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BALANCING NUMBERS : SOME IDENTITIES

A report

submitted by

KABERI PARIDA

Roll No: 412MA2073 for

the partial fulfilment for the award of the degree of

Master of Science in Mathematics

under the supervision of

Dr. GOPAL KRISHNA PANDA

DEPARTMENT OF MATHEMATICS NIT ROURKELA

ROURKELA– 769 008

MAY 2014

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DECLARATION

I hereby declare that the topic β€œBalancing numbers : some identities ” for comple- tion for my master degree has not been submitted in any other institution or university for the award of any other Degree, Fellowship or any other similar titles.

Date:

Place:

Kaberi Parida Roll no: 412MA2073 Department of Mathematics

NIT Rourkela

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NATIONAL INSTITUTE OF TECHNOLOGY,ROURKELA

CERTIFICATE

This is to certify that the project report entitledBalancing numbers : some identities submitted by Kaberi Parida to the National Institute of Technology Rourkela, Odisha for the partial fulfilment of requirements for the degree of master of science in Mathematics is a bonafide record of review work carried out by her under my supervision and guidance.

The contents of this project, in full or in parts, have not been submitted to any other institute or university for the award of any degree or diploma.

May, 2014

Prof. Gopal Krishna Panda Department of Mathematics

NIT Rourkela

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Acknowledgement

I wish to express my deep sense of gratitude to my supervisor Dr.G.K.Panda, Professor, Department of Mathematics, National Institute of Technology, Rourkela for his inspiring, guidance and assistance in the preparation of this project work.

I am grateful to Prof.S.K.Sarangi, Director, National Institute of Technology, Rourkela for providing excellent facilities in the Institute for carrying out this project work.

I owe a lot to the Ph.D. students Mr. Ravi Kumar Daval and Mr. Akshaya kumar Panda for their help during the preparation of this project work.

I am extremely grateful to my parents who are a constant source of inspiration for me.

Kaberi Parida

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Abstract

This paper studies a problem in the theory of figurate numbers identifying and investigat- ing those numbers which are polygonal in two ways - triangular and square. In this report a study of Pell numbers, Associate Pell numbers, Balancing numbers, Lucas Balancing numbers is presented. These numbers can be better computed by means of recurrence relations through Pell’s equation will play a central role.

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Contents

1 Introduction 1

2

Preliminaries 3

2.1 Recurrence Relation . . . 3

2.2 Generating function . . . 3

2.3 Triangular numbers . . . 3

2.4 Diophantine Equation . . . 3

2.5 Binet formula . . . 3

2.6 Pell numbers . . . 3

2.7 Balancing numbers . . . 4

2.8 Associate Pell numbers . . . 4

2.9 Lucas Balancing numbers . . . 4

2.10 Co-Balancing numbers . . . 4

2.11 Some identities of Pell numbers, Balancing numbers and Lucas balancing numbers . . . 4

3 Balancing numbers : some identities 6 3.1 Some identities : part A . . . 10

3.2 Identities: part B . . . 15

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CHAPTER-1

1 Introduction

There is a quote by the famous mathematician Carl Friedrich Gauss (1777βˆ’1855): Math- ematics is the queen of all sciences, and number theory is the queen of mathematics. Num- ber theory, or higher arithmetic is the study of those properties of integers and rational numbers, which go beyond the ordinary manipulations of everyday arithmetic.

In number theory, discovery of number sequences with certain specified properties has been a source of attraction since ancient times.The most beautiful and simplest of all number sequences is the Fibonacci sequence.This sequence was first invented by Leonardo of Pisa (1180βˆ’1250), who was also known as Fibonacci, to describe the growth of a rabbit population.

Other interesting number sequences are the Pell sequence and the associated Pell se- quence. In mathematics, the Pell numbers are infinite sequence of integers that have been known since ancient times.The denominators of the closest rational approximations to the square root of 2. This sequence of approximations begins with 11,32,75,1712,4129; so the sequence of Pell numbers begins with 1,2,5,12,29. The numerators of the same sequence of approximations give the associated Pell sequence.

The concept of balancing numbers was first introduced by Behera and Panda in the year 1999 in connection with a Diophantine equation. It consists of finding a natural number n such that

1 + 2 + 3...+ (nβˆ’1) = (n+ 1) + (n+ 2) +...+ (n+r)

for some natural number r, while they call r, the balancer corresponding to the balancing number n. If the nth triangular number n(n+1)2 is denoted byTn, then the above equation reduces to Tnβˆ’1 +Tn =Tn+r which is the problem of finding two consecutive triangular numbers whose sum is also a triangular number. Since

T5+T6 = 15 + 21 = 36 = T8 6 is a balancing number with balancer 2. Similarly,

T34+T35= 595 + 630 = 1225 =T49

implies that, 35 is also a balancing number with balancer 14.

The balancing numbers, though obtained from a simple Diophantine equation, are very useful for the computation of square triangular numbers. An important result about bal- ancing numbers is that,n is a balancing number if and only if 8n2+ 1 is a perfect square, and the number √

8n2+ 1 is called a Lucas balancing number. The most interesting fact about Lucas-balancing numbers is that, these numbers are associated with balancing numbers in the way Lucas numbers are associated with Fibonacci numbers.

The early investigators of Pell equation were the indian mathematicians Brahmagupta and Bhaskara. In particular Bhaskara studied Pell’s equation for the values d= 8,11,32,61.

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Bhaskar found the solution x = 1776319049, y = 2261590 for d = 61. Fermat was also interested in the Pell’s equation and worked out some of the basic theories regarding Pell’s equation. A special type of Diophantine equation is the Pells equationx2βˆ’dy2 = 1 where d is a natural number which is not a perfect square. Indeed, the english mathematician John Pell (1610βˆ’1685) has nothing to do with this equation.In general Pell’s equation is a Diophantine equation of the formx2βˆ’dy2 = 1. wheredis a positive non square integer.

Cobalancing number n are solutions of the Diophantine equation 1 + 2 + 3...+n= (n+ 1) + (n+ 2) +...+ (n+r)

Where r is called the cobalancer corresponding to the cobalancing numbern.

