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\begin{center}
\vskip 1cm{\LARGE\bf Quasi-Fibonacci Numbers of the \\
\vskip .1in
Seventh Order}
\vskip 1cm
\large
Roman Witu{\l}a, Damian S{\l}ota and Adam Warzy{\'n}ski \\
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 the seventh 
order.  We discover
many surprising relations
and identities, and study some applications to polynomials.
\end{abstract}


\vskip .2in


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\section{Introduction}


Grzymkowski and Witu{\l}a~\cite{grzymkowski}
discovered and studied
the following two identities:
\begin{align}
(1+\xi +\xi ^4)^n&=F_{n+1}+F_n(\xi +\xi ^4),\label{nw-g1}\\
(1+\xi ^2+\xi ^3)^n&=F_{n+1}+F_n(\xi ^2+\xi ^3),\label{nw-g2}
\end{align}
where $F_n$ denote the Fibonacci numbers and $\xi\in\mathbb{C}$ is
a~primitive fifth root of unity (i.e.,~$\xi^5=1$ and
$\xi\neq 1$). These identities make it possible to prove many
classical relations for Fibonacci numbers as well as to generalize
some of them. We may state that  these identities make up an
independent method of proving such relations, which is an
alternative to the methods depending on the application of either
Binet formulas or the generating function of Fibonacci  and Lucas
numbers.

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\subsection{Example of the application of identities~(\ref{nw-g1}) and~(\ref{nw-g2})}


First, we note that
\begin{equation}\label{wzor-a}
(1+\xi +\xi^4)^{u+v}=F_{u+v+1}+ F_{u+v} \, \big( \xi + \xi^4\big)
\end{equation}
and
\begin{multline}\label{wzor-b}
(1+\xi +\xi^4)^{u+v}=(1+\xi +\xi^4)^u\,(1+\xi +\xi^4)^v= \\
=\big( F_{u+1}+F_u\,(\xi +\xi^4)\big)\,\big( F_{v+1}+F_v\,(\xi +\xi^4)\big)= \\
\mbox{(by the identity $1+\xi+\xi^2+\xi^3+\xi^4=0$)}\\
=F_{u+1}\,F_{v+1}+F_u\,F_v+(F_u\,F_{v+1}+F_{u+1}\,F_v-F_u\,F_v)\,(\xi +\xi ^4)=\\
=F_{u+1}\,F_{v+1}+F_u\,F_v+(F_u\,F_{v+1}+F_{u-1}\,F_v)\,(\xi +\xi ^4).
\end{multline}
Replacing $u$ by $u-r$ and $v$ by $v+r$ in~(\ref{wzor-b}) we obtain
\begin{multline}\label{wzor-c}
(1+\xi +\xi^4)^{u+v}=F_{u-r+1}\,F_{v+r+1}+F_{u-r}\,F_{v+r}+ \\
+(F_{u-r}\,F_{v+r+1}+F_{u-r+1}\,F_{v+r}-F_{u-r}\,F_{v+r})(\xi +\xi ^4).
\end{multline}
We note that the numbers~$1$ and $\xi+\xi^4$ are linearly independent over~$\mathbb{Q}$.
Hence comparing the parts without $(\xi +\xi^4)$ of (\ref{wzor-a}) with (\ref{wzor-b})
and (\ref{wzor-b}) with (\ref{wzor-c}) we get two known identities
(see~\cite{graham,koshy})
\[
F_{u+v+1} = F_{u+1}\, F_{v+1} + F_{u}\, F_{v}
\]
and
\begin{align*}
F_{u+1}\,F_{v+1}-F_{u-r+1}\,F_{v+r+1}&=F_{u-r}\,F_{v+r}-F_u\,F_v \\
&\mbox{(after the next $(u-r)$ iterations)} \\
&=(-1)^{u-r+1}(F_r\,F_{v-u+r}-F_0\,F_{v-u+2r}) \\
&=(-1)^{u-r+1}F_r\,F_{v-u+r}.
\end{align*}




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\subsection{The aim of the paper}


In this paper the relations~(\ref{nw-g1}) and~(\ref{nw-g2}) will
be generalized in the following way:
\begin{align*}
(1+\delta (\xi+\xi^6))^n &=
A_n(\delta)+B_n(\delta)(\xi+\xi^6)+C_n(\delta)(\xi^2+\xi^5),\\
(1+\delta (\xi^2 +\xi^5))^n&= A_n(\delta )+B_n(\delta )
(\xi^2+\xi^5)+C_n(\delta )(\xi^3 +\xi^4),
\end{align*}
and
$$
(1+\delta (\xi^3 +\xi^4))^n =
A_n(\delta )+B_n(\delta )(\xi^3 +\xi^4) +C_n(\delta )(\xi +\xi^6),
$$
where $\xi\in\mathbb{C}$
are primitive seventh roots of unity (i.e.,~$\xi^7=1$ and $\xi\neq 1$),
$\delta\in \mathbb{C}$, $\delta\neq 0$.
New families of numbers created by these identities
\begin{equation}\label{pierwszy}
\{A_{n}(\delta) \}_{n=1}^{\infty},\qquad
\{B_{n}(\delta) \}_{n=1}^{\infty},\quad \mbox{and}\quad
\{C_{n}(\delta) \}_{n=1}^{\infty}
\end{equation}
called here ``the quasi-Fibonacci numbers of order $(7;\delta)$'',
$\delta\in \mathbb{C}$, $\delta \neq 0$, are investigated in this
paper. The elements of each of the three
sequences~(\ref{pierwszy}) satisfy the same recurrence relation of
order three
$$
\mathbb{X}_{n+3}+(\delta -3)\mathbb{X}_{n+2}+(3-2\delta
-2\delta^2)\mathbb{X}_{n+1} + (-1+\delta
+2\delta^2-\delta^3)\mathbb{X}_n=0,
$$
which enables a~direct trigonometrical
representation of these numbers (see formulas~(\ref{2.14})--(\ref{2.16})).
In consequence, many surprising
algebraic and trigonometric identities
and summation formulas for these numbers may be generated.
Also the polynomials connected with the numbers $\{C_{n}(\delta) \}_{n=1}^{\infty}$
$$
\sum_{k=1}^{n} C_{k+1}(\delta)\, x^{n-k},\qquad n\in \mathbb{N},
$$
are investigated in this paper.




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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 \cite{watkins}
(see also~\cite{surowski}) 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 $n=2s$, where $T_s(x)$ denotes the $s$-th Chebyshev polynomial of the first kind.
In the sequel, if $n=2s+1$ is a prime number, we obtain
$$
T_{s+1}(x)-T_s(x)=2^s\Psi _1(x)\Psi _n(x).
$$
For example, we have
\begin{align*}
\Psi_7(x) &= \frac{1}{8(x-1)}(T_4(x)-T_3(x)) =\frac{1}{8(x-1)}(8x^4-4x^3-8x^2+3x+1) \\
&=\frac{1}{8(x-1)}8(x-1)\bigg( x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{8}\bigg) =
x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{8},
\end{align*}


\begin{lemma}\label{qf2-lem1.1}
If $n\geqslant 3$ then the roots of $\Psi _n(x)=0$ are $\cos (2\pi k/n)$,
for $0<k\leqslant s$ and $(k,n)=1$ (where $n=2s$ or $n=2s+1$ respectively).
\end{lemma}


Below a~more elementary proof of this result but only for the polynomial
$\Psi_7(x)$, will be presented. Our
proof is based only on the following simple fact:


\begin{lemma}\label{qf2-lem1.2}
Let $q\in\mathbb{Q}$ and $\cos (q\pi )\in\mathbb{Q}$.
Then $\cos (q\pi )\in\Big\{ 0,\pm\frac{1}{2},\pm 1\Big\}$.
\end{lemma}

\begin{proof}
For the elementary proof of this Lemma see, for example,
the Appendix to the Russian translation of Niven's book~\cite{niven}
(written by I.~M.~Yaglom).
\end{proof}

\begin{lemma}\label{qf2-lem1.3}
Let $\xi =\exp (i2\pi /7)$. Then the polynomial
\begin{align*}
p_7(x) &:= (x-\xi -\xi ^6)(x-\xi ^2-\xi ^5)(x-\xi ^3-\xi ^4) \\
&= x^3+x^2(-\xi -\xi ^2-\xi ^3-\xi ^4-\xi ^5-\xi ^6) \\
&\phantom{:=} +x((\xi +\xi ^6)(\xi ^2+\xi ^5)+(\xi +\xi ^6)(\xi ^3+\xi ^4)+
(\xi ^2+\xi ^5)(\xi ^3+\xi ^4)) \\
&\phantom{:=} -(\xi +\xi ^6)(\xi ^2+\xi ^5)(\xi ^3+\xi ^4)=x^3+x^2-2x-1
\end{align*}
is a minimal polynomial of the numbers
\begin{equation}
\xi +\xi^6=2\cos (2\pi /7),\quad
\xi^2+\xi^5=2\cos (4\pi /7),\quad
\xi^3+\xi^4=2\cos (6\pi /7).
\label{1.1}
\end{equation}
Moreover, we have the identity
\begin{equation}
\Psi_7(x)=\frac{1}{8}p_7(2x).
\label{b}
\end{equation}
\end{lemma}


\begin{proof}
Besides Lemma~\ref{qf2-lem1.2}, each
of the numbers in~(\ref{1.1}) is an irrational number,
so the polynomial~$p_7(x)$ is irreducible
 over~$\mathbb{Q}$.
\end{proof}



\begin{corollary}\label{qf2-cor1.3.1}
We have
$$
p_7(x-1)=x^3-2x^2-x+1.
$$
\end{corollary}

See the identity~(\ref{prz1}) below, where the connection of $p_7(x-1)$ with
quasi-Fibonacci numbers of $7$-th order is presented.


\begin{corollary}\label{qf2-cor1.3.2}
The numbers $\cos (2k\pi /7)$, $k=0,1,2$ are linearly independent over~$\mathbb{Q}$.
\end{corollary}


\begin{proof}
If we suppose that
$$
a+b\cos (2\pi /7)+c\cos (4\pi /7)=0
$$
for some $a,b,c\in\mathbb{Q}$, then we also  have
$$
a+b\cos (2\pi /7)+c(2\cos ^2(2\pi /7)-1)=0,
$$
i.e.,
$$
a-c+b\cos (2\pi /7)+2c\cos ^2(2\pi /7)=0
$$
so the degree of $\cos (2\pi /7)$ is $\leqslant 2$,
which by Lemma~\ref{qf2-lem1.3} means that $a=b=c=0$.
\end{proof}


\begin{corollary}\label{qf2-cor1.3.2a}
Every three numbers which belong to the set
$\{1,\xi+\xi^{6},\xi^{2}+\xi^{5},\xi^{3}+\xi^{4}\}$
are linearly independent over~$\mathbb{Q}$.
\end{corollary}

\begin{proof}
It follows from the identity
$1+\xi+\xi^2+\ldots+\xi^{6}=0$ and from Corollary~\ref{qf2-cor1.3.2}.
\end{proof}

\begin{corollary}\label{qf2-cor1.3.3}
The following decomposition holds:
\begin{align*}
f_n(\mathbb{X}) &:= \Big( \mathbb{X} -\Big( 2\cos\frac{2\pi}{7}\Big)^{2^n}\Big)\cdot
\Big(\mathbb{X} -\Big( 2\cos\frac{4\pi}{7}\Big) ^{2^n}\Big)\cdot
\Big(\mathbb{X} -\Big( 2\cos\frac{6\pi}{7}\Big) ^{2^n}\Big) \\
&=\mathbb{X} ^3-\alpha _n\mathbb{X} ^2+\beta _n\mathbb{X} -1,
\end{align*}
where $\alpha _0=-1$, $\beta _0=-2$ and
\begin{equation}\label{gw}
\left\{\begin{array}{ll}
\alpha _{n+1}=\alpha _n^2-2\beta _n, \\
\beta _{n+1}=\beta _n^2-2\alpha _n,\quad n\in\mathbb{N}.
\end{array}\right.
\end{equation}
\end{corollary}

For example, we have
$\alpha_0=-1$, $\beta_0=-2$,
$\alpha _1=5$, $\beta _1=6$, $\alpha _2=13$, $\beta _2=26$,
$\alpha _3=117$ and $\beta _3=650$.
(A~more general decomposition will be presented later, see Lemma~\ref{qf2-lem2.3}~a.)
We note that $13 | \alpha_{n}$ and $13 | \beta_{n}$ for every
$n\in \mathbb{N}$, $n\geq 2$.


\begin{proof}
We have
\begin{multline*}
\big( \mathbb{Y} ^2 -\big( 2\cos\tfrac{2\pi}{7}\big)^{2^{n+1}}\big)\cdot
\big(\mathbb{Y} ^2 -\big( 2\cos\tfrac{4\pi}{7}\big) ^{2^{n+1}}\big)\cdot
\big(\mathbb{Y} ^2 -\big( 2\cos\tfrac{6\pi}{7}\big) ^{2^{n+1}}\big)=\\
=-f_n(\mathbb{Y} )\cdot f_n(-\mathbb{Y} )=(\mathbb{Y} ^3-\alpha _n\mathbb{Y}^2+
\beta _n\mathbb{Y} -1)(\mathbb{Y} ^3+\alpha _n\mathbb{Y} ^2+\beta _n\mathbb{Y} +1)= \\
=(\mathbb{Y} ^3+\beta _n\mathbb{Y} )^2-(\alpha _n\mathbb{Y} ^2+1)^2=
\mathbb{Y} ^6-(\alpha _n^2-2\beta _n)\mathbb{Y} ^4+(\beta _n^2-2\alpha _n)\mathbb{Y} ^2-1.
%\hfill{\Box}
\end{multline*}%\nopagebreak
\end{proof}

