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\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}
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\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 $m\in \mathbb{N}$.

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\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

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\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 $\eta_j=x_{1}^{j}+x_{2}^{j}+\ldots+x_{n}^{j}$,
$\chi_{j} =\sum\limits_{1\leq m_1<m_2<\ldots<m_j\leq n} x_1 x_2\ldots x_{j}$.
Indeed, for $x_m=\tau_{m}^{n}(\delta)$, $m=1,2,\ldots,5$,
from~(\ref{nng}) we get
$$%\begin{multline*}
\chi_{m}=\frac{1}{m!}\,
\left|
\begin{array}{cccccc}
{\cal A}_{n}(\delta) & 1 & 0 & 0 & \ldots & 0 \\
{\cal A}_{2n}(\delta) & {\cal A}_{n}(\delta) & 2 & 0 & \ldots & 0 \\
{\cal A}_{3n}(\delta) & {\cal A}_{2n}(\delta) & {\cal A}_{n}(\delta) & 3 &\ldots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
{\cal A}_{mn}(\delta) & {\cal A}_{(m-1)n}(\delta) & {\cal A}_{(m-2)n}(\delta) &
{\cal A}_{(m-3)n}(\delta) & \ldots & {\cal A}_{n}(\delta) \\
\end{array}
\right|=%{}\\
$$
$${}=
\left\{
\begin{array}{ll}
\frac{1}{2!}\, \big( {\cal A}_{n}^{2}(\delta) - {\cal A}_{2n}(\delta) \big), &
\quad\mbox{for } m=2,\\[1ex]
\frac{1}{3!}\, \big( {\cal A}_{n}^{3}(\delta) + 2\, {\cal A}_{3n}(\delta)
- 3\, {\cal A}_{n}(\delta)\, {\cal A}_{2n}(\delta) \big), &
\quad\mbox{for } m=3,\\[1ex]
\mbox{etc.} & \\
\end{array}
\right.
$$%\end{multline*}
We note that for this case we have
$$
\prod_{m=1}^{5} \big( \mathbb{X} - \tau_{m}^{n}(\delta) \big) =
\mathbb{X}^5 - \chi_{1}\, \mathbb{X}^{4} + \chi_{2}\, \mathbb{X}^{3}
- \chi_{3}\, \mathbb{X}^{2} + \chi_{4}\, \mathbb{X} - \chi_{5}.
$$
Moreover, from~(\ref{w3.22}) we have
$$
\chi_5 = \prod_{m=1}^{5} \tau_{m}(\delta) = -\delta^5+3\, \delta^4 -4\, \delta^2 -\delta+1.
$$

(b)~(\ref{b1}): By~(\ref{qf11-brak}) and~(\ref{114}) we have:
$$
2\, {\cal A}_{n}(\delta) - 11\, A_{n}(\delta) =
2\, \sum_{m=1}^{5} \tau_{m}^{n}(\delta) -
\sum_{m=1}^{5} \big(2-\sigma_{m}\big)\, \tau_{m}^{n}(\delta) =
\sum_{m=1}^{5} \sigma_{5m} \tau_{m}^{n}(\delta).
$$

\noindent
(\ref{b2}): From~(\ref{qf11-brak}) we obtain
$$%\begin{multline*}
\frac{1}{\delta} \big( {\cal A}_{n+1}(\delta) - {\cal A}_{n}(\delta)\big)=
\frac{1}{\delta}\,
\sum_{m=1}^{5} \big( \tau_{m}^{n+1}(\delta) - \tau_{m}^{n}(\delta)\big) =
\sum_{m=1}^{5} \frac{1}{\delta}\, \big( \tau_{m}(\delta) - 1 \big)\, \tau_{m}^{n}(\delta) =
\sum_{m=1}^{5} \sigma_{m} \tau_{m}^{n}(\delta).
$$%\end{multline*}
On the other hand, by~(\ref{113a}) and~(\ref{qf11-brak}) we get
$$%\begin{multline*}
\sum_{m=1}^{5} \sigma_{m} \tau_{m}^{n}(\delta) =
\sum_{m=1}^{5} \sigma_{5m} \tau_{m}^{n}(\delta) +
\sum_{m=1}^{5} \big(\sigma_{m}-\sigma_{5m}\big) \tau_{m}^{n}(\delta) =
11\,  \big( B_{n}(\delta) - A_{n}(\delta) \big) + 2\, {\cal A}_{n}(\delta).
$$%\end{multline*}

Formulas~(\ref{b4}), (\ref{b5}) and~(\ref{b6}) from~(\ref{b1}) and
formulas~($3.2m-1$), $m=2,3,4$ can be easily deduced.

(c)
The following ten simple equalities form the technical
base (by hand calculation) of the proof of identity~(\ref{wwww}):
\begin{align*}
(1+\sigma_{1})(1+\sigma_{3}) &= -\sigma_{5},&
(1+\sigma_{1})(1+\sigma_{2}) &= \sigma_{1} - \sigma_{4} - \sigma_{5}, \\
(1+\sigma_{1})(1+\sigma_{4}) &= -\sigma_{2},&
(1+\sigma_{1})(1+\sigma_{5}) &= \sigma_{5} - \sigma_{2} - \sigma_{3}, \\
(1+\sigma_{2})(1+\sigma_{3}) &= -\sigma_{4}, &
(1+\sigma_{2})(1+\sigma_{4}) &= \sigma_{2} - \sigma_{1} - \sigma_{3}, \\
(1+\sigma_{2})(1+\sigma_{5}) &= -\sigma_{1}, &
(1+\sigma_{3})(1+\sigma_{4}) &= \sigma_{4} - \sigma_{2} - \sigma_{5}, \\
(1+\sigma_{4})(1+\sigma_{5}) &= -\sigma_{3}, &
(1+\sigma_{3})(1+\sigma_{5}) &= \sigma_{3} - \sigma_{1} - \sigma_{4}.
\end{align*}
\end{proof}


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%%%%%%%%%%%%%%   Section 4   %%%%%%%%%%%%%%%%%%
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\section{Summation formulas}




\begin{lemma}\label{qf2-lem3.new.1}
We have:
\begin{align}
\delta\sum\limits_{n=1}^{N}C_n(\delta ) &=D_{N+1}(\delta ),\label{120c}\\
\delta\sum\limits_{n=1}^{N}B_n(\delta ) &=C_{N+1}(\delta )+
\delta \sum_{n=1}^{N} \big( E_{n}(\delta) - D_{n}(\delta) \big)=
C_{N+1}(\delta )-E_{N+1}(\delta ),
\label{120d}\\
\delta \sum_{n=1}^{N} E_{n} (\delta) &
=A_1(\delta)-A_{N+1}(\delta) + 2\delta \sum_{n=1}^{N} B_{n}(\delta)
\label{w3.36a}\\
&= 1-A_{N+1}(\delta)+2C_{N+1}(\delta)-2E_{N+1}(\delta),\label{w3.36b}
\end{align}
and, from~(\ref{adam}) we obtain:
\begin{multline}\label{120a}
\delta ^5\sum\limits_{n=1}^{N}E_n(\delta )=
-E_{N+5}(\delta )+(4-\delta )E_{N+4}(\delta )
+(4\delta ^2+3\delta -6)E_{N+3}(\delta )+ \\
+(3\delta ^3-8\delta ^2-3\delta +4)E_{N+2}(\delta )
+(-3\delta ^4-3\delta ^3+4\delta ^2+\delta -1)E_{N+1}(\delta )+\delta ^4;
\end{multline}
\begin{equation}\label{120b}
\delta\sum\limits_{n=1}^{N}D_n(\delta )=
E_{N+1}(\delta )+\delta\sum\limits_{n=1}^{N}E_n(\delta ),
\end{equation}
\begin{equation}\label{w3.38a}
\delta\sum\limits_{n=0}^{N-1} (1-\delta)^n D_{N-n}(\delta)=
E_{N+1}(\delta)
\end{equation}
or
\begin{equation}\label{w3.38b}
\frac{E_{N}(\delta)}{(1-\delta)^{N-1}}=
\delta\sum\limits_{n=0}^{N-1} \frac{D_{n}(\delta)}{(1-\delta)^n},
\end{equation}
\begin{equation}\label{120e}
\delta\sum\limits_{n=1}^{N}A_n(\delta ) =
-A_{N+1}(\delta )+B_{N+1}(\delta )+2C_{N+1}(\delta )
-D_{N+1}(\delta )-2E_{N+1}(\delta )+1-\delta .
\end{equation}
and
\begin{equation}\label{w3.40new}
\delta^2\sum\limits_{n=1}^{N} (N-n+1)\big(2\, B_n(\delta) -E_{n}(\delta)\big) =
\delta\sum\limits_{n=1}^{N+1} A_n(\delta)  - (N+1)\, \delta.
\end{equation}
\end{lemma}


\begin{proof}

(\ref{120c}):
By~(\ref{adam1}) and Table~\ref{qf2-tab-d2} we get
$$
\delta\sum_{n=1}^{N} C_n(\delta) =
\sum_{n=1}^{N} \big(  D_{n+1}(\delta) - D_{n}(\delta) \big) =
D_{N+1}(\delta) - D_{1}(\delta) = D_{N+1}(\delta).
$$