Panda [13] generalized balancing and cobalancing numbers by introducing sequence bal- ancing and cobalancing numbers, in which, the sequence of natural numbers, used in the definition of balancing and cobalancing numbers is replaced by an arbitrary sequence of real numbers. Thus if {an}∞n=1 is a sequence of real numbers, then ak is called a sequence balancing number if

a1+a2+...+akβˆ’1 =ak+1+ak+2+...+ak+r for some natural number r; ak is called a sequence cobalancing number if

a1+a2+...+ak =ak+1+ak+2+...+ak+r for some natural number r.

The Associated Pell sequence exhausts two sequences generated from balancing and cobalancing numbers,namely, the sequences of Lucas-balancing and the Lucas-cobalancing numbers. Pell and associated Pell numbers also appear as the greatest common divisors of two consecutive balancing numbers or cobalancing numbers or, a pair of balancing and cobalancing numbers of same order.Balancing and cobalancing numbers also arise in the partial sums of even ordered Pell numbers, odd order Pell numbers, even ordered asso- ciated Pell numbers, odd order associated Pell numbers, and in the partial sum of these numbers up to even and odd order.

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CHAPTER - 2

2 Preliminaries

In this chapter we recall some definitions and known results on Pell numbers,associated Pell numbers, triangular numbers, recurrence relations, Binet formula, Diophantine equa- tions including Pells equations. This chapter serves as base and background for the study of subsequent chapters. We shall keep on referring back to it as and when required.

2.1 Recurrence Relation

In mathematics, a recurrence relation is an equation that defines a sequence recursively;

each term of the sequence is defined as a function of the preceding terms.

2.2 Generating function

The generating function of a sequence {xn}∞n=1 of real or complex numbers is given by f(s) = P∞

n=1xnsn. Hence, the nth term of the sequence is obtained as the coefficient of sn in the power series expansion off(s).

2.3 Triangular numbers

A number of the form n(n+ 1)

2 where n ∈Z+ is called a triangular number.The justifi- cations for the name triangular number are many.One such reason may be the fact that the triangular number n(n+ 1)

2 represents the area of a right angled triangle with base n+ 1 and perpendicular n. It is well known that m ∈ Z+ is a triangular number if and only if 8m+ 1 is a perfect square.

2.4 Diophantine Equation

In mathematics, a Diophantine equation is an indeterminate polynomial equation that allows the variables to be integers only. Diophantine problems have fewer equations than unknowns and involve finding integers that work correctly for all the equations.

2.5 Binet formula

While solving a recurrence relation as a difference equation,the nth term of the sequence is obtained in closed form,which is a formula containing conjugate surds of irrational numbers is known as the Binet formula for the particular sequence. These surds are obtained from the auxiliary equation of the recurrence relation for the recursive sequence under consideration.

2.6 Pell numbers

The first two Pell numbers areP1 = 1, P2 = 2 and other terms of the sequence are obtained by means of the

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Recurrence relation : Pn= 2Pnβˆ’1+Pnβˆ’2 for P β‰₯2 Cassini formula : Pnβˆ’1Pn+1βˆ’Pn2 = (βˆ’1)n

Binet formula : Ξ±n1βˆ’Ξ±n2

2√

2 and α1 = 1 +√

2 ,Ξ±2 = 1βˆ’βˆš 2 Where Ξ±1βˆ’Ξ±2 = 2√

2, Ξ±1Ξ±2 =βˆ’1, Ξ±1+Ξ±2 = 2 andPn = 1,2,5,12,29,70...

2.7 Balancing numbers

The first two Balancing numbers are B1 = 1, B2 = 6 and other terms of the sequence are obtained by means of the

Recurrence relation : Bn+1 = 6Bnβˆ’Bnβˆ’1

Cassini formula : Bn2 βˆ’Bn+1Bnβˆ’1 = 1 Binet formula : Ξ±Ξ±n1βˆ’Ξ±n2

1βˆ’Ξ±2 and Ξ±1 = 3 +√

8, Ξ±2 = 3βˆ’βˆš 8 Where Ξ±1βˆ’Ξ±2 = 2√

8, Ξ±1Ξ±2 = 1, Ξ±1+Ξ±2 = 6 andBn= 1,6,35,204,1189...

2.8 Associate Pell numbers

Recurrence relation : Qn = 2Qnβˆ’1+Qnβˆ’2. Cassini formula : Qnβˆ’1Qn+1βˆ’Q2n= (βˆ’1)n Binet formula : Ξ±n1+Ξ±2 n2 and Ξ±1 = 1 +√

2 ,Ξ±2 = 1βˆ’βˆš 2 Where Ξ±1Ξ±2 =βˆ’1, Ξ±1+Ξ±2 = 2 and Qn= 1,3,7,17...

2.9 Lucas Balancing numbers

Recurrence relation : Cn+1 = 6Cnβˆ’Cnβˆ’1

Cassini formula : Cn2βˆ’Cn+1Cnβˆ’1 =βˆ’8 Binet formula : Ξ±n1+Ξ±2 n2 and Ξ±1 = 3 +√

8 ,Ξ±2 = 3βˆ’βˆš 8 Where Ξ±1βˆ’Ξ±2 = 2√

8, Ξ±1Ξ±2 = 1, Ξ±1+Ξ±2 = 6 andCn= 1,3,17,99,577...

2.10 Co-Balancing numbers

Recurrence relation : bn+1 = 6bnβˆ’bnβˆ’1+ 2 Binet formula : bn = Ξ±

2nβˆ’1 1 βˆ’Ξ±2nβˆ’1n

4√

2 and α1 = 1 +√

2,Ξ±2 = 1βˆ’βˆš

2 and bn = 0,2,14,84...

2.11 Some identities of Pell numbers, Balancing numbers and Lucas balancing numbers

(a) Relation between Pell and Associate Pell numbers (1)P1+P2+...+P2nβˆ’1 =Bn+bn.

(2)P1+P3+...+P2nβˆ’1 =bn. (3)P2+P4+...+P2n=bn+1. (4)Q1+Q2+...+Q2n= 2bn+ 1.