It is obvious that
$$
\alpha_n=
\Big( 2 \cos \frac{2\pi}{7} \Big)^{\!2^n}+
\Big( 2 \cos \frac{4\pi}{7} \Big)^{\!2^n}+
\Big( 2 \cos \frac{6\pi}{7} \Big)^{\!2^n}
$$
and
$$
\beta_n=
\Big( 4 \cos \frac{2\pi}{7}\, \cos \frac{4\pi}{7} \Big)^{\!2^n}+
\Big( 4 \cos \frac{2\pi}{7}\, \cos \frac{6\pi}{7} \Big)^{\!2^n}+
\Big( 4 \cos \frac{4\pi}{7}\, \cos \frac{6\pi}{7} \Big)^{\!2^n}
$$
for every $n\in \mathbb{N}$.
Moreover, from~(\ref{gw}), we deduce that
\begin{align*}
\alpha_{n+1}+\beta_{n+1} & = \big(\alpha_n-1\big)^{2} + \big(\beta_n-1\big)^{2} -2,\\
\alpha_{n+1}-\beta_{n+1} & = \big(\alpha_n-1\big)^{2} - \big(\beta_n-1\big)^{2}
=\big( \alpha_n -\beta_n\big) \big(\alpha_n+\beta_n+2\big)=\\
 & =\big( \alpha_n -\beta_n\big) \Big( \big(\alpha_{n-1}-1\big)^{2} +
 \big(\beta_{n-1}-1\big)^{2}\Big),
\end{align*}
which yields
$$
%\begin{multline*}
\beta_{n+1}-\alpha_{n+1}
 = \prod_{k=0}^{n-1}
\Big( \big(\alpha_{k}-1\big)^{2} +
      \big(\beta_{k}-1\big)^{2}\Big)
 = 13 \prod_{k=1}^{n-1}
\Big( \big(\alpha_{k}-1\big)^{2} +
      \big(\beta_{k}-1\big)^{2}\Big)=
%\\
$$
$$
 = \prod_{k=1}^{n}
\big(\alpha_{k} + \beta_{k}+2\big).
$$
%\end{multline*}




The following extension lemma is the basic technical tool to
generating almost all formulas presented in the next sections.

\begin{lemma}\label{qf2-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
$$
\sum_{k=1}^{n} f_k (\delta) \, a_k =
\sum_{k=1}^{n} g_k (\delta) \, a_k
$$
holds 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}


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\section{Quasi-Fibonacci numbers of order~7}


\begin{lemma}\label{qf2-lem2.1}
Let $\delta ,\xi\in\mathbb{C}$, $\xi^7=1$ and $\xi\neq 1$.
Then the following identities hold:
\begin{align}
(1+\delta (\xi+\xi^6))^n &=
a_n(\delta) +b_n(\delta)(\xi^{1\cdot 1} +\xi^{1\cdot 6})
+c_n(\delta)(\xi^{2\cdot 1}+\xi^{2\cdot 6})\nonumber \\
&\phantom{=}+d_n(\delta)(\xi^{3\cdot 1} +\xi^{3\cdot 6}) \nonumber \\
&= a_n(\delta) +b_n(\delta)(\xi +\xi^6) +c_n(\delta)(\xi^2+\xi^5)+
d_n(\delta)(\xi^3 +\xi^4) \nonumber \\
&= A_n(\delta)+B_n(\delta)(\xi+\xi^6)+C_n(\delta)(\xi^2+\xi^5),\label{aaaa}
\end{align}
\begin{align}
(1+\delta (\xi^2 +\xi^5))^n&= a_n(\delta )+b_n(\delta )(\xi^{1\cdot 2} +\xi^{1\cdot 5}) +
c_n(\delta )(\xi^{2\cdot  2}+\xi^{2\cdot 5})\nonumber \\
&\phantom{=}+d_n(\delta )(\xi^{3\cdot 2}+\xi^{3\cdot 5})\nonumber \\
&=a_n(\delta )+b_n(\delta )(\xi^2 +\xi^5) +
c_n(\delta )(\xi^3 +\xi^4) +d_n(\delta )(\xi +\xi^6) \nonumber \\
&= A_n(\delta )+B_n(\delta )(\xi^2 +\xi^5)+C_n(\delta )(\xi^3 +\xi^4),\label{bbbb}\\
(1+\delta (\xi^3 +\xi^4))^n &= a_n(\delta ) +b_n(\delta )(\xi^{1\cdot 3} +\xi^{1\cdot 4})+
c_n(\delta )(\xi^{2\cdot 3}+\xi^{2\cdot 4})\nonumber \\
&\phantom{=}+d_n(\delta )(\xi^{3\cdot 3} +\xi^{3\cdot 4})\nonumber \\
&=a_n(\delta ) +b_n(\delta )(\xi^3 +\xi^4)+
c_n(\delta )(\xi+\xi^6)+d_n(\delta )(\xi^2 +\xi^5) \nonumber \\
&=A_n(\delta )+B_n(\delta )(\xi^3 +\xi^4) +C_n(\delta )(\xi +\xi^6),\label{cccc}
\end{align}
where
\begin{equation}
\left\{\begin{array}{l}
a_0(\delta )=1,\,\,\, b_0(\delta )=c_0(\delta )=d_0(\delta )=0, \\
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 )+(1+\delta )d_n(\delta )
\end{array}\right. \label{pppp}
\end{equation}
for every $n\in\mathbb{N}$ and
\begin{align*}
A_{n+1}(\delta ) &:=a_{n+1}(\delta)-d_{n+1}(\delta)\\
&\phantom{:}= a_n(\delta )+2\delta b_n(\delta )-\delta c_n(\delta )-
(1+\delta )d_n(\delta ) \\
&\phantom{:}=(a_n(\delta )-d_n(\delta ))+2\delta (b_n(\delta )-d_n(\delta ))-
\delta (c_n(\delta )-d_n(\delta )) \\
&\phantom{:}= A_n(\delta )+2\delta B_n(\delta )-\delta C_n(\delta ), \\
B_{n+1}(\delta ) &:=b_{n+1}(\delta)-d_{n+1}(\delta)\\
&\phantom{:}= \delta a_n(\delta )+b_n(\delta )+\delta c_n(\delta )-
\delta c_n(\delta )-(1+\delta )d_n(\delta ) \\
&\phantom{:}= \delta (a_n(\delta )-d_n(\delta ))+b_n(\delta )-d_n(\delta ) \\
&\phantom{:}= \delta A_n(\delta )+B_n(\delta ),\\
%\end{align*}
%\begin{align*}
C_{n+1}(\delta ) &:=c_{n+1}(\delta)-d_{n+1}(\delta)\\
&\phantom{:}= \delta b_n(\delta )+c_n(\delta )+\delta d_n(\delta )-
\delta c_n(\delta )-(1+\delta )d_n(\delta ) \\
&\phantom{:}= \delta (b_n(\delta )-d_n(\delta ))+(1-\delta )(c_n(\delta )-d_n(\delta )) \\
&\phantom{:}= \delta B_n(\delta )+(1-\delta )C_n(\delta ),
\end{align*}
i.e.,
\begin{equation}
\left\{\begin{array}{l}
A_0(\delta )=1,\,\,\, B_0(\delta )=C_0(\delta )=0, \\
A_{n+1}(\delta )=A_n(\delta )+2\delta B_n(\delta )-\delta C_n(\delta ), \\
B_{n+1}(\delta )=\delta A_n(\delta )+B_n(\delta ), \\
C_{n+1}(\delta )=\delta B_n(\delta )+(1-\delta )C_n(\delta )
\end{array}\right.
\label{aa}
\end{equation}
for every $n\in\mathbb{N}$.
\end{lemma}

\begin{proof}
For example, we have
\begin{align*}
(1 &+\delta (\xi +\xi^6))^{n+1}=(1+\delta (\xi +\xi^6))^n(1+\delta (\xi +\xi^6)) \\
&=\big( a_n(\delta )+b_n(\delta )(\xi +\xi^6)+c_n(\delta )(\xi^2+\xi^5)+
  d_n(\delta )(\xi^3+\xi^4)\big) (1+\delta (\xi +\xi^6)) \\
&=a_n(\delta ) +\delta a_n(\delta )(\xi +\xi^6) +b_n(\delta )(\xi +\xi^6) +
  \delta b_n(\delta )(\xi^2 +2 +\xi^5) +c_n(\delta )(\xi^2 +\xi^5) \\
&\phantom{=} +\delta c_n(\delta )(\xi^3+\xi +\xi^6+\xi^4)+d_n(\delta )(\xi^3+\xi^4)+
  \delta d_n(\delta )(\xi^4+\xi^5+\xi^2+\xi^3) \\
&= a_n(\delta )+2\delta b_n(\delta )+(\xi +\xi^6)(\delta a_n(\delta )+b_n(\delta )+
   \delta c_n(\delta )) \\
&\phantom{=} +(\xi^2+\xi^5)(\delta b_n(\delta )+c_n(\delta )+\delta d_n(\delta ))+
    (\xi^3+\xi^4)(\delta c_n(\delta )+(1+\delta )d_n(\delta ))= \\
&\!\!\!\!\!\mbox{(by equality $\xi^3+\xi^4=-1-\xi -\xi^6-\xi^2-\xi^5$)} \\
&= a_n(\delta )+2\delta b_n(\delta )-\delta c_n(\delta )-(1+\delta )d_n(\delta ) \\
&\phantom{=} +(\xi +\xi^6)(\delta a_n(\delta )+b_n(\delta )-(1+\delta )d_n(\delta )) \\
&\phantom{=} +(\xi^2+\xi^5)(\delta b_n(\delta )+(1-\delta )c_n(\delta )-d_n(\delta )).
\end{align*}
\end{proof}



\begin{definition}\label{qf2-rem2.1}
The numbers~(\ref{aa}) will be called
\textit{the quasi-Fibonacci numbers of order~$(7;\delta )$}
(see Table~\ref{qf2-tab-d1} at the end of the paper).
To simplify notation we write $a_n$, $b_n$, $c_n$, $d_n$, $A_n$, $B_n$, $C_n$ instead of
$a_n(1)$, $b_n(1)$, $c_n(1)$, $d_n(1)$, $A_n(1)$,
$B_n(1)$, $C_n(1)$ respectively,
and these numbers will be called
\textit{the quasi-Fibonacci numbers of the seventh order}.
\end{definition}

\begin{remark}
Note that section~$III$ of~\cite{steinbach} contains an explicit reference to the sequences
$A_n$, $B_n$ and $C_n$. This paper is partly based on the properties of $-p_7(-x)$
and $p_7(x-1)$. $A_n$, $B_n$ and $C_n$ also appear in~\cite{kappraff}.
\end{remark}


\begin{corollary}\label{qf2-cor2.1.1}
Adding equalities~(\ref{aaaa}--\ref{cccc}) we obtain the identity
\begin{multline}
(1 + \delta (\xi +\xi^6))^n + (1+ \delta (\xi^2 +\xi^5))^n +
(1 + \delta (\xi^3 +\xi^4))^n =\\
= 3A_n(\delta ) - B_n(\delta ) - C_n(\delta ).
\label{1111111}
\end{multline}
\end{corollary}


\begin{corollary}\label{qf2-cor2.1.2}
There follows from~(\ref{pppp}) and~(\ref{aa}) two special systems (for~$\delta=1$)
\begin{equation}
\left\{\begin{array}{l}
a_0=1,\,\,\,b_0=c_0=d_0=0, \\
a_{n+1}=a_n+2b_n, \\
b_{n+1}=a_n+b_n+c_n, \\
c_{n+1}=b_n+c_n+d_n, \\
d_{n+1}=c_n+2d_n
\end{array}\right.
\label{bb}
\end{equation}
and
\begin{equation}
\left\{\begin{array}{l}
A_0=1,\,\,\, B_0=C_0=0, \\
A_{n+1}=A_n+2B_n-C_n, \\
B_{n+1}=A_n+B_n, \\
C_{n+1}=B_n
\end{array}\right.
\label{cc}
\end{equation}
for every $n\in\mathbb{N}$ (see Table~\ref{qf2-tab1},
\seqnum{A006356} and \seqnum{A006054} in~\cite{sloan}).
\end{corollary}




\begin{corollary}\label{qf2-cor2.1.3}
Another system may also be derived from~(\ref{aa}) for~$\delta=-1$
$$
\left\{\begin{array}{l}
A_0(-1)=1,\,\,\, B_0(-1)=C_0(-1)=0, \\
A_{n+1}(-1)=A_n(-1)-2B_n(-1)+C_n(-1), \\
B_{n+1}(-1)=-A_n(-1)+B_n(-1), \\
C_{n+1}(-1)=-B_n(-1)+2C_n(-1)
\end{array}\right.
$$
for every $n\in\mathbb{N}$
(see Table~\ref{qf2-tab2} and \seqnum{A085810} in~\cite{sloan}).
\end{corollary}


\begin{corollary}\label{qf2-cor2.1.4}
If $\delta\neq 0$ then from~(\ref{aa})  the following identities
can be generated:
\begin{align}
\delta B_n(\delta )&=C_{n+1}(\delta )-(1-\delta )C_n(\delta ),\label{dd}\\
\delta A_n(\delta )&=B_{n+1}(\delta )-B_n(\delta ),\label{ddddddd}\\
\delta^2A_n(\delta )&= \delta B_{n+1}(\delta )-\delta B_n(\delta ) \label{ee} \\
&=C_{n+2}(\delta )-(1-\delta )C_{n+1}(\delta )-\big(C_{n+1}(\delta )-
  (1-\delta )C_n(\delta )\big)\nonumber\\
&=C_{n+2}(\delta )-(2-\delta )C_{n+1}(\delta )+(1-\delta )C_n(\delta ),\nonumber\\
A_{n+1}(\delta )-A_n(\delta )&= 2\delta B_n(\delta )-\delta C_n(\delta ) \label{2.11}\\
&=2C_{n+1}(\delta )-2(1-\delta )C_n(\delta )-\delta C_n(\delta )\nonumber \\
&=2C_{n+1}(\delta )-(2-\delta )C_n(\delta ),\nonumber
\end{align}
hence, we obtain
\begin{equation}
\delta^2A_{n+1} (\delta )-\delta^2A_n(\delta )=
2\delta^2C_{n+1}(\delta )-(2\delta^2-\delta^3)C_n(\delta ).
\label{211}
\end{equation}
On the other hand, by~(\ref{ee}) we obtain
\begin{multline}\label{2111}
\delta^2A_{n+1}(\delta ) -\delta^2A_n(\delta )
=C_{n+3}(\delta ) - (2 - \delta )C_{n+2}(\delta ) + (1 - \delta )C_{n+1}(\delta ) -\\
-C_{n+2}(\delta ) + (2 - \delta )C_{n+1}(\delta ) - (1 - \delta )C_n(\delta )= \\
=C_{n+3}(\delta ) - (3 - \delta )C_{n+2}(\delta ) + (3 - 2\delta )C_{n+1}(\delta )
+ (\delta - 1)C_n(\delta ).
\end{multline}
From~(\ref{211}) and~(\ref{2111}) we obtain the final recurrence
identities for numbers $C_n(\delta )$
\begin{equation}%\begin{multline}
C_{n+3}(\delta )+ (\delta - 3)C_{n+2}(\delta ) +
(3 - 2\delta - 2\delta^2)C_{n+1}(\delta )%+\\
+ ( -1 + \delta  + 2\delta^2 - \delta^3)C_n(\delta ) = 0.
\label{ff}
\end{equation}
%\end{multline}
\end{corollary}