\noindent
(\ref{120d}):
By~(\ref{adam1}) and Table~\ref{qf2-tab-d2} we get
\begin{multline*}
\delta\sum_{n=1}^{N} B_n(\delta) =
\sum_{n=1}^{N} \big( C_{n+1}(\delta) - C_{n}(\delta) \big)
+\delta\sum_{n=1}^{N} \big( E_{n}(\delta) - D_{n}(\delta) \big)
= {}\\
{}=
C_{N+1}(\delta) -C_{1}(\delta) +
\delta\sum_{n=1}^{N} \big( E_{n}(\delta) - E_{n+1}(\delta) \big)
={}\\
{}=
C_{N+1}(\delta) -C_{1}(\delta) + E_{1}(\delta) - E_{N+1}(\delta)
=
C_{N+1}(\delta) - E_{N+1}(\delta).
\end{multline*}


\noindent
(\ref{w3.36a}):  Follows from ~(\ref{wz314p}).
\medskip

\noindent
(\ref{w3.36b}): Can be derived from~(\ref{w3.36a}) and~(\ref{120d}).
The other proof from formulas~(\ref{110}),
(\ref{112}) and~(\ref{114}) follows:
first, by~(\ref{110}) we get
\begin{multline}\label{pomoc1}
\delta\, \sum_{n=1}^{N} E_{n}(\delta) =
\frac{\delta}{11}\,
\sum_{n=1}^{N} \sum_{m=1}^{5} \big( \sigma_{4m}-\sigma_{5m}\big)\, \tau_{m}^{n}(\delta)=
\frac{\delta}{11}\,
\sum_{m=1}^{5} \sum_{n=1}^{N}  \big( \sigma_{4m}-\sigma_{5m}\big)\, \tau_{m}^{n}(\delta)={}\\
{}=
\frac{\delta}{11}\,
\sum_{m=1}^{5} \big( \sigma_{4m}-\sigma_{5m}\big)\,
\frac{\tau_{m}^{N+1}(\delta)-\tau_{m}(\delta)}{\tau_{m}(\delta)-1}=
\frac{1}{11}\,
\sum_{m=1}^{5} \frac{\sigma_{4m}-\sigma_{5m}}{\sigma_m}\,
\big(\tau_{m}^{N+1}(\delta)-\tau_{m}(\delta)\big),
\end{multline}
but we have
\begin{multline*}
\frac{\sigma_{4m}-\sigma_{5m}}{\sigma_m} =
\frac{\sigma_{4m}+\sigma_{5m}}{\sigma_m} - 2\, \frac{\sigma_{5m}}{\sigma_m} =
\sigma_{5m} - 2\, \big( \sigma_{4m} - \sigma_{2m} + 1 \big)={}\\
{}=
2\, \big( \sigma_{2m} - \sigma_{5m} \big) -
2\, \big( \sigma_{4m} - \sigma_{5m} \big) -
\big( 2 - \sigma_{5m} \big),
\end{multline*}
which, by~(\ref{pomoc1}), (\ref{112}),  (\ref{110}) and~(\ref{114}) implies
$$
\delta\, \sum_{n=1}^{N} E_{n}(\delta) =
2\, \big( C_{N+1}(\delta) - C_{1}(\delta) \big) -
2\, \big( E_{N+1}(\delta) - E_{1}(\delta) \big) -
\big( A_{N+1}(\delta) - A_{1}(\delta) \big).
$$
Hence, by~Table~\ref{qf2-tab-d2} the formula~(\ref{w3.36b}) follows.
\medskip


\noindent
(\ref{w3.38a}):
By~(\ref{wz314p}) we obtain
\begin{align*}
\delta\, D_{N}(\delta) &= E_{N+1}(\delta) - (1-\delta)\, E_{N}(\delta),\\
\delta\,(1-\delta)\, D_{N-1}(\delta) &=
(1-\delta)\,E_{N}(\delta) - (1-\delta)^2\, E_{N-1}(\delta),\\
\delta\,(1-\delta)^2\, D_{N-2}(\delta) &=
(1-\delta)^2\,E_{N-1}(\delta) - (1-\delta)^3\, E_{N-2}(\delta),\\
&\vdots  \\
\delta\,(1-\delta)^{N-1}\, D_{1}(\delta) &=
(1-\delta)^{N-1}\,E_{2}(\delta) - (1-\delta)^{N}\, E_{1}(\delta),
\end{align*}
which, after summation implies the formula~(\ref{w3.38a}).
\medskip


\noindent
(\ref{120e}):
By~(\ref{wz314p}) we have
$$
\delta\, \sum_{n=1}^{N} A_{n}(\delta) =
\sum_{n=1}^{N} \big(B_{n+1}(\delta) -B_{n}(\delta)\big) +
\delta\, \sum_{n=1}^{N} E_{n}(\delta) -
\delta\, \sum_{n=1}^{N} C_{n}(\delta).
$$
Hence, by~(\ref{120c}) and~(\ref{w3.36a}) we get the desired formula
$$
= \big(B_{N+1}(\delta) -B_{1}(\delta)\big) +
\big( 1 - A_{N+1}(\delta) +2\, C_{N+1}(\delta)-2\, E_{N+1}(\delta)\big)
-D_{N+1}(\delta).
$$


\noindent
(\ref{w3.40new}):
From~(\ref{wz314p}) we deduce the following system of equalities
\begin{align*}
A_{N+1}(\delta) - A_{N}(\delta) &= 2\, \delta\, B_{N}(\delta) - \delta\, E_{N}(\delta),\\
2\, A_{N}(\delta) - 2\,A_{N-1}(\delta) &=
4\, \delta\, B_{N-1}(\delta) - 2\, \delta\, E_{N-1}(\delta),\\
3\, A_{N-1}(\delta) - 3\,A_{N-2}(\delta) &=
6\, \delta\, B_{N-2}(\delta) - 3\, \delta\, E_{N-2}(\delta),\\
&\vdots  \\
N\, A_{2}(\delta) - N\,A_{1}(\delta) &=
2\, N\, \delta\, B_{1}(\delta) - N\, \delta\, E_{1}(\delta),
\end{align*}
which, after summation implies the desired formulas.
\end{proof}


\begin{corollary}
(See also Corollary~\ref{cor4.2}.)
From~(\ref{w3.38a}) for $\delta=\pm\frac{1}{2}$ we get
$$
2^{N+1}\, E_{N+1}\big(\tfrac{1}{2}\big)=
\sum_{n=0}^{N-1}2^n\, D_{n}\big(\tfrac{1}{2}\big)
$$
and
$$
-2\, \big(\tfrac{2}{3}\big)^{N}\, E_{N+1}\big({-}\tfrac{1}{2}\big)=
\sum_{n=0}^{N-1}\big(\tfrac{2}{3}\big)^{n}\, D_{n}\big({-}\tfrac{1}{2}\big).
$$
\end{corollary}

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%%%%%%%%%%%%%%   Section 5   %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%               %%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Reduction formulas for indices}



\begin{lemma}\label{qf2-cor3.1.1.1}
The following identities hold:
\begin{equation}
\left\{\!\!\begin{array}{l}
A_{m+n} = A_mA_n+2B_mB_n+2C_mC_n+2D_mD_n+2E_mE_n, \\
B_{m+n} = A_mB_n+B_mA_n+B_mC_n+C_mB_n+C_mD_n
         +D_mC_n+D_mE_n+E_mD_n,\\
C_{m+n} = A_mC_n+B_mB_n+B_mD_n+D_mB_n+C_mA_n
   +C_mE_n+E_mC_n, \\
D_{m+n} = A_mD_n+B_mC_n+B_mE_n+D_mA_n+C_mB_n
+E_mb_n+E_mE_n, \\
E_{m+n} = A_mE_n+B_mD_n+E_mA_n+D_mB_n+C_mC_n
+D_mE_n+E_mD_n,
\end{array}\right.
\end{equation}
and (the special cases when $m=n$):
\begin{equation}
\left\{\begin{array}{l}
A_{2n} = A_n^2+2B_n^2+2C_n^2+2D_n^2+2E_n^2, \\
B_{2n} = 2A_nB_n+2B_nC_n+2C_nD_n+2D_nE_n, \\
C_{2n} = 2A_nC_n+2C_nE_n+2D_nB_n+B_n^2, \\
D_{2n} = 2A_nD_n+2C_nB_n+2B_nE_n+E_n^2, \\
E_{2n} = 2A_nE_n+2E_nD_n+2D_nB_n+C_n^2.
\end{array}\right.
\end{equation}
\end{lemma}

%\begin{proof}
The proof runs similar to the proof of
Lemma~3.16 from~\cite{wsw1}.
%\end{proof}
\bigskip


\begin{lemma}
We have the identity:
\begin{multline}
\Delta:=
\left|
\begin{array}{ccc}
A_{k}(\delta) & 2 B_{k}(\delta) - E_{k}(\delta) &
2 C_{k}(\delta) - D_{k}(\delta) - E_{k}(\delta) \\
B_{k}(\delta) &  A_{k}(\delta) + C_{k}(\delta)- E_{k}(\delta) &
B_{k}(\delta) - E_{k}(\delta) \\
C_{k}(\delta) & B_{k}(\delta)+D_{k}(\delta)-E_{k}(\delta) &
A_{k}(\delta)-D_{k}(\delta)  \\
D_{k}(\delta) & C_{k}(\delta) & B_{k}(\delta)-D_{k}(\delta)-E_{k}(\delta) \\
E_{k}(\delta) & D_{k}(\delta)-E_{k}(\delta) & C_{k}(\delta)-D_{k}(\delta)-E_{k}(\delta) \\
\end{array}
\right.\\[1ex]
\left.
\begin{array}{cc}
 -C_{k}(\delta) + D_{k}(\delta) &
-B_{k}(\delta)- C_{k}(\delta)+2 E_{k}(\delta) \\
 -D_{k}(\delta) + E_{k}(\delta) &
-B_{k}(\delta) - C_{k}(\delta) + 2 E_{k}(\delta) \\
 B_{k}(\delta)-C_{k}(\delta)-D_{k}(\delta) &
-B_{k}(\delta) \\
A_{k}(\delta)-C_{k}(\delta)-D_{k}(\delta) & -C_{k}(\delta)+E_{k}(\delta) \\
B_{k}(\delta)-C_{k}(\delta)-D_{k}(\delta)+E_{k}(\delta) &
A_{k}(\delta)-B_{k}(\delta)-C_{k}(\delta)+D_{k}(\delta) \\
\end{array}
\right|=\\[1ex]
=\big( -\delta^5+3\, \delta^4+3\, \delta^3-4\, \delta^2 -\delta+1\big)^k.
\end{multline}
\end{lemma}