(5)Q1+Q2+...+Q2nβˆ’1 = 2Bnβˆ’1.

(11)

(6)P2n+Q2nβˆ’1 =b2n.

(7)Pm+n =PmPn+1+Pmβˆ’1Pn. (8)Pm+n = 2PmQnβˆ’(βˆ’1)nPmβˆ’n. (9)P12 +P22+...+Pn2 = Pn+12Pn. (10)P2n+1 = 12(PnQn+1+QnPn+1).

(11)Qn+1Qnβˆ’1βˆ’Q2n= 6(βˆ’1)nβˆ’1.

(b) Relation between Balancing and Lucas Balancing numbers (1)B2nβˆ’1 =Bn2 βˆ’Bnβˆ’12 .

(2)B2n =Bn(Bn+1βˆ’Bnβˆ’1).

(3)B1+B3+...+B2nβˆ’1 =Bn2. (4)B2+B4+...+Bn =BnBn+1,

(5)Bn+12 = 34Bn2 + 2 (Bn2 βˆ’Bn+1Bnβˆ’1)βˆ’Bnβˆ’12 . (6)Bn+1 = 3Bn+ (3Bnβˆ’Bnβˆ’1).

(7)Bn+1 = 3Bn+p

8Bn2+ 1.

(8)Bnβˆ’1 = 3Bnβˆ’p

8Bn2+ 1.

(9)Cn+12 = (3Cn+ 8Bn).

(10)Bnβˆ’1 +Bn=Cn.

(11)Cmβˆ’n =Bm+1Cnβˆ’BmCn+1.

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CHAPTER - 3

3 Balancing numbers : some identities

The nth polygonal number of order g is the non negative integer such that fg(m) = n

2[{(gβˆ’2)(nβˆ’1)}+ 2], n = 0,1,2, .... Every triangular number taken 8 times and then increased by 1 gives a square i.e, 8n(n+ 1)

2 + 1 = (2n+ 1)2. If a triangular number is polygonal in two ways

(i) Triangular n(n+ 1)

2 .

(ii) Square m2, then it is a Balancing number.

Example : 36 = 8.9

2 = 62 is such a number represented in two ways. Let the nth triangular number Tn = n(n+ 1)

2 = m2 = sm be the mth square. Tn consists of the two consecutive numbers n and n+ 1, one is even and the other is odd. They are necessarily co-prime and so if n is even, thengcd(n

2, n+ 1) = 1, or ifn is odd thengcd(n,n+ 1 2 ) = 1.

Whenever of the two numbers are even, it contains only odd powers of 2. If n is even, then setting n+ 1 =x2 i.e, n =x2βˆ’1, and n

2 =y2 i.e, n= 2y2 and equating n in terms of x and y.

x2βˆ’2y2 = 1 (1)

Alternately, if n is odd then putting n =x2 i.e, n+ 1

2 =y2, n = 2y2βˆ’1 and equating n in terms of xand y

x2βˆ’2y2 =βˆ’1 (2)

In the above cases, Tn = (xy)2 = m2 =sm = Nk. Solutions of Pell equations are given the above two equation would give all Nk and (3,2) is the solution of (1) and (1,1) is the solution of (2) i.e,

1 = 1k = (βˆ’1)2k =

(3βˆ’2√

2)(3 + 2√ 2)

k

=

(1βˆ’2√

2)(1 + 2√ 2)

2k

βˆ’1 = (βˆ’1)2kβˆ’1 =

(1βˆ’2√

2)(1 + 2√

2)2kβˆ’1 Hence, the general solution of (1) and that of (2) respectively is:

xk+yk√

2 = (3 + 2√

2)k= (1 +√

2)2k (3)

xk+yk√

2 = (1 +√

2)2kβˆ’1 (4)

Combining the even and odd power solutions in (3) and (4), we get the single formula for all solutions of (1) and (2):

xk+yk√

2 = (1 +√

2)k (5)

We also have the following explicit formula:

Qk =xk = (1 +√

2)k+ (1βˆ’βˆš 2)k

2 (6)

(13)

Pk =yk = (1 +√

2)kβˆ’(1βˆ’βˆš 2)k 2√

2 (7)

The recurrence relations beginning with x0 = 1, y0 = 0 define them:

Qk+1 =Qk+1 = 2Qk+Qkβˆ’1 (8) Pk+1 =Pk+1 = 2Pk+Pkβˆ’1 (9) Recurrence relations can be very easily translated into generating functions and we have:

f(x) = 1 +x

(1βˆ’2xβˆ’x2) = 1 + 3x+ 7x2+... (10)

f(y) = 1

(1βˆ’2yβˆ’y2) = 1 + 2y+ 5y2+... (11) These Lucas-type formula involving binomial coefficients are adapted from Weisstein [6]:

Qk=

[k2]

X

r=0

2r k

2r

(12)

Pk=

[k+12 ]

X

r=1

2rβˆ’1 k

2rβˆ’1

(13) Having computed the values of Qand P, we can get the required values ofn and m. For the even values of subscript k of n, we use n = x2βˆ’1 = 2y2, and for the odd, we take n =x2 =√

2y2βˆ’1. Form, we simply need to multiply the corresponding values ofxand y. These relations holds forQk, Bn+bn;Pk, Bn:

((Bn+2rβˆ’1 +bn+2rβˆ’1)βˆ’(Bn+bn)) Q2rβˆ’1

= (Bk+2rβˆ’1+Bk) P2rβˆ’1

=Q2k+2rβˆ’1; ((Bn+2r+bn+2r)βˆ’(Bk+bk))

2P2r = (Bk+2r+Bk)

Q2r =P2k+2r (Bk+2rβˆ’1βˆ’Bk)

Q2rβˆ’1

=P2k+2rβˆ’1;(Bk+2rβˆ’Bk)

P2r =Q2k+2r The following explicit formula for Bn+bn and Bn are due to Euler:

(Bn+bn) = nk = (3 + 2√

2)k+ (3βˆ’2√

2)kβˆ’2

4 (14)

Bn=mk = (3 + 2√

2)kβˆ’(3βˆ’2√ 2)k 4√

2 (15)

The following recurrence relations, with initial values 0 and 1, are also given by Euler:

