Double-Recurrence Fibonacci Numbers and Generalizations

TL;DR

This paper introduces and analyzes double-recurrence Fibonacci numbers, extending the classical Fibonacci sequence to two variables, providing formulas and identities, and exploring generalizations of this new sequence.

Contribution

The paper defines a new two-variable Fibonacci extension, derives formulas for its computation, and explores its identities and generalizations, expanding understanding of Fibonacci sequence variants.

Findings

Derived a formula for double-recurrence Fibonacci numbers in terms of standard Fibonacci numbers.

Established identities involving sums of double-recurrence Fibonacci numbers.

Explored the general case of the double-recurrence sequence.

Abstract

Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence given by the recurrence $F_{n+2}=F_{n+1}+F_n$, for $n\geq 0$, where $F_0=0$ and $F_1=1$. There are several generalizations of this sequence and also several interesting identities. In this paper, we investigate a homogeneous recurrence relation that, in a way, extends the linear recurrence of the Fibonacci sequence for two variables, called {\it double-recurrence Fibonacci numbers}, given by ${F(m,n) = F(m-1, n-1)+F (m-2, n-2)}$, for $n,m\geq 2$, where $F (m, 0) = F_m$, $F (m, 1) = F_{m+1}$, $F (0, n) = F_n$ and $F (1, n) = F_{n+1}$. We exhibit a formula to calculate the values of this double recurrence, only in terms of Fibonacci numbers, such as certain identities for their sums are outlined. Finally, a general case is studied.

Tables

Table 1. Table 1: Related sequences.

Initial Condition $[a,b,c,d]$	∑_i=1^n∑_j=0^i-1H(i,j)	∑_j=0^n∑_i=0^jH(i,j)	∑_i,j=0^mH(i,j)
$[0,0,0,1]$	A006478$\left(n\right)$	A001629$\left(n+1\right)$	A006478$\left(n+1\right)$
$[0,0,1,0]$	A002940$\left(n-2\right)$	A006478$\left(n+1\right)$	-
$[0,0,1,1]$	-	A122491$\left(n+2\right)$	-
$[0,1,0,0]$	A001629$\left(n+1\right)$	A006478$\left(n\right)$	A006478$\left(n+1\right)$
$[0,1,0,1]$	A006478$\left(n+1\right)$	A006478$\left(n+1\right)$	A178523$\left(n+1\right)$
$[0,1,1,0]$	A014286$\left(n\right)$	A002940$\left(n-1\right)$	-
$[0,1,1,1]$	-	A178523$\left(n+1\right)$	-
$[1,0,0,0]$	A001629$\left(n\right)$	A010049$\left(n+1\right)$	-
$[1,0,0,1]$	A122491$\left(n+1\right)$	A001629$\left(n+2\right)$	-
$[1,0,1,0]$	A178523$\left(n\right)$	-	-
$[1,0,1,1]$	-	A006478$\left(n+2\right)$	-
$[1,1,0,0]$	A006478$\left(n+1\right)$	A190062$\left(n+1\right)$	-
$[1,1,1,0]$	A002940$\left(n-1\right)$	-	-
$[1,1,1,1]$	-	A014286$\left(n+11\right)$	-

Equations

$$f_{m,n}=f_{m-1,n}+f_{m-2,n},$$

$$f_{m,n}=f_{m-1,n-1}+f_{m-2,n-2},$$

$$f_{0,0}=f_{1,0}=f_{1,1}=f_{2,1}=1.$$

$$F(m,n)=F(m-1,n-1)+F(m-2,n-2),\text{mbox}form,n\ge 2,$$

$$F(m,n)=F_k{F}_{|m-n|+2}+F_{k-1}{F}_{|m-n|}.$$

$$F(m+1,n)$$

$$=F_{k^{\prime }}{F}_{|m-n+1|+2}+F_{k^{\prime }-1}{F}_{|m-n+1|}+F_{k^{\prime }-1}{F}_{|m-n+1|+2}+F_{k^{\prime }-2}{F}_{|m-n+1|},$$

$$F(m+1,n)=F_{k^{\prime }+1}{F}_{|(m+1)-n|+2}+F_{k^{\prime }}{F}_{|(m+1)-n|},$$

$$H(m,n)=H(m-1,n-1)+H(m-2,n-2)$$

$$H(m,0)=H_1(m)$$

$$H(m,1)=H_1^2(m)$$

$$H(0,n)=H_2(n)$$

$$H(1,n)=H_2^1(n)$$

$$H(m,m)=H_1^1(m),\text{mbox}with{H}_1^1(0)=a\text{mbox}and{H}_1^1(1)=c.$$

$$F_a^b(m,n):=b{F}_n{F}_{|m-n|+2}+a{F}_{n-1}{F}_{|m-n|}.$$

$$F_a^b(m+2,n+2)=F_a^b(m+1,n+1)+F_a^b(m,n),$$

$$H(m,n)=F_n{H}_1^2(m-n+1)+F_{n-1}{H}_1(m-n),$$

$$H(m,n)$$

$$=c{F}_n{F}_{m-n+1}+a{F}_{n-1}{F}_{m-n-1}+d{F}_n{F}_{m-n}+b{F}_{n-1}{F}_{m-n}.$$

