Theory and Modern Applications

# On the Smarandache-Pascal derived sequences and some of their conjectures

## Abstract

For any sequence $\left\{{b}_{n}\right\}$, the Smarandache-Pascal derived sequence $\left\{{T}_{n}\right\}$ of $\left\{{b}_{n}\right\}$ is defined as ${T}_{1}={b}_{1}$, ${T}_{2}={b}_{1}+{b}_{2}$, ${T}_{3}={b}_{1}+2{b}_{2}+{b}_{3}$, generally, ${T}_{n+1}={\sum }_{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {b}_{k+1}$ for all $n\ge 2$, where $\left(\genfrac{}{}{0}{}{n}{k}\right)=\frac{n!}{k!\left(n-k\right)!}$ is the combination number. In reference (Murthy and Ashbacher in Generalized Partitions and New Ideas on Number Theory and Smarandache Sequences, 2005), authors proposed a series of conjectures related to Fibonacci numbers and its Smarandache-Pascal derived sequence, one of them is that if $\left\{{b}_{n}\right\}=\left\{{F}_{1},{F}_{9},{F}_{17},\dots \right\}$, then we have the recurrence formula ${T}_{n+1}=49\cdot \left({T}_{n}-{T}_{n-1}\right)$, $n\ge 2$. The main purpose of this paper is using the elementary method and the properties of the second-order linear recurrence sequence to study these problems and to prove a generalized conclusion.

## 1 Introduction

For any sequence $\left\{{b}_{n}\right\}$, we define a new sequence $\left\{{T}_{n}\right\}$ through the following method: ${T}_{1}={b}_{1}$, ${T}_{2}={b}_{1}+{b}_{2}$, ${T}_{3}={b}_{1}+2{b}_{2}+{b}_{3}$, generally, ${T}_{n+1}={\sum }_{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {b}_{k+1}$ for all $n\ge 2$, where $\left(\genfrac{}{}{0}{}{n}{k}\right)=\frac{n!}{k!\left(n-k\right)!}$ is the combination number. This sequence is called the Smarandache-Pascal derived sequence of $\left\{{b}_{n}\right\}$. It was introduced by professor Smarandache in [1] and studied by some authors. For example, Murthy and Ashbacher [2] proposed a series of conjectures related to Fibonacci numbers and its Smarandache-Pascal derived sequence; three of them are as follows.

Conjecture 1 Let $\left\{{b}_{n}\right\}=\left\{{F}_{8n+1}\right\}=\left\{{F}_{1},{F}_{9},{F}_{17},{F}_{25},\dots \right\}$, $\left\{{T}_{n}\right\}$ be the Smarandache-Pascal derived sequence of $\left\{{b}_{n}\right\}$, then we have the recurrence formula

${T}_{n+1}=49\cdot \left({T}_{n}-{T}_{n-1}\right),\phantom{\rule{1em}{0ex}}n\ge 2.$

Conjecture 2 Let $\left\{{b}_{n}\right\}=\left\{{F}_{10n+1}\right\}=\left\{{F}_{1},{F}_{11},{F}_{21},{F}_{31},\dots \right\}$, $\left\{{T}_{n}\right\}$ be the Smarandache-Pascal derived sequence of $\left\{{b}_{n}\right\}$, then we have the recurrence formula

${T}_{n+1}=125\cdot \left({T}_{n}-{T}_{n-1}\right),\phantom{\rule{1em}{0ex}}n\ge 2.$

Conjecture 3 Let $\left\{{b}_{n}\right\}=\left\{{F}_{12n+1}\right\}=\left\{{F}_{1},{F}_{13},{F}_{25},{F}_{37},\dots \right\}$, $\left\{{T}_{n}\right\}$ be the Smarandache-Pascal derived sequence of $\left\{{b}_{n}\right\}$, then we have the recurrence formula

${T}_{n+1}=324\cdot \left({T}_{n}-{T}_{n-1}\right),\phantom{\rule{1em}{0ex}}n\ge 2.$

Regarding these conjectures, it seems that no one has studied them yet; at least, we have not seen any related results before. These conjectures are interesting; they reveal the profound properties of the Fibonacci numbers. The main purpose of this paper is using the elementary method and the properties of the second-order linear recurrence sequence to study these problems and to prove a generalized conclusion. That is, we shall prove the following.

