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

Li, Xiaoxue; Han, Di

doi:10.1186/1687-1847-2013-240

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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

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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).

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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.

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

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Xiaoxue Li
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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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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

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