$$H(m,n)$$

$$=F_{a+b}^c(m-1,n)+F_b^d(m-2,n).$$

$$H(m,n)=F_{a+d}^c(n-1,m)+F_b^d(n-2,m),$$

$$\sum _{i=1}^{n}i{G}_i=n{G}_{n+2}-G_{n+3}+G_3$$

$$\sum _{i=1}^{n}i{G}_i$$

$$=g_0\sum _{i=0}^{n-1}(i+1)F_i+g_1\sum _{i=1}^{n}i{F}_i$$

$$=g_0((n-1)F_{n+1}-F_{n+2}+2+F_{n+1}-1)+g_1(n{F}_{n+2}-F_{n+3}+2)$$

$$=n(g_0{F}_{n+1}+g_1{F}_{n+2})-(g_0{F}_{n+2}+g_1{F}_{n+3})+2g_0+g_1$$

$$=n{G}_{n+2}-G_{n+3}+G_3$$

$$\sum _{\genfrac{}{}{0ex}{}{i,j=0}{j\le i}}^{m}F(i,j)=\frac{2}{5}\left(m{L}_{m+3}-L_{m+4}+2F_{m+2}\right)+2.$$

$$\sum _{i,j=0}^{m}F(i,j)=\frac{4}{5}\left(m{L}_{m+3}-L_{m+4}+2F_{m+2}\right)-F_{m+2}+5.$$

$$\sum _{\genfrac{}{}{0ex}{}{i,j=0}{i\ge j}}^{m}F(i,j)$$

$$=\sum _{i=0}^m\sum _{j=0}^i{F}_j{F}_{i-j+1}+F_{j-1}{F}_{i-j}+F_j{F}_{i-j},$$

$$=\sum _{i=0}^m\sum _{j=0}^i{F}_i+F_j{F}_{i-j}$$

$$=\sum _{i=0}^m\left(\left(i+1\right)F_i+\sum _{j=0}^i{F}_{i-j}{F}_j\right).$$

$$\sum _{\genfrac{}{}{0ex}{}{i,j=0}{i\ge j}}^{m}F(i,j)$$

$$=m\left(\frac{L_{m+2}}{5}+F_{m+2}\right)-\left(\frac{L_{m+3}}{5}+F_{m+3}\right)+\frac{4}{5}{F}_{m+2}+2,$$

$$=\frac{m\left(L_{m+2}+L_{m+1}+L_{m+3}\right)}{5}-\frac{\left(L_{m+4}+L_{m+2}+L_{m+3}\right)}{5}$$

$$+\frac{4}{5}{F}_{m+2}+2$$

$$\therefore \sum _{\genfrac{}{}{0ex}{}{i,j=0}{i\ge j}}^{m}F(i,j)$$

Full text

1 Introduction

Fibonacci numbers are known for their amazing properties, association with geometric figures, among others [7, 4]. Using the usual notation for such numbers, $(F_{n})_{n\geq 0}$, they are given by the following linear recurrence of order two: $F_{n+2}=F_{n+1}+F_{n}$, for $n\geq 0$, where $F_{0}=0$ and $F_{1}=1$. The Fibonacci sequence has been generalized in many ways, some by preserving the initial conditions, and others by studying high order recurrences with similar initial conditions [6, 3].

Our interest relies in a generalization that uses a recurrence for two indices (called a double-recurrence), such as the one studied by Hosoya [2], who defined a set of integers $\{f_{m,n}\}$ satisfying:

[TABLE]

[TABLE]

for all $m\geq 2$, $m\geq n\geq 0$, where

[TABLE]

Those numbers, when arranged triangularly, provide the famous Fibonacci Triangle (also known as Hosoya’s Triangle). One of our goals is to construct an analogue of the Fibonacci Triangle, studying a similar double-recurrence. The set of numbers $\{F(m,n)\}$, will be required to satisfy the following,

[TABLE]

with initial values

\begin{array}[]{l l}F(m,0)=F_{m},&F(1,n)=F_{n+1},\\ F(m,1)=F_{m+1},&F(0,n)=F_{n}\ .\end{array}

The initial conditions above, along with (1), are sufficient to calculate the value of $F(m,n)$ at each $(m,n)\in\mathbb{N}^{2}$. We call the values of the set $\{F(m,n)\}$, double-recurrence Fibonacci numbers. Note that $F(m,n)$ is a symmetric function, since the initial conditions above and below the main diagonal are the same, and that $F(k,i)=F(k,k-i)$ for all $0\leq i\leq\lfloor k/2\rfloor$. Figure 1, displays a few values for $F(m,n)$, considering the bottom left corner as the origin $(0,0)$, and the $(m,n)$ coordinate having the value for $F(m,n)$.

