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

## 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, $T n + 1 = ∑ k = 0 n ( n k ) ⋅ b k + 1$ for all $n≥2$, where $( n k ) = n ! k ! ( n − k ) !$ 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 ${ b n }={ F 1 , F 9 , F 17 ,…}$, then we have the recurrence formula $T n + 1 =49⋅( T n − T n − 1 )$, $n≥2$. 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, $T n + 1 = ∑ k = 0 n ( n k ) ⋅ b k + 1$ for all $n≥2$, where $( n k ) = n ! k ! ( n − k ) !$ is the combination number. This sequence is called the Smarandache-Pascal derived sequence of ${ b n }$. It was introduced by professor Smarandache in  and studied by some authors. For example, Murthy and Ashbacher  proposed a series of conjectures related to Fibonacci numbers and its Smarandache-Pascal derived sequence; three of them are as follows.

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

$T n + 1 =49⋅( T n − T n − 1 ),n≥2.$

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

$T n + 1 =125⋅( T n − T n − 1 ),n≥2.$

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

$T n + 1 =324⋅( T n − T n − 1 ),n≥2.$

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 $n≥1$, where $a 2 +4b>0$. For any positive integer $d≥2$, we define the Smarandache-Pascal derived sequence of ${ X d n + 1 }$ as

$T n + 1 = ∑ k = 0 n ( n k ) ⋅ X d k + 1 .$

Then we have the recurrence formula

$T n + 1 =(2+ A d +b⋅ A d − 2 )⋅ T n − ( 1 + A d + b ⋅ A d − 2 + ( − b ) d ) ⋅ T n − 1 ,$

where the sequence ${ A n }$ is defined as $A 0 =1$, $A 1 =a$, $A n + 1 =a⋅ A n +b⋅ A n − 1$ for all $n≥1$. In fact this time, the general term is

$A n = 1 a 2 + 4 b [ ( a + a 2 + 4 b 2 ) n + 1 − ( a − a 2 + 4 b 2 ) n + 1 ] .$

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 $n≥1$. For any even number $d≥2$, we have the recurrence formula

$T n + 1 =(2+ A d + A d − 2 )⋅( T n − T n − 1 ),n≥2.$

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 $n≥1$. For any odd number $d≥2$, we have the recurrence formula

$T n + 1 =(2+ A d + A d − 2 )⋅ T n −( A d + A d − 2 )⋅ T n − 1 ,n≥2,$

where

$A n = A n (a)= 1 a 2 + 4 [ ( a + a 2 + 4 2 ) n + 1 − ( a − a 2 + 4 2 ) n + 1 ] .$

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 .

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 $n≥1$, 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

$T n + 1 = ∑ k = 0 n ( n k ) ⋅ P 2 d k + 1 ,$

we have the recurrence formula

$T n + 1 =(2+ P 2 d + 1 + P 2 d − 1 )⋅( T n − T n − 1 ),n≥2.$

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 $m≥0$ and $n≥2$. If the sequence ${ X n }$ satisfying the recurrence relations $X n + 2 =a⋅ X n + 1 +b⋅ X n$, $n≥0$, then we have the identity

$X m + n = A n − 1 ⋅ X m + 1 +b⋅ A n − 2 ⋅ X m ,$

where $A n$ is defined as $A 0 =1$, $A 1 =a$ and $A n + 1 =a⋅ A n +b⋅ A n − 1$ for all $n≥1$, or

$A n = 1 a 2 + 4 b [ ( a + a 2 + 4 b 2 ) n + 1 − ( a − a 2 + 4 b 2 ) n + 1 ] .$

Proof Now we prove this lemma by mathematical induction. Note that the recurrence formula $X m + 2 =a⋅ X m + 1 +b⋅ X m$, $A 1 =a$, $A 0 =1$, $A n + 1 =a⋅ A n +b⋅ A n − 1$ for all $n≥1$. So $X m + 2 = A 1 ⋅ X m + 1 +b⋅ A 0 ⋅ X m$. That is, the lemma holds for $n=2$. Since $X m + 3 =a⋅ X m + 2 +b⋅ X m + 1 =a⋅(a⋅ X m + 1 +b⋅ X m )+b⋅ X m + 1 =( a 2 +b)⋅ X m + 1 +ba⋅ X m = A 2 ⋅ X m + 1 +b A 1 ⋅ X m$. That is, the lemma holds for $n=3$. Suppose that for all integers $2≤n≤k$, we have $X m + n = A n − 1 ⋅ X m + 1 +b⋅ A n − 2 ⋅ X m$. Then for $n=k+1$, from the recurrence relations for $X m$ and the inductive hypothesis, we have

$X m + k + 1 = a ⋅ X m + k + b ⋅ X m + k − 1 = a ⋅ ( A k − 1 ⋅ X m + 1 + b ⋅ A k − 2 ⋅ X m ) + b ⋅ ( A k − 2 ⋅ X m + 1 + b ⋅ A k − 3 ⋅ X m ) = ( a ⋅ A k − 1 + b ⋅ A k − 2 ) ⋅ X m + 1 + b ⋅ ( a ⋅ A k − 2 + b ⋅ A k − 3 ) ⋅ X m = A k ⋅ X m + 1 + b ⋅ A k − 1 ⋅ X m − 1 .$

