Balancing polynomials and their derivatives
We study the generalization of balancing numbers with a new sequence of numbers called $k$-balancing numbers. Moreover, by using the Binet formula for $k$-balancing numbers, we obtain the identities including the generating function of these numbers. In addition, the properties of divisibility of th...
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| author | Ray, P. K. Раі, П. К. |
| author_facet | Ray, P. K. Раі, П. К. |
| author_sort | Ray, P. K. |
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| description | We study the generalization of balancing numbers with a new sequence of numbers called $k$-balancing numbers. Moreover,
by using the Binet formula for $k$-balancing numbers, we obtain the identities including the generating function of these
numbers. In addition, the properties of divisibility of these numbers are investigated. Further, balancing polynomials that
are natural extensions of the $k$-balancing numbers are introduced and some relations for the derivatives of these polynomials
in the form of convolution are also proved. |
| first_indexed | 2026-03-24T02:11:14Z |
| format | Article |
| fulltext |
UDC 517.5
P. K. Ray (Sambalpur Univ., India)
BALANCING POLYNOMIALS AND THEIR DERIVATIVES
БАЛАНСУЮЧI ПОЛIНОМИ ТА ЇХ ПОХIДНI
We study the generalization of balancing numbers with a new sequence of numbers called k-balancing numbers. Moreover,
by using the Binet formula for k-balancing numbers, we obtain the identities including the generating function of these
numbers. In addition, the properties of divisibility of these numbers are investigated. Further, balancing polynomials that
are natural extensions of the k-balancing numbers are introduced and some relations for the derivatives of these polynomials
in the form of convolution are also proved.
Вивчається узагальнення балансуючих чисел з новою послiдовнiстю чисел, що називаються k-балансуючими чис-
лами. Бiльш того, за допомогою формули Бiне для k-балансуючих чисел отримано тотожностi, що включають
породжуючу функцiю для цих чисел. Вивчено властивостi подiльностi цих чисел. Крiм того, ми вводимо балансуючi
полiноми, що є природним узагальненням k-балансуючих чисел, i доводимо деякi спiввiдношення для похiдних
цих полiномiв у формi згорток.
1. Introduction. There is a huge interest of many number theorists in the study of a newly developed
number sequence which is popularly known as balancing numbers. According to Behera et al., a
positive integer n is called a balancing number, if it is the solution of a simple Diophantine equation
1 + 2 + . . .+ (n - 1) = (n+ 1) + (n+ 2) + . . .+ (n+ r),
where r is the balancer corresponding to the balancing number n. The balancing number Bn is the
n th term of the sequence \{ 0, 1, 6, 35, 204, . . .\} beginning with the values B0 = 0 and B1 = 1 and
having recurrence relation
Bn+1 = 6Bn - Bn - 1, n \geq 1. (1.1)
A number sequence closely associated with the sequence of balancing numbers is the sequence of
Lucas-balancing numbers \{ Cn\} , where the n th Lucas-balancing number Cn is given by the relation
Cn =
\sqrt{}
8B2
n + 1 with n \geq 0 [6]. The recurrence relation of Lucas-balancing numbers is same
as that of balancing numbers but with different initials, that is, Cn+1 = 6Cn - Cn - 1, n \geq 1 with
C0 = 1 and C1 = 3. The closed form of balancing numbers popularly known as Binet formula is
given by
Bn =
\lambda n
1 - \lambda n
2
\lambda 1 - \lambda 2
,
where \lambda 1 = 3 +
\surd
8 and \lambda 2 = 3 -
\surd
8 are the roots of the equation \lambda 2 - 6\lambda + 1 = 0 [1]. On the
other hand, the Binet formula for Lucas-balancing numbers is the expression Cn =
\lambda n
1 + \lambda n
2
2
[6].
Balancing and Lucas-balancing numbers can also be extended negatively, in particular, B - n = - Bn
and C - n = Cn [9]. Ray, in [9], has shown some interesting properties relating to balancing numbers
using matrices. Among those properties, one important identity was the bilinear index reduction
formula for balancing numbers which is given by the relation
BaBb - BcBd = Ba - 1Bb - 1 - Bc - 1Bd - 1,
c\bigcirc P. K. RAY, 2017
550 ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 551
whenever a+ b = c+ d [9].