Theorem Let $\left\{{X}_{n}\right\}$ be a second-order linear recurrence sequence with ${X}_{0}=u$, ${X}_{1}=v$, ${X}_{n+1}=a{X}_{n}+b{X}_{n-1}$ for all $n\ge 1$, where ${a}^{2}+4b>0$. For any positive integer $d\ge 2$, we define the Smarandache-Pascal derived sequence of $\left\{{X}_{dn+1}\right\}$ as

${T}_{n+1}=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {X}_{dk+1}.$

Then we have the recurrence formula

${T}_{n+1}=\left(2+{A}_{d}+b\cdot {A}_{d-2}\right)\cdot {T}_{n}-\left(1+{A}_{d}+b\cdot {A}_{d-2}+{\left(-b\right)}^{d}\right)\cdot {T}_{n-1},$

where the sequence $\left\{{A}_{n}\right\}$ is defined as ${A}_{0}=1$, ${A}_{1}=a$, ${A}_{n+1}=a\cdot {A}_{n}+b\cdot {A}_{n-1}$ for all $n\ge 1$. In fact this time, the general term is

${A}_{n}=\frac{1}{\sqrt{{a}^{2}+4b}}\left[{\left(\frac{a+\sqrt{{a}^{2}+4b}}{2}\right)}^{n+1}-{\left(\frac{a-\sqrt{{a}^{2}+4b}}{2}\right)}^{n+1}\right].$

Now we take $b=1$, then from our theorem, we may immediately deduce the following three corollaries.

Corollary 1 Let $\left\{{X}_{n}\right\}$ be a second-order linear recurrence sequence with ${X}_{0}=u$, ${X}_{1}=v$, ${X}_{n+1}=a{X}_{n}+{X}_{n-1}$ for all $n\ge 1$. For any even number $d\ge 2$, we have the recurrence formula

${T}_{n+1}=\left(2+{A}_{d}+{A}_{d-2}\right)\cdot \left({T}_{n}-{T}_{n-1}\right),\phantom{\rule{1em}{0ex}}n\ge 2.$

Corollary 2 Let $\left\{{X}_{n}\right\}$ be a second-order linear recurrence sequence with ${X}_{0}=u$, ${X}_{1}=v$, ${X}_{n+1}=a{X}_{n}+{X}_{n-1}$ for all $n\ge 1$. For any odd number $d\ge 2$, we have the recurrence formula

${T}_{n+1}=\left(2+{A}_{d}+{A}_{d-2}\right)\cdot {T}_{n}-\left({A}_{d}+{A}_{d-2}\right)\cdot {T}_{n-1},\phantom{\rule{1em}{0ex}}n\ge 2,$

where

${A}_{n}={A}_{n}\left(a\right)=\frac{1}{\sqrt{{a}^{2}+4}}\left[{\left(\frac{a+\sqrt{{a}^{2}+4}}{2}\right)}^{n+1}-{\left(\frac{a-\sqrt{{a}^{2}+4}}{2}\right)}^{n+1}\right].$

It is clear that ${F}_{n+1}\left(a\right)={A}_{n}\left(a\right)$ is a polynomial of a; sometimes, it is called a Fibonacci polynomial, because ${F}_{n}\left(1\right)={F}_{n}$ is Fibonacci number, see [35].

