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

Li, Xiaoxue; Han, Di

doi:10.1186/1687-1847-2013-240

Research
Open access
Published: 08 August 2013

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

Xiaoxue Li¹ &
Di Han¹

Advances in Difference Equations volume 2013, Article number: 240 (2013) Cite this article

1747 Accesses
1 Citations
Metrics details

Abstract

For any sequence ${b_{n}}$ , the Smarandache-Pascal derived sequence ${T_{n}}$ of ${b_{n}}$ is defined as $T_{1} = b_{1}$ , $T_{2} = b_{1} + b_{2}$ , $T_{3} = b_{1} + 2 b_{2} + b_{3}$ , generally, $T_{n + 1} = \sum_{k = 0}^{n} (\binom{n}{k}) \cdot b_{k + 1}$ for all $n \geq 2$ , where $(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$ 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 ${b_{n}} = {F_{1}, F_{9}, F_{17}, \dots}$ , then we have the recurrence formula $T_{n + 1} = 49 \cdot (T_{n} - T_{n - 1})$ , $n \geq 2$ . 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 ${b_{n}}$ , we define a new sequence ${T_{n}}$ through the following method: $T_{1} = b_{1}$ , $T_{2} = b_{1} + b_{2}$ , $T_{3} = b_{1} + 2 b_{2} + b_{3}$ , generally, $T_{n + 1} = \sum_{k = 0}^{n} (\binom{n}{k}) \cdot b_{k + 1}$ for all $n \geq 2$ , where $(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$ is the combination number. This sequence is called the Smarandache-Pascal derived sequence of ${b_{n}}$ . 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 ${b_{n}} = {F_{8 n + 1}} = {F_{1}, F_{9}, F_{17}, F_{25}, \dots}$ , ${T_{n}}$ be the Smarandache-Pascal derived sequence of ${b_{n}}$ , then we have the recurrence formula

T_{n + 1} = 49 \cdot (T_{n} - T_{n - 1}), n \geq 2 .

Conjecture 2 Let ${b_{n}} = {F_{10 n + 1}} = {F_{1}, F_{11}, F_{21}, F_{31}, \dots}$ , ${T_{n}}$ be the Smarandache-Pascal derived sequence of ${b_{n}}$ , then we have the recurrence formula

T_{n + 1} = 125 \cdot (T_{n} - T_{n - 1}), n \geq 2 .

Conjecture 3 Let ${b_{n}} = {F_{12 n + 1}} = {F_{1}, F_{13}, F_{25}, F_{37}, \dots}$ , ${T_{n}}$ be the Smarandache-Pascal derived sequence of ${b_{n}}$ , then we have the recurrence formula

T_{n + 1} = 324 \cdot (T_{n} - T_{n - 1}), n \geq 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 ${X_{n}}$ 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 $n \geq 1$ , where $a^{2} + 4 b > 0$ . For any positive integer $d \geq 2$ , we define the Smarandache-Pascal derived sequence of ${X_{d n + 1}}$ as

T_{n + 1} = \sum_{k = 0}^{n} (\binom{n}{k}) \cdot X_{d k + 1} .

Then we have the recurrence formula

T_{n + 1} = (2 + A_{d} + b \cdot A_{d - 2}) \cdot T_{n} - (1 + A_{d} + b \cdot A_{d - 2} + {(- b)}^{d}) \cdot T_{n - 1},

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

A_{n} = \frac{1}{\sqrt{a^{2} + 4 b}} [{(\frac{a + \sqrt{a^{2} + 4 b}}{2})}^{n + 1} - {(\frac{a - \sqrt{a^{2} + 4 b}}{2})}^{n + 1}] .

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

Corollary 1 Let ${X_{n}}$ 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 $n \geq 1$ . For any even number $d \geq 2$ , we have the recurrence formula

T_{n + 1} = (2 + A_{d} + A_{d - 2}) \cdot (T_{n} - T_{n - 1}), n \geq 2 .

Corollary 2 Let ${X_{n}}$ 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 $n \geq 1$ . For any odd number $d \geq 2$ , we have the recurrence formula

T_{n + 1} = (2 + A_{d} + A_{d - 2}) \cdot T_{n} - (A_{d} + A_{d - 2}) \cdot T_{n - 1}, n \geq 2,

where

A_{n} = A_{n} (a) = \frac{1}{\sqrt{a^{2} + 4}} [{(\frac{a + \sqrt{a^{2} + 4}}{2})}^{n + 1} - {(\frac{a - \sqrt{a^{2} + 4}}{2})}^{n + 1}] .

