Download A New Integer Sequence

A New Integer Sequence
M.A.Gopalan1, S.Vidhyalaksmi2, J.Shanthi3
1
Professor, Department of Mathematics, Shrimati Indira Gandhi College,Tamil nadu,India,
Professor,Department of Mathematics, Shrimati Indira Gandhi College, Tamil nadu, India,
3
Asst.Professor,Department of Mathematics,Shrimati Indira Gandhi College,Tamil nadu, India,
2
Abstract— In this paper, a new integer sequence is developed by defining the recurrence relation
Jn2  Jn1  (k 2  k)Jn with the initial conditions J 0  0, J1  1. Various interesting relations
among these numbers are exhibited. Also, Diophantine quadruples with property D(k 2n ) are
constructed. Some numerical examples are given.
Keywords: Integer sequence, Binet’s formula
I.
INTRODUCTION
It is well known that the Fibonacci sequence is famous for its wonderful and amazing properties.
Fibonacci composed a number text in which he did important work in number theory and the
solution of algebraic equations. The equation of rabbit problem posed by Fibonacci is known as the
first mathematical model for population growth. From the statement of rabbit problem, the famous
Fibonacci numbers can be derived. This sequence of Fibonacci numbers is extremely fruitful and
apperrs in different areas in mathematics and science.
The Fibonacci sequence, Lucas sequence, Pell sequence, Pell-Lucas sequence, Jacobsthal sequence
and Jacobsthal –Lucas sequence are most prominent examples of recursive sequences.
The Fibonacci sequence [3] is defined by the recurrence relation Fk  Fk 1  Fk  2 , k  2
with F0  0, F1  1. The Lucas sequence [3] is defined by the recurrence relation L k  Lk 1  Lk 2 , k  2
with L 0  2, L1  1.
The second order recurrence sequence has been generalized in two ways mainly, first by preserving
the initial conditions and second by preserving the recurrence relation. In this context, one may refer
[9].
D.
Kalman
and
R.Mena[2]
by Fn  aFn 1  bFn  2 , n  2 with F0  0, F1  1.
generalized
the
Fibonacci
sequence
A. F. Horadam[1] defined generalized the Fibonacci sequence {Hn } by H n  H n1  H n2 , n  3 with
H1  pH2  p  q , where p and q are arbitrary integers.
B. Singh, O. Sikhwal and S. Bhatnagar [12], defined Fibonacci like sequence by recurrence relation
Sk  S k 1  S k 2 , k  3 with S0  2, S1  2. The associated initial conditions S 0 a n dS1 are the sum of
the Fibonacci and Lucas sequence respectively. i.e, S0  F0  L 0 and S1  F1  L1.
L.R. Natividad [11], Deriving a formula in solving Fibonacci like sequence. He found missing terms
in Fibonacci like sequence and solved by standard formula.
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V.K. Gupta, V.Y. Panwar and O. Sikhwal [7], defined generalized Fibonacci sequences and derivd
its identies connection formulae and other results. V.K. Gupta, V.Y. Panwar and N.Gupta[6], stated
and derived identies for Fibonacci like sequence. Also, described and derived connection formulae
and negation formualae for Fibonacci like sequence. B.Singh, V.K.Gupta and V.Y.Panwar [4],
present many combination of higher powers of Fibonacci like sequence.
The k-Fibonacci numbers defined by Falco ‘n’ Plaza.A [5], depending only an one integer parameter
k as follows, for any positive real number k, the Fibonacci sequence is defined recurrently by
Fn,k  kFk,n1  Fk,n2 , n  2 with Fk, 0  0, Fk,1 1.
In [10], A.D. Godase and M.B. Dhakne have presented some properties of k-Fibonacci and k-Lucas
numbers by using matrices.
In [8], Yashwant , K.Panwar, G.P. Rathore and Richa Chawla have established some interesting
properties of k-Fibonacci like numbers.
The above results motivated us to search for new integer sequences with suitable initial conditions.
In this communication, a new integer sequence is developed by defining the recurrence relation
Jn2  Jn1  (k2  k)Jn with the initial condition J 0  0, J1  1. Various interesting relations among
these numbers are exhibited. Also, Diophantine quadruples with property D(k 2n ) are constructed.
Some numerical examples are given.
II.
Consider a sequence {Jn } defined by
METHOD OF ANALYSIS
Jn2  Jn1  (k2  k)Jn
with the initial condition J 0  0, J 1  1.
The auxiliary equation associated with the recurrence relation (1) is given by
(1)
m 2  m  (k 2  k )  0
whose roots are
    1,   -(k 2  k )
Thus the general solution of (1) is J n  A n  B n
From the initial conditions we infer that
A B  0, A  B 1
Solving for A and B, we get
 1 
1