(Bk+1+bk+1) =nk+1 = 6(Bk+bk)βˆ’(Bkβˆ’1+bkβˆ’1) + 2 (16) Bk+1 =mk = 6Bkβˆ’Bkβˆ’1, kβ‰₯1 (17) These generating functions define them:

f(u) = 1 +u

(1βˆ’u)(1βˆ’6u+u2) = 1 + 8u+ 49u2+... (18)

(14)

f(v) = 1

(1βˆ’6v+v2) = 1 + 6v+ 35v2+... (19) These Lucas-type formula involving binomial coefficients from (12) and (13):

(Bn+bn) = nk =

n

X

r=1

2rβˆ’1 2n

2r

(20)

Bn =mk=

n

X

r=1

2rβˆ’2 2n

2rβˆ’1

(21) The following summation formula holds:

2

n

X

r=1

(Br+br) = ( n

X

r=1

2rβˆ’1

2n+ 1 2r

)

βˆ’n (22)

2

n

X

r=1

Br = ( n

X

r=1

2rβˆ’1

2n+ 1 2r+ 1

)

+n (23)

Now, the triangular square gives the formula:

Nk =m2k=Bn2 = (3 + 2√

2)kβˆ’(3βˆ’2√ 2)k 4√

2

!2

= (17 + 12√

2)kβˆ’(17βˆ’12√

2)kβˆ’2

32 . (24)

Bn+12 =Nk+1 = 34Bn2 βˆ’Bnβˆ’12 + 2 (25) Generating function for the square triangular numbers:

f(z) = 1 +z

(1βˆ’z)(1βˆ’34z+z2) = 1 + 36z+ 1225z2+... (26) A product formula for the kth triangular square is recorded by Weisstein[6]:

Bn2 =Nk= 22nβˆ’5

2n

Y

k=1

3 +coskΟ€ k

(27) Since cos(Ο€βˆ’ΞΈ) = cos(Ο€+ΞΈ) and cosΟ€ =βˆ’1,cos2Ο€= 1 then

2n

Y

k=n+1

3 +coskΟ€ k

= 2

n

Y

k=1

3 +coskΟ€ k

Hence,

Bn2 =Nk = 2kβˆ’2

n

Y

k=1

3 +coskΟ€ k

2!

Bn= 2kβˆ’2

n

Y

k=1

3 +coskΟ€ k

(28)

(15)

The above-noted formula give these values of Qk, Pk, TBn+bn, sBk and the associated square triangular numbers Bn2:

K Qn Pn Tnk SBn Bn2

0 1 0 T0 S0 0

1 1 1 T1 S1 1

2 3 2 T8 S6 36

3 7 5 T49 S35 1225

4 17 12 T288 S204 41616

5 41 29 T1681 S1189 1413721

6 99 70 T9800 S6930 48024900

7 239 169 T57121 S40391 1631432881

8 577 408 T332928 S235416 55420693056 9 1393 985 T1940449 S1372105 1882672131025 10 3363 2378 T11309768 S7997214 63955431761796

The values assumed by Pk are known as Pell Numbers which are related to square trian- gular numbers:

(Pk(Pk+Pkβˆ’1))2 =

(Pk+Pkβˆ’1)2βˆ’(βˆ’1)k βˆ—(Pk+Pkβˆ’1)2

2 , k β‰₯1

There exist infinitely many primitive Pythagorean triples < a, b, c > of positive integers satisfying a2 +b2 = c2.These are given by a = 2st, b = s2 βˆ’t2, c = s2 +t2; s, t ∈ Z+. Hatch studied special triples with | a βˆ’b | to uncover connection between them and triangular squares. Then, s2βˆ’t2βˆ’2st=Β±1β‡’(s+t)2βˆ’2s2 =βˆ“1. One can verify that if < g, g+ 1, h >is a special primitive Pythagorean triple, then so is <3g+ 2h+ 1,3g+ 2h+ 2,4g + 3h+ 2 >. Beginning with the primitive triple < 0,1,1 >, we successively derive < 3,4,5 >, < 20,21,29 > , < 119,120,169 >,... This construction exhausts all such triples. They can also be obtained from any of the following four relations:

(QkQkβˆ’1)2+

QkQkβˆ’1+ (βˆ’1)k 2 =

Q2kβˆ’1+Q2kβˆ’2

2

2

, kβ‰₯1 (29) (2PkPkβˆ’1)2 +

PkPkβˆ’1+ (βˆ’1)k 2 = (P2kβˆ’1)2 (30) ((Bn+bn)βˆ’(Bnβˆ’1+bnβˆ’1)βˆ’1)

2

2

+

((Bn+bn)βˆ’(Bnβˆ’1+bnβˆ’1) + 1) 2

2

=

((Bn+bn) + (Bnβˆ’1+bnβˆ’1) + 1) 2

2

;k β‰₯1 (31)

(Bnβˆ’Bnβˆ’1)2 =

7Bnβˆ’1βˆ’Bnβˆ’2βˆ’1 2

2

+

7Bnβˆ’1βˆ’Bnβˆ’2+ 1 2

2

, kβ‰₯1, Bβˆ’1 =βˆ’1 (32) It can also be shown that there exist infinitely many primitive Pythagorean triples having the property a = Tk,b = Tk+1 and c = T(k+1)2. We know Tx2 is a perfect square for infinitely many values of n = x2. In fact, if < g, g+ 1, h > forms a Pythagorean triple, then so does < T2g, T2g+1,(2g + 1)h > proving the infinity of triangular squares as on putting n=hg1, m=ghβˆ’1

2 , we get n(n+ 1)

2 =m2.