While searching of balancing numbers, Liptai, in [3], has found that the only balancing number
in the sequence of Fibonacci numbers is 1. He has also shown that there is no Lucas-balancing
number in the sequence of Fibonacci numbers [4]. In [7], Panda has established many fascinating
results for balancing and Lucas-balancing numbers. For example, the identities that resemble the
trigonometry identities \mathrm{s}\mathrm{i}\mathrm{n}(x \pm y) = \mathrm{s}\mathrm{i}\mathrm{n}x \mathrm{c}\mathrm{o}\mathrm{s} y \pm \mathrm{c}\mathrm{o}\mathrm{s}x \mathrm{s}\mathrm{i}\mathrm{n} y are Bm\pm n = BmCn \pm BnCm . Later,
Panda et al. (see [6]) have linked balancing numbers with well known Pell and associated Pell
numbers by showing that the n th balancing number Bn is the product of n th Pell number Pn and
n th associated Pell number Qn . As usual, Pell and associated Pell numbers are recursively defined
as Pn+1 = 2Pn + Pn - 1, n \geq 2 with P1 = 1, P2 = 2 and Qn+1 = 2Qn + Qn - 1, n \geq 2 with
Q1 = 1, Q2 = 3 respectively.
Different forms of generalization for balancing numbers are available in literature [2, 5, 8, 12]. In
[5], Liptai et al. have generalized the concept of balancing numbers in the following way. Let y, k, l
be fixed positive integers with y \geq 4. A positive integer x with x \leq y - 2 is called a (k, l)-power
numerical center for y if
1k + . . .+ (x - 1)k = (x+ 1)l + . . .+ (y - 1)l.
Also, in [5], several effective and ineffective finiteness results were proved for (k, l)-power numerical
centers using certain Baker-type Diophantine results and Bilu – Tichy theorem, respectively. For
example, they proved that, for a fixed positive integer k with k \geq 1 and l \in \{ 1, 3\} , if (k, l) \not = (1, 1),
then there are only finitely many (k, l)-balancing numbers, and these balancing numbers are bounded
by an effectively computable constant depending only on k . In [12], Szakács has studied a further
generalization of balancing numbers by introducing multiplying balancing numbers. A positive
integer n is called a multiplying balancing number if
1 \cdot 2 . . . (n - 1) = (n+ 1)(n+ 2) . . . (n+ r),
for some positive integer r which is called as multiplying balancer corresponding to the multiplying
balancing number n. He has proved that the only multiplying balancing number is n = 7 with
multiplying balancer r = 3. As a generalization of the notion of balancing numbers, in [2], Bérczes
et al. called R = \{ Ri\} \infty i=0 = R(A,B,R0, R1) a second order linear recurrence sequence if the
recurrence relation
Ri = ARi - 1 +BRi - 2, i \geq 2,
holds, where A,B \not = 0, R0, R1 are fixed rational integers and | R0| + | R1| > 0. They proved that
any sequence Ri = R(A,B, 0, R1) with D = A2 + 4B > 0, (A,B) \not = (0, 1) is not a balancing
sequence.
The present paper is organized as follows. In Section 2, the sequence of k-balancing numbers is
considered and some identities concerning these numbers are derived. Moreover, some divisibility
properties of k-balancing numbers are investigated. In Section 3, balancing polynomials which are
the natural extension of k-balancing numbers are introduced and it can be seen that many of their
properties admit straightforward proofs. We also present the derivatives of the balancing polynomials
in the form of convolution of these polynomials in Section 4.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
552 P. K. RAY
2. \bfitk -Balancing numbers.
Definition 2.1. If k is any positive number, the sequence of k-balancing numbers \{ Bk,n\} \infty n=1
recursively defined as
Bk,n+1 = 6kBk,n - Bk,n - 1, n \geq 1, (2.1)
where Bk,n is the n th k-balancing number with Bk,0 = 0 and Bk,1 = 1.
Some initial k-balancing numbers are
Bk,0 = 0,
Bk,1 = 1,
Bk,2 = 6k,
Bk,3 = 36k2 - 1,
Bk,4 = 216k3 - 12k,
Bk,5 = 1296k4 - 108k2 + 1,
. . . . . . . . . . . . . . . . . . . . . . . . . . .
Noting that, the sequence \{ B1,n\} \infty n=1 is the sequence of balancing numbers.