\begin{remark}
We note that the elements of each of the sequences $\{A_{n}(\delta)\}$ and $\{B_{n}(\delta)\}$
also satisfy this recurrence relation (which follows from~(\ref{dd}) and~(\ref{ddddddd})).
\end{remark}


\begin{lemma}\label{qf2-lem2707}
The characteristic polynomial $p_7({\mathbb X};\delta)$ of~(\ref{ff})
has the following decomposition:
\begin{multline}\label{2.13}
p_7({\mathbb X};\delta):=\mathbb{X}^3+(\delta -3)\mathbb{X}^2+
(3-2\delta -2\delta^2)\mathbb{X} +
(-1+\delta +2\delta^2-\delta^3)=  \\
=(\mathbb{X} -1-\delta (\xi +\xi^6))(\mathbb{X} -1-\delta (\xi^2+\xi^5))
(\mathbb{X} -1-\delta (\xi^3 +\xi^4))
\end{multline}
where $\xi =\exp (i2\pi /7)$. So the relation
$$
C_n(\delta )=\alpha (1+\delta (\xi +\xi^6))^n+\beta (1+\delta (\xi^2+\xi^5))^n+
\gamma (1+\delta (\xi^3+\xi^4))^n
$$
holds for every $n\in\mathbb{N}$ and for some $\alpha, \beta, \gamma\in\mathbb{R}$.
\end{lemma}

\begin{remark}
To find $\alpha, \beta, \gamma$, from~Lemma~\ref{qf2-lem2707},
it is sufficient to solve the following linear system:
$$
\left\{\!\!\!\begin{array}{l}
C_1(\delta ) = 0 = \alpha (1 +\delta (\xi +\xi^6)) +\beta (1 +\delta (\xi^2+\xi^5))
 +\gamma (1 +\delta (\xi^3+\xi^4)), \\
C_2(\delta ) = \delta^2 = \alpha (1 +\delta (\xi +\xi^6))^2\!
  +\beta (1 +\delta (\xi^2+\xi^5))^2\! +\gamma (1 +\delta (\xi^3+\xi^4))^2, \\
C_3(\delta ) = 3\delta^2\!-\delta^3 = \alpha (1 +\delta (\xi +\xi^6))^3\!
  +\beta (1 +\delta (\xi^2+\xi^5))^3\! +\gamma (1 +\delta (\xi^3+\xi^4))^3.
\end{array}\right.
$$
\end{remark}

After some calculations we obtain the identities
\begin{align}\label{2.14}
7C_n(\delta )&=
 (\xi^{2\cdot 1}+\xi^{2\cdot 6}-\xi^{3\cdot 1}-\xi^{3\cdot 6})(1+\delta (\xi +\xi^6))^n\\
&\phantom{=}+(\xi^{2\cdot 2}+\xi^{2\cdot 5}-\xi^{3\cdot 2}-\xi^{3\cdot 5})
(1+\delta (\xi^2+\xi^5))^n\nonumber\\
&\phantom{=}+(\xi^{2\cdot 3}+\xi^{2\cdot 4}-\xi^{3\cdot 3}-\xi^{3\cdot 4})
(1+\delta (\xi^3+\xi^4))^n\nonumber\\
&=(\xi^2+\xi^5-\xi^3-\xi^4)(1+\delta (\xi +\xi^6))^n
%\nonumber\\
%&\phantom{=}
+(\xi^3+\xi^4-\xi -\xi^6)\nonumber\\
&\phantom{=}\times(1+\delta (\xi^2+\xi^5))^n
+(\xi +\xi^6-\xi^2-\xi^5)(1+\delta (\xi^3+\xi^4))^n\nonumber\\
&=2\Big(\cos\frac{4}{7}\pi +\cos\frac{\pi}{7}\Big)
\Big( 1+2\delta\cos\frac{2\pi}{7}\Big)^n
%\nonumber\\
%&\phantom{=}
-2\Big(\cos\frac{\pi}{7} +\cos\frac{2}{7}\pi\Big)\nonumber\\
&\phantom{=}
\times\Big( 1+2\delta\cos\frac{4\pi}{7}\Big)^n
+2\Big(\cos\frac{2\pi}{7}-\cos\frac{4\pi}{7}\Big)
  \Big( 1+2\delta\cos\frac{6\pi}{7}\Big)^n.\nonumber
\end{align}
Hence, by (\ref{dd}) we get
\begin{align}
7B_n(\delta ) &=\frac{1}{\delta}(7C_{n+1}(\delta )
-7(1 -\delta )C_n(\delta ))\label{2.15}\\
&= (\xi^{1\cdot 1}+\xi^{1\cdot 6}-\xi^{3\cdot 1}-\xi^{3\cdot 6})
(1+\delta (\xi +\xi^6))^n\nonumber\\
&\phantom{=}+(\xi^{1\cdot 2}+\xi^{1\cdot 5}-\xi^{3\cdot 2}-\xi^{3\cdot 5})
(1+\delta (\xi^2+\xi^5))^n\nonumber\\
&\phantom{=}+(\xi^{1\cdot 3}+\xi^{1\cdot 4}-\xi^{3\cdot 3}-\xi^{3\cdot 4})
(1+\delta (\xi^3+\xi^4))^n\nonumber\\
&=(\xi +\xi^6 -\xi^3 -\xi ^4)(1 + \delta (\xi +\xi^6))^n
%\nonumber\\
%&\phantom{=}
+(\xi^2+\xi^5-\xi -\xi^6) \nonumber\\
&\phantom{=}
\times(1+\delta (\xi^2+\xi^5))^n
+(\xi^3+\xi^4-\xi^2-\xi^5)(1+\delta (\xi^3+\xi^4))^n \nonumber\\
&=2\,\Big(\cos\frac{2\pi}{7} +\cos\frac{\pi}{7}\Big)
\Big( 1+2\delta\cos\frac{2\pi}{7}\Big)^n
%\nonumber\\
%&\phantom{=}
+2\,\Big(\cos\frac{4\pi}{7} -\cos\frac{2\pi}{7}\Big)\nonumber \\
&\phantom{=}
\times\Big( 1+2\delta\cos\frac{4\pi}{7}\Big)^n
-2\,\Big(\cos\frac{\pi}{7}+\cos\frac{4\pi}{7}\Big)
\Big( 1+2\delta\cos\frac{6\pi}{7}\Big)^n\nonumber
\end{align}
and by (\ref{ddddddd})
\begin{multline}\label{2.16}
7A_n(\delta )=\frac{1}{\delta}(7B_{n+1}(\delta )-7B_n(\delta ))%=\\
= (\xi^{0\cdot 1}+\xi^{0\cdot 6}-\xi^{3\cdot 1}-\xi^{3\cdot 6})
(1+\delta (\xi +\xi^6))^n+\\
+(\xi^{0\cdot 2}+\xi^{0\cdot 5}-\xi^{3\cdot 2}-\xi^{3\cdot 5})
(1+\delta (\xi^2+\xi^5))^n+\\
+(\xi^{0\cdot 3}+\xi^{0\cdot 4}-\xi^{3\cdot 3}-\xi^{3\cdot 4})
(1+\delta (\xi^3+\xi^4))^n=\\
=(2-\xi ^3-\xi ^4)(1+\delta (\xi +\xi ^6))^n
+(2-\xi -\xi ^6)(1+\delta (\xi ^2+\xi ^5))^n +\\
+(2-\xi ^2-\xi ^5)(1+\delta (\xi ^3+\xi ^4))^n= \\
%\end{multline}
%\begin{multline*}
=2\Big( 1+\cos\frac{\pi}{7}\Big)\Big( 1+2\delta\cos\frac{2\pi}{7}\Big)^n+
2\Big( 1-\cos\frac{2\pi}{7}\Big)\Big( 1+2\delta\cos\frac{4\pi}{7}\Big) ^n+ \\
+2\Big( 1-\cos\frac{4\pi}{7}\Big)\Big( 1+2\delta\cos\frac{6\pi}{7}\Big)^n.
\end{multline}

\begin{remark}\label{qf2-rem2.1.1}
In the sequel, for $\delta =1$, we obtain from~(\ref{ff})
\begin{equation}
C_{n+3}-2C_{n+2}-C_{n+1}+C_n=0
\label{gg}
\end{equation}
for every $n\in\mathbb{N}$. Of course,
the elements of the sequences $\{ B_n\}$ and $\{ A_n\}$ satisfy
the identity~(\ref{gg}).
The characteristic polynomial of~(\ref{gg}) has the following decomposition:
(see Corollary~\ref{qf2-cor1.3.1})
$$
\mathbb{X} ^3-2\mathbb{X} ^2-\mathbb{X} +1=
(\mathbb{X} -1-\xi -\xi ^6)(\mathbb{X} -1-\xi ^2-\xi ^5)(\mathbb{X} -1-\xi ^3-\xi ^4),
$$
where $\xi =\exp (i2\pi /7)$. Moreover,
the decompositions of $C_n$ (there are cyclic
transformations of the sums $\xi +\xi ^6$, $\xi ^2+\xi ^5$ and $\xi ^3+\xi ^4$
in elements of the first equality)  follows immediately
from~(\ref{2.14})
\begin{multline}\label{2.18a}
7C_n=(\xi ^2+\xi ^5-\xi ^3-\xi ^4)(1+\xi +\xi ^6)^n +\\
+(\xi ^3+\xi ^4-\xi -\xi ^6)(1+\xi ^2+\xi ^5)^n+
(\xi +\xi ^6-\xi ^2-\xi ^5)(1+\xi ^3+\xi ^4)^n= \\
=(\xi +\xi ^6-\xi ^3-\xi ^4)(1+\xi +\xi ^6)^{n-1}+
(\xi ^2+\xi ^5-\xi -\xi ^6)(1+\xi ^2+\xi ^5)^{n-1}+ \\
+(\xi ^3+\xi ^4-\xi ^2-\xi ^5)(1+\xi ^3+\xi ^4)^{n-1}
\end{multline}
for every $n=1,2,3,\ldots$
\end{remark}







\begin{lemma}\label{qf2-lem2.2}
The following summation formulas for the elements
of the sequences $\{ C_n(\delta )\}$, $\{ B_n(\delta )\}$
and $\{ A_n(\delta )\}$ hold:
\begin{align}
\delta\sum\limits_{n=1}^{N}A_n(\delta )&=B_{N+1}(\delta ) -\delta;\label{2.24a}\\
\delta\sum\limits_{n=1}^{N}B_n(\delta )&=C_{N+1}(\delta )
+\delta\sum\limits_{n=1}^{N}C_n(\delta );
\end{align}
hence, from~(\ref{2.11}) we obtain
\begin{align}
A_{N+1}(\delta)-A_{1}(\delta) &=
2\delta\sum\limits_{n=1}^{N}B_n(\delta )-\delta\sum\limits_{n=1}^{N}C_n(\delta )
=2C_{N+1}(\delta )+\delta\sum\limits_{n=1}^{N}C_n(\delta ),
\end{align}
i.e.,
\begin{equation}
\delta\sum\limits_{n=1}^{N}C_n(\delta )=-1+A_{N+1}(\delta )-2C_{N+1}(\delta );
\label{1001}
\end{equation}
\begin{align}
7\delta\sum\limits_{n=1}^{N}C_n(\delta )&=
-7-7C_{N+1}(\delta )-7B_{N+1}(\delta )\label{101}\\
&+\Big( 3-\frac{2}{\delta}\Big) A_{N+1}(\delta )
+\frac{2}{\delta} A_{N+2}(\delta );\nonumber
\end{align}
\begin{multline}\label{102}
\delta ^3\sum\limits_{n=1}^{N}C_n(\delta ) =
C_{N+3}(\delta ) + (\delta  -2)C_{N+2}(\delta ) +
(1 - \delta - 2\delta^2)C_{N+1}(\delta ) - \delta^2 = \\
= C_{N+2}(\delta ) + (\delta  - 2)C_{N+1}(\delta ) +
(\delta ^3 - 2\delta ^2 - \delta  + 1)C_N(\delta ) - \delta ^2;
\end{multline}
\begin{equation}\label{2.22a}
\delta\sum\limits_{n=1}^{N}B_n(\delta )=
-1+A_{N+1}(\delta )-C_{N+1}(\delta )
\end{equation}
and
\begin{equation}
\delta ^3\sum\limits_{n=1}^{N}B_n(\delta )=
C_{N+3}(\delta ) + (\delta -2)C_{N+2}(\delta ) +
(1-\delta -\delta ^2)C_{N+1}(\delta ) - \delta ^2.
\label{2.23a}
\end{equation}
\end{lemma}


\begin{definition}\label{qf2-def2.1}
We define $\mathcal{A} _n(\delta ):=3A_n(\delta )-B_n(\delta )-C_n(\delta )$,
$\delta\in\mathbb{C}$, $n\in\mathbb{N}$.
Moreover, we set $\mathcal{A}_n:=\mathcal{A} _n(1)$
(see Table~\ref{qf2-tab3a} and \seqnum{A033304} in~\cite{sloan}).
\end{definition}


\begin{lemma}\label{qf2-lem2.3}
From identities~(\ref{2.14}--\ref{2.16}) the following special identities
can be deduced:

