Open Access

On the Smarandache-Pascal derived sequences of generalized Tribonacci numbers

Advances in Difference Equations20132013:284

https://doi.org/10.1186/1687-1847-2013-284

Received: 8 August 2013

Accepted: 29 August 2013

Published: 7 November 2013

Abstract

For any sequence recurrence formula, the Smarandache-Pascal derived sequence { T n } of { b n } is defined by T n + 1 = k = 0 n ( n k ) b k + 1 for all n 2 , where ( n k ) = n ! k ! ( n k ) ! denotes the combination number. The recurrence formula of { T n } is obtained by the properties of the third-order linear recurrence sequence.

Keywords

Smarandache-Pascal derived sequenceTribonacci numberscombination numberelementary method

1 Introduction

For any sequence { b n } , a new sequence { T n } is defined by 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 the Smarandache-Pascal derived sequence, one of them is as follows.

Conjecture 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 .

Li and Han [3] studied these problems and proved a generalized conclusion as follows.

Proposition 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 .

It is clear that if we take b = 1 , then X n is the Fibonacci polynomials, see [47].

The main purpose of this paper is, using the elementary method and the properties of the third-order linear recurrence sequence, to unify the above results by proving the following theorem.

Theorem Let { X n } be a third-order linear recurrence sequence X n + 3 = a X n + 2 + b X n + 1 + c X n with the initial values X 0 = u , X 1 = v and X 2 = w for all n 1 , where a, b and c are positive integers. 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 = g 1 g 3 g 1 g 6 + g 7 g 3 g 6 T n + g 3 g 4 g 2 g 6 g 1 g 7 g 3 g 6 T n 1 + g 3 g 5 g 2 g 7 g 3 g 6 T n 2 ,
where
g 1 = f 1 + f 2 + c A d f 5 , g 2 = f 3 A d + 1 f 1 f 2 + c A d f 6 , g 3 = c A d + 1 2 c A d f 2 + c A d f 4 , g 4 = f 3 A d + 1 f 1 f 2 + c A d ( f 5 + f 6 ) , g 5 = c A d f 6 , g 6 = c ( A d + 1 A d f 2 + A d f 4 A d ) , g 7 = c A d f 4
and
f 1 = b A d + c A d 1 + 1 , f 2 = 1 + A d + 2 , f 3 = b A d 1 + c A d , f 4 = 1 + c A d 1 c A d 2 A d + 1 , f 5 = A d A d + 1 , f 6 = b A d 1 + c A d 2 b A d 2 + c A d 1 A d + A d A d + 1 ,

the sequence { A n } is defined by A n + 3 = a A n + 2 + b A n + 1 + c A n with the initial values A 1 = 0 , A 2 = 1 and A 3 = a for all n 1 .

From our theorem we know that if { b n } is a third-order linear recurrence sequence, then its Smarandache-Pascal derived sequence { T n } is also a third-order linear recurrence sequence.

2 Proof of the theorem

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

Lemma Let integers m 0 and n 3 . If the sequence { X n } satisfies the recurrence relations X n + 3 = a X n + 2 + b X n + 1 + c X n , n 0 , then we have the identity
X m + n = A n X m + 2 + ( b A n 1 + c A n 2 ) X m + 1 + c A n 1 X m ,

where A n is defined by A n + 3 = a A n + 2 + b A n + 1 + c A n with the initial values A 1 = 0 , A 2 = 1 and A 3 = a for all n 1 .