The proof of this lemma runs like the proof of Lemma~3.21 from~\cite{wsw1}.
\bigskip

\begin{lemma}
The following identities hold:
$$
%\begin{align*}
A_{n-k} (\delta) = \frac{\Delta_1}{\Delta},\quad
B_{n-k} (\delta) = \frac{\Delta_2}{\Delta},\quad
C_{n-k} (\delta) = \frac{\Delta_3}{\Delta},\quad
%$$
%$$
D_{n-k} (\delta) = \frac{\Delta_4}{\Delta},\quad
E_{n-k} (\delta) = \frac{\Delta_5}{\Delta},
%\end{align*}
$$
where $\Delta_i$ arise from the determinant $\Delta$ by replacing
its i-th column by the vector $(A_{n}, B_{n},C_{n},$ $D_{n},E_{n})^{T}$
(the Cramers rule for respective system of equations).
\end{lemma}


The proof is similar to the proof of Lemma~3.20 from~\cite{wsw1}.
%\bigskip
%\bigskip

\begin{remark}
After multiplication we obtain
\begin{multline}
\big( \tau_{2}(1) \big)^{2n}  \big( \tau_{5}(1) \big)^{2n} =
\big( 2+\sigma_{2}\big)^{n}   =
2^n \big( \tau_{2}\big(\tfrac{1}{2}\big) \big)^{n}
\stackrel{\rm by~(\ref{3.4a})}{=} {}\\
{} =  2^n\big( A_{n}\big(\tfrac{1}{2}\big)
+B_{n}\big(\tfrac{1}{2}\big)\, \sigma_{2}
+C_{n}\big(\tfrac{1}{2}\big)\, \sigma_{4}
+D_{n}\big(\tfrac{1}{2}\big)\, \sigma_{5}
+E_{n}\big(\tfrac{1}{2}\big)\, \sigma_{3}\big).
\end{multline}
On the other hand, by~(\ref{3.4a}) and~(\ref{3.10a}) we get:
\begin{multline*}
\big( \tau_{2}(1) \big)^{2n}  \big( \tau_{5}(1) \big)^{2n} = {}\\
{}=
\big( A_{2n} +
B_{2n}\, \sigma_{2}+
C_{2n}\, \sigma_{4} +
D_{2n}\, \sigma_{5}+
E_{2n}\, \sigma_{3} \big)
\big( A_{2n} +
B_{2n}\, \sigma_{5}+
C_{2n}\, \sigma_{1}+
D_{2n}\, \sigma_{4}+
E_{2n}\, \sigma_{2}
\big)={}\\
{}=
\sigma_{2}\,
\big(A_{2n}\, \big(B_{2n}-C_{2n}+E_{2n}\big)-B_{2n}\, \big(C_{2n}-E_{2n}\big)
+E_{2n}\, \big(2\, C_{2n}-D_{2n}-E_{2n}\big)\big) +{}\\
{}+
\sigma_{3}\,
\big( C_{2n} \, \big( C_{2n}-A_{2n}-B_{2n}+D_{2n}\big)+B_{2n}\, \big(B_{2n}-D_{2n}\big)
 + E_{2n}\, \big(A_{2n} +B_{2n} \big)-D_{2n}^{2}-E_{2n}^{2}\big)+{}\\
{}+
\sigma_{4}\,
\big( B_{2n}\, \big(B_{2n}-2\, C_{2n}-D_{2n}+E_{2n}\big)+C_{2n}\, \big(D_{2n}+E_{2n}\big)
+D_{2n}\, \big(A_{2n}-D_{2n}+E_{2n}\big)-E_{2n}^{2}\big)+{}\\
{}+
\sigma_{5}\,
\big( A_{2n}\,  \big( B_{2n}+D_{2n}-C_{2n}\big)
+C_{2n}\, \big(C_{2n}-2\, B_{2n}+D_{2n}+E_{2n}\big)-D_{2n}\, \big(D_{2n}+E_{2n}\big)\big)+{}\\
{}+
\big(A_{2n}\, \big(A_{2n}-C_{2n}\big)-B_{2n}\, \big(2\, C_{2n}-D_{2n}-2\, E_{2n}\big)
+D_{2n}\, \big(2\, C_{2n}-D_{2n}-E_{2n}\big)-E_{2n}^{2}\big).
\end{multline*}
Hence, comparing the coefficients in the expression $\sigma_{m}$,
$m=0,2,3,4,5$ we obtain,
by Lemma~\ref{qf11-lem2.20},
the following interesting identities:
\begin{align}
2^n\, A_{n} \big( \tfrac{1}{2}\big) &=
A_{2n}\, \big(A_{2n}-C_{2n}\big)-B_{2n}\, \big(2\, C_{2n}-D_{2n}-2\, E_{2n}\big)\\
&\phantom{=}+D_{2n}\, \big(2\, C_{2n}-D_{2n}-E_{2n}\big)-E_{2n}^{2},\nonumber\\
%\end{align}
%\begin{align}
2^n\,  B_{n} \big( \tfrac{1}{2}\big) &=
A_{2n}\, \big(B_{2n}-C_{2n}+E_{2n}\big)-B_{2n}\, \big(C_{2n}-E_{2n}\big)\\
&\phantom{=}+E_{2n}\, \big(2\, C_{2n}-D_{2n}-E_{2n}\big),\nonumber\\
2^n\,  C_{n} \big( \tfrac{1}{2}\big) &=
B_{2n}\, \big(B_{2n}-2\, C_{2n}-D_{2n}+E_{2n}\big)+C_{2n}\, \big(D_{2n}+E_{2n}\big)\\
&\phantom{=}+D_{2n}\, \big(A_{2n}-D_{2n}+E_{2n}\big)-E_{2n}^{2},\nonumber\\
2^n \, D_{n} \big( \tfrac{1}{2}\big) &=
A_{2n}\,  \big( B_{2n}+D_{2n}-C_{2n}\big) \\
&\phantom{=}+C_{2n}\, \big(C_{2n}-2\, B_{2n}+D_{2n}+E_{2n}\big)-D_{2n}\, \big(D_{2n}+E_{2n}\big),\nonumber\\
2^n\,  E_{n} \big( \tfrac{1}{2}\big) &=
C_{2n} \, \big( C_{2n}-A_{2n}-B_{2n}+D_{2n}\big)+B_{2n}\, \big(B_{2n}-D_{2n}\big) \\
&\phantom{=} + E_{2n}\, \big(A_{2n} +B_{2n} \big)-D_{2n}^{2}-E_{2n}^{2}.\nonumber
\end{align}
\end{remark}
%\bigskip

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\section{Some new recurrence formulas}\label{roz6}

The following recurrence formulas determined the coefficients of the polynomials
$A_{n}(\Delta)$, $B_{n}(\Delta)$, \ldots, $E_{n}(\Delta)$.


\begin{lemma}
Let $\Delta\in \mathbb{C}\setminus \{0\}$, $n\in \mathbb{N}$. Then the following recurrence
formulas hold
\begin{align}
(-1)^{n}\, & A_{n} (\Delta) =
\Delta^{n}\, \big( A_{n}\, (A_{n}-B_{n}-D_{n}) +
B_{n}\, (D_{n}+2\, E_{n}) +
C_{n}\, (D_{n}-C_{n}-E_{n}) - E_{n}^{2} \big) \nonumber\\
&+\sum_{k=0}^{n-1} \binom{n}{k}\, (-1)^{k+1}\, A_{k}(\Delta);\label{d1}%\\
\end{align}
\begin{align}
(-1)^{n}\, & B_{n} (\Delta) =
\Delta^{n}\, \big( C_{n}\, (A_{n}+2\,B_{n}-C_{n}-D_{n}-E_{n}) +
D_{n}\, (D_{n}+ E_{n}-A_{n}) - A_{n}\,B_{n} \big) \nonumber\\
&+\sum_{k=0}^{n-1} \binom{n}{k}\, (-1)^{k+1}\, B_{k}(\Delta);\label{d2}\\
(-1)^{n}\, & C_{n} (\Delta) =
\Delta^{n}\, \big( C_{n}\, (B_{n}-C_{n}-D_{n}+E_{n}) +
D_{n}\, (D_{n}-A_{n}) +E_{n}\, (A_{n}+B_{n}-E_{n}) \big) \nonumber\\
&+\sum_{k=0}^{n-1} \binom{n}{k}\, (-1)^{k+1}\, C_{k}(\Delta);\label{d3}\\
(-1)^{n}\, & D_{n} (\Delta) =
\Delta^{n}\, \big( B_{n}\, (B_{n}+C_{n}+E_{n}-A_{n}-D_{n}) +
E_{n}\, (A_{n}+D_{n} - C_{n}-E_{n}) - A_{n}\,D_{n} \big) \nonumber\\
&+\sum_{k=0}^{n-1} \binom{n}{k}\, (-1)^{k+1}\, D_{k}(\Delta);\label{d4}\\
(-1)^{n}\, & E_{n} (\Delta) =
\Delta^{n}\, \big( B_{n}\, (B_{n}+E_{n}-A_{n}-D_{n}) +
C_{n}\, (A_{n}-C_{n}) +E_{n}\, (2\, D_{n}-E_{n}) \big) \nonumber\\
&+\sum_{k=0}^{n-1} \binom{n}{k}\, (-1)^{k+1}\, E_{k}(\Delta).\label{d5}
\end{align}
\end{lemma}

