Open Access

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

Advances in Difference Equations20132013:240

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

Received: 9 June 2013

Accepted: 23 July 2013

Published: 8 August 2013

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.

Keywords

Smarandache-Pascal derived sequenceFibonacci numbercombination numberelementary methodconjecture

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 [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 , } , { 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 + 4 b > 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 [35].

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 + b a 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).

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, Northwest University, Xi’an, Shaanxi, China

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Copyright

© Li and Han; licensee Springer 2013

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