We observe that, the equation (2.1) is a second order difference equation having characteristic
equation \alpha 2 = 6k\alpha - 1, whose roots are indeed \alpha 1 = 3k +
\surd
9k2 - 1 and \alpha 2 = 3k -
\surd
9k2 - 1
with 9k2 - 1 \geq 0. Clearly, \alpha 2 is the conjugate root of \alpha 1 and also notice that, the sum, the product
and the difference of these roots are respectively given as 6k, 1 and 2
\surd
9k2 - 1.
The following are some identities involving k-balancing numbers.
As \alpha 1 = 3k +
\surd
9k2 - 1 and \alpha 2 = 3k -
\surd
9k2 - 1 are roots of \alpha 2 = 6k\alpha - 1, we have
\alpha 2
1 = 6k\alpha 1 - 1, and \alpha 2
2 = 6k\alpha 2 - 1. Multiplying \alpha n
1 and \alpha n
2 to both these equations respectively,
we obtain the following result.
Lemma 2.1. For any integer n \geq 1, \alpha n+2
1 = 6k\alpha n+1
1 - \alpha n
1 and \alpha n+2
2 = 6k\alpha n+1
2 - \alpha n
2 .
Binet formulas for certain sequences such as Fibonacci sequence, Lucas sequence, balancing
sequence, etc. are useful to establish many of their identities. With the help of mathematical
induction, it is easy to derive the Binet formula for k-balancing numbers and is given by the identity
Bk,n =
\alpha n
1 - \alpha n
2
\alpha 1 - \alpha 2
, (2.2)
with \alpha 1 = 3k +
\surd
9k2 - 1 and \alpha 2 = 3k -
\surd
9k2 - 1. An application of this Binet formula gives a
combinatorial identity of k-balancing numbers as follows.
Lemma 2.2. Let
\biggl(
n
j
\biggr)
denote the usual notation for combination. Then for any integer n \geq 0,
we have
n\sum
j=0
( - 1)n+j
\biggl(
n
j
\biggr)
6jkjBk,j = Bk,2n.
Proof. Using Binet formula (2.2), the left-hand side of the identity reduces to
n\sum
j=0
( - 1)n+j
\biggl(
n
j
\biggr)
6jkj
\Biggl[
\alpha j
1 - \alpha j
2
\alpha 1 - \alpha 2
\Biggr]
,
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 553
which on simplification gives
1
\alpha 1 - \alpha 2
\left[ n\sum
j=0
( - 1)n+j
\biggl(
n
j
\biggr)
(6k\alpha 1)
j -
n\sum
j=0
( - 1)n+j
\biggl(
n
j
\biggr)
(6k\alpha 2)
j
\right] .
With some algebraic manipulations, the expression further simplifies to
1
\alpha 1 - \alpha 2
[(6k\alpha 1 - 1)n - (6k\alpha 2 - 1)n] =
\alpha 2n
1 - \alpha 2n
2
\alpha 1 - \alpha 2
= Bk,2n,
which ends the proof.
The k-balancing numbers are also extended negatively like balancing numbers. Some further
applications of Binet formula show the following facts.
Proposition 2.1. For n \geq 1, Bk, - n = - Bk,n.
Proposition 2.2. For any natural numbers p, q and r, Bk,p+q - 1 = Bk,pBk,q - Bk,p - 1Bk,q - 1.
Proposition 2.3. For any natural numbers p, q and r, Bk,p+q - 2 =
1
6k
[Bk,pBk,q - Bk,p - 2Bk,q - 2] .
Proposition 2.4. For any natural numbers p, q and r,
Bk,p+q+r - 3 =
1
6k
[Bk,pBk,qBk,r - 6kBk,p - 1Bk,q - 1Bk,r - 1 +Bk,p - 2Bk,q - 2Bk,r - 2] .
Since \alpha 1 + \alpha 2 = 6k and \alpha 1\alpha 2 = 1, we have \alpha 2
1 = 6k\alpha 1 - 1 = \alpha 1Bk,2 - Bk,1, \alpha 3
1 =
= \alpha 1(6k\alpha 1 - 1) = 6k\alpha 2
1 - \alpha 1 = (36k2 - 1)\alpha 1 - 6k = \alpha 1Bk,3 - Bk,2 and so on. Continuing this
way similarly for \alpha 2, we obtain the following result.