(16)

3.1 Some identities : part A

Balancing numbers and the associated numbers discovered this relation between neigh- bouring Balancing numbers that are perfect squares:

(Bn+1+bn+1) = 3(Bn+bn) + 2p

2(Bn+bn)((Bn+bn) + 1) + 1 (33) (Bnβˆ’1+bnβˆ’1) = 3(Bn+bn)βˆ’2p

2(Bn+bn)((Bn+bn) + 1) + 1 (34) The following relations holds:

2

((Bn+bn)βˆ’(Bnβˆ’1+bnβˆ’1))2+ 1 ={((Bn+bn) + (Bnβˆ’1+bnβˆ’1) + 1)}2 (35) The following relation holds between two consecutive squares that are Balancing numbers:

Bn+1 = 3Bn+p

8Bn2+ 1 (36)

Bnβˆ’1 = 3Bnβˆ’p

8Bn2 + 1 (37)

from (37) the following relation:

2 (Bnβˆ’Bnβˆ’1)2 = (Bn+Bnβˆ’1)2+ 1 (38) The general recurrence relations regarding (Bn+bn) and Bn:

(Bn+d+bn+d) = {(Bd+1βˆ’Bdβˆ’1) (Bn+bn)} βˆ’ {(Bnβˆ’d+bnβˆ’d)βˆ’2 (Bd+bd)}, dβ‰₯1 (39) Bn+d={(Bd+1βˆ’Bdβˆ’1)Bn} βˆ’Bnβˆ’d, dβ‰₯1 (40) The identities involving 2k+ 1 number of consecutive values of (Bn+bn) and Bn:

2k

X

j=0

(Bi+j +bi+j) (βˆ’1)j =Qk+1Β·Pk(2Bi+k+bi+k) + (βˆ’1)k(Bi+k+bi+k)βˆ’r (41) r = 1 if 2k+ 1≑3(mod4) ,r= 0 if 2k+ 1≑1(mod4);iβ‰₯1, k β‰₯1.

2k

X

j=0

(Bi+j+bi+j) = (Bn+Bn+1) (Bi+k+bi+k) + 2

k

X

j=1

(Bj+bj), iβ‰₯1, k β‰₯1 (42)

2k

X

j=0

Bi+j(βˆ’1)j = 2Qk+1Β·Pk(Bi+k) + (βˆ’1)k(Bi+k), iβ‰₯1, k β‰₯1 (43)

2k

X

j=0

Bi+j = (Bk+Bk+1)(Bi+k), iβ‰₯1, k β‰₯1 (44) Putting i= 1 in (42) and (43),we get these sum formula:

2k

X

j=0

(B1+j +b1+j) = (Bn+Bn+1) (B1+k+b1+k) + 2

k

X

j=1

(Bj+bj), kβ‰₯1 (45)

2k

X

j=0

B1+j = (Bk+Bk+1)(B1+k), kβ‰₯1 (46)

(17)

The general formula for 2k number of consecutive values of (Bn+bn) andBn:

2kβˆ’1

X

j=0

(Bi+j +bi+j) (βˆ’1)j+1 =Bk((Bi+k+bi+k)βˆ’(Bi+kβˆ’1+bi+kβˆ’1)), iβ‰₯1, k β‰₯1 (47)

2kβˆ’1

X

j=0

(Bi+j+bi+j) = Bk((Bi+kβˆ’1+bi+kβˆ’1) + (Bi+k+bi+k) + 1)βˆ’k, iβ‰₯1, kβ‰₯1 (48)

2kβˆ’1

X

j=0

Bi+j(βˆ’1)j+1 =Bk((Bi+k)βˆ’(Bi+kβˆ’1)), iβ‰₯1, kβ‰₯1 (49)

2kβˆ’1

X

j=0

Bi+j =Bk((Bi+kβˆ’1) + (Bi+k)), iβ‰₯1, kβ‰₯1 (50) Putting i= 1 in (48) and (49),we get these sum formula:

2kβˆ’1

X

j=0

(B1+j +b1+j) =Bk((Bn+bn) + (Bn+1+bn+1) + 1)βˆ’k, k β‰₯1 (51)

2kβˆ’1

X

j=0

B1+j =Bk(Bk+Bk+1), kβ‰₯1 (52) We could get (52) by combining the next two results:

Identity: 1.

k

X

j=1

B2jβˆ’1 = (Bk)2, kβ‰₯1 (53) Proof. We can prove it by induction. The identity is obviously true for n = 1. Suppose the identity is true for n=k. Then,Pk

j=1B2jβˆ’1 = (Bk)2 Hence

k+1

X

j=1

B2jβˆ’1 =

k

X

j=1

B2jβˆ’1+B2k+1 = (Bk)2+B2k+1

= (3 + 2√

2)kβˆ’(3βˆ’2√ 2)k 4√

2

!2

+(3 + 2√

2)2k+1βˆ’(3βˆ’2√ 2)2k+1 4√

2

= (3 + 2√

2)2kβˆ’2 + (3βˆ’2√ 2)2k 4√

2 +(3 + 2√

2)2k+1βˆ’(3βˆ’2√ 2)2k+1 4√

2

= (3 + 2√

2)2k+ 4√

2(3 + 2√ 2)2k+1

32 + (3βˆ’2√

2)2kβˆ’4√

2(3βˆ’2√ 2)2k+1

32 βˆ’ 2

32

= (3 + 2√ 2)2k

1 + 4√

2(3 + 2√ 2)

32 +(3βˆ’2√

2)2k

1βˆ’4√

2(3βˆ’2√ 2)

32 βˆ’ 2

32

= (17 + 12√ 2)k

17 + 12√ 2

32 +(17βˆ’12√

2)k

17βˆ’12√ 2

32 βˆ’ 2

32 (17 + 12√

2)k+1+ (17βˆ’12√

2)k+1βˆ’2

32 =Bk+12 = (Bk+1)2

(18)

Thus if it is true for n=k then it is true for n=k+ 1 also.Hence true for all n.