It is clear that $F_{n + 1} (a) = A_{n} (a)$ is a polynomial of a; sometimes, it is called a Fibonacci polynomial, because $F_{n} (1) = F_{n}$ is Fibonacci number, see [3–5].

If we take $a = 1$ , $X_{0} = 0$ , $X_{1} = 1$ in Corollary 1, then ${X_{n}} = {F_{n}}$ 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 $n \geq 1$ , 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} (\binom{n}{k}) \cdot P_{2 d k + 1},

we have the recurrence formula

T_{n + 1} = (2 + P_{2 d + 1} + P_{2 d - 1}) \cdot (T_{n} - T_{n - 1}), n \geq 2 .

On the other hand, from our theorem, we know that if ${b_{n}}$ is a second-order linear recurrence sequence, then its Smarandache-Pascal derived sequence ${T_{n}}$ 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 $m \geq 0$ and $n \geq 2$ . If the sequence ${X_{n}}$ satisfying the recurrence relations $X_{n + 2} = a \cdot X_{n + 1} + b \cdot X_{n}$ , $n \geq 0$ , 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 $n \geq 1$ , or

A_{n} = \frac{1}{\sqrt{a^{2} + 4 b}} [{(\frac{a + \sqrt{a^{2} + 4 b}}{2})}^{n + 1} - {(\frac{a - \sqrt{a^{2} + 4 b}}{2})}^{n + 1}] .

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 $n \geq 1$ . 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 $X_{m + 3} = a \cdot X_{m + 2} + b \cdot X_{m + 1} = a \cdot (a \cdot X_{m + 1} + b \cdot X_{m}) + b \cdot X_{m + 1} = (a^{2} + b) \cdot X_{m + 1} + b a \cdot X_{m} = A_{2} \cdot X_{m + 1} + b A_{1} \cdot X_{m}$ . That is, the lemma holds for $n = 3$ . Suppose that for all integers $2 \leq n \leq k$ , 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 (A_{k - 1} \cdot X_{m + 1} + b \cdot A_{k - 2} \cdot X_{m}) + b \cdot (A_{k - 2} \cdot X_{m + 1} + b \cdot A_{k - 3} \cdot X_{m}) \\ = & (a \cdot A_{k - 1} + b \cdot A_{k - 2}) \cdot X_{m + 1} + b \cdot (a \cdot A_{k - 2} + b \cdot A_{k - 3}) \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 $(\binom{n}{k})$ , we have

\begin{array}{rcl} (\binom{n - 1}{k}) + (\binom{n - 1}{k - 1}) & = & \frac{(n - 1)!}{k! (n - 1 - k)!} + \frac{(n - 1)!}{(k - 1)! (n - k)!} \\ = & \frac{(n - 1)!}{(k - 1)! (n - k - 1)!} (\frac{1}{k} + \frac{1}{n - k}) = (\binom{n}{k}) \end{array}

(1)

and

\begin{array}{rcl} T_{n + 1} & = & \sum_{k = 0}^{n} (\binom{n}{k}) \cdot X_{d k + 1} \\ = & X_{1} + X_{d n + 1} + \sum_{k = 1}^{n - 1} ((\binom{n - 1}{k}) + (\binom{n - 1}{k - 1})) X_{d k + 1} \\ = & \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \cdot X_{d k + 1} + \sum_{k = 0}^{n - 2} (\binom{n - 1}{k}) \cdot X_{d k + d + 1} + X_{d n + 1} \\ = & T_{n} + \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) X_{d k + d + 1} . \end{array}

(2)

From the lemma, we have $X_{d k + d + 1} = A_{d} \cdot X_{d k + 1} + b \cdot A_{d - 1} \cdot X_{d k}$ , 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} (\binom{n - 1}{k}) \cdot (A_{d} \cdot X_{d k + 1} + b \cdot A_{d - 1} \cdot X_{d k}) \\ = & T_{n} + A_{d} \cdot \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \cdot X_{d k + 1} + b \cdot A_{d - 1} \cdot \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \cdot X_{d k} \\ = & (1 + A_{d}) \cdot T_{n} + b \cdot A_{d - 1} \cdot \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \cdot X_{d k} . \end{array}

(3)