\noindent
a)
\begin{align*}
\mathcal{A} _n^2&=\mathcal{A} _{2n}+2(\xi +\xi ^6)^n+2(\xi ^2+\xi ^5)^n+
2(\xi ^3+\xi ^4)^n \\
&=\mathcal{A} _{2n}+2\Big( 2\cos\frac{2\pi}{7}\Big)^n+2\Big( 2\cos\frac{4\pi}{7}\Big)^n+
2\Big( 2\cos\frac{6\pi}{7}\Big) ^n
\end{align*}
or
$$
\Big( 2\cos\frac{2\pi}{7}\Big) ^n+\Big( 2\cos\frac{4\pi}{7}\Big) ^n+
\Big( 2\cos\frac{6\pi}{7}\Big) ^n=\frac{1}{2}(\mathcal{A} _n^2-\mathcal{A} _{2n}),
$$
and
\begin{multline*}
\Big( 4\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\Big) ^n+
\Big( 4\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}\Big) ^n+
\Big( 4\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}\Big) ^n= \\
=(-1)^n\big[ (1+\xi +\xi ^6)^n+(1+\xi ^2+\xi ^5)^n+
(1+\xi ^3+\xi ^4)^n\big] =(-1)^n\mathcal{A} _n.
\end{multline*}
Consequently, we get the following decomposition:
\begin{multline*}
\Big(\mathbb{X} -\Big( 2\cos\frac{2\pi}{7}\Big) ^n\Big)
\Big(\mathbb{X} -\Big( 2\cos\frac{4\pi}{7}\Big) ^n\Big)
\Big(\mathbb{X} -\Big( 2\cos\frac{6\pi}{7}\Big) ^n\Big)=\\
=\mathbb{X} ^3-\frac{1}{2}(\mathcal{A} _n^2-\mathcal{A} _{2n})
\mathbb{X} ^2+(-1)^n\mathcal{A} _n\mathbb{X} -1;
\end{multline*}

\noindent
b)
$$
2^{-n}\mathcal{A} _n\Big(\frac{1}{2}\Big) =
\Big(\cos\frac{\pi}{7}\Big) ^{2n}+\Big(\cos\frac{2\pi}{7}\Big) ^{2n}+
\Big(\cos\frac{3\pi}{7}\Big) ^{2n};
$$

\noindent
c)
\begin{align*}
\mathcal{A} _n^3&=\mathcal{A} _{3n}+6(-1)^n+3\mathcal{A} _n(-1)+
3(\xi +\xi ^6-\xi ^2-\xi ^5-1)^n \\
&\phantom{=}+3(\xi ^3+\xi ^4-\xi -\xi ^6-1)^n+3(\xi ^2+\xi ^5-\xi ^3-\xi ^4-1)^n;
\end{align*}

\noindent
d)
\begin{multline*}
7C_{n+1}\mathcal{A} _n -7C_n\mathcal{A} _{n+1}=
(-1 + \xi ^3 + \xi ^4 - 4(\xi ^2 + \xi ^5))(\xi ^3 +\xi ^4)^n+ \\
+(-1 + \xi ^2 +\xi ^5 - 4(\xi +\xi ^6))(\xi ^2 +\xi ^5)^n +
(-1 +\xi +\xi ^6 - 4(\xi ^3 +\xi ^4))(\xi  +\xi ^6)^n= \\
=\Big(-1 + 2\cos\frac{6\pi}{7}-8\cos\frac{4\pi}{7}\Big)
\Big( 2\cos\frac{6\pi}{7}\Big)^n +\\
+\Big(-1 + 2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}\Big)
\Big( 2\cos\frac{4\pi}{7}\Big)^n+ %\\
\end{multline*}
$$%\begin{multline*}
+\Big(  - 1 + 2\cos\frac{2\pi}{7}-8\cos\frac{6\pi}{7}\Big)
\Big( 2\cos\frac{2\pi}{7}\Big) ^n
=\left\{\begin{array}{lcl}
7; &  \mbox{\it for} & n=2,4; \\
-7; &  \mbox{\it for} & n=6; \\
-70; &  \mbox{\it for} & n=8; \\
14; &  \mbox{\it for} & n=1; \\
7\cdot A_{(n+1)/2}; & \mbox{\it for} & n=3,5,7; \\
\end{array}\right.
$$%\end{multline*}

\noindent
e)
\begin{multline*}
7(B_{n+1} + C_{n+1} - 2A_{n+1})(A_n + B_n - C_n) - 7(B_n + C_n - 2A_n)(A_{n+1} +
B_{n+1} - C_{n+1})= \\
=(8 + 3(\xi ^2 +\xi ^5))(\xi +\xi ^6)^n + (8 + 3(\xi +\xi ^6))
(\xi ^3 +\xi ^4)^n +(8 + 3(\xi ^3 +\xi ^4))(\xi ^2 +\xi ^5)^n= \\
=\Big( 8+6\cos\frac{4\pi}{7}\Big)\Big( 2\cos\frac{2\pi}{7}\Big)^n+
\Big( 8+6\cos\frac{2\pi}{7}\Big)\Big( 2\cos\frac{6\pi}{7}\Big) ^n+ \\
+\Big( 8+6\cos\frac{6\pi}{7}\Big)\Big( 2\cos\frac{4\pi}{7}\Big) ^n
=\left\{\begin{array}{lcl}
-14; &  \mbox{\it for} & n=1; \\
49; & \mbox{\it for} & n=2; \\
-56; & \mbox{\it for} & n=3,
\end{array}\right.
\end{multline*}

\noindent
and a~more general identity

\noindent
f)
\begin{align*}
(\alpha 7A_{n+1} &+ \beta 7B_{n+1}+\gamma 7C_{n+1})
(\varepsilon 7A_n+\omega 7B_n+\varphi 7C_n)- \\
&\phantom{=} -(\alpha 7A_n+\beta 7B_n+\gamma 7C_n)
 (\varepsilon 7A_{n+1}+\omega 7B_{n+1}+\varphi 7C_{n+1})= \\
&= \big((\alpha _0 +\beta _0(\xi +\xi ^6)+
\gamma _0(\xi ^2+\xi ^5))(1+\xi +\xi ^6)^{n+1}+ \\
&\phantom{=} +(\alpha _0 +\beta _0(\xi ^2+\xi ^5)+
  \gamma _0(\xi ^3+\xi ^4))(1+\xi ^2+\xi ^5)^{n+1}+ \\
&\phantom{=} +(\alpha _0 +\beta _0(\xi ^3+\xi ^4)+
  \gamma _0(\xi +\xi ^6))(1+\xi ^3+\xi ^4)^{n+1}\big) \times \\
  &\phantom{=} \times \big((\varepsilon _0 +\omega _0(\xi +\xi ^6)+
  \varphi _0(\xi ^2+\xi ^5))(1+\xi +\xi ^6)^n+%\\
\end{align*}
\begin{align*}
&\phantom{=} +(\varepsilon _0 +\omega _0(\xi ^2+\xi ^5)+
   \varphi _0(\xi ^3+\xi ^4))(1+\xi ^2+\xi ^5)^n+ \\
&\phantom{=} +(\varepsilon _0 +\omega _0(\xi ^3+\xi ^4)+
    \varphi _0(\xi +\xi ^6))(1+\xi ^3+\xi ^4)^n\big)- \\
&\phantom{=} -\big((\alpha _0 +\beta _0(\xi +\xi ^6)+
   \gamma _0(\xi ^2+\xi ^5))(1+\xi +\xi ^6)^n+ \\
&\phantom{=} +(\alpha _0 +\beta _0(\xi ^2+\xi ^5)+
   \gamma _0(\xi ^3+\xi ^4))(1+\xi ^2+\xi ^5)^n+ \\
&\phantom{=} +(\alpha _0 +\beta _0(\xi ^3+\xi ^4)+
   \gamma _0(\xi +\xi ^6))(1+\xi ^3+\xi ^4)^n\big) \times \\
&\phantom{=} \times \big((\varepsilon _0 +\omega _0(\xi +\xi ^6)+
   \varphi _0(\xi ^2+\xi ^5))(1+\xi +\xi ^6)^{n+1}+ \\
&\phantom{=} +(\varepsilon _0 +\omega _0(\xi ^2+\xi ^5)+
  \varphi _0(\xi ^3+\xi ^4))(1+\xi ^2+\xi ^5)^{n+1}+ \\
&\phantom{=} +(\varepsilon _0 +\omega _0(\xi ^3+\xi ^4)+
  \varphi _0(\xi +\xi ^6))(1+\xi ^3+\xi ^4)^{n+1}\big)= \\
&= \big(\mathcal{A} +(\xi +\xi ^6)\mathcal{B} +
(\xi ^2+\xi ^5)\mathcal{C} \big)(\xi +\xi ^6)^n+ \\
&\phantom{=} +\big(\mathcal{A} +(\xi +\xi ^6)\mathcal{C} +
  (\xi ^3+\xi ^4)\mathcal{B}\big) (\xi ^3+\xi ^4)^n+ \\
&\phantom{=} +\big(\mathcal{A} +(\xi ^2+\xi ^5)\mathcal{B} +
  (\xi ^3+\xi ^4)\mathcal{C}\big) (\xi ^3+\xi ^5)^n= \\
%\end{align*}
%\begin{align*}
&= \Big( \mathcal{A} +2\cos\frac{2\pi}{7}\mathcal{B} +
  2\cos\frac{4\pi}{7}\mathcal{C}\Big)\Big( 2\cos\frac{2\pi}{7}\Big) ^n+ \\
&\phantom{=} +\Big( \mathcal{A} -2\cos\frac{\pi}{7}\mathcal{B} +
  2\cos\frac{2\pi}{7}\mathcal{C}\Big)\Big( -2\cos\frac{\pi}{7}\Big) ^n+ \\
&\phantom{=} +\Big( \mathcal{A} +2\cos\frac{4\pi}{7}\mathcal{B} -
  2\cos\frac{\pi}{7}\mathcal{C}\Big)\Big( 2\cos\frac{4\pi}{7}\Big) ^n,
\end{align*}
where
\begin{align*}
\mathcal{A} &:= 5\varepsilon _0\beta _0-5\alpha _0\omega _0+
   \alpha_0\varphi_0-\varepsilon_0\gamma_0+2\omega_0\gamma_0-2\beta_0\varphi_0, \\
\mathcal{B} &:= 5\alpha _0\omega _0-\varepsilon _0\beta _0+
  3\varepsilon _0\gamma _0-3\alpha _0\varphi _0+\gamma _0\omega _0-\beta _0\varphi _0, \\
\mathcal{C} &:= 2\varepsilon _0\beta _0-2\alpha _0\omega _0+
2\beta _0\varphi _0-2\gamma _0\omega _0+\varepsilon _0\gamma _0-\alpha _0\varphi _0,
\end{align*}
\begin{align*}
\alpha _0 &:=3\alpha +\beta +\gamma, &\beta _0&:=\alpha +2\beta +\gamma ,
&\gamma _0&:=\alpha +\beta +2\gamma ,\\
\varepsilon_0&:=3\varepsilon +\omega +\varphi, &\omega_0&:=\varepsilon +2\omega +\varphi,
&\varphi _0&:=\varepsilon +\omega +2\varphi .
\end{align*}
\end{lemma}

\begin{proof}
The identity \textit{b)} follows from~(\ref{1111111}).
\end{proof}
\bigskip

\begin{remark}
The sequence $\{ \tfrac{1}{2} (\mathcal{A}_{n}^{2}-\mathcal{A}_{2n})\}_{n=1}^{\infty}$
(see Table~\ref{qf2-tab3a}) is an accelerator sequence
for Catalan's constant (see~\cite{bradley}).
The generating function of these numbers has the form
$$
\frac{3+2\, x-2\, x^2}{1+x-2\, x^2-x^3}
$$
(see also  \seqnum{A094648} in~\cite{sloan}).
\end{remark}


Some applications of identities~(\ref{aaaa}--\ref{cccc}) will be presented now.