Consider the value of the coordinate $(7,4)$, given by $F(7,4)=19$, and then draw a parallel to the antidiagonal from this point towards the axis, where the interactions begin with initial values $F(3,0)=F_{3}$ and $F(4,1)=F_{5}$. This means that, in order to determine $F(7,4)$, we only needed the pair $F_{3}$ and $F_{5}$, in other words, only Fibonacci numbers. The following proposition, asserts that this property is true for all $F(m,n)$, meaning that these values can be obtained using only Fibonacci numbers.

subgriddiv=1,griddots=7,gridlabels=0pt(7,7) 13211819191821138131112111382158778513183545381119233257121912134711181123581321011235813

Proposition 1**.**

Let  $m,n\in\mathbb{N}$, and $F(m,n)$ be a double-recurrence Fibonacci number, with $k:=\min\{m,n\}$. Then,

[TABLE]

Proof.

We proceed by the induction principle for two variables. It Is straightforward that $F(0,0)=F_{0}=F_{0}F_{2}+F_{-1}F_{0}$. So, supposing that (2) holds for all $\ i\leq m$ and $j\leq n$, we have

[TABLE]

where $k^{\prime}=\min\{m,n-1\}\Rightarrow k^{\prime}-1=\min\{m-1,n-2\}$. Therefore,

[TABLE]

and since $k^{\prime}+1=\min\{m+1,n\}$, the identity holds in this case. Analogously, following the same steps, the identity also holds for $F(m,n+1)$, which completes the proof. ∎

In the homogeneous double-recurrence (1), one could replace the initial conditions by a general linear recurrence sequence of order two, or even arithmetic functions. In other words, we have the following:

Definition 2**.**

Let $m,n\in\mathbb{N}$. The function $H(m,n)$ satisfying

[TABLE]

for all $m,n\geq 2$, where the following initial conditions are given

\begin{array}[]{ll}H(m,0)=H_{1}(m),&\;\;H(0,n)=H_{2}(n),\\ H(m,1)=H_{1}^{2}(m),&\;\;H(1,n)=H_{2}^{1}(n),\end{array}

with $H_{1}$, $H_{2}$, $H_{1}^{2}$ and $H_{2}^{1}$ arithmetic functions, is called a double-recurrence function. If $H_{1}$, $H_{2}$, $H_{1}^{2}$ and $H_{2}^{1}$ are linear recurrence sequences of order two, the function satisfying (3) is called a spin Function.

In this way, double-recurrence Fibonacci numbers are values of a spin Function, such as every Fibonacci and Lucas numbers. Now, let $H(m,n)$ be a spin Function, where

[TABLE]

and if $m=n$, we have a linear recurrence sequence of order two, given by:

[TABLE]

The motivation for the term spin function, relies on the way that we can reach, from the initial terms, all pairs of $(m,n)\in\mathbb{N}^{2}$, where the function is evaluated, using every secondary diagonal on it, that we refer as strings. A graphical representation of it, can be seen next.

subgriddiv=1,griddots=7,gridlabels=0pt(4,4) H_{1}(m)$$a$$b$$H_{1}^{2}(m)$$c$$d$$H_{1}(n)$$H_{1}^{2}(n)$$H_{1}^{1}(n)

subgriddiv=1,griddots=7,gridlabels=0pt(4,4) $F_{m}$[math]1$$F_{m+1}$$1$$1$$F_{n}$$F_{n+1}$$F_{n}

2 Properties and Identities

Among several generalizations for Fibonacci numbers, we now consider the ones that satisfies the Fibonacci recurrence relation, but with arbitrary initial conditions.

Definition 3**.**

Let $(G_{n})_{n}$ a linear recurrence sequence of order two, where $G_{1}=a$, $G_{2}=b$ and $G_{n+2}=G_{n+1}+G_{n}$, $n\geq 1$. The ensuing sequence is called a generalized Fibonacci sequence (GFS).

The following, is a classical result, that can be easily proved by induction, which states that every term on a GFS, can be written only in terms of Fibonacci numbers and their initial conditions.

Theorem 4**.**

Let $G_{n}$ denote the nth term of the GFS. Then $G_{n+2}=bF_{n+1}+aF_{n}$, $n\geq 1$.

Proof.

