{n} {}\nonumber\\ &\phantom{=} + 2 \big( {-}k_1 - k_4 \big) \big( 1 - 2\, \delta\, k_3 \big)^{n} + 2 \big( k_4 - k_6 \big) \big( 1 - 2\, \delta\, k_1 \big)^{n},\nonumber \end{align} which implies, the following identities for $\delta =\frac{1}{2}$: \begin{align} \frac{11}{2^{n+1}}E_n\big(\tfrac{1}{2}\big) &= \big( k_1 - k_3 \big)\, k_{1}^{2n} +\big( k_6 - k_2 \big)\, k_{2}^{2n} {} \\ &\phantom{=} +\big( k_2 + k_3 \big)\, k_{3}^{2n} -\big( k_1 + k_4 \big)\, k_{4}^{2n} +\big( k_4 - k_6 \big)\, k_{5}^{2n} \nonumber \\ &= 2\, s_1\, s_2 \, k_{1}^{2n} - 2\, s_2\, s_4 \, k_{2}^{2n} + 2\, s_3\, s_5 \, k_{3}^{2n} - 2\, s_3\, s_4 \, k_{4}^{2n} + 2\, s_1\, s_5 \, k_{5}^{2n}, \nonumber \end{align} and for $\delta =-\frac{1}{2}$ (by~(\ref{A})): \begin{equation} \frac{11}{2^{n+2}}\, E_n\big({-}\tfrac{1}{2}\big) = s_2\, s_{1}^{2n+1} - s_4\, s_{2}^{2n+1} + s_5\, s_{3}^{2n+1} - s_3\, s_{4}^{2n+1} + s_1\, s_{5}^{2n+1}. \end{equation} Now, from~(\ref{adam1}), (\ref{110}), (\ref{a1}) and~(\ref{a2}) the following decomposition may be generated: \begin{align} 11\, D_n(\delta)&= \frac{1}{\delta} \big( \big(11\,E_{n+1}(\delta)\big)-(1-\delta )\, \big(11\,E_n(\delta)\big)\big) = {}\label{111a}\\ &= \frac{1}{\delta}\, \sum_{m=1}^{5} \big( \sigma_{4m} - \sigma_{5m} \big) \big(\tau_{m}(\delta)-1+\delta\big)\, \tau_{m}^{n}(\delta)\nonumber\\ &= \sum_{m=1}^{5} \big( \sigma_{4m} - \sigma_{5m} \big) \big(\sigma_{m}+1\big)\, \tau_{m}^{n}(\delta)%\nonumber\\ %& = \sum_{m=1}^{5} \big( \sigma_{3m} - \sigma_{5m} \big) \tau_{m}^{n}(\delta)\nonumber\\ &= \big( \sigma_{3} - \sigma_{5} \big) \, \tau_{1}^{n}(\delta) + \big( \sigma_{5} - \sigma_{1} \big) \, \tau_{2}^{n}(\delta) + \big( \sigma_{2} - \sigma_{4} \big) \, \tau_{3}^{n}(\delta) {}\nonumber\\ &\phantom{=} {} + \big( \sigma_{1} - \sigma_{2} \big) \, \tau_{4}^{n}(\delta) + \big( \sigma_{4} - \sigma_{3} \big) \, \tau_{5}^{n}(\delta) \nonumber \\ &= 2\, \big( k_1 + k_6 \big)\, \big( 1 + 2\, \delta\, k_{2} \big)^{n} - 2\, \big( k_1 + k_2 \big)\, \big( 1 + 2\, \delta\, k_{4} \big)^{n} + 2\, \big( k_3 + k_4 \big)\, \big( 1 + 2\, \delta\, k_{6} \big)^{n} {}\nonumber\\ &\phantom{=} {} + 2\, \big( k_2 - k_4 \big)\, \big( 1 - 2\, \delta\, k_{3} \big)^{n} - 2\, \big( k_3 + k_6 \big)\, \big( 1 - 2\, \delta\, k_{1} \big)^{n},\nonumber \end{align} which implies the following identity for $\delta =\frac{1}{2}$: \begin{align} \frac{11}{2^{n+1}}D_n\big(\tfrac{1}{2}\big)&= \big( k_1 - k_5 \big)\, k_{1}^{2n} - \big( k_1 + k_2 \big)\, k_{2}^{2n} + \big( k_3 + k_4 \big)\, k_{3}^{2n} {}\\ &\phantom{=} {} + \big( k_2 - k_4 \big)\, k_{4}^{2n} - \big( k_3 + k_6 \big)\, k_{5}^{2n} \nonumber \\ &= 2\, s_2 \, s_3 \, k_{1}^{2n} - 2\, s_4 \, s_6 \, k_{2}^{2n} + 2\, s_2 \, s_6 \, k_{3}^{2n} + 2\, s_1 \, s_3 \, k_{4}^{2n} - 2\, s_1 \, s_4 \, k_{5}^{2n},\nonumber \end{align} and for $\delta =-\frac{1}{2}$ (by~(\ref{A})): \begin{equation} \frac{11}{2^{n+2}}\, D_n\big({-}\tfrac{1}{2}\big) = s_2\, s_3\, s_{1}^{2n} - s_4\, s_6\, s_{2}^{2n} + s_2\, s_6\, s_{3}^{2n} + s_1\, s_3\, s_{4}^{2n} - s_1\, s_4\, s_{5}^{2n}. \end{equation} Next, by~(\ref{adam1}), (\ref{111a}), (\ref{a1}) and~(\ref{a2}), we obtain: \begin{align} 11\, C_n(\delta ) &= \frac{11}{\delta} \big( D_{n+1}(\delta ) - D_n(\delta ) \big) = {}\label{112}\\ &= \sum_{m=1}^{5} \big( \sigma_{3m} - \sigma_{5m} \big)\, \sigma_{m}\, \tau_{m}^{n}(\delta)%\nonumber\\ = \sum_{m=1}^{5} \big( \sigma_{2m} - \sigma_{5m} \big) \tau_{m}^{n}(\delta)\nonumber\\ &= \big( \sigma_{2} - \sigma_{5} \big) \, \tau_{1}^{n}(\delta) + \big( \sigma_{4} - \sigma_{1} \big) \, \tau_{2}^{n}(\delta) + \big( \sigma_{5} - \sigma_{4} \big) \, \tau_{3}^{n}(\delta) {}\nonumber\\ &\phantom{=} {} + \big( \sigma_{3} - \sigma_{2} \big) \, \tau_{4}^{n}(\delta) + \big( \sigma_{1} - \sigma_{3} \big) \, \tau_{5}^{n}(\delta) \nonumber\\ &= 2\, \big( k_1 + k_4 \big)\, \big( 1 + 2\, \delta\, k_{2} \big)^{n} - 2\, \big( k_2 + k_3 \big)\, \big( 1 + 2\, \delta\, k_{4} \big)^{n} + 2\, \big( k_3 - k_1 \big)\, \big( 1 + 2\, \delta\, k_{6} \big)^{n} {}\nonumber\\ &\phantom{=} {} + 2\, \big( k_6 - k_4 \big)\, \big( 1 - 2\, \delta\, k_{3} \big)^{n} + 2\, \big( k_2 - k_6 \big)\, \big( 1 - 2\, \delta\, k_{1} \big)^{n},\nonumber \end{align} which implies the following identity for $\delta =\frac{1}{2}$: \begin{align} \frac{11}{2^{n+1}}C_n\big(\tfrac{1}{2}\big) &= \big( k_1 + k_4 \big)\, k_{1}^{2n} - \big( k_2 + k_3 \big)\, k_{2}^{2n} {}\\ &\phantom{=} {} + \big( k_3 - k_1 \big)\, k_{3}^{2n} + \big( k_6 - k_4 \big)\, k_{4}^{2n} + \big( k_2 - k_6 \big)\, k_{5}^{2n} \nonumber \\ &= 2\, s_3 \, s_4 \, k_{1}^{2n} - 2\, s_3 \, s_5 \, k_{2}^{2n} - 2\, s_1 \, s_2 \, k_{3}^{2n} - 2\, s_1 \, s_5 \, k_{4}^{2n} + 2\, s_2 \, s_4 \, k_{5}^{2n},\nonumber \end{align} and for $\delta =-\frac{1}{2}$ (by~(\ref{A})): \begin{equation} \frac{11}{2^{n+2}}\, C_n\big({-}\tfrac{1}{2}\big) = s_3\, s_4\, s_{1}^{2n} - s_3\, s_5\, s_{2}^{2n} - s_1\, s_2\, s_{3}^{2n} - s_1\, s_5\, s_{4}^{2n} + s_2\, s_4\, s_{5}^{2n}. \end{equation} Now, by~(\ref{adam1}), (\ref{110}), (\ref{111a}), (\ref{112}), (\ref{a1}) and~(\ref{a2}), we may generate the formula: \begin{align} 11\, B_n(\delta ) &= \frac{11}{\delta} \big( C_{n+1}(\delta) - C_n(\delta) \big)+ 11\, \big( E_n(\delta )- D_n(\delta) \big) = {}\label{113a} \\ &= \sum_{m=1}^{5} \Big(\big(\sigma_{2m}-\sigma_{5m}\big)\, \sigma_{m} +\sigma_{4m}-\sigma_{5m}-\sigma_{3m}+\sigma_{5m} \Big)\, \tau_{m}^{n}(\delta)%\nonumber\\ = \sum_{m=1}^{5} \big( \sigma_{m} - \sigma_{5m} \big) \tau_{m}^{n}(\delta)\nonumber\\ &= 2\, \big( k_1 + k_2 \big)\, \big( 1 + 2\, \delta\, k_{2} \big)^{n} + 2\, \big( k_4 - k_2 \big)\, \big( 1 + 2\, \delta\, k_{4} \big)^{n} {}\nonumber\\ &\phantom{=} {} + 2\, \big( k_3 + k_6 \big)\, \big( 1 + 2\, \delta\, k_{6} \big)^{n} - 2\, \big( k_3 + k_4 \big)\, \big( 1 - 2\, \delta\, k_{3} \big)^{n} - 2\, \big( k_1 + k_6 \big)\, \big( 1 - 2\, \delta\, k_{1} \big)^{n},\nonumber \end{align} hence, for $\delta =\frac{1}{2}$ we obtain: \begin{align} \frac{11}{2^{n+1}}B_n\big(\tfrac{1}{2}\big) &= \big( k_1 + k_2 \big)\, k_{1}^{2n} + \big( k_4 - k_2 \big)\, k_{2}^{2n} {}\\ &\phantom{=} {} + \big( k_3 + k_6 \big)\, k_{3}^{2n} - \big( k_3 + k_4 \big)\, k_{4}^{2n} - \big( k_1 + k_6 \big)\, k_{5}^{2n},\nonumber \end{align} and for $\delta =-\frac{1}{2}$ (by~(\ref{A})): \begin{equation} \frac{11}{2^{n+2}}\, B_n\big({-}\tfrac{1}{2}\big) = s_4\, s_5\, s_{1}^{2n} - s_1\, s_3\, s_{2}^{2n} + s_1\, s_4\, s_{3}^{2n} - s_2\, s_5\, s_{4}^{2n} - s_2\, s_3\, s_{5}^{2n}. \end{equation} Immediately from Corollary~\ref{qf2-cor3.1.1} we derive the identity: \begin{align} 55\, A_n(\delta ) &= 11\, \Big( \sum_{m=1}^{5} \tau_{m}^{n}(\delta) + B_{n}(\delta) + C_{n}(\delta) + D_{n}(\delta) + E_{n}(\delta) \Big)\label{114} \\ & = \sum_{m=1}^{5} \big( 11+\sigma_{m}+\sigma_{2m}+\sigma_{3m}+\sigma_{4m}-4\,\sigma_{5m} \big)\, \tau_{m}^{n}(\delta) \nonumber\\ & = \sum_{m=1}^{5} \big( \underbrace{(1+\sigma_{m}+\sigma_{2m}+\sigma_{3m}+\sigma_{4m}+\sigma_{5m})}_{=0}+ 10-5\,\sigma_{5m} \big)\, \tau_{m}^{n}(\delta) \nonumber\\ & = 5\, \sum_{m=1}^{5} \big( 2 - \sigma_{5m} \big) \tau_{m}^{n}(\delta),\nonumber \end{align} and thus we obtain the following trigonometrical form of $A_n(\delta )$: \begin{multline} 11\, A_n(\delta ) = s_{5}^{2}\, \big( 1 + 2\, \delta\, k_{2} \big)^{n} + s_{1}^{2}\, \big( 1 + 2\, \delta\, k_{4} \big)^{n} + s_{4}^{2}\, \big( 1 + 2\, \delta\, k_{6} \big)^{n} + {}\\ {}+ s_{2}^{2}\, \big( 1 - 2\, \delta\, k_{3} \big)^{n} + s_{3}^{2}\, \big( 1 - 2\, \delta\, k_{1} \big)^{n}.