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

\subsection{Reduction formulas for indices}

\begin{lemma}\label{qf2-lem2.1.1}
The following identities hold:
\begin{align}
A_{m+n}(\delta)&=A_m(\delta)A_n(\delta)+2B_m(\delta)B_n(\delta)+
C_m(\delta)C_n(\delta)\\
&-B_m(\delta)C_n(\delta)-B_n(\delta)C_m(\delta),\nonumber \\
B_{m+n}(\delta)&=-C_m(\delta)C_n(\delta)+A_m(\delta)B_n(\delta)+A_n(\delta)B_m(\delta), \\
C_{m+n}(\delta)&=B_m(\delta)B_n(\delta)-C_m(\delta)C_n(\delta)+
A_m(\delta)C_n(\delta)\\
&+A_n(\delta)C_m(\delta)-B_m(\delta)C_n(\delta)-B_n(\delta)C_m(\delta).\nonumber
\end{align}
\end{lemma}


\begin{proof}
First, we note that
$$
(1+\delta(\xi +\xi ^6))^{m+n}=A_{m+n}(\delta)+B_{m+n}(\delta)(\xi +\xi ^6)+
C_{m+n}(\delta)(\xi ^2+\xi ^5).
$$
On the other hand, we have
\begin{align*}
(1+& \delta(\xi +\xi ^6))^{m+n} = (1+\delta(\xi +\xi ^6))^m(1+\delta(\xi +\xi ^6))^n= \\
&=\Big( A_m(\delta)+B_m(\delta)(\xi +\xi ^6)+C_m(\delta)(\xi ^2+\xi ^5)\Big)\times\\
&\phantom{=}\times \Big( A_n(\delta)+B_n(\delta)(\xi +\xi ^6)+
C_n(\delta)(\xi ^2 +\xi ^5)\Big)% \\
\end{align*}
\begin{align*}
&=A_m(\delta)A_n(\delta)+B_m(\delta)B_n(\delta)(\xi ^2+\xi ^5+2)+
C_m(\delta)C_n(\delta)(\xi ^4+\xi ^3+2) \\
&\phantom{=}+A_m(\delta)B_n(\delta)(\xi +\xi ^6)+A_m(\delta)C_n(\delta)(\xi ^2+\xi ^5)+
  A_n(\delta)B_m(\delta)(\xi +\xi ^6)\\
&\phantom{=} +A_n(\delta)C_m(\delta)(\xi ^2+\xi ^5)+
B_m(\delta)C_n(\delta)(\xi +\xi ^6)(\xi ^2+\xi ^5)\\
&\phantom{=}+B_n(\delta)C_m(\delta)(\xi +\xi ^6)(\xi ^2+\xi ^5)\\
&\!\!\!\!\!\!\mbox{(we have $(\xi +\xi ^6)(\xi ^2+\xi ^5)=
              \xi ^3+\xi ^6+\xi +\xi ^4=-1-\xi ^2-\xi ^5$)} \\
&=A_m(\delta)A_n(\delta)+2B_m(\delta)B_n(\delta)+C_m(\delta)C_n(\delta)
-B_m(\delta)C_n(\delta)-B_n(\delta)C_m(\delta) \\
&\phantom{=}+(\xi +\xi ^6)(A_m(\delta)B_n(\delta)+A_n(\delta)B_m(\delta)-
C_m(\delta)C_n(\delta)) \\
%\end{align*}
%\begin{align*}
&\phantom{=}+(\xi ^2+\xi ^5)(B_m(\delta)B_n(\delta)-C_m(\delta)C_n(\delta)+
A_m(\delta)C_n(\delta)\\
&\phantom{=}+A_n(\delta)C_m(\delta)-B_n(\delta)C_m(\delta)-B_m(\delta)C_n(\delta))
\end{align*}
hence by linear independence of~$1$, $\xi +\xi ^6=2\cos\frac{2\pi}{7}$,
$\xi ^2+\xi ^5=2\cos\frac{4\pi}{7}$ over $\mathbb{Q}$
and by Lemma~\ref{qf2-lem2.20} the
reduction formulas follow.
\end{proof}



\begin{corollary}\label{qf2-cor2.1.1.1}
We have
\begin{align}
A_{2n}(\delta)&=A_n^2(\delta)+B_n^2(\delta)+(B_n(\delta)-C_n(\delta))^2, \\
B_{2n}(\delta)&=2A_n(\delta)B_n(\delta)-C_n^2(\delta), \\
C_{2n}(\delta)&=B_n^2(\delta)-C_n^2(\delta)+2C_n(\delta)(A_n(\delta)-B_n(\delta)).
\end{align}
\end{corollary}

\begin{remark}\label{qf2-rem2.1.1.1}
(Which comes from~\cite{sloan}, sequence \seqnum{A006356})
Let $u(k)$, $v(k)$, $w(k)$ be defined by $u(1)=1$, $v(1)=0$, $w(1)=0$ and
\begin{align*}
u(k+1) &= u(k)+v(k)+w(k), \\
v(k+1) &= u(k)+v(k), \\
w(k+1) &= u(k).
\end{align*}
Then we have
$$
u(n+1)=A_n,\quad v(n+1)=B_n,\quad w(n+1)=B_n-C_n;
$$
for every $n\in\mathbb{N}\cup \{0\}$. So, according to Table~\ref{qf2-tab1} we get
\begin{align*}
\{ u(n)\} &=1,1,3,6,14,31,\ldots ,\\
\{ v(n)\} &=0,1,2,5,11,25,\ldots ,
\end{align*}
and
$$
\{ w(n)\} =\{ u(n)\}
$$
(Benoit Cloitre (abcloitre@wanadoo.fr), Apr~05~2002). Moreover
$$
\big(u(k)\big)^2+\big(v(k)\big)^2+\big(w(k)\big)^2=u(2k)
\quad\mbox{\rm (Gary Adamson, Dec 23 2003)}.
$$
The $n$-th term of the sequence $\{u(n)\}$ is the number of paths for a ray of light that enters
two layers of glass, and then is reflected exactly $n$ times before leaving the
layers of glass. One such path (with~$2$ plates of glass and~$3$ reflections) might be
\smallskip

\mbox{\null}\hfill%
\begin{minipage}{120pt}
\begin{verbatim}
...\........./...
-----------------
....\/\..../.....
-----------------
........\/.......
-----------------
\end{verbatim}
\end{minipage}
\hfill\mbox{\null}
\end{remark}


\begin{corollary}\label{qf2-cor2.1.1.2}
We have
\begin{align}
A_{3n}(\delta) &= A_n(\delta)A_{2n}(\delta)-B_n(\delta)C_{2n}(\delta)-
B_{2n}(\delta)C_n(\delta)\\
&\phantom{=}+2B_n(\delta)B_{2n}(\delta)+C_n(\delta)C_{2n}(\delta)\nonumber \\
&= A_n^3(\delta)-B_n^3(\delta)+6A_n(\delta)B_n^2(\delta)+3A_n(\delta)C_n^2(\delta)
-3B_n(\delta)C_n^2(\delta)\nonumber\\
&\phantom{=}+3B_n^2(\delta)C_n(\delta)-6A_n(\delta)B_n(\delta)C_n(\delta),\nonumber \\
%\end{align}
%\begin{align}
B_{3n}(\delta) &= A_{2n}(\delta)B_n(\delta)+A_n(\delta)B_{2n}(\delta)-
C_n(\delta)C_{2n}(\delta) \\
&= 2B_n^3(\delta)+C_n^3(\delta)+3A_n^2(\delta)B_n(\delta)+
3A_n(\delta)C_n^2(\delta)\nonumber\\
&\phantom{=}+3B_n(\delta)C_n^2(\delta)-3B_n^2(\delta)C_n(\delta),\nonumber \\
C_{3n}(\delta) &= B_nvB_{2n}(\delta)+A_n(\delta)C_{2n}(\delta)+A_{2n}(\delta)C_n(\delta)
-C_n(\delta)C_{2n}(\delta)\\
&\phantom{=}-B_n(\delta)C_{2n}(\delta)-B_{2n}(\delta)C_n(\delta)\nonumber \\
&= 3C_n^3(\delta)-B_n^3(\delta)+3A_n^2(\delta)C_n(\delta)+3A_n(\delta)B_n^2(\delta)
+3B_n^2(\delta)C_n(\delta)\nonumber\\
&\phantom{=}-6A_n(\delta)B_n(\delta)C_n(\delta).\nonumber
\end{align}
\end{corollary}



\begin{lemma}\label{qf2-lem3.18}
The following identities are satisfied:
\begin{align}
A_{n-k} (\delta) &= \frac{1}{\Delta}
\left|
\begin{array}{ccc}
A_{n}(\delta) & 2 B_{k}(\delta) - C_{k}(\delta) & C_{k}(\delta) - B_{k}(\delta) \\
B_{n}(\delta) &  A_{k}(\delta) & -C_{k}(\delta) \\
C_{n}(\delta) &  B_{k}(\delta) - C_{k}(\delta) &
A_{k}(\delta) - B_{k}(\delta) - C_{k}(\delta) \\
\end{array}
\right|,\label{kropka1}\\
B_{n-k} (\delta) &= \frac{1}{\Delta}
\left|
\begin{array}{ccc}
A_{k}(\delta) &  A_{n}(\delta) & C_{k}(\delta) - B_{k}(\delta) \\
B_{k}(\delta) &  B_{n}(\delta) & -C_{k}(\delta) \\
C_{k}(\delta) &  C_{n}(\delta) & A_{k}(\delta) - B_{k}(\delta) - C_{k}(\delta) \\
\end{array}
\right|,\label{kropka2}\\
\intertext{and}
C_{n-k} (\delta) &= \frac{1}{\Delta}
\left|
\begin{array}{ccc}
A_{k}(\delta) & 2 B_{k}(\delta) - C_{k}(\delta) & A_{n}(\delta) \\
B_{k}(\delta) &  A_{k}(\delta) & B_{n}(\delta) \\
C_{k}(\delta) &  B_{k}(\delta) - C_{k}(\delta) & C_{n}(\delta) \\
\end{array}
\right|,\label{kropka3}
\end{align}
where
$$
\Delta:=
\left|
\begin{array}{ccc}
A_{k}(\delta) & 2 B_{k}(\delta) - C_{k}(\delta) & C_{k}(\delta) - B_{k}(\delta) \\
B_{k}(\delta) &  A_{k}(\delta) & -C_{k}(\delta) \\
C_{k}(\delta) &  B_{k}(\delta) - C_{k}(\delta) &
A_{k}(\delta) - B_{k}(\delta) - C_{k}(\delta) \\
\end{array}
\right|.
$$
\end{lemma}

\begin{proof}
First we note that
$$
\bigl( 1+\delta (\xi+\xi^6)\bigr)^{n-k} =
A_{n-k}(\delta) +B_{n-k}(\delta) (\xi+\xi^6) +C_{n-k}(\delta) (\xi^2+\xi^5).
$$
On the other hand we obtain
$$
%\begin{multline*}
\bigl( 1+\delta (\xi+\xi^6)\bigr)^{n-k} =
\frac{\bigl( 1+\delta (\xi+\xi^6)\bigr)^{n}}{\bigl( 1+\delta (\xi+\xi^6)\bigr)^{k}} =
\frac{A_{n}(\delta) +B_{n}(\delta) (\xi+\xi^6) +
C_{n}(\delta) (\xi^2+\xi^5)}{A_{k}(\delta) +B_{k}(\delta) (\xi+\xi^6) +
C_{k}(\delta) (\xi^2+\xi^5)}.
$$
%\end{multline*}
The final form of formulas~(\ref{kropka1}),~(\ref{kropka2}) and~(\ref{kropka3})
from the following identity could be derived:
$$
\frac{a +b (\xi+\xi^6) +c (\xi^2+\xi^5)}{d +e (\xi+\xi^6) +f (\xi^2+\xi^5)} =
\alpha +\beta (\xi+\xi^6) +\gamma (\xi^2+\xi^5),
$$
where
\begin{align*}
&\alpha := \frac{1}{D}
\left|
\begin{array}{ccc}
a & 2 e -f  & f-e \\
b &  d & -f \\
c &  e-f & d-e-f \\
\end{array}
\right|, & &
\beta := \frac{1}{D}
\left|
\begin{array}{ccc}
d & a & f-e \\
e & b & -f \\
f & c & d-e-f \\
\end{array}
\right|,\\
&\gamma := \frac{1}{D}
\left|
\begin{array}{ccc}
d & 2 e -f  & a \\
e &  d & b \\
f &  e-f & c \\
\end{array}
\right|
%\end{align*}
\qquad  \mbox{ and}
%$$
& &
D :=
\left|
\begin{array}{ccc}
d & 2 e -f  & f-e \\
e &  d & -f \\
f &  e-f & d-e-f \\
\end{array}
\right|.
%$$
\end{align*}
\end{proof}


\begin{lemma}
The following identity for the determinant $\Delta$ from Lemma~\ref{qf2-lem3.18}
holds:
\begin{multline}\label{duzeA}
\Delta=
\left|
\begin{array}{ccc}
A_{k}(\delta) & 2 B_{k}(\delta) - C_{k}(\delta) & C_{k}(\delta) - B_{k}(\delta) \\
B_{k}(\delta) &  A_{k}(\delta) & -C_{k}(\delta) \\
C_{k}(\delta) &  B_{k}(\delta) - C_{k}(\delta) &
A_{k}(\delta) - B_{k}(\delta) - C_{k}(\delta) \\
\end{array}
\right|=\\[1ex]
=\big(A_{k}(\delta)\big)^{\!3} +\big(B_{k}(\delta)\big)^{\!3} +
\big(C_{k}(\delta)\big)^{\!3} +
3\,A_{k}(\delta)\,B_{k}(\delta)\,C_{k}(\delta)
-\big(A_{k}(\delta)\big)^{\!2} \big(B_{k}(\delta) + C_{k}(\delta)\big)-\\
-2\, A_{k}(\delta) \Big(\big(B_{k}(\delta)\big)^{\!2}+ \big(C_{k}(\delta)\big)^{\!2}\Big)
+ 3\, \big(B_{k}(\delta)\big)^{\!2}\, C_{k}(\delta)
-4\,B_{k}(\delta)\, \big(C_{k}(\delta)\big)^{\!2}=\\
=\big( \delta^3-2\, \delta^2-\delta+1 \big)^{k}.
\end{multline}
\end{lemma}