See [5, Th. 7.1]. ∎

Note that, Proposition 1 can be seen as a generalization of Theorem 4 for double-recurrence Fibonacci numbers. Our immediate purpose is to show that an analogous result also holds for spin functions. In order to do so, we introduce a double-recurrence function that will play the same role as Fibonacci numbers on Theorem 4. Let $m,n,a,b\in\mathbb{N}$. Then, define

[TABLE]

It is easy to see that $F_{a}^{b}(m,n)$ is a double-recurrence function, but not necessarily a spin Function, i.e.,

[TABLE]

but the functions on the initial conditions are not necessarily linear recurrence sequences of order two. For that, we have the following result.

Proposition 5**.**

Let $m,n\in\mathbb{N}$ and the spin function $H(m,n)$, such as on Definition 2. Then,

1. i.

If $n\leq m-1$, then $H(m,n)=F_{a+b}^{c}(m-1,n)+F_{b}^{d}(m-2,n)$. 2. ii.

If $m-1<n$, then $H(m,n)=F_{a+d}^{c}(n-1,m)+F_{d}^{b}(n-2,m)$.

Proof.

Let $H(m,n)$ be a spin function for $n\leq m-1$, with functions $H_{1}^{2}$ and $H_{1}$ given by the initial conditions described previously. Similarly to the Proposition 1, we have

[TABLE]

and since $H_{1}^{2}$ and $H_{1}$ are linear recurrence sequences, using Theorem 4, we get

[TABLE]

Using that $bF_{n-1}F_{m-n}=b\cdot(F_{n-1}F_{m-n-2}+F_{n-1}F_{m-n-1})$, we obtain

[TABLE]

Analogously, for $m-1<n$, considering $H_{2}^{1}$ and $H_{2}$, we get

[TABLE]

which completes the proof. ∎

subgriddiv=1,griddots=7,gridlabels=0pt(4,4) H_{1}(m)$$a$$b$$H_{1}^{2}(m)$$d$$c$$H_{2}(n)$$H_{2}^{1}(n)$$H_{1}^{1}(n)

subgriddiv=1,griddots=7,gridlabels=0pt(4,4) H_{1}(m)$$a$$b$$H_{1}^{2}(m)$$d$$c$$H_{2}(n)$$H_{2}^{1}(n)$$H_{1}^{1}(n)

Now, we return our attention to sums of double-recurrence Fibonacci numbers. But first, we recall an interesting identity for Generalized Fibonacci Numbers [9], giving an alternative proof for it.

Proposition 6**.**

Let $(G_{n})_{n}$ be a GFS, where $G_{n}=G_{n-1}+G_{n-2}$ with initial conditions $G_{0}=g_{0}$ and $G_{1}=g_{1}$. Then

[TABLE]

Proof.

Straightforward from Theorem 4, we have $G_{n}=g_{0}F_{n-1}+g_{1}F_{n}$. Thus,

[TABLE]

Where, from (8) to (9), the identity $\sum_{i=1}^{n}iF_{i}=nF_{n+2}-F_{n+3}+2$, [8, p.16, Ex.10], is used. ∎

The following proposition, consists of a closed form to calculate the sums of Double-Fibonacci numbers, where the indices are in $\{1,\ldots,m\}^{2}$.

Proposition 7**.**

Let $F(i,j)$ be Double-Fibonacci Numbers, where $i,j\in\{0,1,\ldots,m\}$. Then,

1. i.

The sum of all Double-Fibonacci Numbers with indices below the main diagonal, including it, is given by

[TABLE] 2. ii.

The sum of all Double-Fibonacci Numbers, with indices on the square $m\times m$, is

[TABLE]

Proof.

First, we proceed to prove (i), and use it to prove (ii). Rewriting (10), and using the closed form on Proposition 1, we have

[TABLE]

completing the proof for (i). For (ii), we use the symmetry satisfied by double-recurrence Fibonacci Numbers, $F(m,n)=F(n,m)$, giving us that the sum on (ii) is two times the sum on (i), minus the sum for indices on the main diagonal:

[TABLE]

∎

Out of curiosity, equation (10) happens to be the same formula for the path length of the Fibonacci tree of order $n$. (A178523 of [1])

3 Acknowledgements

During the preparation of this paper, Ana Paula Chaves was supported in part by CNPq Universal 01/2016 - 427722/2016-0 grant, and Carlos Alirio Rico Acevedo was fully supported by a Masters Scholarship from CNPq.

Appendix

The following table explicit some interesting sequences founded on [1], that can be obtained from the sum of the terms of $H(i,j)$, with initial conditions $a,b,c$ and $d$, considering $0\leq j<i\leq n$, $0\leq i\leq j\leq n$, and all $i,j\in\{0,1,\ldots,n\}^{2}$.