If we take $a=1$, ${X}_{0}=0$, ${X}_{1}=1$ in Corollary 1, then $\left\{{X}_{n}\right\}=\left\{{F}_{n}\right\}$ is a Fibonacci sequence. Note that ${A}_{n}={F}_{n+1}$, $2+{A}_{8}+{A}_{6}=2+{F}_{9}+{F}_{7}=2+34+13=49$, $2+{A}_{10}+{A}_{8}=2+{F}_{11}+{F}_{9}=2+89+34=125$, $2+{A}_{12}+{A}_{10}=2+{F}_{13}+{F}_{11}=2+233+89=324$; from Corollary 1, we may immediately deduce that the three conjectures above are true.

If we take $a=2$, ${X}_{0}={P}_{0}=0$, ${X}_{1}={P}_{1}=1$ and ${P}_{n+1}=2{P}_{n}+{P}_{n-1}$ for all $n\ge 1$, then ${P}_{n}$ are the Pell numbers. From Corollary 1, we can also deduce the following.

Corollary 3 Let ${P}_{n}$ be the Pell number. Then for any positive integer d and

${T}_{n+1}=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {P}_{2dk+1},$

we have the recurrence formula

${T}_{n+1}=\left(2+{P}_{2d+1}+{P}_{2d-1}\right)\cdot \left({T}_{n}-{T}_{n-1}\right),\phantom{\rule{1em}{0ex}}n\ge 2.$

On the other hand, from our theorem, we know that if $\left\{{b}_{n}\right\}$ is a second-order linear recurrence sequence, then its Smarandache-Pascal derived sequence $\left\{{T}_{n}\right\}$ is also a second-order linear recurrence sequence.

## 2 Proof of the theorem

To complete the proof of our theorem, we need the following.

Lemma Let integers $m\ge 0$ and $n\ge 2$. If the sequence $\left\{{X}_{n}\right\}$ satisfying the recurrence relations ${X}_{n+2}=a\cdot {X}_{n+1}+b\cdot {X}_{n}$, $n\ge 0$, then we have the identity

${X}_{m+n}={A}_{n-1}\cdot {X}_{m+1}+b\cdot {A}_{n-2}\cdot {X}_{m},$

where ${A}_{n}$ is defined as ${A}_{0}=1$, ${A}_{1}=a$ and ${A}_{n+1}=a\cdot {A}_{n}+b\cdot {A}_{n-1}$ for all $n\ge 1$, or

${A}_{n}=\frac{1}{\sqrt{{a}^{2}+4b}}\left[{\left(\frac{a+\sqrt{{a}^{2}+4b}}{2}\right)}^{n+1}-{\left(\frac{a-\sqrt{{a}^{2}+4b}}{2}\right)}^{n+1}\right].$

Proof Now we prove this lemma by mathematical induction. Note that the recurrence formula ${X}_{m+2}=a\cdot {X}_{m+1}+b\cdot {X}_{m}$, ${A}_{1}=a$, ${A}_{0}=1$, ${A}_{n+1}=a\cdot {A}_{n}+b\cdot {A}_{n-1}$ for all $n\ge 1$. So ${X}_{m+2}={A}_{1}\cdot {X}_{m+1}+b\cdot {A}_{0}\cdot {X}_{m}$. That is, the lemma holds for $n=2$. Since ${X}_{m+3}=a\cdot {X}_{m+2}+b\cdot {X}_{m+1}=a\cdot \left(a\cdot {X}_{m+1}+b\cdot {X}_{m}\right)+b\cdot {X}_{m+1}=\left({a}^{2}+b\right)\cdot {X}_{m+1}+ba\cdot {X}_{m}={A}_{2}\cdot {X}_{m+1}+b{A}_{1}\cdot {X}_{m}$. That is, the lemma holds for $n=3$. Suppose that for all integers $2\le n\le k$, we have ${X}_{m+n}={A}_{n-1}\cdot {X}_{m+1}+b\cdot {A}_{n-2}\cdot {X}_{m}$. Then for $n=k+1$, from the recurrence relations for ${X}_{m}$ and the inductive hypothesis, we have