\begin{proof}
By~(\ref{2.13}) we obtain
\begin{multline}\label{duzeB}
\Big[ \big( 1+\delta (\xi+\xi^6)\big)
\big( 1+\delta (\xi^2+\xi^5)\big)
\big( 1+\delta (\xi^3+\xi^4)\big)
\Big]^{k}
=\big( \delta^3-2\, \delta^2-\delta+1 \big)^{k}.
\end{multline}
On the other hand, by~(\ref{aaaa})--(\ref{cccc}) we get
\begin{multline}\label{duzeC}
\Big( \big( 1+\delta (\xi+\xi^6)\big)
\big( 1+\delta (\xi^2+\xi^5)\big)
\big( 1+\delta (\xi^3+\xi^4)\big)
\Big)^{k}=\\
=\big( 1+\delta (\xi+\xi^6)\big)^k
\big( 1+\delta (\xi^2+\xi^5)\big)^k
\big( 1+\delta (\xi^3+\xi^4)\big)^k=\\
=\big( A_k(\delta) + B_k (\delta) ( \xi + \xi^6 ) + C_k (\delta) (\xi^2 + \xi^5 ) \big)
\big( A_k(\delta) + B_k (\delta) ( \xi^2 + \xi^5 ) +
C_k (\delta) (\xi^3 + \xi^4 ) \big)\times\\
\times\big( A_k(\delta) + B_k (\delta) ( \xi^3 + \xi^4 ) +
C_k (\delta) (\xi + \xi^6 ) \big)=\\
=\big(A_{k}(\delta)\big)^3 +a\, \big(B_{k}(\delta)\big)^3 +
a\, \big(C_{k}(\delta)\big)^3 +
b\, \,A_{k}(\delta)\,B_{k}(\delta)\,C_{k}(\delta)+\qquad\qquad{}\\
+c\,\big(A_{k}(\delta)\big)^2 \big(B_{k}(\delta) + C_{k}(\delta)\big)
+d\,\, A_{k}(\delta) \Big(\big(B_{k}(\delta)\big)^2+ \big(C_{k}(\delta)\big)^2\Big)+\\
+ e\, \big(B_{k}(\delta)\big)^2\, C_{k}(\delta)
+f\,B_{k}(\delta)\, \big(C_{k}(\delta)\big)^2,
\end{multline}
where (see~Lemma~\ref{qf2-lem1.3})
\begin{align*}
a&= ( \xi + \xi^6 )(\xi^2 + \xi^5 )(\xi^3 + \xi^4 )=1,\\
b&= ( \xi + \xi^6 )(\xi^2 + \xi^5 )+( \xi + \xi^6 )(\xi^3 + \xi^4 )+
(\xi^2 + \xi^5 )(\xi^3 + \xi^4 )+\\
 &\ \ \ +( \xi + \xi^6 )^2+(\xi^2 + \xi^5 )^2+
(\xi^3 + \xi^4 )^2=3,\\
c&= \xi +\xi^2 + \xi^3 + \xi^4 + \xi^5 +\xi^6 =-1,\\
d&=( \xi + \xi^6 )(\xi^2 + \xi^5 )+( \xi + \xi^6 )(\xi^3 + \xi^4 )+
(\xi^2 + \xi^5 )(\xi^3 + \xi^4 )=-2,\\
e&=( \xi + \xi^6 )^2(\xi^2 + \xi^5 )+( \xi + \xi^6 )(\xi^3 + \xi^4 )^2+
(\xi^2 + \xi^5 )^2(\xi^3 + \xi^4 )=3,\\
f&=( \xi + \xi^6 )(\xi^2 + \xi^5 )^2+( \xi + \xi^6 )^2(\xi^3 + \xi^4 )+
(\xi^2 + \xi^5 )(\xi^3 + \xi^4 )^2=-4.
\end{align*}
Hence~(\ref{duzeA}) follows from~(\ref{duzeB}) and~(\ref{duzeC}).
\end{proof}


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


\subsection{Reduction formulas for $\delta$-arguments}


Next, our aim will be to prove some more general identities
connecting the quasi-Fibo\-na\-cci numbers
with different $\delta$'s ($\delta\neq 2$).
Let us start with the following formula:
\begin{multline}\label{100a}
\big(2 + \delta (\xi ^2 +\xi ^5)+\delta (\xi ^3 +\xi ^4)\big)^n =
\big((2 -\delta ) -\delta (\xi +\xi ^6)\big)^n=\\
=(2 -\delta )^n\bigg( 1 +\frac{\delta}{\delta -2}(\xi  +\xi ^6)\bigg)^n
= (2 - \delta )^n A_n \bigg(\frac{\delta}{\delta -2}\bigg) +\\
+(2 -\delta )^nB_n \bigg(\frac{\delta}{\delta  - 2}\bigg) (\xi  +\xi ^6)  +
(2 -\delta )^nC_n \bigg(\frac{\delta}{\delta  - 2}\bigg) (\xi ^2 + \xi ^5).
\end{multline}
But, we can deduce another equivalent formula
\begin{align}\label{101a}
\Big(2&+\delta (\xi ^2 + \xi ^5)+\delta (\xi ^3+\xi ^4)\Big)^n=
\Big(\big(1+\delta (\xi ^2+\xi ^5)\big)+\big(1+\delta (\xi ^3+\xi ^4)\big)\Big)^n \\
&= \sum\limits_{k=0}^{n}\binom{n}{k} (1+\delta (\xi ^2+\xi ^5))^k
    (1+\delta (\xi ^3+\xi ^4))^{n-k}\nonumber \\
&= \sum\limits_{k=0}^{n}\binom{n}{k}
  \Big( A_k(\delta )+B_k(\delta )(\xi ^2+\xi ^5)+
  C_k(\delta )(\xi ^3+\xi ^4)\Big)\times\nonumber \\
&\phantom{=}\times \Big( A_{n-k}(\delta )+
  B_{n-k}(\delta )(\xi ^3+\xi ^4)+C_{n-k}(\delta )(\xi +\xi ^6)\Big)\nonumber \\
&= \sum\limits_{k=0}^{n}\binom{n}{k}
   \Big( A_k(\delta )A_{n-k}(\delta )+A_k(\delta )
     B_{n-k}(\delta )(\xi ^3+\xi ^4)+A_k(\delta )C_{n-k}(\delta )(\xi +\xi ^6)\nonumber \\
&\phantom{=}+A_{n-k}(\delta )B_k(\delta )(\xi ^2+\xi ^5)+
   B_k(\delta )B_{n-k}(\delta )(\xi ^5+\xi ^6+\xi +\xi ^2)\nonumber \\
&\phantom{=}+B_k(\delta )C_{n-k}(\delta )(\xi ^3+\xi +\xi ^6+\xi ^4)+
   A_{n-k}(\delta )C_k(\delta )(\xi ^3+\xi ^4)\nonumber \\
&\phantom{=}+B_{n-k}(\delta )C_k(\delta )(\xi +\xi ^6+2)+
  C_k(\delta )C_{n-k}(\delta )(\xi ^4+\xi ^5+\xi ^2+\xi ^3)\Big)=\nonumber
\end{align}
\begin{align*}
&= \sum\limits_{k=0}^{n}\binom{n}{k}(A_k(\delta )A_{n-k}(\delta )-
  A_k(\delta )B_{n-k}(\delta )-B_k(\delta )C_{n-k}(\delta )-
  A_{n-k}(\delta )C_k(\delta ) \\
&\phantom{=} +2B_{n-k}(\delta )C_k(\delta )-C_k(\delta )C_{n-k}(\delta )) \\
&\phantom{=}+(\xi +\xi ^6)\sum\limits_{k=0}^{n}\binom{n}{k}
(-A_k(\delta )B_{n-k}(\delta )+
  A_k(\delta )C_{n-k}(\delta )+B_k(\delta )B_{n-k}(\delta ) \\
&\phantom{=} -A_{n-k}(\delta )C_k(\delta )+B_{n-k}(\delta )C_k(\delta )-
   C_k(\delta )C_{n-k}(\delta )) \\
&\phantom{=}+(\xi ^2+\xi ^5)\sum\limits_{k=0}^{n}\binom{n}{k}
(-A_k(\delta )B_{n-k}(\delta )+
  A_{n-k}(\delta )B_k(\delta ) \\
&\phantom{=} +B_k(\delta )B_{n-k}(\delta )-B_k(\delta )C_{n-k}(\delta )-
  A_{n-k}(\delta )C_k(\delta )).
\end{align*}
From~(\ref{100a}) and~(\ref{101a}),
again by linear independence of~$1$, $\xi +\xi ^6=2\cos\frac{2}{7}\pi$ and
$\xi ^2+\xi ^5=2\cos\frac{4}{7}\pi$ over $\mathbb{Q}$
and by Lemma~\ref{qf2-lem2.20}
the identities below can be generated:

\begin{lemma}
We have
\begin{align}
(2 &-\delta )^n A_n\Big(\frac{\delta}{\delta - 2}\Big)=
\sum\limits_{k=0}^{n}\binom{n}{k}\Big( A_k(\delta )A_{n-k}(\delta ) -
  A_k(\delta )B_{n-k}(\delta )  \label{2.25a}\\
&-B_k(\delta )C_{n-k}(\delta ) -A_{n-k}(\delta )C_k(\delta ) +
2B_{n-k}(\delta )C_k(\delta ) -
 C_k(\delta )C_{n-k}(\delta )\Big),\nonumber %\\
\end{align}
\begin{align}
(2 &-\delta )^n B_n\Big(\frac{\delta}{\delta - 2}\Big) =
\sum\limits_{k=0}^{n}\binom{n}{k}\Big(  - A_k(\delta )B_{n-k}(\delta )  +
 A_k(\delta )C_{n-k}(\delta )  \label{2.26a} \\
&+B_k(\delta )B_{n-k}(\delta )-A_{n-k}(\delta )C_k(\delta ) +
B_{n-k}(\delta )C_k(\delta ) -
C_k(\delta )C_{n-k}(\delta )\Big)\nonumber\\
\intertext{and}
(2 &-\delta )^n C_n\bigg(\frac{\delta}{\delta - 2}\bigg) =
\sum\limits_{k=0}^{n}\binom{n}{k}\Big(  - A_k(\delta )B_{n-k}(\delta ) \label{2.27a} \\
&+A_{n-k}(\delta )B_k(\delta )+B_k(\delta )B_{n-k}(\delta ) -
B_k(\delta )C_{n-k}(\delta ) -
A_{n-k}(\delta )C_k(\delta )\Big).\nonumber
\end{align}
\end{lemma}

In the sequel, from the above formulas the following
special identities can be obtained:

\noindent
the case $\delta =1$
\begin{align}
A_n(-1)=&\sum\limits_{k=0}^{n}\binom{n}{k}\big(A_kA_{n-k}-A_kB_{n-k}-
   B_kC_{n-k}-A_{n-k}C_k\label{2.28a} \\
&+2B_{n-k}C_k-C_kC_{n-k}\big),\nonumber\\
B_n(-1)=&\sum\limits_{k=0}^{n}\binom{n}{k}\big(-A_kB_{n-k}+A_kC_{n-k}+
   B_kB_{n-k}-A_{n-k}C_k\label{2.29a} \\
&+B_{n-k}C_k-C_kC_{n-k}\big),\nonumber%\\
\end{align}
\begin{align}
C_n(-1)=&\sum\limits_{k=0}^{n}\binom{n}{k}
\big(-A_kB_{n-k}+A_{n-k}B_k+B_kB_{n-k}\label{2.30a} \\
&-B_kC_{n-k}-A_{n-k}C_k\big);\nonumber
\end{align}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% przypadek \delta=3

\noindent
the case $\delta=3$
\begin{align}
(&-1)^n A_n(3)=\sum\limits_{k=0}^{n}\binom{n}{k}\big(A_k(3)A_{n-k}(3)-A_k(3)B_{n-k}(3)-
   \label{d3a} \\
&-B_k(3)C_{n-k}(3)-A_{n-k}(3)C_k(3)+2B_{n-k}(3)C_k(3)-C_k(3)C_{n-k}(3)\big),\nonumber\\
(&-1)^n B_n(3)=\sum\limits_{k=0}^{n}\binom{n}{k}\big(-A_k(3)B_{n-k}(3)+A_k(3)C_{n-k}(3)+
   \label{d3b} \\
&+B_k(3)B_{n-k}(3)-A_{n-k}(3)C_k(3)+B_{n-k}(3)C_k(3)-C_k(3)C_{n-k}(3)\big),\nonumber\\
(&-1)^n C_n(3)=\sum\limits_{k=0}^{n}\binom{n}{k}\big(-A_k(3)B_{n-k}(3)+A_{n-k}(3)B_k(3)+
\label{d3c} \\
&+B_k(3)B_{n-k}(3)-B_k(3)C_{n-k}(3)-A_{n-k}(3)C_k(3)\big).\nonumber
\end{align}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% przypadek \delta=2

The case $\delta=2$ must be treated in another way
\begin{align*}
\big(2&+2 (\xi^2 +\xi^5) +2 (\xi^3 +\xi^4)\big)^{2n}=
\big(-2(\xi +\xi^6 ) \big)^{2n}=\\
&=2^{2n}\big(2+(\xi^2 +\xi^5 ) \big)^{n}=
2^{3n}\big(1+\tfrac{1}{2}(\xi^2 +\xi^5 ) \big)^{n}=\\
&= 2^{3n} \Big( A_{n} \big( \tfrac{1}{2} \big) +
B_{n} \big( \tfrac{1}{2} \big) (\xi^2+\xi^5)+
C_{n} \big( \tfrac{1}{2} \big) (\xi^3+\xi^4) \Big)=\\
&= 2^{3n} \Big( A_{n} \big( \tfrac{1}{2} \big) +
B_{n} \big( \tfrac{1}{2} \big) (\xi^2+\xi^5)+
C_{n} \big( \tfrac{1}{2} \big) (-1-\xi-\xi^6-\xi^2-\xi^5) \Big)=\\
&= 2^{3n} \Big( A_{n} \big( \tfrac{1}{2} \big) - C_{n} \big( \tfrac{1}{2} \big)+
\big( B_{n}\big( \tfrac{1}{2} \big) - C_{n} \big( \tfrac{1}{2} \big) \big) (\xi^2+\xi^5)-
C_{n} \big( \tfrac{1}{2} \big) (\xi+\xi^6) \Big),
\end{align*}
which, by~(\ref{101a}), implies the next five identities
\begin{align}
2^{3n}&\big( A_{n} \big( \tfrac{1}{2} \big) - C_{n} \big( \tfrac{1}{2} \big) \big)=
\sum\limits_{k=0}^{2n}\binom{2n}{k}\big(A_k(2)A_{2n-k}(2)-A_k(2)B_{2n-k}(2)-
   \label{d2a} \\
&-B_k(2)C_{2n-k}(2)-A_{2n-k}(2)C_k(2)+2B_{2n-k}(2)C_k(2)-
C_k(2)C_{2n-k}(2)\big),\nonumber\\
2^{3n}&\big( B_{n} \big( \tfrac{1}{2} \big) - C_{n} \big( \tfrac{1}{2} \big) \big)=
\sum\limits_{k=0}^{2n}\binom{2n}{k}\big(-A_k(2)B_{2n-k}(2)-A_{2n-k}(2)B_{k}(2)+
   \label{d2b} \\
&+B_k(2)B_{2n-k}(2)-B_{k}(2)C_{2n-k}(2)-A_{2n-k}(2)C_{k}(2)\big),\nonumber\\
-2^{3n}& C_{n} \big( \tfrac{1}{2} \big) =
\sum\limits_{k=0}^{2n}\binom{2n}{k}\big(-A_k(2)B_{2n-k}(2)+A_{k}(2)C_{2n-k}(2)+
   \label{d2c} \\
&+B_k(2)B_{2n-k}(2)-A_{2n-k}(2)C_{k}(2)-B_{2n-k}(2)C_{k}(2)-
C_k(2) C_{2n-k} (k) \big),\nonumber
\end{align}
hence, after some manipulations, we obtain
\begin{align}
2^{3n} A_{n} \big( \tfrac{1}{2} \big) &=
\sum\limits_{k=0}^{2n}\binom{2n}{k}\big(A_k(2)A_{2n-k}(2)-A_k(2)C_{2n-k}(2)-
   \label{d2aa} \\
&-B_k(2)B_{2n-k}(2)-B_{k}(2)C_{2n-k}(2)+B_{2n-k}(2)C_k(2)\big),\nonumber\\
2^{3n}B_{n} \big( \tfrac{1}{2} \big) &=
\sum\limits_{k=0}^{2n}\binom{2n}{k}\big(-A_k(2)C_{2n-k}(2)+A_{2n-k}(2)B_{k}(2)-
   \label{d2bb} \\
&-B_k(2)C_{2n-k}(2)-B_{2n-k}(2)C_{k}(2)+C_{k}(2)C_{2n-k}(2)\big).\nonumber
\end{align}