Lemma 2.3. For any integer n \geq 1, \alpha n
1 = \alpha 1Bk,n - Bk,n - 1, \alpha
n
2 = \alpha 2Bk,n - Bk,n - 1.
In [7], Panda has shown that two consecutive balancing numbers are relatively prime, that is
(Bn - 1, Bn) = 1, where (a, b) denotes the greatest common divisor of a and b. The following is a
similar result concerning k-balancing numbers.
Lemma 2.4. Any two consecutive k-balancing numbers are relatively prime,
i.e., (Bk,n, Bk,n - 1) = 1.
Proof. The proof of this result is easy and based on Euclidean algorithm where Bk,n is the
dividend and Bk,n - 1 is the divisor and we have
Bk,n = 6kBk,n - 1 - Bk,n - 2,
Bk,n - 1 = 6kBk,n - 2 - Bk,n - 3,
. . . . . . . . . . . . . . . . . . . . . . . . . . .
Bk,3 = 6kBk,2 - Bk,1,
Bk,2 = 6kBk,1 - Bk,0,
consequently, (Bk,n, Bk,n - 1) = Bk,1 = 1.
Panda has also shown certain divisibility properties for balancing numbers in [7]. Later, Ray,
in [10], has investigated some more of these properties using congruences. The following are some
results concerning divisibility properties for k-balancing numbers.
Lemma 2.5. Let n and m be any positive integers, then Bk,m divides Bk,mn.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
554 P. K. RAY
Proof. The proof is based on induction on n. The result is true for n = 1. Assume that, it
is true for all r \geq 1, that is, Bk,m divides Bk,mj , for every j lies between 1 and r. Therefore, by
Proposition 2.2, Bk,m(r+1) = Bk,mrBk,m+1 - Bk,mr - 1Bk,m. Consequently, Bk,m divides Bk,m(r+1)
by hypothesis and the result follows.
Lemma 2.6. For every positive integers m and t, (Bk,mt - 1, Bk,m) = 1.
Proof. Let d = (Bk,mt - 1, Bk,m). It follows that, d divides Bk,mt - 1 and d divides Bk,m . As
Bk,m divides Bk,mt by Lemma 2.5, d divides Bk,mt. Further, d divides Bk,mt - 1 and d divides Bk,mt
and as Bk,mt - 1 and Bk,mt are relatively prime, by division algorithm, d divides 1 which follows
that d = 1.
Lemma 2.7. Let m, n, s and t are positive integers. Then for m = sn + t, (Bk,m, Bk,n) =
= (Bk,n, Bk,t).
Proof. Since m = sn + t, the greatest common divisor of Bk,m and Bk,n is given by
(Bk,m, Bk,n) = (Bk,sn+t, Bk,n). Using Proposition 2.2, the right-hand side expression reduces to
(Bk,snBk,t+1 - Bk,sn - 1Bk,t, Bk,n). Further simplification gives (Bk,m, Bk,n) = (Bk,sn - 1Bk,t, Bk,n).
Therefore, the proof of the result follows as Bk,sn - 1 and Bk,n are relatively prime by Lemma 2.6.
The greatest common divisor of any two k-balancing numbers is again a k-balancing number.
The following result shows this fact.
Lemma 2.8. For any two positive integers m and n, (Bk,m, Bk,n) = Bk,(m,n).
Proof. Let m and n are any two positive integers with m \geq n. Then by Euclidean algorithm,
m = q0n+ r1, 0 \leq r1 < n,
n = q1r1 + r2, 0 \leq r2 < r1,
. . . . . . . . . . . . . . . . . . . . . . . . . . .
rn - 2 = qn - 1rn - 1 + rn, 0 \leq rn < rn - 1,
rn - 1 = qnrn + 0.
It follows that, (m,n) = rn . By virtue of Lemma 2.7, we have
(Bk,m, Bk,n) = (Bk,n, Bk,r1) = (Bk,r1 , Bk,r2) = . . . = (Bk,rn - 1 , Bk,rn).
On the other hand as rn| rn - 1, by Lemma 2.5, Bk,rn | Bk,rn - 1 . It follows that (Bk,rn - 1 , Bk,rn) =
= Bk,rn . Consequently, (Bk,m, Bk,n) = Bk,rn = Bk,(m,n) .