Proof Now we prove this lemma by mathematical induction. Note that the recurrence formula X m + 3 = a X m + 2 + b X m + 1 + c X m = A 3 X m + 2 + ( b A 2 + c A 1 ) X m + 1 + c A 2 X m for all n 1 . That is, the lemma holds for n = 3 since
X m + 4 = a ( a X m + 2 + b X m + 1 + c X m ) + b X m + 2 + c X m + 1 = ( a 2 + b ) X m + 2 + ( a b + c ) X m + 1 + a c X m = A 4 X m + 2 + ( b A 3 + c A 2 ) X m + 1 + c A 3 X m .
That is, the lemma holds for n = 4 . Suppose that for all integers 2 n k , we have X m + n = A n X m + 2 + ( b A n 1 + c A n 2 ) X m + 1 + c A n 1 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 + c X m + k 2 = a ( A k X m + 2 + ( b A k 1 + c A k 2 ) X m + 1 + c A k 1 X m ) + b ( A k 1 X m + 2 + ( b A k 2 + c A k 3 ) X m + 1 + c A k 2 X m ) + c ( A k 2 X m + 2 + ( b A k 3 + c A k 4 ) X m + 1 + c A k 3 X m ) = ( a A k + b A k 1 + c A k 2 ) X m + 2 + ( a b A k 1 + ( a c + b 2 ) A k 2 + 2 b c A k 3 + c 2 A k 4 ) X m + 1 + c ( a A k 1 + b A k 2 + c A k 3 ) X m = A k + 1 X m + 2 + ( a b A k 1 + b 2 A k 2 + b c A k 3 + c A k 1 ) X m + 1 + c A k X m = A k + 1 X m + 2 + ( b A k + c A k 1 ) X m + 1 + c A k X m .