We have also the identity
$$
\big( 1 + \xi + \xi^6 \big)
\big( 1 + \delta \big( \xi^2 + \xi^5\big) \big) =
\big( 1 + \xi + \xi^6 \big) -\delta,
$$
i.e.,
$$
\big( 1 + \xi + \xi^6 \big)^{n}
\big( 1 + \delta \big( \xi^2 + \xi^5\big) \big)^{n} =
\Big(\big( 1 + \xi + \xi^6 \big) -\delta\Big)^{\!n},
$$
which, by~(\ref{aaaa}) and~(\ref{bbbb}), yields
\begin{multline*}
\Big(A_n+B_n \big( \xi + \xi^6\big) +C_n \big( \xi^2 + \xi^5 \big) \Big)
\Big(A_n(\delta)+B_n(\delta) \big( \xi^2 + \xi^5\big) +C_n(\delta)
\big( \xi^3 + \xi^4 \big) \Big)=\\
= \sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k} \big( 1 + \xi + \xi^6 \big)^{k}
= \sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k}A_k +\\
+\big( \xi + \xi^6\big)\sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k} B_k
+\big( \xi^2 + \xi^5\big) \sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k}C_k.
\end{multline*}
Hence, by Lemma~\ref{qf2-lem2.20}, the following three identities
can be discovered:
\begin{align}
A_n A_n (\delta) &- A_n C_n(\delta) - B_n B_n(\delta) -B_n C_n(\delta)
+ C_n B_n(\delta)=\\
&=\sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k} A_k,\nonumber\\
-A_n C_n (\delta) &+ B_n A_n(\delta) - B_n C_n(\delta) - C_n B_n(\delta)
+ C_n C_n(\delta)=\\
&=\sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k} B_k,\nonumber\\
A_n B_n (\delta) &- A_n C_n(\delta) - B_n B_n(\delta) + C_n A_n(\delta)
- C_n B_n(\delta) + C_n C_n(\delta)=\\
&=\sum_{k=0}^{n} (-\delta)^{n-k} \binom{n}{k} C_k.\nonumber
\end{align}


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

\subsection{Polynomials associated with quasi-Fibonacci
numbers\protect{\newline} of order~7}

Directly from equation~(\ref{aaaa}) we obtain
(for $x_k:=1+2\delta\cos\frac{2k\pi}{7}$, $k=1,2,3$)
\begin{align*}
x^{n}_{k} &=
  A_n(\delta )+B_n(\delta )\bigg( 2\cos\frac{2k\pi}{7}\bigg)+
  C_n(\delta )\bigg( 2\cos\frac{4k\pi}{7}\bigg) \\
&= A_n(\delta ) + \frac{B_n(\delta )}{\delta}\, x_{k}  -
 \frac{B_n(\delta )}{\delta} + \frac{C_n(\delta )}{\delta ^2}
 \bigg( 2\delta ^2\bigg( 2\cos ^2\frac{2k\pi}{7} -1\bigg)\bigg) \\
&= A_n(\delta ) + \frac{B_n(\delta )}{\delta}\, x_{k}
    -  \frac{B_n(\delta )}{\delta}
  + \frac{C_n(\delta )}{\delta ^2}
   \Big(x_{k}^{2} - 2\,x_{k} +1-2\delta ^2\Big) \\
&= A_n(\delta ) - \frac{B_n(\delta )}{\delta} -
\frac{C_n(\delta )}{\delta^2}(2\delta^2-1) +
   \frac{B_n(\delta )}{\delta} x_{k} - 2\frac{C_n(\delta )}{\delta^2} x_{k} +
   \frac{C_n(\delta )}{\delta ^2} x_{k}^{2},
\end{align*}
i.e.,
\begin{multline*}
W_{n,7}(x;\delta):=\delta ^2x^n-C_n(\delta )x^2+(2C_n(\delta )-\delta B_n(\delta ))x+\\
+\delta B_n(\delta )+(2\delta ^2-1) C_n(\delta )-\delta ^2A_n(\delta )=0
\end{multline*}
for $x=x_k$, $k=1,2,3$.
So, by identity~(\ref{2.13}) the polynomial
$p_7(\mathbb{X};\delta)$ is a~divisor of the polynomial
$W_{n,7}(\mathbb{X};\delta)$
(for every $n\geqslant 3$). In the sequel we obtain (for $\delta =1$)
\begin{equation}\label{qf2-w3.46}
(x^3-2x^2-x+1)\big| (x^n-C_nx^2+(2C_n-B_n)x+B_n+C_n-A_n)
\end{equation}
for every $n\in\mathbb{N}$, $n\geqslant 3$.
More precisely, the following decomposition holds:
\begin{equation}\label{prz}
p_7(\mathbb{X};\delta)
\Big(\sum\limits_{k=1}^{n-2}C_{k+1}(\delta )\mathbb{X} ^{n-2-k}\Big) =
W_{n,7}(\mathbb{X};\delta),
\end{equation}
hence, for $\delta =1$, we obtain
%\begin{multline}
\begin{equation}
(x^3 - 2x^2 - x + 1)\bigg(\sum\limits_{k=1}^{n-2}B_kx^{n-2-k}\bigg)  %=\\
= x^n - C_nx^2 + (2C_n - B_n)x + B_n + C_n - A_n.
\label{prz1}
\end{equation}
%\end{multline}

\begin{remark}
Equation~(\ref{prz})
after differentiating and some manipulating
enables us to generate
the sums formulas of the following form:
$$
\Big(p_7(\mathbb{X};\delta)\Big)^{r+1}
\bigg(\sum\limits_{k=r}^{n}k^rC_k(\delta )\mathbb{X} ^{n-k-r}\bigg)=
\begin{array}{l}
\mbox{polynomial depending}\\
\mbox{on $W_{n,7}(\mathbb{X};\delta)$, $p_7(\mathbb{X};\delta)$}\\
\mbox{and their derivatives}
\end{array}
$$
For example, the identity~(\ref{prz}) implies the identity~(\ref{1001})
(for $\mathbb{X} =1$).
\end{remark}


\begin{remark}
There exist other ways to obtain the relation~(\ref{qf2-w3.46}).
For example, by~(\ref{aa}) we have
\begin{multline}\label{gwiazda346}
\alpha A_{n+1} (\delta) + \beta B_{n+1} (\delta) + \gamma C_{n+1} (\delta) =\\
=\big( \alpha + \beta\, \delta \big) A_{n} (\delta)
+\big( 2\,\alpha\, \delta + \beta +\gamma\, \delta \big) B_{n} (\delta)
+ \big( -\delta\, \alpha + (1-\delta)\gamma \big) C_{n} (\delta).
\end{multline}
We are interested when the following system of equations holds:
$$
\frac{\alpha+\beta\, \delta}{\alpha} =
\frac{2\,\alpha\, \delta + \beta +\gamma\, \delta}{\beta} =
\frac{-\delta\, \alpha + (1-\delta)\gamma}{\gamma},
$$
which is equivalent to the following one:
$$
\left\{
\begin{array}{l}
\gamma=\frac{1}{\alpha} \Big( \beta^2-2\, \alpha^2\Big),\\
\Big( \frac{\beta}{\alpha}\Big)^3 +
\Big( \frac{\beta}{\alpha}\Big)^2 -
2 \Big( \frac{\beta}{\alpha}\Big) -1 =0.\\
\end{array}
\right.
$$
Hence, by Remark~\ref{qf2-rem2.1.1}, it follows that
$$
\frac{\beta}{\alpha} \in \Big\{ 1+2 \cos \frac{2k\pi}{7}:\ k=1,2,3 \Big\}.
$$
So the equation~(\ref{gwiazda346}) then has the following form:
\begin{align*}
&A_{n+1}(\delta) + 2 \cos \frac{2k\pi}{7} B_{n+1}(\delta) -
\Big( 1+2 \cos \frac{2k\pi}{7}\big)^{-1} C_{n+1}(\delta)=\\
&= \Big( 1+ 2\, \delta \cos \frac{2k\pi}{7}\big)
\Big(A_{n}(\delta) + 2 \cos \frac{2k\pi}{7} B_{n}(\delta) -
\Big( 1+2 \cos \frac{2k\pi}{7}\big)^{-1} C_{n}(\delta)\Big)=\\
&= \Big( 1+ 2\, \delta \cos \frac{2k\pi}{7}\big)^{n}
\Big(A_{1}(\delta) + 2 \cos \frac{2k\pi}{7} B_{1}(\delta) -
\Big( 1+2 \cos \frac{2k\pi}{7}\big)^{-1} C_{1}(\delta)\Big)=\\
&\hspace*{-4mm}\mbox{(by~(\ref{aa}))}\\
&= \Big( 1+ 2\, \delta \cos \frac{2k\pi}{7}\big)^{n+1}
\end{align*}
or
$$
\delta^2\, x^{n+1} = \delta \big( x+\delta -1 \big) \, A_{n}(\delta) +
\big( x+\delta -1 \big) \big( x-1\big)\,B_{n}(\delta)
-\delta^2 C_{n}(\delta),
$$
i.e.,
\begin{multline*}
\delta^2\, x^{n+1}
- x^2\, B_{n}(\delta) + x \big( (2-\delta) \, B_{n}(\delta) -
\delta\, A_{n}(\delta)\big)+\\
+\delta\, (1-\delta) \, A_{n}(\delta) +(\delta-1) \, B_{n}(\delta) +
\delta^2 \, C_{n}(\delta) = 0
\end{multline*}
for $x:=1+2\delta\cos \frac{2k\pi}{7}$, $k=1,2,3$, which is equivalent to
$$
W_{n,7}(\mathbb{X};\delta)=0.
$$
\end{remark}





\begin{lemma}
Immediately from~(\ref{prz1}) we obtain the following identities:
\begin{multline}
(x^3-2x^2-x+1)^2\cdot\Big(\sum\limits_{k=1}^{n-2}B_kx^{n-2-k}\Big)'= \label{33a} \\
=(x^3-2x^2-x+1)^2\cdot\sum\limits_{k=1}^{n-3}(n-2-k)B_kx^{n-3-k}= \\
=(n-3)x^{n+2}-2(n-2)x^{n+1}-(n-1)x^n+nx^{n-1}+C_nx^4+ \\
+2(B_n-2C_n)x^3+(3A_n-5B_n+2C_n)x^2+(-4A_n+4B_n+2C_n)x-A_n+3C_n,
\end{multline}
and
\begin{multline}
(x^3-2x^2-x+1)^3\cdot\Big(\sum\limits_{k=1}^{n-2}B_kx^{n-2-k}\Big)'' \label{34a}
=(n^2-7n+12)x^{n+4}+\\
+(-4n^2+24n-32)x^{n+3}
+(2n^2-10n+18)x^{n+2}
+(6n^2-24n+6)x^{n+1}+\\
+(-3n^2+9n+6)x^{n}
+(2n-2n^2)x^{n-1}
+(n^2-n)x^{n-2}
-2C_nx^6+\\
+(-6B_n+12C_n)x^5
+(-12A_n+24B_n-18C_n)x^4
+(32A_n-42B_n+6C_n)x^3+\\
+(-18A_n+30B_n-18C_n)x^2
+(-6A_n-6B_n+30C_n)x
-6A_n+4B_N+8C_n.
\end{multline}
\end{lemma}

\begin{corollary}\label{qf2-cor2.2.1}
We have
\begin{align*}
\sum\limits_{k=1}^{n-2}&B_k=A_n-2C_n-1,\\
\sum\limits_{k=1}^{n-2}&(-1)^k B_k=-1+(-1)^n \big(A_n-2B_n+2C_{n}\big),\\
\sum\limits_{k=1}^{n-3}&(n-2-k)B_k=-2(A_n-2C_n-1)-n+B_n,\\
\sum\limits_{k=1}^{n-2}&kB_k=nA_n-B_n-2nC_n,\\
\sum\limits_{k=1}^{n-3}&(n-2-k) (-1)^k B_k=6-n+(-1)^{n-1} \big(6A_n-11B_n+8C_{n}\big),\\
\intertext{and}
\sum\limits_{k=1}^{n-3}&(-1)^k k B_k=-4+(-1)^n \Big( 6A_n-11B_n+8C_{n}+\\
&\qquad\qquad\qquad+(n-2)\big(A_n-2B_n+2C_{n}- B_{n-2}\big)\Big).
\end{align*}
\end{corollary}

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

\subsection{Numerical remarks about the zeros of polynomials~$\mathcal{B} _n(x)$}

Let us set
$$
\mathcal{B} _n(x):=\sum\limits_{k=1}^{n}B_kx^{n-k},\qquad n\in\mathbb{N}.
$$
Then,
$$
\mathcal{B} _1(x)\equiv 1,\quad \mathcal{B} _2(x)=x+2,\quad \mathcal{B} _3(x)=x^2+2x+5,
$$
and, thus, we have the recurrence relations (see~(\ref{gg}))
$$
\mathcal{B} _{n+3}(x)=x^{n+2}+2\mathcal{B} _{n+2}(x)+
\mathcal{B} _{n+1}(x)-\mathcal{B} _n(x),
\quad n\geqslant 1.
$$
All polynomials $\mathcal{B} _{n}(x)$ for $n=2k+1$,
$k\in\mathbb{N}$, have an even degree, for example
\begin{align*}
\mathcal{B}_5(x)&=x^4+2x^3+5x^2+11x+25, \\
\mathcal{B}_7(x)&=x^6+2x^5+5x^4+11x^3+25x^2+56x+126.
\end{align*}
As follows from calculations, these polynomials
have no real zeros for $n\leqslant 200$.