$\begin{array}{rcl}{X}_{m+k+1}& =& a\cdot {X}_{m+k}+b\cdot {X}_{m+k-1}\\ =& a\cdot \left({A}_{k-1}\cdot {X}_{m+1}+b\cdot {A}_{k-2}\cdot {X}_{m}\right)+b\cdot \left({A}_{k-2}\cdot {X}_{m+1}+b\cdot {A}_{k-3}\cdot {X}_{m}\right)\\ =& \left(a\cdot {A}_{k-1}+b\cdot {A}_{k-2}\right)\cdot {X}_{m+1}+b\cdot \left(a\cdot {A}_{k-2}+b\cdot {A}_{k-3}\right)\cdot {X}_{m}\\ =& {A}_{k}\cdot {X}_{m+1}+b\cdot {A}_{k-1}\cdot {X}_{m-1}.\end{array}$

That is, the lemma also holds for $n=k+1$. This completes the proof of our lemma by mathematical induction. □

Now, we use this lemma to complete the proof of our theorem. For any positive integer d, from the definition of ${T}_{n}$ and the properties of the binomial coefficient $\left(\genfrac{}{}{0}{}{n}{k}\right)$, we have

$\begin{array}{rcl}\left(\genfrac{}{}{0}{}{n-1}{k}\right)+\left(\genfrac{}{}{0}{}{n-1}{k-1}\right)& =& \frac{\left(n-1\right)!}{k!\left(n-1-k\right)!}+\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}\\ =& \frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k-1\right)!}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\left(\genfrac{}{}{0}{}{n}{k}\right)\end{array}$
(1)

and

$\begin{array}{rcl}{T}_{n+1}& =& \sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {X}_{dk+1}\\ =& {X}_{1}+{X}_{dn+1}+\sum _{k=1}^{n-1}\left(\left(\genfrac{}{}{0}{}{n-1}{k}\right)+\left(\genfrac{}{}{0}{}{n-1}{k-1}\right)\right){X}_{dk+1}\\ =& \sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk+1}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk+d+1}+{X}_{dn+1}\\ =& {T}_{n}+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right){X}_{dk+d+1}.\end{array}$
(2)

From the lemma, we have${X}_{dk+d+1}={A}_{d}\cdot {X}_{dk+1}+b\cdot {A}_{d-1}\cdot {X}_{dk}$, by (2) and the definition of ${T}_{n}$, we may deduce that

$\begin{array}{rcl}{T}_{n+1}& =& {T}_{n}+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot \left({A}_{d}\cdot {X}_{dk+1}+b\cdot {A}_{d-1}\cdot {X}_{dk}\right)\\ =& {T}_{n}+{A}_{d}\cdot \sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk+1}+b\cdot {A}_{d-1}\cdot \sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk}\\ =& \left(1+{A}_{d}\right)\cdot {T}_{n}+b\cdot {A}_{d-1}\cdot \sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk}.\end{array}$
(3)

On the other hand, from the lemma, we also have ${X}_{dk+d}={A}_{d-1}\cdot {X}_{dk+1}+b\cdot {A}_{d-2}\cdot {X}_{dk}$, from this and formula (1), we have

$\begin{array}{rcl}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk}& =& {X}_{0}+{X}_{d\left(n-1\right)}+\sum _{k=1}^{n-2}\left(\genfrac{}{}{0}{}{n-1}{k}\right)\cdot {X}_{dk}\\ =& {X}_{0}+{X}_{d\left(n-1\right)}+\sum _{k=1}^{n-2}\left(\left(\genfrac{}{}{0}{}{n-2}{k}\right)+\left(\genfrac{}{}{0}{}{n-2}{k-1}\right)\right)\cdot {X}_{dk}\\ =& \sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk}+\sum _{k=0}^{n-3}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk+d}+{X}_{d\left(n-1\right)}\\ =& \sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot \left({A}_{d-1}\cdot {X}_{dk+1}+b\cdot {A}_{d-2}\cdot {X}_{dk}\right)\\ =& \left(1+b\cdot {A}_{d-2}\right)\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk}+{A}_{d-1}\cdot {T}_{n-1}.\end{array}$
(4)