Identity: 2.

k

X

j=1

B2j =BkBk+1, kβ‰₯1 (54) Proof. We prove it by induction again.The result is true for n = 1. Suppose it is true for n =k. Then,Pk

j=1B2j =BkBk+1. Hence

k+1

X

j=1

B2j =

k

X

j=1

B2j+B2k+2 =BkBk+1+B2k+2

= (3 + 2√

2)kβˆ’(3βˆ’2√ 2)k 4√

2 βˆ— (3 + 2√

2)k+1βˆ’(3βˆ’2√ 2)k+1 4√

2 +(3 + 2√

2)2k+2βˆ’(3βˆ’2√ 2)2k+2 4√

2

= (3 + 2√

2)2k+1βˆ’(3βˆ’2√

2)βˆ’(3 + 2√

2) + (3βˆ’2√ 2)2k+1 32

+(3 + 2√

2)2k+2βˆ’(3βˆ’2√ 2)2k+2 4√

2 (3 + 2√

2)2k+1βˆ’6 + (3βˆ’2√ 2)2k+1

32 + (3 + 2√

2)2k+2βˆ’(3βˆ’2√ 2)2k+2 4√

2

= (3 + 2√

2)2k+1+ 4√

2(3 + 2√

2)2k+2βˆ’6 32

+(3βˆ’2√

2)2k+1βˆ’4√

2(3βˆ’2√ 2)2k+2 32

= (3 + 2√

2)2k+1

1 + 4√

2(3 + 2√ 2)

βˆ’6 32

+(3βˆ’2√

2)2k+1

1βˆ’4√

2(3βˆ’2√ 2) 32

= (3 + 2√

2)2k+1

17 + 12√ 2

βˆ’6 + (3βˆ’2√

2)2k+1

17βˆ’12√ 2 32

= (3 + 2√

2)2k+3βˆ’6 + (3βˆ’2√ 2)2k+3 32

= (3 + 2√

2)k+1βˆ’(3βˆ’2√ 2)k+1 4√

2 βˆ— (3 + 2√

2)k+2βˆ’(3βˆ’2√ 2)k+2 4√

2

=Bk+1Bk+2.

Thus if it is true for n = k then it is true for n =k+ 1 also. Hence true for all n. The interesting sum formula holds:

6

[kβˆ’12 ]

X

j=0

B2kβˆ’4jβˆ’1 =BkBk+1, kβ‰₯1 (55)

[] is the greatest integer function.

6

k

X

r=1

B2r = 36

[k+12 ]

X

r=1

B2kβˆ’4r+3 =B2k+B2k+1βˆ’1 (56)

(19)

The identities expressing n and m in terms of all preceding values:

(Bi+bi) = 5(Biβˆ’1 +biβˆ’1) + 4

iβˆ’2

X

j=1

(Bj +bj) + (2iβˆ’1), iβ‰₯1. (57)

Bi = 5Biβˆ’1+ 4

iβˆ’2

X

j=1

Bj + 1, iβ‰₯1 (58)

This identity is more interesting.

Identity: 3.

B2rβˆ’1 = (B2r)2βˆ’(Brβˆ’1)2 (59) Proof. We can establish it by simply manipulating the defining formula.

R.H.S = (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

!2

βˆ’ (3 + 2√

2)rβˆ’1βˆ’(3βˆ’2√ 2)rβˆ’1 4√

2

!2

= (3 + 2√

2)2rβˆ’2 + (3βˆ’2√ 2)2r

32 βˆ’(3 + 2√

2)2rβˆ’2βˆ’2 + (3βˆ’2√ 2)2rβˆ’2 32

= (3 + 2√

2)2rβˆ’(3 + 2√

2)2rβˆ’2+ (3βˆ’2√

2)2rβˆ’(3βˆ’2√ 2)2rβˆ’2 32

= (3 + 2√

2)2rβˆ’2

(3 + 2√

2)2βˆ’1

+ (3βˆ’2√

2)2rβˆ’2

(3βˆ’2√

2)2βˆ’1 32

= (3 + 2√

2)2rβˆ’2(16 + 12√

2) + (3βˆ’2√

2)2rβˆ’2(16βˆ’12√ 2) 32

= (3 + 2√

2)2rβˆ’2(4 + 3√

2) + (3βˆ’2√

2)2rβˆ’2(4βˆ’3√ 2) 8

= (3 + 2√

2)2rβˆ’2(2√

2 + 3) + (3βˆ’2√

2)2rβˆ’2(2√ 2βˆ’3) 4√

2

= (3 + 2√

2)2rβˆ’1βˆ’(3βˆ’2√ 2)2rβˆ’1 4√

2 =B2rβˆ’1

=L.H.S Identity: 4.

B2r =Br(Br+1βˆ’Brβˆ’1) (60) Proof. We establish it with the help of the method employed above.

R.H.S = (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

(3 + 2√

2)r+1βˆ’(3βˆ’2√ 2)r+1 4√

2 βˆ’ (3 + 2√

2)rβˆ’1βˆ’(3βˆ’2√ 2)rβˆ’1 4√

2

!

= (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

(3 + 2√ 2)rβˆ’1

(3 + 2√

2)2βˆ’1 4√

2 βˆ’ (3βˆ’2√

2)rβˆ’1

(3βˆ’2√

2)2βˆ’1 4√

2

= (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

(3 + 2√

2)rβˆ’1(4 + 3√

√ 2)

2 βˆ’(3βˆ’2√

2)rβˆ’1(4βˆ’3√

√ 2) 2

= (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2 [(3 + 2√

2)rβˆ’1(2√

2 + 3)βˆ’(3βˆ’2√

2)rβˆ’1(2√

2βˆ’3)]

= (3 + 2√

2)2rβˆ’(3βˆ’2√ 2)2r 4√

2 =B2r

=L.H.S

(20)

Identity: 5.

(B2rβˆ’1+b2rβˆ’1) = ((Br+br)βˆ’(Brβˆ’1+brβˆ’1))2 (61) Proof.

R.H.S = (3 + 2√

2)r+ (3βˆ’2√

2)rβˆ’2

4 βˆ’(3 + 2√

2)rβˆ’1+ (3βˆ’2√

2)rβˆ’1βˆ’2 4

!2

= (3 + 2√

2)rβˆ’(3 + 2√ 2)rβˆ’1

4 + (3βˆ’2√

2)rβˆ’(3βˆ’2√ 2)rβˆ’1 4

!2

= (3 + 2√

2)rβˆ’1(3 + 2√ 2βˆ’1)

4 + (3βˆ’2√

2)rβˆ’1(3βˆ’2√ 2βˆ’1) 4

!2

= (3 + 2√

2)rβˆ’1(1 +√ 2)

2 +(3βˆ’2√

2)rβˆ’1(1βˆ’βˆš 2) 2

!2

= (3 + 2√

2)2rβˆ’2(1 +√ 2)2

4 +(3βˆ’2√

2)2rβˆ’2(1βˆ’βˆš 2)2 4

+2βˆ— (3 + 2√

2)rβˆ’1(1 +√ 2)

2 βˆ— (3βˆ’2√

2)rβˆ’1(1 +√ 2) 2

= (3 + 2√

2)2rβˆ’1+ (3βˆ’2√ 2)2rβˆ’1

4 = (B2rβˆ’1+b2rβˆ’1)

=L.H.S Identity: 6.