\label{115a} \end{multline} Hence, for $\delta =\frac{1}{2}$ we obtain: \begin{equation} \frac{11}{2^{n+2}} A_n\big(\tfrac{1}{2}\big) = s_{5}^{2}\, k_{1}^{2n} + s_{1}^{2}\, k_{2}^{2n} + s_{4}^{2}\, k_{3}^{2n} + s_{2}^{2}\, k_{4}^{2n} + s_{3}^{2}\, k_{5}^{2n}, \end{equation} and for $\delta =-\frac{1}{2}$ (by~(\ref{A})): \begin{equation} \frac{11}{2^{n+2}}\, A_n\big({-}\tfrac{1}{2}\big) = s_{5}^{2}\, s_{1}^{2n} + s_{1}^{2}\, s_{2}^{2n} + s_{4}^{2}\, s_{3}^{2n} + s_{2}^{2}\, s_{4}^{2n} + s_{3}^{2}\, s_{5}^{2n}. \end{equation} %%%% The next lemma is an attempt at generalizing the Lemma~3.13 from~\cite{wsw1}. In view of an extensive form of the formulas, only some identities will be presented here. \begin{lemma}\label{qf2-lem3.2} The following identities hold: \noindent a) (decomposition using Newton's formulas for elementary symmetric polynomials): \begin{align} \prod_{m=1}^{5} \big( \mathbb{X} - \tau_{m}^{n}(\delta) \big) &= \mathbb{X}^5 - {\cal A}_{n} (\delta)\, \mathbb{X}^4 +\frac{1}{2}\, \big( {\cal A}_{n}^{2} (\delta) - {\cal A}_{2n} (\delta) \big)\, \mathbb{X}^3 {}\label{w3.40}\\ &- \frac{1}{6}\, \big({\cal A}_{n}^{3} (\delta) +2\, {\cal A}_{3n} (\delta) - 3\, {\cal A}_{2n} (\delta)\, {\cal A}_{n} (\delta) \big)\, \mathbb{X}^2\nonumber\\ &+ \frac{1}{24}\, \big({\cal A}_{n}^{4} (\delta) - 6\, {\cal A}_{2n} (\delta)\,{\cal A}_{n}^{2} (\delta) + 8\, {\cal A}_{3n} (\delta)\, {\cal A}_{n} (\delta) - 6\, {\cal A}_{4n} (\delta) + 3\, {\cal A}_{2n}^{2} (\delta) \big)\, \mathbb{X}\nonumber\\ &+ \big( \delta^5 -3\, \delta^4 -3\, \delta^3 +4\, \delta^2 +\delta -1 \big)^n.\nonumber \end{align} \noindent b) \begin{equation} 2\, {\cal A}_{n}(\delta) - 11\, A_{n}(\delta) = \sum_{m=1}^{5} \sigma_{5m} \tau_{m}^{n}(\delta), \label{b1} \end{equation} \begin{equation}\label{b2} 11\, \big( B_{n}(\delta) - A_{n}(\delta) \big) + 2\, {\cal A}_{n}(\delta) = \sum_{m=1}^{5} \sigma_{m} \tau_{m}^{n}(\delta) = \frac{1}{\delta}\, \big( {\cal A}_{n+1}(\delta) - {\cal A}_{n}(\delta) \big), \end{equation} which implies also the recurrence