\begin{proof}
%Let $\xi\in \mathbb{C}$, $\xi \neq 1$ and~$\xi^{11}=1$.
We have the equality:
$$
\tau_{1}(1)\, \tau_{3}(1) = -\sigma_{5}.
$$
Hence we get ($\delta\in \mathbb{C}\setminus\{0\}$):
$$%\begin{multline*}
\tau_{1}^{n}(1)\, \tau_{3}^{n}(1) =
\big( \delta - (\delta + \sigma_{5}) \big)^{n} =
\sum_{k=0}^{n} \binom{n}{k}\, (-1)^{k}\, \delta^{n-k}\, (\delta+\sigma_{5})^k=
\delta^{n}\, \sum_{k=0}^{n} \binom{n}{k}\, (-1)^{k}\,
\tau_{5}^{k} \big(\tfrac{1}{\delta}\big),
$$%\end{multline*}
which by~(\ref{2.2}), (\ref{3.6a}) and~(\ref{3.10a}), implies the identity
\begin{multline*}
\big(
A_{n}
+ B_{n}\, \sigma_{1}
+ C_{n}\, \sigma_{2}
+ D_{n}\, \sigma_{3}
+ E_{n}\, \sigma_{4}
\big)%\times \\
%\times
\big(
A_{n}
+ B_{n}\, \sigma_{3}
+ C_{n}\, \sigma_{5}
+ D_{n}\, \sigma_{2}
+ E_{n}\, \sigma_{1}
\big)= \\
=
\delta^{n} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k}
\big(
A_{n}(\tfrac{1}{\delta})
+ B_{n}(\tfrac{1}{\delta})\, \sigma_{5}
+ C_{n}(\tfrac{1}{\delta})\, \sigma_{1}
+ D_{n}(\tfrac{1}{\delta})\, \sigma_{4}
+ E_{n}(\tfrac{1}{\delta})\, \sigma_{2}
\big).
\end{multline*}
After the calculation of the left side of the above identity and the following substitution:
$$
\sigma_{3} =
-1 - \sigma_{1} - \sigma_{2} - \sigma_{4} - \sigma_{5}
$$
and making use of the linear independence over $\mathbb{Q}$
of the numbers $1$, $\sigma_{1}$,
$\sigma_{2}$, $\sigma_{4}$, $\sigma_{5}$
(see Corollary~\ref{qf2-cor1.4.2a}), we obtain:
\begin{multline*}
\delta^{n}\, \sum_{k=0}^{n} \binom{n}{k}\, (-1)^{k}\, A_{k} (\tfrac{1}{\delta}) =
A_{n}\, (A_{n}-B_{n}-D_{n}) +
B_{n}\, (D_{n}+2\, E_{n}) + \\
+ C_{n}\, (D_{n}-C_{n}-E_{n}) - E_{n}^{2}
\end{multline*}
etc., which, after the substitution of $\Delta:=1/\delta$ and minor transformations,
implies identities~(\ref{d1})--(\ref{d5}).
\end{proof}


\begin{corollary}
If $\deg \big( A_{n}(\Delta)\big) = n$ then
$$
\coeff \big( A_{n}(\Delta); \Delta^n \big) =
(-1)^n
\big(
A_{n}\, \big( A_{n} - B_{n} - D_{n} \big) +
B_{n}\, \big( D_{n} + 2\, E_{n} \big) +
C_{n}\, \big( D_{n} - C_{n} - E_{n} \big) -
E_{n}^{2}
\big)
$$
and
$$
\coeff \big( B_{n}(\Delta); \Delta^n \big) =
(-1)^n
\big(
C_{n}\, \big( A_{n} + 2\, B_{n} - C_{n} - D_{n} - E_{n} \big) +
D_{n}\, \big( D_{n} + E_{n} - A_{n} \big) -
A_{n}\, B_{n}
\big).
$$
\end{corollary}
\bigskip

\noindent
\textbf{Problem.}
\textit{
Is it true that
$$
\deg \big( A_{n}(\Delta)\big) = \deg \big( B_{n}(\Delta)\big) =
\deg \big( C_{n}(\Delta)\big) = \deg \big( D_{n}(\Delta)\big) =
\deg \big( E_{n}(\Delta)\big) = n
$$
for every $n=5,6,\ldots$?
}

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\section{Some convolution type identities}


\begin{lemma}
We have
$$
\tau_{1}(1)\,
\tau_{2}(1)\,
\tau_{3}(1)\,
=
1+\sigma_{1}+\sigma_{2}.
$$
Hence, for $\delta\in \mathbb{C}$, $\delta\, (1-\delta)\neq 0$, we obtain
\begin{multline*}
\tau_{1}^{n}(1)\,
\tau_{2}^{n}(1)\,
\tau_{3}^{n}(1)\,
=
\Big(
\big( 1-\delta +\sigma_{1} \big) +
\big( \delta +\sigma_{2} \big)
\Big)^{\! n} = {}\\
{} =
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\tau_{1}^{k} \big( 1/(1-\delta)\big)\,
\tau_{2}^{n-k} \big( 1/\delta \big),
\end{multline*}
which by~(\ref{2.2}), (\ref{3.4a}) and~(\ref{3.6a}),
implies the identity:
\begin{multline*}
\big(
A_n +
B_n\, \sigma_{1} +
C_n\, \sigma_{2} +
D_n\, \sigma_{3} +
E_n\, \sigma_{4}
\big) \,
\big(
A_n +
B_n\, \sigma_{2} +
C_n\, \sigma_{4} +
D_n\, \sigma_{5} +
E_n\, \sigma_{3}
\big) \times {} \\
{}\times
\big(
A_n +
B_n\, \sigma_{3} +
C_n\, \sigma_{5} +
D_n\, \sigma_{2} +
E_n\, \sigma_{1}
\big) = {}\\
{}=
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\Big(
A_k\big(\tfrac{1}{1-\delta}\big) +
B_k\big(\tfrac{1}{1-\delta}\big)\, \sigma_{1} +
C_k\big(\tfrac{1}{1-\delta}\big)\, \sigma_{2} +
D_k\big(\tfrac{1}{1-\delta}\big)\, \sigma_{3} {} \\
{} +
E_k\big(\tfrac{1}{1-\delta}\big)\, \sigma_{4}
\Big)\,
\Big(
A_{n-k}\big(\tfrac{1}{\delta}\big) +
B_{n-k}\big(\tfrac{1}{\delta}\big)\, \sigma_{2} + %{} \\
%{}+
C_{n-k}\big(\tfrac{1}{\delta}\big)\, \sigma_{4} +
D_{n-k}\big(\tfrac{1}{\delta}\big)\, \sigma_{5} +
E_{n-k}\big(\tfrac{1}{\delta}\big)\, \sigma_{3}
\Big).
\end{multline*}
Hence, after some calculation, by comparing the absolute terms and the coefficients
in the expressions $\sigma_{1}$, $\sigma_{2}$, $\sigma_{3}$ and $\sigma_{4}$
we get the following five independent identities:
\begin{multline}
A_{n}^{3} + B_{n}^{3} + 2\, C_{n}^{3} + D_{n}^{3} + E_{n}^{3}
- A_{n}^{2}\, \big( C_{n} + D_{n} \big)
- B_{n}^{2}\, \big( A_{n} + D_{n} \big)
+ C_{n}^{2}\, \big( D_{n} - A_{n} - B_{n} - 2\, E_{n}  \big) + {} \\
{}+ D_{n}^{2}\, \big( 2\, E_{n} - A_{n} - B_{n} - 3\, C_{n} \big)
- E_{n}^{2}\, \big( A_{n} + 2\, B_{n} \big)
+ A_{n}\, B_{n}\, \big( D_{n} + 2\, E_{n} - C_{n} \big) + {} \\
{}+ A_{n}\, D_{n}\, \big( 3\, C_{n} - 2\, E_{n} \big)
+ B_{n}\, C_{n}\, \big( 2\, D_{n} + E_{n} \big)
+ D_{n}\, E_{n}\, \big( B_{n} - 2\, C_{n} \big) = {} \\
{}=
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\Big[
A_{k}\big(\tfrac{1}{1-\delta}\big)\,
\Big( A_{n-k}\big(\tfrac{1}{\delta}\big) -
D_{n-k}\big(\tfrac{1}{\delta}\big) \Big) + {} \\
{}+
C_{k}\big(\tfrac{1}{1-\delta}\big)\,
\Big( 2\, B_{n-k}\big(\tfrac{1}{\delta}\big) -
 C_{n-k}\big(\tfrac{1}{\delta}\big) -
 E_{n-k}\big(\tfrac{1}{\delta}\big)\Big)
- B_{k}\big(\tfrac{1}{1-\delta}\big)\,
\Big(
D_{n-k}\big(\tfrac{1}{\delta}\big) +
C_{n-k}\big(\tfrac{1}{\delta}\big)
\Big) +{}\\
{}+D_{k}\big(\tfrac{1}{1-\delta}\big)\,
\Big(E_{n-k}\big(\tfrac{1}{\delta}\big) -
B_{n-k}\big(\tfrac{1}{\delta}\big) \Big) +
E_{k}\big(\tfrac{1}{1-\delta}\big)\,
\Big( 2\,  C_{n-k}\big(\tfrac{1}{\delta}\big) -
 B_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
\Big],
\end{multline}