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 $( n k )$, we have

$( n − 1 k ) + ( n − 1 k − 1 ) = ( n − 1 ) ! k ! ( n − 1 − k ) ! + ( n − 1 ) ! ( k − 1 ) ! ( n − k ) ! = ( n − 1 ) ! ( k − 1 ) ! ( n − k − 1 ) ! ( 1 k + 1 n − k ) = ( n k )$
(1)

and

$T n + 1 = ∑ k = 0 n ( n k ) ⋅ X d k + 1 = X 1 + X d n + 1 + ∑ k = 1 n − 1 ( ( n − 1 k ) + ( n − 1 k − 1 ) ) X d k + 1 = ∑ k = 0 n − 1 ( n − 1 k ) ⋅ X d k + 1 + ∑ k = 0 n − 2 ( n − 1 k ) ⋅ X d k + d + 1 + X d n + 1 = T n + ∑ k = 0 n − 1 ( n − 1 k ) X d k + d + 1 .$
(2)

From the lemma, we have$X d k + d + 1 = A d ⋅ X d k + 1 +b⋅ A d − 1 ⋅ X d k$, by (2) and the definition of $T n$, we may deduce that

$T n + 1 = T n + ∑ k = 0 n − 1 ( n − 1 k ) ⋅ ( A d ⋅ X d k + 1 + b ⋅ A d − 1 ⋅ X d k ) = T n + A d ⋅ ∑ k = 0 n − 1 ( n − 1 k ) ⋅ X d k + 1 + b ⋅ A d − 1 ⋅ ∑ k = 0 n − 1 ( n − 1 k ) ⋅ X d k = ( 1 + A d ) ⋅ T n + b ⋅ A d − 1 ⋅ ∑ k = 0 n − 1 ( n − 1 k ) ⋅ X d k .$
(3)

On the other hand, from the lemma, we also have $X d k + d = A d − 1 ⋅ X d k + 1 +b⋅ A d − 2 ⋅ X d k$, from this and formula (1), we have

$∑ k = 0 n − 1 ( n − 1 k ) ⋅ X d k = X 0 + X d ( n − 1 ) + ∑ k = 1 n − 2 ( n − 1 k ) ⋅ X d k = X 0 + X d ( n − 1 ) + ∑ k = 1 n − 2 ( ( n − 2 k ) + ( n − 2 k − 1 ) ) ⋅ X d k = ∑ k = 0 n − 2 ( n − 2 k ) ⋅ X d k + ∑ k = 0 n − 3 ( n − 2 k ) ⋅ X d k + d + X d ( n − 1 ) = ∑ k = 0 n − 2 ( n − 2 k ) ⋅ X d k + ∑ k = 0 n − 2 ( n − 2 k ) ⋅ ( A d − 1 ⋅ X d k + 1 + b ⋅ A d − 2 ⋅ X d k ) = ( 1 + b ⋅ A d − 2 ) ⋅ ∑ k = 0 n − 2 ( n − 2 k ) ⋅ X d k + A d − 1 ⋅ T n − 1 .$
(4)

From (3), we can also deduce that

$T n =(1+ A d )⋅ T n − 1 +b⋅ A d − 1 ⋅ ∑ k = 0 n − 2 ( n − 2 k ) ⋅ X d k .$
(5)

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

$T n + 1 = ( 1 + A d ) ⋅ T n + b ⋅ A d − 1 ⋅ ( ( 1 + b ⋅ A d − 2 ) ⋅ ∑ k = 0 n − 2 ( n − 2 k ) ⋅ X d k + A d − 1 ⋅ T n − 1 ) = ( 1 + A d ) ⋅ T n + b ⋅ A d − 1 2 ⋅ T n − 1 + ( 1 + b ⋅ A d − 2 ) ⋅ ( T n − ( 1 + A d ) ⋅ T n − 1 )$

or equivalent to

$T n + 1 = ( 2 + A d + b ⋅ A d − 2 ) ⋅ T n − ( 1 + A d + b ⋅ A d − 2 + b ⋅ A d ⋅ A d − 2 − b ⋅ A d − 1 2 ) ⋅ T n − 1 = ( 2 + A d + b ⋅ A d − 2 ) ⋅ T n − ( 1 + A d + b ⋅ A d − 2 + ( − b ) d ) ⋅ T n − 1 ,$
(6)

where we have used the identity

$A d ⋅ A d − 2 − A d − 1 2 = − ( − b ) d − 1 a 2 + 4 b [ ( a + a 2 + 4 b 2 ) 2 + ( a − a 2 + 4 b 2 ) 2 ] + 2 ( − b ) d a 2 + 4 b = − ( − b ) d − 1 ⋅ a 2 + 2 b + 2 b a 2 + 4 b = − ( − b ) d − 1 .$

Now, our theorem follows from formula (6).

## References

1. 1.

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

2. 2.

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

3. 3.

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

4. 4.

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

5. 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 