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1. Fibonacci sequences

A Fibonacci sequence is a sequence of integers where every number is the sum of the two previous ones. Such sequences are called ``secondary Fibonacci sequences" to distinguish them from the tertiary Fibonacci sequences, where every term is the sum of the three previous terms. Beginning with $F(0)=1; F(1)=1,$ we get

\begin{displaymath}
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...  ,
\end{displaymath} (1.1)

where $F(n + 1) = F(n) + F(n - 1).$ One may extend the notion of secondary Fibonacci sequences, originating the so-called generalized secondary Fibonacci sequence (GSFS) $G(n)$:
\begin{displaymath}
a, b, pb + qa, p(pb + qa) + qb,...
\end{displaymath} (1.2)

that satisfies relations of the type
\begin{displaymath}
G(n+1) = p G(n) + q G(n - 1),\;n=1,2,...
\end{displaymath} (1.3)

where $a,b,p,$ and $q$ are real numbers.

The following assertion holds:


If the limit

\begin{displaymath}
x=\lim_{n\rightarrow
\infty}\frac{G(n+1)}{G(n)}
\end{displaymath}

exists then it is a root of equation
\begin{displaymath}
x^2 - px - q = 0,
\end{displaymath} (1.4)

i.e., equals
\begin{displaymath}
\sigma=\frac{p+\sqrt{p^2+4q}}{2}\quad \mbox{ or } \quad
\tau=\frac{p-\sqrt{p^2+4q}}{2}
\end{displaymath} (1.5)

and is independent of $a, b.$


Indeed, from (1.3) we get

\begin{displaymath}
\frac{G(n+1)}{G(n)} = p+ q\frac{G(n-1)}{G(n)},
\end{displaymath}

and passing to the limit gives $x=p+\frac{q}{x}$, which is equivalent to (1.4), (1.5).


Theorem 1.1   Let a GSFS % latex2html id marker 939
$(\ref{q12})$, % latex2html id marker 941
$(\ref{q13})$ be such that $a,b,p$, and $q$ are real numbers such that $p>0,\;p^2+4q>0.$ Then there exists a limit

\begin{displaymath}
\lim_{n \rightarrow \infty}\frac{G(n+1)}{G(n)},
\end{displaymath}

which is a real number.


Proof. To find an expression for the $n$th term of the GSFS, let us write a matrix equation equivalent to (1.3)

\begin{displaymath}
\left(
\begin{array}{c}
G(n+1) G(n)
\end{array}\right)= A
\left(
\begin{array}{c}
G(n) G(n-1)
\end{array}\right),
\end{displaymath}

where
\begin{displaymath}
A =\left(\begin{array}{cc} p & q 1 & 0
\end{array}\right).
\end{displaymath} (1.6)

It is easily seen that

\begin{displaymath}
\left(
\begin{array}{c}
G(n+1) G(n)
\end{array}\right)= A^...
...\right)= A^n
\left(
\begin{array}{c}
b a
\end{array}\right).
\end{displaymath}

Thus, the problem is reduced to finding the $n$th power of the matrix $A$. The characteristic equation of $A$ is

\begin{displaymath}
\left\vert
\begin{array}{cc}
p-\lambda & q\\
1 & -\lambda
\end{array}\right\vert=\lambda^2-p\lambda-q=0,
\end{displaymath}

with the two eigenvalues $\sigma$ and $\tau$ (see above). To diagonalize $A$ and convert it into

\begin{displaymath}
A_d=
\left(\begin{array}{cc}
\sigma & 0\\
0 & \tau
\end{array}\right),
\end{displaymath}

we shall use the matrix

\begin{displaymath}
P=
\left(\begin{array}{cc}
\sigma & \tau\\
1 & 1
\end{array}\right).
\end{displaymath}

Set

\begin{displaymath}
D=
\left\vert\begin{array}{cc}
\sigma & \tau\\
1 & 1
\end{array}\right\vert=\sigma-\tau.
\end{displaymath}

It is easy to check that

\begin{displaymath}
P^{-1}=\frac{1}{D^2}
\left(\begin{array}{cc}
1 & -\tau\\
-1 & \sigma
\end{array}\right)
\end{displaymath}

and that $A=PA_dP^{-1}$. The $n$th power of $A$ is calculated by applying the similarity transformation

\begin{displaymath}
A^n=(PA_dP^{-1})^n=PA_dP^{-1}PA_dP^{-1}\ldots
PA_dP^{-1}=PA_d^nP^{-1}.
\end{displaymath}

Further we have

\begin{displaymath}
A_d^n=
\left(\begin{array}{cc}
\sigma^n & 0\\
0 & \tau^n
\end{array}\right),
\end{displaymath}

so that

\begin{displaymath}
A^n=
\left(\begin{array}{cc}
\sigma & \tau\\
1 & 1
\end{arr...
...(\begin{array}{cc}
1 & -\tau\\
-1 & \sigma
\end{array}\right)
\end{displaymath}


\begin{displaymath}
=\frac{1}{D^2}
\left(\begin{array}{cc}
\sigma^{n+1}- \tau^{n...
...\tau^{n} & -\sigma^{n}\tau+ \tau^{n}\sigma
\end{array}\right).
\end{displaymath}

So

\begin{displaymath}
\left(
\begin{array}{c}
G(n+1) G(n)
\end{array}\right)=\fr...
...{n}) b +(-\sigma^{n}\tau+ \tau^{n}\sigma) a
\end{array}\right)
\end{displaymath}

and

\begin{displaymath}
G(n+1)=\frac{1}{D}\{
(\sigma^{n+1}- \tau^{n+1}) b-\sigma\tau(\sigma^{n}- \tau^{n})a\},
\end{displaymath}


\begin{displaymath}
G(n)=\frac{1}{D}\{(\sigma^{n}- \tau^{n}) b +(-\sigma^{n}\tau+
\tau^{n}\sigma) a\}.
\end{displaymath} (1.7)

It follows that

\begin{displaymath}
\frac{G(n+1)}{G(n)}=\frac{
(\sigma^{n+1}- \tau^{n+1}) b-\sig...
...(\sigma^{n}- \tau^{n})b +(-\sigma^{n}\tau+ \tau^{n}\sigma)a} .
\end{displaymath}

Setting $\varepsilon=\tau/\sigma$ and taking into account that $\vert\varepsilon\vert<1$, we obtain

\begin{displaymath}
\frac{G(n+1)}{G(n)}=\frac{(\sigma-\tau\varepsilon^n)b-\sigma...
...epsilon^n)a}{(1-\varepsilon^n)b+(-\tau+\sigma\varepsilon^n)a},
\end{displaymath}

which tends to $\sigma$ if $b\ne \tau a$ (otherwise, $G(n+1)=\tau^na$, a geometric progression, and the fraction tends to $\tau$). The theorem is proved.


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Next: 2 The family of Up: SPINADEL Previous: SPINADEL
2003-06-05