(B2r+b2r) = (2(Br+br) + 1)2βˆ’1, rβ‰₯1 (62) Proof.

R.H.S = (2(Br+br) + 1)2βˆ’1

= 2(3 + 2√

2)r+ (3βˆ’2√

2)rβˆ’2

4 + 1

!2

βˆ’1

= (3 + 2√

2)r+ (3βˆ’2√

2)rβˆ’2 + 2 2

!2

βˆ’1

= (3 + 2√

2)2r+ (3βˆ’2√

2)2rβˆ’2

4 = (B2r+b2r) =L.H.S

(B3r+b3r) = (Br+br) (4 (Br+br) + 3)2 (63) Identity: 7.

B3r =Br 25Br2+ 3

(64) Proof.

R.H.S = (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

ο£±

ο£²

ο£³

32 (3 + 2√

2)rβˆ’(3βˆ’2√ 2)r 4√

2

!2

+ 3

ο£Ό

ο£½

ο£Ύ

= 1

4√ 2

n

(3 + 2√

2)rβˆ’(3βˆ’2√ 2)ro n

(3 + 2√

2)2r+ (3βˆ’2√

2)2r+ 1o

= 1

4√ 2

n

(3 + 2√

2)3rβˆ’(3βˆ’2√ 2)3ro

=B3r =L.H.S

(21)

The general identity for B(2n+1)r is

B(2n+1)r = 25Br2n+1+ 25(nβˆ’1)(2n+ 1)Br2nβˆ’1 +

"

(2n+ 1) (nβˆ’1

X

k=1

25(nβˆ’kβˆ’1) Pk

(k+ 1)!Br2(nβˆ’k)βˆ’1 )#

where the product

Pk =

k

Y

j=1

2(nβˆ’k) +j βˆ’1 (65)

some interesting relations between n and m:

(Bi +bi) + (Biβˆ’2j+1+biβˆ’2j+1) = 2 (Bjβˆ’Bjβˆ’1) (Biβˆ’j+1βˆ’Biβˆ’j)βˆ’1, j β‰₯1, iβ‰₯2j (66) Bi+Biβˆ’2j+1 = (Bjβˆ’Bjβˆ’1) ((Biβˆ’j+1+biβˆ’j+1)βˆ’(Biβˆ’j+biβˆ’j))j β‰₯0, iβ‰₯2j (67) ((Bi+j +bi+j) + (Biβˆ’j +biβˆ’j)) = (2(Bj+bj) + 1) (3Biβˆ’Biβˆ’1)βˆ’1 (68) ((Bi+j+bi+j)βˆ’(Biβˆ’j +biβˆ’j)) = 8BiBj;i > j >0 (69) Bi+2jβˆ’1βˆ’Biβˆ’2j+1 = 2B2jβˆ’1(2(Bi+bi) + 1), j β‰₯0, iβ‰₯2j (70)

(Bi+bi) = Bi+ 2

iβˆ’1

X

j=1

Bi, iβ‰₯1 (71)

3.2 Identities: part B

We consider the triangular squares. Neighbouring triangular squares can be expressed in terms of each other:

Bn+12 = 6p

Bn2(8Bn2 + 1) + 17Bn2+ 1 (72) B2nβˆ’1 = 17Bn2βˆ’6p

Bn2(8B2n+ 1) + 1 (73) As Bn2 is both a square and triangular number, the quantity under the square root above is an integer.equation (72) and equation (73), obtained from (2), can be proved otherwise.

The relation from (72):

Identity: 1.

n 9

8 Bn2βˆ’Bnβˆ’12 2+ 1o

=

8(B2n+Bnβˆ’12 ) + 1 2 (74)

(22)

Proof. We establish it by using definition and the identity to come later with proof.

R.H.S = 8(17 + 12√

2)k+ (17βˆ’12√

2)kβˆ’2

32 + 8(17 + 12√

2)kβˆ’1+ (17βˆ’12√

2)kβˆ’1βˆ’2

32 + 1

!

= (17 + 12√

2)kβˆ’1(17 + 12√ 2 + 1)

4 +(17βˆ’12√

2)kβˆ’1(17βˆ’12√

2 + 1)kβˆ’1 4

!2

= 9 (17 + 12√

2)kβˆ’1(3 + 2√ 2)

2 + (17βˆ’12√

2)kβˆ’1(3βˆ’2√ 2) 2

!2

= 9 (3 + 2√ 2)2kβˆ’1

2 +(3βˆ’2√ 2)2kβˆ’1 2

!2

= 9

((3 + 2√

2)2(2kβˆ’1)+ 2 + (3βˆ’2√

2)2(2kβˆ’1) 4

)

= 9 (

8(3 + 2√

2)2(2kβˆ’1)βˆ’2 + (3βˆ’2√

2)2(2kβˆ’1)

32 + 1

)

= 9 (

8(17 + 12√

2)2kβˆ’1+ (17βˆ’12√

2)2kβˆ’1βˆ’2

32 + 1

)

= 9

8B2kβˆ’12 + 1 = 9n

8 Bk2βˆ’Bkβˆ’12 2

+ 1o

=L.H.S This relation can be verified easily using (55) and (56):

Bn2 +Bnβˆ’12 = 6βˆ—q

Bn2Bnβˆ’12 + 1 (75)

The general recurrence relation:

Bn+d2 =

(B2d+1βˆ’B2d+1)B2n βˆ’Bnβˆ’d2 + 2Bd2, dβ‰₯1 (76) The following relation holds between four consecutive triangular squares:

Bn+12 βˆ’B2nβˆ’2 = 35 Bn2βˆ’Bnβˆ’12

(77) We now have identities involving many more consecutive triangular squares:

4kβˆ’2

X

j=0

Bi+j2 (βˆ’1)j =Bn 16Bi+k2 + 1

+Bi+k2 ;i, k β‰₯1 (78)

Bn0 = 2Pk

i=1B4iβˆ’3; alternatively, Bn0 == 2

B2n2 βˆ’Pk

i=1B4iβˆ’3

.