%3
\begin{multline}
B_{n}^3 - C_{n}^3 - D_{n}^3 + E_{n}^3
+ A_{n}^2\,\big( C_{n} + D_{n} - E_{n} \big)
+ B_{n}^2\, \big( C_{n} - A_{n} \big)
+ C_{n}^2 \, \big( A_{n} + D_{n} - E_{n} \big) + {}\\
{}+ D_{n}^2\, \big( C_{n} - B_{n} + E_{n} \big)
- E_{n}^{2}\, \big(2\, B_{n} + 2\, D_{n} \big)
+ A_{n}\, B_{n}\, \big( E_{n} - 2\, C_{n} - D_{n} \big) + {} \\
{} + D_{n}\, E_{n}\, \big( A_{n} + 3\, B_{n} + C_{n} \big)
- C_{n}\, \big( A_{n}\, E_{n} + B_{n}\, D_{n}\big)
= {} \\
{}=
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\Big[
A_{k}\big(\tfrac{1}{1-\delta}\big)\, B_{n-k}\big(\tfrac{1}{\delta}\big)
+ B_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( E_{n-k}\big(\tfrac{1}{\delta}\big) -
A_{n-k}\big(\tfrac{1}{\delta}\big) - B_{n-k}\big(\tfrac{1}{\delta}\big) \Big) + {} \\
{} + C_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( A_{n-k}\big(\tfrac{1}{\delta}\big) +
C_{n-k}\big(\tfrac{1}{\delta}\big) - E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
+ D_{k}\big(\tfrac{1}{1-\delta}\big)\, \big( D_{n-k}\big(\tfrac{1}{\delta}\big) -
B_{n-k}\big(\tfrac{1}{\delta}\big) - C_{n-k}\big(\tfrac{1}{\delta}\big) \Big) + {}\\
{}+ E_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( B_{n-k}\big(\tfrac{1}{\delta}\big) -
E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
\Big],
\end{multline}

%4
\begin{multline}
E_{n}^3 - D_{n}^3
+ A_{n}^2\, D_{n} +
B_{n}^2\, \big( 3\, C_{n} - A_{n}  - 2\, E_{n} \big)
- C_{n}^2\, B_{n}
+ D_{n}^2\, \big( A_{n} + B_{n} + C_{n} + E_{n} \big) + {}\\
{}+ A_{n}\, E_{n}\, \big( C_{n} - 2\, D_{n} \big)
-E_{n}^{2} \,  \big( 3\, C_{n} + 2\, D_{n} \big)
+ B_{n}\, E_{n}\, \big( 2\, C_{n} + 4\, D_{n} - A_{n} \big)
+ C_{n}\, D_{n}\, \big( 2\, E_{n} - 5\, B_{n} \big)
= {} \\
{}=
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\Big[
A_{k}\big(\tfrac{1}{1-\delta}\big)\, E_{n-k}\big(\tfrac{1}{\delta}\big)
+ B_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( C_{n-k}\big(\tfrac{1}{\delta}\big) -
A_{n-k}\big(\tfrac{1}{\delta}\big) \Big) + {} \\
{}+ C_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( D_{n-k}\big(\tfrac{1}{\delta}\big) -
E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
+ D_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( A_{n-k}\big(\tfrac{1}{\delta}\big) -
B_{n-k}\big(\tfrac{1}{\delta}\big) - C_{n-k}\big(\tfrac{1}{\delta}\big) +
D_{n-k}\big(\tfrac{1}{\delta}\big) \Big) +{} \\
{}+ E_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( C_{n-k}\big(\tfrac{1}{\delta}\big) -
D_{n-k}\big(\tfrac{1}{\delta}\big) - E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
\Big],
\end{multline}


%5
\begin{multline}
B_{n}^3
+ A_{n}^2\, \big( C_{n} - B_{n} \big)
+ B_{n}^2\, \big( C_{n} - A_{n} - D_{n} - E_{n} \big)
+ C_{n}^2\, \big( D_{n} - 3\, B_{n} \big)
+ D_{n}^2\, \big( E_{n} - C_{n} \big) + {} \\
{}+ E_{n}^2\, \big( A_{n} - 3\, D_{n} \big)
+ A_{n}\, C_{n}\, \big( 2\, B_{n} + D_{n} - 3\, E_{n} \big)
+ B_{n}\, C_{n}\, \big( E_{n} - D_{n} \big) + {} \\
+ D_{n}\, E_{n}\, \big(5\, B_{n} + C_{n} \big)
=
\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
\Big[
A_{k}\big(\tfrac{1}{1-\delta}\big)\, C_{n-k}\big(\tfrac{1}{\delta}\big) + {} \\
{} + B_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( D_{n-k}\big(\tfrac{1}{\delta}\big) -
A_{n-k}\big(\tfrac{1}{\delta}\big) - B_{n-k}\big(\tfrac{1}{\delta}\big) +
E_{n-k}\big(\tfrac{1}{\delta}\big) \Big) + {} \\
{} + C_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( B_{n-k}\big(\tfrac{1}{\delta}\big) +
D_{n-k}\big(\tfrac{1}{\delta}\big) - E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
- D_{k}\big(\tfrac{1}{1-\delta}\big)\, B_{n-k}\big(\tfrac{1}{\delta}\big) + {} \\
{} + E_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( A_{n-k}\big(\tfrac{1}{\delta}\big) -
D_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
\Big],
\end{multline}
etc.

%6
%\begin{multline}
%B_{n}^3 - C_{n}^3 + E_{n}^3
%+ A_{n}^2\, \big( C_{n} - B_{n} + D_{n} - E_{n} \big)
%+ B_{n}^2\, \big( C_{n} - A_{n} - 2\, D_{n} - E_{n} \big)
%+ C_{n}^2\, \big( 2\, E_{n} - D_{n} \big) + {} \\
%{} + D_{n}^2\, \big( B_{n} + C_{n} - 2\, E_{n} \big)
%- E_{n}^2\, \big( B_{n} + C_{n} + 2\ D_{n} \big)
%- B_{n}\, C_{n}\, \big( 2\, D_{n}  E_{n} \big)
%- A_{n}\, C_{n}\, \big( D_{n} + E_{n} \big) + {} \\
%{}+ A_{n}\, B_{n}\, \big( C_{n} + D_{n} \big)
%+ D_{n}\, E_{n}\, \big( 2\, A_{n}  + 4\, B_{n} + 3\, C_{n} \big)
%= {} \\
%{}=
%\sum_{k=0}^{n} \binom{n}{k} (1-\delta)^{k}\, \delta^{n-k}\,
%\Big[
%A_{k}\big(\tfrac{1}{1-\delta}\big)\, D_{n-k}\big(\tfrac{1}{\delta}\big) + {} \\
%{}+ B_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( C_{n-k}\big(\tfrac{1}{\delta}\big) -
%A_{n-k}\big(\tfrac{1}{\delta}\big) - B_{n-k}\big(\tfrac{1}{\delta}\big) +
%D_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
%+ C_{k}\big(\tfrac{1}{1-\delta}\big)\, C_{n-k}\big(\tfrac{1}{\delta}\big) + {} \\
%{}+ D_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( E_{n-k}\big(\tfrac{1}{\delta}\big) -
%C_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
%+ E_{k}\big(\tfrac{1}{1-\delta}\big)\, \Big( B_{n-k}\big(\tfrac{1}{\delta}\big) -
%D_{n-k}\big(\tfrac{1}{\delta}\big) - E_{n-k}\big(\tfrac{1}{\delta}\big) \Big)
%\Big].
%\end{multline}
\end{lemma}



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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Polynomials associated with quasi-Fibonacci\protect{\newline}
 numbers of order~11}

By taking into account the Chebyshev polynomials $T_n(x)$  again we obtain,
for $\tau_{k}=\tau_{k}(\delta)$,  $k=1,2,\ldots,5$:
\begin{equation}\label{wz2.47}
\left\{\!\!
\begin{array}{ll}
\sigma_{k}&\!\!\!\!= 2\cos\frac{2k\pi}{11}=\frac{1}{\delta}(\tau_{k}-1), \\[1ex]
\sigma_{2k} &\!\!\!\!= 2\cos\frac{4k\pi}{11}=2T_2\big(\cos\frac{2k\pi}{11}\big)=
2T_2\big( \frac{1}{2\delta}(\tau_{k}-1) \big)
=\frac{1}{\delta ^2}(\tau_{k}^2-2\tau_{k}+1-2\delta ^2), \\[1ex]
\sigma_{3k} &\!\!\!\!= 2\cos\frac{6k\pi}{11}=2T_3\big( \cos\frac{2k\pi}{11}\big)=
2T_3\big( \frac{1}{2\delta}(\tau_{k}-1) \big)=
\\[1ex]
&\!\!\!\!=
\frac{1}{\delta ^3}\big(\tau_{k}^3-3\tau_{k}^2+(3-3\delta ^2)\tau_{k}-1+3\delta ^2 \big), \\[1ex]
\sigma_{4k}  &\!\!\!\!= 2\cos\frac{8k\pi}{11}=2T_4\big( \cos\frac{2k\pi}{11}\big)=
2T_4\big( \frac{1}{2\delta}(\tau_{k}-1) \big)= \\[1ex]
&\!\!\!\!= \frac{1}{\delta ^4}\big(\tau_{k}^4-4\tau_{k}^3+(6-4\delta ^2)\tau_{k}^2+
(-4+8\delta ^2)\tau_{k}+1-4\delta ^2+2\delta ^4\big), \\[1ex]
\sigma_{5k}  &\!\!\!\!= 2\cos\frac{10k\pi}{11}=2T_5\big(\cos\frac{2k\pi}{11}\big)
= \frac{1}{\delta ^5}\big(\tau_{k}^5-5\tau_{k}^4+(10-5\delta ^2)\tau_{k}^3+\\[1ex]
&\phantom{=}\!\!\!\!+(-10+15\delta ^2)\tau_{k}^2+
(10-30\delta ^2+5\delta ^4)\tau_{k}-6+20\delta ^2-5\delta ^4\big).
\end{array}
\right.
\end{equation}