A
, B  
 -
   
Thus a notable sequence {Jn } whose terms are given below is obtained.
(k  1) n  (k ) n
Jn 
2k  1
(2)
The first ten numbers are given below:
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Jn  0
J1  1
J2 1
J3 1  k  k 2
J 4  1  2k  2k 2
J 5  1  3k  4k 2  2k 3  k 4
J 6  1  4k  7k 2  6k 3  3k 4
J 7  1  5k  11k 2  13k 3  9k 4  3k 5  k 6
J 8  1  6k  16k 2  24k 3  22k 4  12k 5  4k 6
J 9  1  7k  22k 2  40k 3  46k 4  34k 5  16k 6  4k 7  k 8
J 1 0  1  8k  29k 2  62k 3  86k 4  80k 5  50k 6  20k 7  5k 8
The new sequence {Jn } is found to satisfy the following relations:
Relations:
3[J 2n 1  k (k  1) J 2 n1  (2k  1) J n ] is a Nasty Number.
1.
Proof:
1
J n1  k (k 1) J n1 
[( k 1)( k 1) n  (k )( k ) n  k (k 1) n  (k 1)( k ) n ]
(2k 1)
= ( k  1) n  (  k ) n
= (2k 1)Jn  2(k)n
from (2)
= 2(k 1)n  (2k 1)Jn
Replacing n by 2n in the above equation, we get
J2n1  k(k 1)J2n1  2(k 1)2n (2k 1)J2n
 3[J2n1 k(k 1)J2n1 (2k 1)Jn ]  6(k 1)2n
Hence, 3[J 2 n1  k (k  1) J 2n1  (2k  1) J n ] is a Nasty Number.
2.
Jmn (2k 1)JmJn  (k 1)m Jn (k 1)n Jm
Proof:
J mn  (2k  1) J m J n 
=
1
[2(k  1) mn  (k  1) m (k ) n  (k  1) n (k ) m ]
(2k  1)
1
[( k 1) m (k 1) n  (k ) n  (k 1) n (k 1) m  (k ) m ]
(2k 1)
= (k 1)m Jn  (k 1)n Jm

Jmn (2k 1)JmJn  (k 1)m Jn (k 1)n Jm.
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Jmn (2k 1)JmJn  (k)m Jn (k)n Jm
3.
Proof:
J m  n  ( 2k  1) J m J n 
1
[( k 1) m (k ) n  (k 1) n (k ) m  2(k ) mn ]
(2k 1)
 (k)m Jn (k)n Jm
 Jmn (2k 1)JmJn  (k)m Jn  (k)n Jm.
J 2n1  k 2 (k 1) 2 J n21  J 2n
4.
Proof:
J 2n 1  k 2 (k  1) 2 J n21 

1
(k  1) 2n2  (k ) 2n2  2(k  1) n1 (k ) n1
2
(2k  1)

=
1
(2k 1)
1
(2k  1)
= J 2n
2
2

[k 2 (k 1) 2n  (k 1) 2 (k) 2n  2(k 1) n1 (k) n1 ]
(k  1)
2n

(2k  1)  (k ) 2n (2k  1)
 J2n1  k2(k 1)2 Jn21  J2n.
J
5.