2kβˆ’1

X

j=0

B2i+j(βˆ’1)j+1 =B2n Bi+k2 βˆ’Bi+kβˆ’12

, iβ‰₯1, kβ‰₯2 (79)

B2n = B2nβˆ’2P[

kβˆ’2 2 ]

j=0 B2kβˆ’4jβˆ’3 ;[ ] denotes the greatest integer function;k β‰₯ 2. Alterna- tively, B2n=

Pk

j=1B2jβˆ’1(βˆ’1)j

;kβ‰₯2.

(23)

We next calculate the even order Balancing numbers. We shall prove B2k : B2k+2 = B2kβˆ’2+ 32B2n+ 2;k β‰₯ 1. Observe thatB4kβˆ’2βˆ’1 =B2kβˆ’2(B2k+1βˆ’B2kβˆ’1) ;k β‰₯2;B4k = B2kβˆ—(B2k+1βˆ’B2kβˆ’1) 2;k β‰₯2

Puttingi= 1 in (79) gives the difference of the first 2k triangular squares:

2kβˆ’1

X

j=0

B1+j2 (βˆ’1)j+1 =B2n B1+k2 βˆ’Bk2

, iβ‰₯1, kβ‰₯2 (80) We have these general sum formula:

2k

X

j=0

Bi+j2 =B2k+1Bi+k2 + 2

k

X

j=1

B2j, iβ‰₯1, k β‰₯1 (81)

2kβˆ’1

X

j=0

Bi+j2 =B2k Bi+kβˆ’12 +Bi+k

+ 4Bkβˆ’12 , iβ‰₯1, k β‰₯2 (82) Puttingi= 1 in the preceding identities gives the sum of first 2k+1/2ktriangular squares:

2k

X

j=0

B1+j2 =B2k+1B1+k2 + 2

k

X

j=1

Bj2, kβ‰₯1 (83)

2kβˆ’1

X

j=0

B1+j2 =

ο£±

ο£²

ο£³

Bn2βˆ’2

[kβˆ’22 ]

X

j=0

B2kβˆ’4jβˆ’3

ο£Ό

ο£½

ο£Ύ

Bn2+Bn+12

+ 4Bnβˆ’12 , kβ‰₯2 (84)

2kβˆ’1

X

j=0

B1+j2 =

k

X

j=1

B2jβˆ’1(βˆ’1)j

Bn2 +Bn+12

+ 4B2nβˆ’1, k β‰₯2 (85) I found this relation expressing N in terms of all preceding values:

Bn2 = 33Bnβˆ’12 + 32

kβˆ’2

X

j=1

Bj2+ (2kβˆ’1), kβ‰₯1. (86)

Identity:2.

B2rβˆ’12 = Br2βˆ’Brβˆ’12 2

(87) Proof.

R.H.S = (17 + 12√

2)r+ (17βˆ’12√

2)rβˆ’2

32 βˆ’(17 + 12√

2)rβˆ’1+ (17βˆ’12√

2)rβˆ’1 βˆ’2 32

!2

= (17 + 12√

2)rβˆ’1(17 + 12√ 2βˆ’1)

32 βˆ’ (17βˆ’12√

2)rβˆ’1(17βˆ’12√ 2βˆ’1) 32

!2

= (17 + 12√

2)rβˆ’1(4 + 3√ 2)

8 βˆ’ (17βˆ’12√

2)rβˆ’1(4βˆ’3√ 2) 8

!2

= (17 + 12√

2)2rβˆ’1+ (17βˆ’12√

2)2rβˆ’1βˆ’2

32 =B22rβˆ’1 =L.H.S.

(24)

Identity:3.

B2r2 = 4Br2 8B2r + 1

, r≀1. (88)

Proof.

R.H.S = 32Br2+ 4Br2

= 32 (17 + 12√

2)rβˆ’(17βˆ’12√

2)rβˆ’2 32

!2

+ 4 (17 + 12√

2)rβˆ’(17βˆ’12√

2)rβˆ’2 32

!

= (17 + 12√

2)2r+ (17βˆ’12√

2)2r+ 4 + 2(17 + 12√

2)r(17βˆ’12√ 2)r 32

βˆ’4(17βˆ’12√

2)rβˆ’4(17 + 12√ 2)r

32 + (17 + 12√

2)r+ (17βˆ’12√

2)rβˆ’2 8

!

= (17 + 12√

2)2r+ (17βˆ’12√

2)2rβˆ’2 32

!

βˆ’ (17 + 12√

2)r+ (17βˆ’12√

2)rβˆ’2 8

!

+ (17 + 12√

2)r+ (17βˆ’12√

2)rβˆ’2 8

!

=B2r2 =L.H.S

(25)

References

[1] A. Behera and G. K. Panda, On the square roots of triangular numbers, Fib. Quart., 37(2) (1999), 98105.

[2] Z. Cerin, Properties of odd and even terms of the Fibonacci sequence, Demonstratio Math., 39(1) (2000), 55-60.

[3] Z. Cerin, Some alternating sums of Lucas numbers, Cent. Eur. J. Math., 3(1) (2005), 1-13.

[4] Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas numbers, Integers, 6 (2006).

[5] Z. Cerin and G. M. Gianella, Formulas for sums of squares and products of Pell numbers, Acc. Sc. Torino-Atti Sc. Fis., 140 (2006), 113-122

[6] Weisstein, Eric W., Square Triangular Number, From MathWorld-A Wolfram Web Resource,

[7] G. K. Panda and P. K. Ray, Cobalancing numbers and cobalancers, Internat. J. Math.

Math. Sci. 8 (2005), 11891200.

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References

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