formula: \begin{equation}\label{b3} {\cal A}_{n+1}(\delta) = (2\, \delta+1)\, {\cal A}_{n}(\delta) +11\, \delta\, \big( B_{n}(\delta) - A_{n}(\delta) \big); \end{equation} \begin{equation} 11\, \big( C_{n}(\delta) - A_{n}(\delta) \big) + 2\, {\cal A}_{n}(\delta) = \sum_{m=1}^{5} \sigma_{2m} \tau_{m}^{n}(\delta), \label{b4} \end{equation} \begin{equation} 11\, \big( D_{n}(\delta) - A_{n}(\delta) \big) + 2\, {\cal A}_{n}(\delta) = \sum_{m=1}^{5} \sigma_{3m} \tau_{m}^{n}(\delta), \label{b5} \end{equation} \begin{equation} 11\, \big( E_{n}(\delta) - A_{n}(\delta) \big) + 2\, {\cal A}_{n}(\delta) = \sum_{m=1}^{5} \sigma_{4m} \tau_{m}^{n}(\delta). \label{b6} \end{equation} \noindent c) (we set here ${\cal A}_n\equiv {\cal A}_n(1)$ and $E_n\equiv E_n(1)$) \begin{align} 11&\big(E_{n+1}\,{\cal A}_n-E_{n}\,{\cal A}_{n+1}\big)= \sum_{m=1}^{5} \big( 5\, \sigma_{5m} + 3\, (\sigma_{m} + \sigma_{4m})\big) (-\sigma_{m})^n {}\label{wwww} \\ &+\sum_{m=1}^{5} \big( 2 + \sigma_{3m} - \sigma_{2m} - \sigma_{4m})\big) \big( \sigma_{m} - \sigma_{4m} - \sigma_{5m} \big)^n \nonumber\\ =& 2^{n+1}\, \Big( ( -5\, k_1 + 3\, k_2 - 3\, k_3 )\, k_{9}^{n} + ( 5\, k_2 + 3\, k_4 + 3\, k_6 )\, k_{7}^{n} + ( -5\, k_3 + 3\, k_2 + 3\, k_6 )\, k_{5}^{n} \nonumber\\ &{}+ ( 5\, k_4 - 3\, k_3 - 3\, k_1 )\, k_{3}^{n} + ( 5\, k_6 + 3\, k_4 - 3\, k_1 )\, k_{1}^{n} + \big( 1 + k_6\, ( 1 - 2\, k_2) \big) \big( k_2\, (1+2\, k_1) \big)^{n} \nonumber\\ &+ \big( 1 - k_1\, ( 1 - 2\, k_4) \big) \big( k_4\, (1-2\, k_2) \big)^{n} + \big( 1 + k_4\, ( 1 + 2\, k_5) \big) \big( k_6\, (1+2\, k_3) \big)^{n} \nonumber\\ &+ \big( 1 + k_2\, ( 1 + 2\, k_3) \big) \big( k_8\, (1-2\, k_4) \big)^{n} + \big( 1 - k_3\, ( 1 + 2\, k_1) \big) \big( k_{10}\, (1-2\, k_5) \big)^{n} \Big).\nonumber \end{align} \end{lemma} \begin{proof} (a) The decompositions~(\ref{w3.40}) from Vieta's formulas, formula~(\ref{w3.22}) and the following (version) of Newton's formula for elementary symmetric polynomials~\cite{mostowski} follows: \begin{equation}\label{nng} \chi_{m}=\frac{1}{m!}\, \left| \begin{array}{lllll} \eta_1 & 1 & 0 & \ldots & 0 \\ \eta_2 & \eta_1 & 2 & \ldots & 0 \\ \eta_3 & \eta_2 & \eta_1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \eta_m & \eta_{m-1} & \eta_{m-2} & \ldots & \eta_1 \\ \end{array} \right| \end{equation} where $\eta_j=x_{1}^{j}+x_{2}^{j}+\ldots+x_{n}^{j}$, $\chi_{j} =\sum\limits_{1\leq m_1