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. From 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)
For any positive integer d, from the lemma we have X d k + d + 1 = A d + 1 X d k + 2 + ( b A d + c A d 1 ) X d k + 1 + c A d X d k . By the definition of T n , we may deduce that
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 = T n + k = 0 n 1 ( n 1 k ) ( A d + 1 X d k + 2 + ( b A d + c A d 1 ) X d k + 1 + c A d X d k ) = ( b A d + c A d 1 + 1 ) T n + A d + 1 k = 0 n 1 ( n 1 k ) X d k + 2 + c A d k = 0 n 1 ( n 1 k ) X d k .
For convenience, we let f 1 ( A k ) = b A d + c A d 1 + 1 (briefly f 1 ), then the above identity implies that
T n + 1 = f 1 T n + A d + 1 k = 0 n 1 ( n 1 k ) X d k + 2 + c A d k = 0 n 1 ( n 1 k ) X d k .
(2)
From this identity, we can also deduce
T n = f 1 T n 1 + A d + 1 k = 0 n 2 ( n 2 k ) X d k + 2 + c A d k = 0 n 2 ( n 2 k ) X d k
and
T n 1 = f 1 T n 2 + A d + 1 k = 0 n 3 ( n 3 k ) X d k + 2 + c A d k = 0 n 3 ( n 3 k ) X d k .
They are equivalent to
k = 0 n 2 ( n 2 k ) X d k + 2 = 1 A d + 1 ( T n f 1 T n 1 c A d k = 0 n 2 ( n 2 k ) X d k )
(3)
and
k = 0 n 3 ( n 3 k ) X d k + 2 = 1 A d + 1 ( T n 1 f 1 T n 2 c A d k = 0 n 3 ( n 3 k ) X d k ) .
(4)
On the other hand, from the lemma we also deduce X d k + d + 2 = A d + 2 X d k + 2 + ( b A d + 1 + c A d ) X d k + 1 + c A d + 1 X d k . Then we have
k = 0 n 1 ( n 1 k ) X d k + 2 = X 2 + X d n d + 2 + k = 1 n 2 ( n 1 k ) X d k + 2 = X 2 + X d n d + 2 + k = 1 n 2 ( ( n 2 k ) + ( n 2 k 1 ) ) X d k + 2 = k = 0 n 2 ( n 2 k ) X d k + d + 2 + k = 0 n 2 ( n 2 k ) X d k + 2 = k = 0 n 2 ( n 2 k ) ( A d + 2 X d k + 2 + ( b A d + 1 + c A d ) X d k + 1 + c A d + 1 X d k ) + k = 0 n 2 ( n 2 k ) X d k + 2 = ( 1 + A d + 2 ) k = 0 n 2 ( n 2 k ) X d k + 2 + c A d + 1 k = 0 n 2 ( n 2 k ) X d k + ( b A d + 1 + c A d ) T n 1 .
(5)
Similarly, applying formula (1) and identity (3), we have
k = 0 n 1 ( n 1 k ) X d k = X 0 + X d n d + k = 1 n 2 ( n 1 k ) X d k = X 0 + X d n d + k = 1 n 2 ( ( n 2 k ) + ( n 2 k 1 ) ) X d k = k = 0 n 2 ( n 2 k ) ( A d X d k + 2 + ( b A d 1 + c A d 2 ) X d k + 1 + c A d 1 X d k ) + k = 0 n 2 ( n 2 k ) X d k = A d k = 0 n 2 ( n 2 k ) X d k + 2 + ( 1 + c A d 1 ) k = 0 n 2 ( n 2 k ) X d k + ( b A d 1 + c A d 2 ) T n 1 = A d A d + 1 ( T n ( b A d + c A d 1 + 1 ) T n 1 c A d k = 0 n 2 ( n 2 k ) X d k ) + ( 1 + c A d 1 ) k = 0 n 2 ( n 2 k ) X d k + ( b A d 1 + c A d 2 ) T n 1 = A d A d + 1 T n + ( b A d 1 + c A d 2 b A d 2 + c A d 1 A d + A d A d + 1 ) T n 1 + ( 1 + c A d 1 c A d 2 A d + 1 ) k = 0 n 2 ( n 2 k ) X d k .
(6)
For convenience, we let
f 2 = 1 + A d + 2 , f 3 = b A d + 1 + c A d , f 4 = 1 + c A d 1 c A d 2 A d + 1 , f 5 = A d A d + 1 , f 6 = b A d 1 + c A d 2 b A d 2 + c A d 1 A d + A d A d + 1 ,
then identities (5) and (6) imply that
k = 0 n 1 ( n 1 k ) X d k + 2 = f 2 k = 0 n 2 ( n 2 k ) X d k + 2 + c A d + 1 k = 0 n 2 ( n 2 k ) X d k + f 3 T n 1 ,
(7)
k = 0 n 1 ( n 1 k ) X d k = f 4 k = 0 n 2 ( n 2 k ) X d k + f 5 T n + f 6 T n 1 .
(8)
Combining (2), (3), (7) and (8), we deduce
T n + 1 = ( f 1 + f 2 + c A d f 5 ) T n + ( f 3 A d + 1 f 1 f 2 + c A d f 6 ) T n 1 + ( c A d + 1 2 c A d f 2 + c A d f 4 ) k = 0 n 2 ( n 2 k ) X d k .
(9)
Applying formula (1), we deduce
( n k ) = ( n 1 k ) + ( n 1 k 1 ) = ( n 2 k ) + ( n 2 k 1 ) + ( n 1 k 1 ) .
(10)
From this and identities (3) and (4), note that X d k + d = A d X d k + 2 + ( b A d 1 + c A d 2 ) X d k + 1 + c A d 1 X d k , we have
k = 0 n 1 ( n 1 k ) X d k = X 0 + X d n d + ( n 1 n 2 ) X d n 2 d + k = 1 n 3 ( n 1 k ) X d k = X 0 + X d n d + ( n 1 ) X d n 2 d + k = 1 n 3 ( ( n 3 k ) + ( n 3 k 1 ) + ( n 2 k 1 ) ) X d k = X 0 + X d n d + ( n 1 ) X d n 2 d + k = 0 n 3 ( n 3 k ) X d k X 0 + k = 0 n 3 ( n 3 k ) X d k + d X d n 2 d + k = 0 n 2 ( n 2 k ) X d k + d ( n 2 ) X d n 2 d X d n d = k = 0 n 3 ( n 3 k ) X d k + k = 0 n 3 ( n 3 k ) X d k + d + k = 0 n 2 ( n 2 k ) X d k + d = A d k = 0 n 2 ( n 2 k ) X d k + 2 + A d k = 0 n 3 ( n 3 k ) X d k + 2 + c A d 1 k = 0 n 2 ( n 2 k ) X d k + ( c A d 1 + 1 ) k = 0 n 3 ( n 3 k ) X d k + ( b A d 1 + c A d 2 ) ( T n 1 + T n 2 ) = A d A d + 1 T n + ( b A d 1 + c A d 2 b A d 2 + c A d 1 A d A d + 1 ) T n 1 + ( b A d 1 + c A d 2 b A d 2 + c A d 1 A d + A d A d + 1 ) T n 2 + ( c A d 1 c A d 2 A d + 1 ) k = 0 n 2 ( n 2 k ) X d k + ( c A d 1 + 1 c A d 2 A d + 1 ) k = 0 n 3 ( n 3 k ) X d k = f 5 T n + ( f 5 + f 6 ) T n 1 + f 6 T n 2 + ( f 4 1 ) k = 0 n 2 ( n 2 k ) X d k + f 4 k = 0 n 3 ( n 3 k ) X d k .
(11)
Combining (2), (3), (7) and (11), we deduce
T n + 1 = ( f 1 + f 2 + c A d f 5 ) T n + ( f 3 A d + 1 f 1 f 2 + c A d ( f 5 + f 6 ) ) T n 1 + c A d f 6 T n 2 + c ( A d + 1 A d f 2 + A d f 4 A d ) k = 0 n 2 ( n 2 k ) X d k + c A d f 4 k = 0 n 3 ( n 3 k ) X d k .
(12)
From identity (9) we can also deduce
T n = ( f 1 + f 2 + c A d f 5 ) T n 1 + ( f 3 A d + 1 f 1 f 2 + c A d f 6 ) T n 2 + ( c A d + 1 2 c A d f 2 + c A d f 4 ) k = 0 n 3 ( n 3 k ) X d k .
(13)
For convenience, we let
g 1 = f 1 + f 2 + c A d f 5 , g 2 = f 3 A d + 1 f 1 f 2 + c A d f 6 , g 3 = c A d + 1 2 c A d f 2 + c A d f 4 , g 4 = f 3 A d + 1 f 1 f 2 + c A d ( f 5 + f 6 ) , g 5 = c A d f 6 , g 6 = c ( A d + 1 A d f 2 + A d f 4 A d ) , g 7 = c A d f 4 .
Inserting (9) and (13) into (12), we deduce
T n + 1 = g 1 g 3 g 1 g 6 + g 7 g 3 g 6 T n + g 3 g 4 g 2 g 6 g 1 g 7 g 3 g 6 T n 1 + g 3 g 5 g 2 g 7 g 3 g 6 T n 2 .
(14)

