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

$$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... ,$$ (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) $G(n)$:

$$a, b, pb + qa, p(pb + qa) + qb,...$$ (1.2)

that satisfies relations of the type

$$G(n+1) = p G(n) + q G(n - 1),\;n=1,2,...$$ (1.3)

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

The following assertion holds:

If the limit

$$x=\lim_{n\rightarrow \infty}\frac{G(n+1)}{G(n)}$$

exists then it is a root of equation

$$x^2 - px - q = 0,$$ (1.4)

i.e., equals

$$\sigma=\frac{p+\sqrt{p^2+4q}}{2}\quad \mbox{ or } \quad \tau=\frac{p-\sqrt{p^2+4q}}{2}$$ (1.5)

and is independent of $a, b.$

Indeed, from (1.3) we get

$$\frac{G(n+1)}{G(n)} = p+ q\frac{G(n-1)}{G(n)},$$

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 $(\ref{q12})$, $(\ref{q13})$ be such that $a,b,p$, and $q$ are real numbers such that $p>0,\;p^2+4q>0.$ Then there exists a limit

$$\lim_{n \rightarrow \infty}\frac{G(n+1)}{G(n)},$$

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)

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

where

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

It is easily seen that

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

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

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

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

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

we shall use the matrix

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

Set

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

It is easy to check that

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

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

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

Further we have

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

so that

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

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

So

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

and

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

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

It follows that

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

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

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

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.