2
2
2 2
2
n1  k (k  1) J n1 (2k  1)
 (2k 2  2k  1)J 2n1  k(k  1)J 2n1  4(k  1) n1(k) n1
Proof:
J

2
2
2 2
2
n1  k (k 1) J n1 (2k 1)

 (k 1) 2n2  (k) 2n2  2(k 1) n1(k) n1

+ [k 2 (k  1) 2 n  (k  1) 2 (k ) 2 n  2(k  1) n 1 (k ) n 1 ]
= (2k 2  2k 1)J2n1  k(k 1)J2n1 4(k 1)n1(k)n1
Hence,
J

2
2
2 2
2
n1  k (k  1) J n1 (2k  1)
 (2k 2  2k  1)J 2n1  k(k  1)J 2n1  4(k  1)n1(k)n1.
6.
JmJn1  Jn Jm1  (k)n Jm (k)m Jn
Proof:
1
J m J n1  J n J m1 
(k  1) n (k ) m (k  k  1)  (k  1) m (k ) n (2k  1)
(2k  1) 2
1
(k  1) m (k ) n  (k  1) n (k ) m
=
(2k  1)
1
(k ) n (k  1) m  (k ) m  (k ) mn  (k ) m (k  1) n  (k ) n  (k ) mn
=
(2k  1)










= (k)n Jm (k)m Jn
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Hence, JmJn1  JnJm1  (k)n Jm (k)mJn.
7.
(2k 1)2 Jn1Jn1  J2n1  k(k 1)J 2n1(k 1)n1(k)n1(2k 2  2k 1)
Proof:


(2k 1)2 Jn1Jn1  (k 1)n1 (k)n1 (k 1)n1 (k)n1
= (k  1)  (k  1)
2n
n 1
(k )
n 1
 (k  1)
n 1

(k ) n1  (k ) 2 n
k 
k  1

= (k  1) 2n  (k ) 2n  (k  1) n (k ) n 
k  1
 k
= J2n1  k(k 1)J2n1 (k 1)n1(k)n1(2k 2  2k 1)
Hence, (2k 1)2 Jn1Jn1  J2n1  k(k 1)J 2n1(k 1)n1(k)n1(2k 2  2k 1)
(2k 1)J n21  2(k) n1 J n1  J 2n2
8.
Proof:
(2k 1)J n21  (2k 1)J 2n2  2(k) 2n2  2(k 1)
n1
(k) n1
n 1


= (2k  1) J 2n 2  2(k )n1 (k  1) (k )n1


= (2k 1)J2n2 2(k)n1(2k 1)Jn1
(2k 1) 2 J n21  2(k) n1 J n1  (2k 1)J 2n2
Hence, (2k 1)J n21  2(k) n1 J n1  J 2n2 .
(2k 1)Jnr1Jnr1  J2n (k)nr Jnr (k)nr Jnr
9.
Proof:
(2k 1)Jnr1Jnr1  (k 1)2n  (k)2n (k 1)nr (k)nr (k 1)nr (k)nr
= (k  1) 2 n  (k ) 2 n  (k ) nr (k  1) nr  (k ) nr  (k ) 2 n
- ( k ) n  r ( k  1) n r  ( k ) n r  (  k ) 2 n
= (k  1)  (k ) 2n  (k ) nr (2k  1) J nr  (k ) nr (2k  1) J nr
2n
Hence, (2k 1)Jnr 1Jnr 1  J2n (k)nr Jnr (k)nr Jnr .
n1
J
1
S n   J i  n1
k (k  1)
i 0
10.
Proof:
n1
Let S n   J i
i 0
=

1 n1
 (k  1)i  (k )i
(2k  1) i0
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=
n
 (k  1) n  1 
1
 (k )  1




(2k  1) 
k

k

1





=
1  (k  1) n1  (k  1)  (k ) n1  k 


(2k  1) 
k (k  1)

=
J n1 1
k(k  1)
n1
J
1
Hence, S n   J i  n1
.
k
(
k

1
)
i 0
11.
4(J3n  J n J 2n1  k(k  1)J n J 2n1)  J n21  k 2 (k  1) 2 J n21  2k(k  1)J n1J n1  (2k  1) 2 J n2
Proof:
Replacing n by 3n in (2), we have
1
J 3n 
(k  1)3n  (k )3n
(2k  1)
1