From (3), we can also deduce that

${T}_{n}=\left(1+{A}_{d}\right)\cdot {T}_{n-1}+b\cdot {A}_{d-1}\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk}.$
(5)

Now, combining (3), (4) and (5), we may immediately get

$\begin{array}{rcl}{T}_{n+1}& =& \left(1+{A}_{d}\right)\cdot {T}_{n}+b\cdot {A}_{d-1}\cdot \left(\left(1+b\cdot {A}_{d-2}\right)\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n-2}{k}\right)\cdot {X}_{dk}+{A}_{d-1}\cdot {T}_{n-1}\right)\\ =& \left(1+{A}_{d}\right)\cdot {T}_{n}+b\cdot {A}_{d-1}^{2}\cdot {T}_{n-1}+\left(1+b\cdot {A}_{d-2}\right)\cdot \left({T}_{n}-\left(1+{A}_{d}\right)\cdot {T}_{n-1}\right)\end{array}$

or equivalent to

$\begin{array}{rcl}{T}_{n+1}& =& \left(2+{A}_{d}+b\cdot {A}_{d-2}\right)\cdot {T}_{n}-\left(1+{A}_{d}+b\cdot {A}_{d-2}+b\cdot {A}_{d}\cdot {A}_{d-2}-b\cdot {A}_{d-1}^{2}\right)\cdot {T}_{n-1}\\ =& \left(2+{A}_{d}+b\cdot {A}_{d-2}\right)\cdot {T}_{n}-\left(1+{A}_{d}+b\cdot {A}_{d-2}+{\left(-b\right)}^{d}\right)\cdot {T}_{n-1},\end{array}$
(6)

where we have used the identity

$\begin{array}{rcl}{A}_{d}\cdot {A}_{d-2}-{A}_{d-1}^{2}& =& \frac{-{\left(-b\right)}^{d-1}}{{a}^{2}+4b}\left[{\left(\frac{a+\sqrt{{a}^{2}+4b}}{2}\right)}^{2}+{\left(\frac{a-\sqrt{{a}^{2}+4b}}{2}\right)}^{2}\right]+\frac{2{\left(-b\right)}^{d}}{{a}^{2}+4b}\\ =& -{\left(-b\right)}^{d-1}\cdot \frac{{a}^{2}+2b+2b}{{a}^{2}+4b}=-{\left(-b\right)}^{d-1}.\end{array}$

Now, our theorem follows from formula (6).

## References

1. Smarandache F: Only Problems, Not Solutions. Xiquan Publishing House, Chicago; 1993.

2. Murthy A, Ashbacher C: Generalized Partitions and New Ideas on Number Theory and Smarandache Sequences. Hexis, Phoenix; 2005:79.

3. Rong M, Wenpeng Z: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 2007, 45: 164–170.

4. Yuan Y, Wenpeng Z: Some identities involving the Fibonacci polynomials. Fibonacci Q. 2002, 40: 314–318.

5. Tingting W, Wenpeng Z: Some identities involving Fibonacci, Lucas polynomials and their applications. Bull. Math. Soc. Sci. Math. Roum. 2012, 55(1):95–103.

## Acknowledgements

The authors would like to thank the referee for carefully examining this paper and providing a number of important comments. This work is supported by the N.S.F. (11071194, 11001218) of P.R. China and G.I.C.F. (YZZ12062) of NWU.

## Author information

Authors

### Corresponding author

Correspondence to Di Han.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

XL studied the Smarandache-Pascal derived sequences and proved a generalized conclusion. DH participated in the research and summary of the study.

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Li, X., Han, D. On the Smarandache-Pascal derived sequences and some of their conjectures. Adv Differ Equ 2013, 240 (2013). https://doi.org/10.1186/1687-1847-2013-240

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• DOI: https://doi.org/10.1186/1687-1847-2013-240

### Keywords

• Smarandache-Pascal derived sequence
• Fibonacci number
• combination number
• elementary method
• conjecture