On the other hand, from the lemma, we also have $X_{d k + d} = A_{d - 1} \cdot X_{d k + 1} + b \cdot A_{d - 2} \cdot X_{d k}$ , from this and formula (1), we have

\begin{array}{rcl} \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \cdot X_{d k} & = & X_{0} + X_{d (n - 1)} + \sum_{k = 1}^{n - 2} (\binom{n - 1}{k}) \cdot X_{d k} \\ = & X_{0} + X_{d (n - 1)} + \sum_{k = 1}^{n - 2} ((\binom{n - 2}{k}) + (\binom{n - 2}{k - 1})) \cdot X_{d k} \\ = & \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot X_{d k} + \sum_{k = 0}^{n - 3} (\binom{n - 2}{k}) \cdot X_{d k + d} + X_{d (n - 1)} \\ = & \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot X_{d k} + \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot (A_{d - 1} \cdot X_{d k + 1} + b \cdot A_{d - 2} \cdot X_{d k}) \\ = & (1 + b \cdot A_{d - 2}) \cdot \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot X_{d k} + A_{d - 1} \cdot T_{n - 1} . \end{array}

(4)

From (3), we can also deduce that

T_{n} = (1 + A_{d}) \cdot T_{n - 1} + b \cdot A_{d - 1} \cdot \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot X_{d k} .

(5)

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

\begin{array}{rcl} T_{n + 1} & = & (1 + A_{d}) \cdot T_{n} + b \cdot A_{d - 1} \cdot ((1 + b \cdot A_{d - 2}) \cdot \sum_{k = 0}^{n - 2} (\binom{n - 2}{k}) \cdot X_{d k} + A_{d - 1} \cdot T_{n - 1}) \\ = & (1 + A_{d}) \cdot T_{n} + b \cdot A_{d - 1}^{2} \cdot T_{n - 1} + (1 + b \cdot A_{d - 2}) \cdot (T_{n} - (1 + A_{d}) \cdot T_{n - 1}) \end{array}

or equivalent to

\begin{array}{rcl} T_{n + 1} & = & (2 + A_{d} + b \cdot A_{d - 2}) \cdot T_{n} - (1 + A_{d} + b \cdot A_{d - 2} + b \cdot A_{d} \cdot A_{d - 2} - b \cdot A_{d - 1}^{2}) \cdot T_{n - 1} \\ = & (2 + A_{d} + b \cdot A_{d - 2}) \cdot T_{n} - (1 + A_{d} + b \cdot A_{d - 2} + {(- b)}^{d}) \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{- {(- b)}^{d - 1}}{a^{2} + 4 b} [{(\frac{a + \sqrt{a^{2} + 4 b}}{2})}^{2} + {(\frac{a - \sqrt{a^{2} + 4 b}}{2})}^{2}] + \frac{2 {(- b)}^{d}}{a^{2} + 4 b} \\ = & - {(- b)}^{d - 1} \cdot \frac{a^{2} + 2 b + 2 b}{a^{2} + 4 b} = - {(- b)}^{d - 1} . \end{array}

Now, our theorem follows from formula (6).

References

Smarandache F: Only Problems, Not Solutions. Xiquan Publishing House, Chicago; 1993.
Google Scholar
Murthy A, Ashbacher C: Generalized Partitions and New Ideas on Number Theory and Smarandache Sequences. Hexis, Phoenix; 2005:79.
Google Scholar
Rong M, Wenpeng Z: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 2007, 45: 164–170.
Google Scholar
Yuan Y, Wenpeng Z: Some identities involving the Fibonacci polynomials. Fibonacci Q. 2002, 40: 314–318.
Google Scholar
Tingting W, Wenpeng Z: Some identities involving Fibonacci, Lucas polynomials and their applications. Bull. Math. Soc. Sci. Math. Roum. 2012, 55(1):95–103.
MathSciNet Google Scholar

Download references

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 and Affiliations

Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R, China
Xiaoxue Li & Di Han

Authors

Xiaoxue Li
View author publications
You can also search for this author in PubMed Google Scholar
Di Han
View author publications
You can also search for this author in PubMed Google Scholar

Corresponding author

Correspondence to Di Han.

Additional information

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.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and permissions

About this article

Cite this article

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

Download citation

Received: 09 June 2013
Accepted: 23 July 2013
Published: 08 August 2013
DOI: https://doi.org/10.1186/1687-1847-2013-240

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

Abstract

1 Introduction

2 Proof of the theorem

References

Acknowledgements

Author information

Authors and Affiliations

Corresponding author

Additional information

Competing interests

Authors’ contributions

Rights and permissions

About this article

Cite this article

Share this article

Keywords