Therefore, immediately from equation (\ref{2.1}) we obtain the equality:
\begin{align*}
\tau_{k}^{n} =& A_n(\delta )+\frac{1}{\delta}B_n(\delta )(\tau_{k}-1)+
\frac{1}{\delta ^2}C_n(\delta )(\tau_{k}^{2}-2\tau_{k}+1-2\delta ^2) \\
&+\frac{1}{\delta ^3}D_n(\delta )(\tau_{k}^{3}-3\tau_{k}^{2}+(3-3\delta ^2)\tau_{k}-1+3\delta ^2) \\
&+\frac{1}{\delta ^4}E_n(\delta )(\tau_{k}^{4}-4\tau_{k}^{3}+(6-4\delta ^2)\tau_{k}^{2}+
(-4+8\delta ^2)\tau_{k}+1-4\delta ^2+2\delta ^4),
\end{align*}
i.e.,
\begin{align*}
W_{n,11}&(\tau_{k};\delta ) := \tau_{k}^{n}-\tfrac{1}{\delta ^4}E_n(\delta )\tau_{k}^{4}+
  \big[ -\tfrac{1}{\delta ^3}D_n(\delta )+\tfrac{4}{\delta ^4}E_n(\delta )\big] \tau_{k}^{3} \\
&+\big[ \tfrac{1}{\delta ^4}(-6+4\delta ^2)E_n(\delta )+\tfrac{3}{\delta ^3}D_n(\delta )-
 \tfrac{1}{\delta ^2}C_n(\delta )\big]\tau_{k}^{2} \\
&+\big[\tfrac{1}{\delta ^4}(4-8\delta ^2)E_n(\delta )+
 \tfrac{1}{\delta ^3}(-3+3\delta ^2)D_n(\delta )+\tfrac{2}{\delta ^2}C_n(\delta )-
 \tfrac{1}{\delta}B_n(\delta )\big] \tau_{k} \\
&+\big[\tfrac{1}{\delta ^4}(-1+4\delta ^2-2\delta ^4)E_n(\delta )+
 \tfrac{1}{\delta ^3}(1-3\delta ^2)D_n(\delta )+
 \tfrac{1}{\delta ^2}(-1+2\delta ^2)C_n(\delta ) \\
&
+\tfrac{1}{\delta}B_n(\delta )-A_n(\delta )\big] =0,
\end{align*}
which means that:
$$
p_{11}(x;\delta )\Big| W_{n,11}(x;\delta ),\quad n\geqslant 5.
$$
For example for $\delta=1$, we have the relation:
\begin{multline*}
(x^5-4x^4+2x^3+5x^2-2x-1)\bigg|\Big[ x^n-E_nx^4+(4E_n-D_n)x^3+ \\
+(-2E_n+3D_n-C_n)x^2+(-4E_n+2C_n-B_n)x+E_n-2D_n+C_n-A_n\Big]
\end{multline*}
for all $n\geqslant 5$.

More precisely, the following decomposition can be deduced:

\begin{lemma}
We have:
\begin{equation}
p_{11}(x;\delta )\Big(\sum\limits_{k=1}^{n-4}E_k(\delta )x^{n-4-k}\Big) =W_{n,11}(x;\delta ).
\end{equation}
For $\delta=1$ we obtain special decomposition:
\begin{multline}\label{1000}
(x^5-4x^4+2x^3+5x^2-2x-1)\bigg(\sum\limits_{k=3}^{n-2}D_kx^{n-2-k}\bigg)= \\
=x^n-E_nx^4+(4E_n-D_n)x^3+(-2E_n+3D_n-C_n)x^2+ \\
+(-4E_n+2C_n-B_n)x+E_n-2D_n+C_n+B_n-A_n,
\end{multline}
and, after differentiating of~(\ref{1000}):
\begin{multline}
(5x^4-16x^3+6x^2+10x-2)\bigg(\sum\limits_{k=3}^{n-2}D_k x^{n-2-k}\bigg)+\\
+(x^5-4x^4+2x^3+5x^2-2x-1)\bigg(\sum\limits_{k=3}^{n-2}D_k x^{n-2-k}\bigg)'= \\
=n\, x^{n-1}-4E_nx^3+3(4E_n-D_n)x^2+2(-C_n+3D_n-2E_n)x%- \\
-B_n+2C_n-4E_n.
\end{multline}
Hence, by applying~(\ref{1000}), we get
\begin{multline}
(x^5-4x^4+2x^3+5x^2-2x-1)^2\bigg(\sum\limits_{k=3}^{n-3}(n-2-k)D_k x^{n-3-k}\bigg)= \\
=\big(x^5-4x^4+2x^3+5x^2-2x-1\big)\Big(n\, x^{n-1}-4E_nx^3+3(4E_n-D_n)x^2+ \\
+2(-C_n+3D_n-2E_n)x-B_n+2C_n-4E_n\Big)+\\
+\big({-}5x^4+16x^3-6x^2-10x+2\big)\Big( x^n-E_nx^4+(4E_n-D_n)x^3+(-2E_n+3D_n-C_n)x^2+ \\
+(-4E_n+2C_n-B_n)x+E_n-2D_n+C_n+B_n-A_n\Big).
\end{multline}
\end{lemma}

\begin{corollary}\label{cor4.2}
We have:
\begin{align}
&\sum_{k=3}^{n-2} D_k = -A_{n}+2C_{n}-2E_{n}+1,\\
(-1)^{n+1}&\sum_{k=3}^{n-2} (-1)^k D_k = (-1)^n -A_{n}+2B_{n}-2C_{n}+2D_{n}-2E_{n},\\
%\end{align}
%\begin{align}
&\sum_{k=3}^{n-3} (n-2-k) D_k = n -3 +3A_{n}-B_{n}-6C_{n}+3D_{n}+6E_{n},\\
(-1)^{n+1}&\sum_{k=3}^{n-3} (-1)^k (n-2-k) D_k = (-1)^n (n-15) +15A_{n}-29B_{n}+\\
&\phantom{===}+26C_{n}-21D_{n}+14E_{n},\nonumber\\
\intertext{and}
&\sum_{k=3}^{n-2} k D_k = 1 -(n+1)A_{n}+B_{n}+2(n+1)C_{n}-3D_{n}-2(n+1)E_{n}.
\end{align}
\end{corollary}



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%%%%%%%%%%%%%%               %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%   Section 9   %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%               %%%%%%%%%%%%%%%%%%
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\section{Some properties of zeros of polynomials $\mathcal{D}_n(x)$}


Let us set:
$$
\mathcal{D}_n(x)=\sum\limits_{k=1}^{n}D_kx^{n-k},\quad n\in
{\mathbb N}.
$$
Then,
\begin{align*}
\mathcal{D}_1(x)&\equiv 0,&
\mathcal{D}_2(x)&\equiv 0,&
\mathcal{D}_3(x)&\equiv 1,\\
\mathcal{D}_4(x)&=x+4,&
\mathcal{D}_5(x)&=x^2+4x+14,
\end{align*}
and we obtain the recurrence relations (see~(\ref{2.4})):
\begin{multline*}
\null\quad\mathcal{D}_{n+5}(x)=x^{n+2}+4\mathcal{D}_{n+4}(x)-2\mathcal{D}_{n+3}(x)
-5\mathcal{D}_{n+2}(x)+\\
+2\mathcal{D}_{n+1}(x)+\mathcal{D}_{n}(x), \qquad n\geqslant 1.\quad\null
\end{multline*}
All polynomials $\mathcal{D}_{n}(x)$ for $n=2k+1$, $k\in\mathbb{N}$,
$k\geqslant 2$, have
an even degree, for example:
\begin{align*}
\mathcal{D}_7(x)&=x^4+4x^3+14x^2+43x+126, \\
\mathcal{D}_9(x)&=x^6+4x^5+14x^4+43x^3+126x^2+357x+993.
\end{align*}
As follows from calculation, these polynomials have no real
roots for $n\leqslant 200$.

