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

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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 = k = 0 n ( n k ) b k + 1 for all n2, 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 ), n2. 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 n2, 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 ),n2.

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 ),n2.

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 ),n2.

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 n1, where a 2 +4b>0. For any positive integer d2, 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 n1. 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 n1. For any even number d2, we have the recurrence formula

T n + 1 =(2+ A d + A d 2 )( T n T n 1 ),n2.

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 n1. For any odd number d2, we have the recurrence formula

T n + 1 =(2+ A d + A d 2 ) T n ( A d + A d 2 ) T n 1 ,n2,

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 n1, 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 ),n2.

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 m0 and n2. If the sequence { X n } satisfying the recurrence relations X n + 2 =a X n + 1 +b X n , n0, 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 n1, 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 n1. 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 2nk, 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

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    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.

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    Tingting W, Wenpeng Z: Some identities involving Fibonacci, Lucas polynomials and their applications. Bull. Math. Soc. Sci. Math. Roum. 2012, 55(1):95–103.

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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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Correspondence to Di Han.

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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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Keywords

  • Smarandache-Pascal derived sequence
  • Fibonacci number
  • combination number
  • elementary method
  • conjecture