This completes the proof of our theorem.

Remark In fact, using the above formulas, we can also obtain the recurrence formula of the Smarandache-Pascal derived sequence { T n } of { u n } , where { u n } denotes the m th-order linear recursive sequences as follows:
u n = a 1 u n 1 + a 2 u n 2 + + a m 1 u n m + 1 + a m u n m ,

with initial values u i N for n > m and 0 i < m .

Declarations

Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11071194, 11001218) of P.R. China and the G.I.C.F. (YZZ12062) of NWU.

Authors’ Affiliations

(1)
Department of Mathematics, Northwest University
(2)
College of Science, Xi’an University of Technology

References

  1. Smarandache F: Only Problems, Not Solutions. Xiquan Publishing House, Chicago; 1993.MATHGoogle Scholar
  2. Murthy A, Ashbacher C: Generalized Partitions and New Ideas on Number Theory and Smarandache Sequences. Hexis, Phoenix; 2005.MATHGoogle Scholar
  3. Li X, Han D: On the Smarandache-Pascal derived sequences and some of their conjectures. Adv. Differ. Equ. 2013., 2013: Article ID 240Google Scholar
  4. Ma R, Zhang W: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 2007, 45: 164-170.MATHGoogle Scholar
  5. Yi Y, Zhang W: Some identities involving the Fibonacci polynomials. Fibonacci Q. 2002, 40: 314-318.MATHMathSciNetGoogle Scholar
  6. Wang T, Zhang W: Some identities involving Fibonacci, Lucas polynomials and their applications. Bull. Math. Soc. Sci. Math. Roum. 2012, 55: 95-103.Google Scholar
  7. Riordan J: Combinatorial Identities. Wiley, New York; 1968.MATHGoogle Scholar

Copyright

© Wu et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.