However, all polynomials $\mathcal{D}_{n}(x)$ for $n=2k$, $k\in\mathbb{N}$,
$k\geqslant 2$, have an odd degree, for example:
\begin{align*}
\mathcal{D}_6(x)&=x^3+4x^2+14x+43, \\
\mathcal{D}_8(x)&=x^5+4x^4+14x^3+43x^2+126x+357.
\end{align*}


\begin{remark}
Polynomials $\mathcal{D}_{2k}(x)$ have exactly one real root $s_k$, which is less than
zero (for $k\leqslant 100$). The roots~$s_k$ form an increasing sequence from
$s_{2}=-4$ to $s_{100}=-2.699746$ (by numerical calculation).
All numerical calculations were performed using 
\textit{Mathematica}.\footnote{\textit{Mathematica\/} is registered trademark of
Wolfram Research~Inc.}
\end{remark}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%                %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%   Section 10   %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%                %%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Jordan decomposition}\label{roz-jordan}


For sequences $A_{n}(\delta)$, $B_{n}(\delta)$,
$C_{n}(\delta)$, $D_{n}(\delta)$ and $E_{n}(\delta)$
we have the identity:
\begin{equation}
\left[
\begin{array}{l}
A_{n+1}(\delta) \\
B_{n+1}(\delta) \\
C_{n+1}(\delta) \\
D_{n+1}(\delta) \\
E_{n+1}(\delta) \\
\end{array}
\right]
=
\mathcal{W}(\delta)
\left[
\begin{array}{l}
A_{n}(\delta) \\
B_{n}(\delta) \\
C_{n}(\delta) \\
D_{n}(\delta) \\
E_{n}(\delta) \\
\end{array}
\right],
\qquad n\in \mathbb{N},
\end{equation}
where
\begin{equation}
\mathcal{W}(\delta)=
\left[
\begin{array}{ccccc}
1 & 2\, \delta & 0 & 0 & -\delta \\
\delta & 1 & \delta & 0 & - \delta \\
0 & \delta & 1 & \delta & -\delta \\
0 & 0 & \delta & 1 & 0 \\
0 & 0 & 0 & \delta & 1-\delta \\
\end{array}
\right].
\end{equation}
Matrix $\mathcal{W}(\delta)$ is diagonalizable, because of the following decomposition:
\begin{equation}
\mathcal{W}(\delta)=
A\cdot
\diag \big[
\tau_{1}(\delta),
\tau_{2}(\delta),
\ldots,
\tau_{5}(\delta)
\big]
\cdot A^{-1},
\end{equation}
where:
\begin{equation*}
A =
\big[ K_1, K_2, K_3, K_4, K_5 \big]
\end{equation*}
and
\begin{equation*}
K_m= \big[
\sigma_{m}^{-4} + 2\, \sigma_{m}^{-3} - 2\, \sigma_{m}^{-2} -2\, \sigma_{m}^{-1},
\sigma_{m}^{-2}-\sigma_{m}^{-1}-1,
-\sigma_{m}^{-1}-\sigma_{m}^{-2},
-\sigma_{m}^{-3}-\sigma_{m}^{-2},
-\sigma_{m}^{-3}
\big]^{T},
\end{equation*}
for every $m=1,2,\ldots,5$. We note, that:
\begin{equation*}
A^{-1} = \frac{1}{11}\,
\big[ L_1, L_2, L_3, L_4, L_5 \big]
\cdot
\left[
\begin{array}{rrrrr}
11 & -3 & -20 & 8 & 16 \\[1ex]
 -2 & 2 & 7 & -7 & -11 \\[1ex]
 -13 & 7 & 22 & -15 & -17 \\[1ex]
 1 & -1 & -3 & 3 & 4 \\[1ex]
 3 & -2 & -5 & 4 & 4
\end{array}
\right],
\end{equation*}
where
\begin{equation*}
L_m=
\big[
\sigma_{1}^{m+3},
\sigma_{2}^{m+3},
\sigma_{3}^{m+3},
\sigma_{4}^{m+3},
\sigma_{5}^{m+3}
\big]^{T},
\end{equation*}
for $m=1,2,\ldots,5$.

Notice, that the characteristic polynomial $w(\lambda)$ of the matrix $\mathcal{W}(\delta)$
has the form:
$$
w(\lambda) =  p_{11} \Big(\lambda; \delta \Big).
$$

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Final remarks}\label{finalsection}



Quasi-Fibonacci numbers of order ($7$, $\delta$) and
($11$, $\delta$)
constitute a special case
of the so-called quasi-Fibonacci numbers of order ($k$, $\delta$),
where $k$ is an odd positive integer, $k\geq 5$ and $\delta\in \mathbb{C}$.
These numbers are defined to be the elements of the following sequences
of polynomials:
$$
\big\{ a_{n,i}(\delta)\big\}_{n\in \mathbb{N}}\subset \mathbb{Z}[\delta],
\qquad i=1,2,\ldots,\frac{1}{2}\, \varphi(k),
$$
where $\varphi$ is the Euler's totient function,
which are determined by the following relations:
\begin{equation}\label{w-ig}
\big( 1+\delta\, (\xi^l + \xi^{k-l}) \big)^{n} =
a_{n,1}(\delta) +
\sum_{\genfrac{}{}{0pt}{}{i\in \{2,3,\ldots,(k-1)/2\}}{(i,k)=1}}
a_{n,f(i)}(\delta)\, (\xi^{i\, l} + \xi^{i\,(k-l)})
\end{equation}
for $l=1,2,\ldots, \frac{k-1}{2}$, $(l,k)=1$, $n\in \mathbb{N}$,
where $\xi\in \mathbb{C}$ is a~primitive root of unity of order~$k$
and where $f$ is the increasing bijection of the set
$\{i\in \mathbb{N}:\ 2\leq i\leq (k-1)/2\ \mbox{ and }\ (i,k)=1\}$
onto the set $\{2,3,\ldots,\varphi(k)/2\}$.
In the sequel, if $k$ is a prime number, relations~(\ref{w-ig})
have the following simpler form:
\begin{equation}\label{w-ig2}
\big( 1+\delta\, (\xi^l + \xi^{k-l}) \big)^{n} =
a_{n,1}(\delta) +
\sum_{i=1}^{(k-3)/2}
a_{n,i+1}(\delta)\, (\xi^{i\, l} + \xi^{i\,(k-l)})
\end{equation}
for $l=1,2,\ldots, \frac{k-1}{2}$ and $n\in \mathbb{N}$.

Quasi-Fibonacci numbers of order ($k$, $\delta$) possess many interesting properties
and natural applications. We remark that these are for the most cases
derived immediately from relations~(\ref{w-ig}).
For example the quasi-Fibonacci numbers enable us to describe
the coefficients of the following polynomials
 in the pure algebraic language:
$$
\prod_{\genfrac{}{}{0pt}{}{l\in \{1,2,\ldots,k\}}{(l,2k+1)=1}}
\Big ( \mathbb{X} - \cos^n \big( \tfrac{2\, l\, \pi}{2\, k+1}\big) \Big),\qquad
\prod_{\genfrac{}{}{0pt}{}{l\in \{1,2,\ldots,k\}}{(l,2k+1)=1}}
\Big ( \mathbb{X} - \sin \big( \tfrac{2\, l\, \pi}{2\, k+1}\big)\,
\cos^n \big( \tfrac{2\, l\, \pi}{2\, k+1}\big) \Big),
$$
etc. (see for example~\cite{wsw1,ws2}).
It is possible also to do it for
many other trigonometric sums,
for example, in~\cite{ws2} the following identity was derived
(plus twelve others like this one):
$$
2\,\cos ( \tfrac{2\, \pi}{7})\, \sqrt[3]{2\cos ( \tfrac{4\, \pi}{7})} +
2\,\cos ( \tfrac{4\, \pi}{7})\, \sqrt[3]{2\cos ( \tfrac{8\, \pi}{7})} +
2\,\cos ( \tfrac{8\, \pi}{7})\, \sqrt[3]{2\cos ( \tfrac{2\, \pi}{7})} =
\sqrt[3]{-2-3\,\sqrt[3]{49}};
$$
%\begin{multline*}
%2\, \cos \big( \tfrac{2\, \pi}{7} \big)\,
%\sqrt[3]{2\, \cos \big( \tfrac{8\, \pi}{7} \big)\,} +
%2\, \cos \big( \tfrac{4\, \pi}{7} \big)\,
%\sqrt[3]{2\, \cos \big( \tfrac{2\, \pi}{7} \big)\,} +
%2\, \cos \big( \tfrac{8\, \pi}{7} \big)\,
%\sqrt[3]{2\, \cos \big( \tfrac{4\, \pi}{7} \big)\,} = {} \\
%{}=
%\sqrt[3]{5-3\, \sqrt[3]{7}\, \big(\sqrt[3]{7}-1\big)\,},
%\end{multline*}
which is a~variation of the known identity of Ramanujan (see~\cite{qf2-lit3}):
$$
\sqrt[3]{\cos \big( \tfrac{2\, \pi}{7} \big)\,}+
\sqrt[3]{\cos \big( \tfrac{4\, \pi}{7} \big)\,}+
\sqrt[3]{\cos \big( \tfrac{8\, \pi}{7} \big)\,}=
\sqrt[3]{\tfrac{1}{2}\, \big( 5-3\, \sqrt[3]{7}\big)\,}.
$$




It seems almost definite that some types of the identities will be characteristic
of only a certain kind of the quasi-Fibonacci numbers
of order ($k$, $\delta$),
with respect to the odd $k\in \mathbb{N}$, $k\geq 5$. Accordingly, it is relevant
to analyze these numbers separately and independently, for different values of~$k$.

We note that quasi-Fibonacci numbers of order ($5$, $1$) are equal
to the classical Fibonacci numbers.
Moreover, for simplicity of notation,
the quasi-Fibonacci numbers of order ($11$, $\delta$) discussed
in our paper are denoted by
(see the relations~(\ref{w-ig2}) above for $k=11$ and the relations ($3.2m$) for $m=1,2,3,4,5$
in Section~\ref{roz3}):
$$
\begin{array}{lll}
A_{n}(\delta):=a_{n,1}(\delta)  & \qquad
B_{n}(\delta):=a_{n,2}(\delta)  & \qquad
C_{n}(\delta):=a_{n,3}(\delta)   \\
D_{n}(\delta):=a_{n,4}(\delta)  & \qquad
E_{n}(\delta):=a_{n,5}(\delta).  &   \\
\end{array}
$$

\section{Acknowledgements}

The authors wish to thank the reviewer for his valuable criticisms
and suggestions, leading to the present improved version of our
paper.