(k 1)n  (k )n (k 1)2n  (k )2n  (k 1)n (k )n
(2k 1)






 J n1  k (k  1) J n1  (2k  1) J n 

 J 2n1  k (k  1) J 2n1  
2



= Jn
  J n1  k (k  1) J n1  (2k  1) J n 

* 


2
 


Hence,
4(J 3n  J n J 2n1  k(k 1)J n J 2n1)  J n21  k 2 (k 1) 2 J n21  2k(k 1)J n1J n1  (2k 1) 2 J n2.
J 4 n  J 2 n ( J 2 n 1  k (k  1) J 2 n 1 ) .
12.
Proof:
Replacing n by 4n in (2), we have
1
J 4n 
(k 1) 4n  (k ) 4n
(2k 1)
1

(k 1)2n  (k )2n (k 1)2n  (k )2n
(2k 1)
Hence, J 4n  J 2n ( J 2n 1  k (k  1) J 2n 1 ) .





J 4 n  J 2 n J n1  k (k  1) J n1 J 2 n1  k (k  1) J 2 n1 
13.
Proof:
Replacing n by 4n in (2), we have
1
J 4n 
(k 1) 4n  (k ) 4n
(2k 1)

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International Journal of Recent Trends in Engineering & Research (IJRTER)
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


1
(k 1)n  (k )n (k 1)n  (k )n (k 1)2n  (k )2n
(2k 1)
Hence, J 4 n  J 2 n J n1  k (k  1) J n1 J 2 n1  k (k  1) J 2 n1  .

14.
Jmn  (  )JmJn   mJ n  n Jm.
Proof:
From (2), we have
1
Jm 
(k  1) m  (k ) m
(2k  1)
1
J mn 
(k 1) mn  (k ) mn
(2k 1)
1
 J mn  (   ) J m J n 
2 mn   m  n   n  m
(   )










1
 m ( n   n )   n ( m   m )
(   )
Hence, Jmn  (  )JmJn  mJn  nJm.
Construction of Diophantine quadruple with property D(k 2n ) :
Let a  (2k 1)Jn  (k 1)n (k)n,b  (Jn1  k(k 1)Jn1  (k 1)n  (k)n
it is noted that ab  k 2n  (k  1) n   p 2 (say ) .
2
Therefore the pair (a,b) is Diophantine -2-tuple with property D(k 2n )
Let c  a  b  2 p  4(k  1) n
it is noted that the triple (a, b, c) is Diophantine-3-tuple with property D(k 2n ) .
By Euler’s formula the fourth tuple ‘d’ is represented by
2
d  a  b  c  2 n (abc  pqr)
k
2
d  6(k  1) n  2 n 8(k  1) 3n  5(k ) 2 n (k  1) n
k
After performing calculation it is seen that (a, b, c, d) is Diophantine quadruple with property


D(k 2n )
It is worth mentioning that the value of ‘d’ is represented by a rational number for particular values
of k and n. A few examples are given in a tabular form below.
Table : Examples
n
k
Quadruple (a, b, c,
d)
Property D(k 2n )
1
1
(3, 1, 8, 120)
D(1)
2
1
(3, 5, 16, 1008)
D(1)
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International Journal of Recent Trends in Engineering & Research (IJRTER)
Volume 02, Issue 06; June - 2016 [ISSN: 2455-1457]
1
2
1
3
(7, 1, 16,
880
)
9
D(9)
1
4
(9, 1, 20, 105)
D(16)
1
5
(11, 1, 24,
2856
)
25
D(25)
4
(9, 41, 100,
14025
)
16
D(256)
develop
III.
a new
2
In
this
paper,
we
(5, 1, 12, 96)
D(4)
CONCLUSION
integer sequence
using
the
recurrence
relation
J n2  J n1  (k  k)J n with the initial condition J 0  0, J1  1, and provided a few interesting
2
relations among its members. To conclude, one may attempt to obtain a new integer sequences
employing different choices for the recurrence relations with suitable initial conditions.
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