However, all polynomials $\mathcal{B}_{n}(x)$ for $n=2k$,
$k\in\mathbb{N}$, have an odd degree, for example
\begin{align*}
\mathcal{B}_4(x)&=x^3+2x^2+5x+11, \\
\mathcal{B}_6(x)&=x^5+2x^4+5x^3+11x^2+25x+56.
\end{align*}
These polynomials have exactly one real zero, which is less than zero
(for $n\leqslant 200$). The zeros~$s_n$
of the consecutive polynomials~$\mathcal{B}_n$ make a~decreasing sequence from
$s_{2}=-2$ to $s_{200}=-2.2442546$ (by numerical calculations).

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

\begin{remark}
All numerical calculations were made in
\textit{Mathematica}.\footnote{\textit{Mathematica\/} is registered trademark of
Wolfram Research~Inc.}
\end{remark}

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

\subsection*{Acknowledgment}

The authors wish to express their gratitude to the referee
for several helpful comments and suggestions
concerning the first version of our paper.


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

\renewcommand{\arraystretch}{1.3}

\begin{table}[H]
{\small %scriptsize
\caption{}\label{qf2-tab-d1}
\medskip

\mbox{\null}\hfill
\begin{tabular}{||c|l||} \hline\hline
      & $A_n(\delta )$                                               \\ \hline\hline
$n=0$ & $1$                             \\ \hline
$n=1$ & $1$                                                                           \\ \hline
$n=2$ & $2\delta ^2+1$                                                                \\ \hline
$n=3$ & $-\delta ^3+6\delta ^2+1$                                                     \\ \hline
$n=4$ & $5\delta ^4-4\delta ^3+12\delta ^2+1$                                         \\ \hline
$n=5$ & $-2\delta ^5+22\delta ^4-7\delta ^3+20\delta ^2+1$                            \\ \hline
$n=6$ & $14\delta ^6-27\delta ^5+72\delta ^4-17\delta ^3+30\delta ^2+1$               \\ \hline
$n=7$ & $-13\delta ^7+92\delta ^6-96\delta ^5+172\delta ^4-32\delta ^3+42\delta ^2+1$ \\ \hline\hline
      &  $B_n(\delta )$      \\ \hline\hline
$n=0$ &  $0$  \\ \hline
$n=1$ &  $\delta$                              \\ \hline
$n=2$ &  $2\delta$                             \\ \hline
$n=3$ &  $2\delta ^3+3\delta$                   \\ \hline
$n=4$ &  $-\delta ^4+8\delta ^3+4\delta$           \\ \hline
$n=5$ &  $5\delta ^5-5\delta ^4+20\delta ^3+5\delta$   \\ \hline
$n=6$ &  $-2\delta ^6+27\delta ^5-12\delta ^4+40\delta ^3+6\delta$  \\ \hline
$n=7$ &  $14\delta ^7-29\delta ^6+99\delta ^5-29\delta ^4+70\delta ^3+7\delta$  \\ \hline\hline
      &  $C_n(\delta )$ \\ \hline\hline
$n=0$ &  $0$ \\ \hline
$n=1$ &  $0$ \\ \hline
$n=2$ &  $\delta ^2$ \\ \hline
$n=3$ &  $-\delta ^3+3\delta ^2$ \\ \hline
$n=4$ &  $3\delta ^4-4\delta ^3+6\delta ^2$ \\ \hline
$n=5$ &  $-4\delta ^5+15\delta ^4-10\delta ^3+10\delta ^2$ \\ \hline
$n=6$ &  $9\delta ^6-24\delta ^5+45\delta ^4-20\delta ^3+15\delta ^2$ \\ \hline
$n=7$ &  $-11\delta ^7+60\delta ^6-81\delta ^5+105\delta ^4-35\delta ^3+21\delta ^2$ \\ \hline\hline
\end{tabular}
\hfill\mbox{\null}

}
\end{table}


\begin{table}[H]
\mbox{\null}\hfill
\begin{minipage}{135pt}
\caption{}\label{qf2-tab1}
\medskip

\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & $A_n$ & $B_n$ & $C_n$ \\ \hline\hline
$0 $ & $1   $ & $0   $ & $0   $ \\ \hline
$1 $ & $1   $ & $1   $ & $0   $ \\ \hline
$2 $ & $3   $ & $2   $ & $1   $ \\ \hline
$3 $ & $6   $ & $5   $ & $2   $ \\ \hline
$4 $ & $14  $ & $11  $ & $5   $ \\ \hline
$5 $ & $31  $ & $25  $ & $11  $ \\ \hline
$6 $ & $70  $ & $56  $ & $25  $ \\ \hline
$7 $ & $157 $ & $126 $ & $56  $ \\ \hline
$8 $ & $353 $ & $283 $ & $126 $ \\ \hline
$9 $ & $793 $ & $636 $ & $283 $ \\ \hline
$10$ & $1782$ & $1429$ & $636 $ \\ \hline
$11$ & $4004$ & $3211$ & $1429$ \\ \hline\hline
\end{tabular}
\end{minipage}
%\end{table}
\hfill
%\begin{table}[htb]
\begin{minipage}{180pt}
\caption{}\label{qf2-tab2}
\medskip

\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & $A_n(-1)$ & $B_n(-1)$ & $C_n(-1)$ \\ \hline\hline
$0 $ & $1    $ & $0     $ & $0    $ \\ \hline
$1 $ & $1    $ & $-1    $ & $0    $ \\ \hline
$2 $ & $3    $ & $-2    $ & $1    $ \\ \hline
$3 $ & $8    $ & $-5    $ & $4    $ \\ \hline
$4 $ & $22   $ & $-13   $ & $13   $ \\ \hline
$5 $ & $61   $ & $-35   $ & $39   $ \\ \hline
$6 $ & $170  $ & $-96   $ & $113  $ \\ \hline
$7 $ & $475  $ & $-266  $ & $322  $ \\ \hline
$8 $ & $1329 $ & $-741  $ & $910  $ \\ \hline
$9 $ & $3721 $ & $-2070 $ & $2561 $ \\ \hline
$10$ & $10422$ & $-5791 $ & $7192 $ \\ \hline
$11$ & $29196$ & $-16213$ & $20175$ \\ \hline\hline
\end{tabular}
\end{minipage}
\hfill\mbox{\null}
\end{table}





\begin{table}[H]
\mbox{\null}\hfill
\begin{minipage}{155pt}
\caption{}\label{qf2-tabd1-2}
\medskip

\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & $A_n(\tfrac{1}{2})$ & $B_n(\tfrac{1}{2})$ & $C_n(\tfrac{1}{2})$ \\ \hline\hline
$0 $ & $1                  $ & $0                  $ & $0                $  \\ \hline
$1 $ & $1                  $ & $\frac{1}{2}        $ & $0                 $  \\ \hline
$2 $ & $\frac{3}{2}        $ & $1                  $ & $\frac{1}{4}       $  \\ \hline
$3 $ & $\frac{19}{8}       $ & $\frac{7}{4}        $ & $\frac{5}{8}       $  \\ \hline
$4 $ & $\frac{61}{16}      $ & $\frac{47}{16}      $ & $\frac{19}{16}     $  \\ \hline
$5 $ & $\frac{197}{32}     $ & $\frac{155}{32}     $ & $\frac{33}{16}     $  \\ \hline
$6 $ & $\frac{319}{32}     $ & $\frac{507}{64}     $ & $\frac{221}{64}    $  \\ \hline
$7 $ & $\frac{2069}{128}   $ & $\frac{413}{32}     $ & $\frac{91}{16}     $  \\ \hline
$8 $ & $\frac{3357}{128}   $ & $\frac{5373}{256}   $ & $\frac{595}{64}    $  \\ \hline
$9 $ & $\frac{10897}{256}  $ & $\frac{4365}{128}   $ & $\frac{7753}{512}  $  \\ \hline
$10$ & $\frac{70755}{1024} $ & $\frac{28357}{512}  $ & $\frac{25213}{1024}$  \\ \hline
$11$ & $\frac{229725}{2048}$ & $\frac{184183}{2048}$ & $\frac{81927}{2048}$  \\ \hline\hline
\end{tabular}
\end{minipage}
%\end{table}
\hfill
%\begin{table}[htb]
\begin{minipage}{170pt}
\caption{}\label{qf2-tabd2}
\medskip

\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & $A_n(2)$ & $B_n(2)$ & $C_n(2)$ \\ \hline\hline
$0 $ & $1     $  & $0     $  & $0     $ \\ \hline
$1 $ & $1     $  & $2     $  & $0     $ \\ \hline
$2 $ & $9     $  & $4     $  & $4     $ \\ \hline
$3 $ & $17    $  & $22    $  & $4     $ \\ \hline
$4 $ & $97    $  & $56    $  & $40    $ \\ \hline
$5 $ & $241   $  & $250   $  & $72    $ \\ \hline
$6 $ & $1097  $  & $732   $  & $428   $ \\ \hline
$7 $ & $3169  $  & $2926  $  & $1036  $ \\ \hline
$8 $ & $12801 $  & $9264  $  & $4816  $ \\ \hline
$9 $ & $40225 $  & $34866 $  & $13712 $ \\ \hline
$10$ & $152265$  & $115316$  & $56020 $ \\ \hline
$11$ & $501489$  & $419846$  & $174612$ \\ \hline\hline
\end{tabular}
\end{minipage}
\hfill\mbox{\null}
\end{table}




\begin{table}[H]
\mbox{\null}\hfill
\begin{minipage}{199pt}
\caption{}\label{qf2-tabd3}
\medskip

\begin{tabular}{||c|c|c|c||}\hline\hline
$n$ & $A_n(3)$ & $B_n(3)$ & $C_n(3)$ \\ \hline\hline
$0 $ & $1       $ & $0       $  & $0      $  \\ \hline
$1 $ & $1       $ & $3       $  & $0      $  \\ \hline
$2 $ & $19      $ & $6       $  & $9      $  \\ \hline
$3 $ & $28      $ & $63      $  & $0      $  \\ \hline
$4 $ & $406     $ & $147     $  & $189    $  \\ \hline
$5 $ & $721     $ & $1365    $  & $63     $  \\ \hline
$6 $ & $8722    $ & $3528    $  & $3969   $  \\ \hline
$7 $ & $17983   $ & $29694   $  & $2646   $  \\ \hline
$8 $ & $188209  $ & $83643   $  & $83790  $  \\ \hline
$9 $ & $438697  $ & $648270  $  & $83349  $  \\ \hline
$10$ & $4078270 $ & $1964361 $  & $1778112$  \\ \hline
$11$ & $10530100$ & $14199171$  & $2336859$  \\ \hline\hline
\end{tabular}
\end{minipage}
%\end{table}
\hfill
\begin{minipage}{183pt}
\caption{}\label{qf2-tab3a}
\medskip

\begin{tabular}{||c|c|c|c||} \hline\hline
$n$ & $\mathcal{A} _n$ &
$\frac{1}{2}(\mathcal{A} _n^2-\mathcal{A} _{2n})$ &
$\mathcal{A} _n(\frac{1}{2})$  \\ \hline\hline
$0 $ & $3   $ & $3   $ & $3$  \\ \hline
$1 $ & $2   $ & $-1  $ & $\frac{5}{2}$  \\ \hline
$2 $ & $6   $ & $ 5  $ & $\frac{13}{4}$  \\ \hline
$3 $ & $11  $ & $-4  $ & $\frac{19}{4}$ \\ \hline
$4 $ & $26  $ & $13  $ & $\frac{117}{16}$ \\ \hline
$5 $ & $57  $ & $-16 $ & $\frac{185}{16}$ \\ \hline
$6 $ & $129 $ & $38  $ & $\frac{593}{32}$ \\ \hline
$7 $ & $289 $ & $-57 $ & $\frac{3827}{128}$ \\ \hline
$8 $ & $650 $ & $117 $ & $\frac{12389}{256}$ \\ \hline
$9 $ & $1460$ & $-193$ & $\frac{40169}{512}$ \\ \hline
$10$ & $3281$ & $370 $ & $\frac{65169}{512}$ \\ \hline
$11$ & $7372$ & $-639$ & $\frac{423065}{2048}$ \\ \hline\hline
\end{tabular}
\end{minipage}
\hfill\mbox{\null}
\end{table}






\renewcommand{\arraystretch}{1}


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\end{thebibliography}


\bigskip
\hrule
\bigskip

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

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


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A006054},
\seqnum{A006356},
\seqnum{A033304},
\seqnum{A085810}, and
\seqnum{A094648}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received January 25 2006;
revised version received August 21 2006; September 5 2006.
Published in {\it Journal of Integer Sequences}, September 6 2006.

\bigskip
\hrule
\bigskip

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


\end{document}