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\section{Tables}

\begin{table}[H]
\begin{center}
{\small \caption{}\label{qf2-tab-d2}
\medskip

\begin{tabular}{||c||l||l||} \hline\hline
$n$ & $A_n(\delta )=A_{11,n}(\delta)$
& $D_n(\delta )=D_{11,n}(\delta)$ \\ \hline\hline $0$ & $1$
& $0$                                     \\ \hline $1$ & $1$
& $0$                                     \\ \hline $2$ & $2\delta
^2+1$
& $0$                                     \\ \hline $3$ & $6\delta
^2+1$
& $\delta ^3$                             \\ \hline $4$ & $6\delta
^4+12\delta ^2+1$
& $4\delta ^3$                            \\ \hline $5$ & $-\delta
^5+30\delta ^4+20\delta ^2+1$
& $4\delta ^5+10\delta ^3$                \\ \hline $6$ & $19\delta
^6-6\delta ^5+90\delta ^4+30\delta ^2+1$
& $-\delta ^6+24\delta ^5+20\delta ^3$    \\ \hline $7$ & $-7\delta
^7+133\delta ^6-21\delta ^5+210\delta ^4+42\delta ^2+1$
& $14\delta ^7-7\delta ^6+84\delta ^5+35\delta ^3$   \\ \hline\hline
$n$ & $B_n(\delta )=B_{11,n}(\delta)$
& $E_n(\delta )=E_{11,n}(\delta)$ \\ \hline\hline $0$ & $0$
& $0$                                      \\ \hline $1$ & $\delta$
& $0$                                      \\ \hline $2$ & $2\delta$
& $0$                                      \\ \hline $3$ & $3\delta
^3+3\delta$
& $0$                                      \\ \hline $4$ & $12\delta
^3+4\delta$
& $\delta ^4$                              \\ \hline $5$ & $9\delta
^5+30\delta ^3+5\delta$
& $-\delta ^5+5\delta ^4$                  \\ \hline $6$ & $-\delta
^6+54\delta ^5+60\delta ^3+6\delta$
& $5\delta ^6-6\delta ^5+15\delta ^4$      \\ \hline $7$ & $28\delta
^7-7\delta ^6+189\delta ^5+105\delta ^3+7\delta$
& $-6\delta ^7+35\delta ^6-21\delta ^5+35\delta ^4$   \\
\hline\hline $n$ &  \multicolumn{2}{l||}{$C_n(\delta
)=C_{11,n}(\delta)$}      \\ \hline\hline $0$ &
\multicolumn{2}{l||}{$0$}
\\ \hline $1$ & \multicolumn{2}{l||}{$0$}
\\ \hline $2$ & \multicolumn{2}{l||}{$\delta ^2$}
\\ \hline $3$ & \multicolumn{2}{l||}{$3\delta ^2$}
\\ \hline $4$ & \multicolumn{2}{l||}{$4\delta ^4+6\delta ^2$}
\\ \hline $5$ & \multicolumn{2}{l||}{$-\delta ^5+20\delta
^4+10\delta ^2$}                            \\ \hline $6$ &
\multicolumn{2}{l||}{$14\delta ^6-6\delta ^5+60\delta ^4+15\delta
^2$}                \\ \hline $7$ & \multicolumn{2}{l||}{$-7\delta
^7+98\delta ^6-21\delta ^5+140\delta ^4+21\delta ^2$}  \\
\hline\hline
\end{tabular}
}
\end{center}
\end{table}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{table}[H]

%\mbox{\null}\hfill%
\begin{minipage}{219pt}
\caption{}\label{qf2-tab3}
\smallskip

{\small
\begin{tabular}{||c|c|c|c|c|c||}\hline\hline
$n$ & $A_n$ & $B_n$ & $C_n$ & $D_n$ & $E_n$ \\ \hline\hline
0  & 1     & 0     & 0     & 0    & 0 \\ \hline
1  & 1     & 1     & 0     & 0    & 0 \\ \hline
2  & 3     & 2     & 1     & 0    & 0 \\ \hline
3  & 7     & 6     & 3     & 1    & 0 \\ \hline
4  & 19    & 16    & 10    & 4    & 1 \\ \hline
5  & 50    & 44    & 29    & 14   & 4 \\ \hline
6  & 134   & 119   & 83    & 43   & 14 \\ \hline
7  & 358   & 322   & 231   & 126  & 43 \\ \hline
8  & 959   & 868   & 636   & 357  & 126 \\ \hline
9  & 2569  & 2337  & 1735  & 993  & 357 \\ \hline
10 & 6886  & 6284  & 4708  & 2728 & 993 \\ \hline
11 & 18461 & 16885 & 12727 & 7436 & 2728 \\ \hline\hline
\end{tabular}
}
\end{minipage}
\hfill
\begin{minipage}{229pt}
\caption{}\label{qf2-tab3b}
\smallskip

{\small
\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & ${\cal A}_n(1)$ & ${\cal A}_n(1/2)$ & ${\cal A}_n(-1/2)$ \\ \hline\hline
0  &         5   &   5           & 5              \\ \hline
1  &         4   &   9/2         & 11/2           \\ \hline
2  &         12  &   25/4        & 33/4           \\ \hline
3  &         25  &   39/4        & 55/4           \\ \hline
4  &         64  &   257/16      & 385/16         \\ \hline
5  &         159 &   437/16      & 693/16         \\ \hline
6  &         411 &   1517/32     & 2541/32        \\ \hline
7  &         1068&   2671/32     & 4719/32        \\ \hline
8  &         2808&   38017/256   & 70785/256      \\ \hline
9  &         7423&   68169/256   & 133705/256     \\ \hline
10 &        19717&  245935/512   & 508079/512     \\ \hline
11 &        52529&  1782735/2048 & 3879865/2048   \\ \hline\hline
\end{tabular}
}
\end{minipage}
%\hfill\mbox{\null}%

\end{table}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{table}[H]
\begin{center}
{\small
\caption{}\label{qf2-tab4}
\medskip

\begin{tabular}{||c|c|c|c|c|c||}\hline\hline
$n$ & $A_n(1/2)$ & $B_n(1/2)$ & $C_n(1/2)$ & $D_n(1/2)$ & $E_n(1/2)$ \\ \hline\hline
0  & 1           & 0           & 0           & 0           & 0          \\ \hline
1  & 1           & 1/2         & 0           & 0           & 0          \\ \hline
2  & 3/2         & 1           & 1/4         & 0           & 0          \\ \hline
3  & 5/2         & 15/8        & 3/4         & 1/8         & 0          \\ \hline
4  & 35/8        & 7/2         & 7/4         & 1/2         & 1/16       \\ \hline
5  & 251/32      & 209/32      & 119/32      & 11/8        & 9/32       \\ \hline
6  & 911/64      & 779/64      & 241/32      & 207/64      & 53/64      \\ \hline
7  & 3327/128    & 1449/64     & 1897/128    & 7           & 65/32      \\ \hline
8  & 6095/128    & 1345/32     & 229/8       & 3689/256    & 289/64     \\ \hline
9  & 5593/64     & 19941/256   & 27949/512   & 7353/256    & 4845/512   \\ \hline
10 & 164407/1024 & 36903/256   & 105641/1024 & 57361/1024  & 19551/1024 \\ \hline
11 & 604487/2048 & 545721/2048 & 12397/64    & 220363/2048 & 4807/128   \\ \hline\hline
\end{tabular}
}
\end{center}
\end{table}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{table}[H]
\begin{center}
{\small
\caption{}\label{qf2-tab5}
\medskip

\begin{tabular}{||c|c|c|c|c|c||}\hline\hline
$n$ & $A_n(-1/2)$ & $B_n(-1/2)$ & $C_n(-1/2)$ & $D_n(-1/2)$ & $E_n(-1/2)$ \\ \hline\hline
0  & 1           & 0            & 0           & 0            & 0          \\ \hline
1  & 1           & -1/2         & 0           & 0            & 0          \\ \hline
2  & 3/2         & -1           & 1/4         & 0            & 0          \\ \hline
3  & 5/2         & -15/8        & 3/4         & -1/8         & 0          \\ \hline
4  & 35/8        & -7/2         & 7/4         & -1/2         & 1/16       \\ \hline
5  & 253/32      & -209/32      & 121/32      & -11/8        & 11/32      \\ \hline
6  & 935/64      & -781/64      & 253/32      & -209/64      & 77/64      \\ \hline
7  & 3509/128    & -1463/64     & 2079/128    & -231/32      & 55/16      \\ \hline
8  & 6655/128    & -1375/32     & 33          & -3927/256    & 561/64     \\ \hline
9  & 3179/32     & -20757/256   & 34067/512   & -8151/256    & 10659/512  \\ \hline
10 & 195415/1024 & -39325/256   & 136609/1024 & -66671/1024  & 48279/1024 \\ \hline
11 & 753709/2048 & -598345/2048 & 136367/512  & -269951/2048 & 52877/512  \\ \hline\hline
\end{tabular}
}
\end{center}
\end{table}








\bibliographystyle{amsplain}
\begin{thebibliography}{10}

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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83, 11A07; Secondary 12D10, 39A10.

\noindent \emph{Keywords:} Fibonacci numbers, primitive roots of
unity, recurrence relation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence \seqnum{A062883}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in} \noindent Received December 9 2006.  Revised version 
received August 3 2007.
Published in {\it Journal of Integer Sequences}, August 5 2007.

\bigskip
\hrule
\bigskip

\noindent Return to \htmladdnormallink{Journal of Integer Sequences
home page}{http://www.math.uwaterloo.ca/JIS/}. \vskip .1in


\end{document}