Corollary 2.1. For any two positive integers m and n, if (m,n) = 1, then Bk,mBk,n divides
Bk,mn.
Proof. By virtue of Lemma 2.5, Bk,m divides Bk,mn and Bk,n divides Bk,mn. Therefore, their
least common multiple [Bk,m, Bk,n] divides Bk,mn . Indeed, (Bk,m, Bk,n) = Bk,mn = Bk,1 = 1,
and using the fact that ab = (a, b)[a, b], we get [Bk,m, Bk,n] = Bk,mBk,n, and the result follows.
The matrix used to represent the recurrence relation (2.1) is M =
\biggl(
6k - 1
1 0
\biggr)
, which
for k = 1 reduces to balancing QB matrix studied in [9]. It is easy to notice that Mn =
=
\biggl(
Bk,n+1 - Bk,n
Bk,n Bk,n - 1
\biggr)
. Consider the matrix I - sM =
\biggl(
1 - 6ks - s
s 1
\biggr)
, where I be
the identity matrix same order as M. Since the determinant of the matrix I - sM is 1 - 6ks + s2,
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 555
its inverse will be
(I - sM) - 1 =
1
1 - 6ks+ s2
\biggl(
1 - s
s 1 - 6ks
\biggr)
.
Let g(s) =
\sum \infty
n=0
snMn = (I - sM) - 1 be the generating function for the sequence of k-balancing
numbers. Then
s0M0 + s1M1 + s2M2 + . . . =
\left(
1
1 - 6ks+ s2
- s
1 - 6ks+ s2
s
1 - 6ks+ s2
1 - 6s
1 - 6ks+ s2
\right) .
By equating (2.1) entry from both sides, we get
Bk,0 + sBk,1 + s2Bk,2 + . . . = g(s) =
s
1 - 6ks+ s2
,
and we have the following result.
Lemma 2.9. The generating function of k-balancing numbers is given by
\infty \sum
n=0
Bk,ns
n =
s
1 - 6ks+ s2
.
The following combinatorial identity straightforwardly obtain from the generating function for
k-balancing numbers.
Lemma 2.10. If Bk,n denotes the n th k-balancing number, then
Bk,n =
\lfloor n - 1
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j - 1
j
\biggr)
(6k)n - 2j - 1.
Proof. The defining generating function of the Chebyshev polynomials of the second kind Un(t)
is
\infty \sum
n=0
Un(t)z
n =
1
1 - 2tz + z2
=
\infty \sum
n=0
\left[ \lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(2t)n - 2j
\right] zn.
Setting t = 3k and z = s, where i2 = - 1, we get
s
1 - 6ks+ s2
=
\infty \sum
n=0
\left[ \lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(6k)n - 2j
\right] sn+1. (2.3)
From (2.3) and Lemma 2.9, it is observed that
Bk,n+1 =
\lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(6k)n - 2j =
=
\lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(6k)n - 2j ,
which ends the proof.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
556 P. K. RAY
3. Balancing polynomials. In this section, the sequence of k-balancing numbers \{ Bk,n\} is
extended to the sequence of balancing polynomials \{ Bx,n\} by replacing k with a real variable x. It
is also observed that many of the properties of balancing polynomials admit straightforward proofs.
Definition 3.1. The sequence of balancing polynomials is recursively defined as follows:
Bn+1(x) =
\left\{
1, if n = 0,
6x, if n = 1,
6xBn(x) - Bn - 1(x), if n > 1.
(3.1)
Some initial balancing polynomials are
B0(x) = 0,
B1(x) = 1,
B2(x) = 6x,
B3(x) = 36x2 - 1,
B4(x) = 216x3 - 12x,
B5(x) = 1296x4 - 108x2 + 1.
It is observed that, Bj(1) = Bj for each j .
Looking into the identity from Lemma 2.10, one can easily obtain the general term of balancing
polynomials as, for n \geq 0,
Bn+1(x) =
\lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(6x)n - 2j . (3.2)
Solving the auxiliary equation of (3.1), the roots are \lambda 1(x) = 3x +
\surd
9x2 - 1 and its conjugate
\lambda 2(x) = 3x -
\surd
9x2 - 1, where 9x2 - 1 \geq 0. Notice that, \lambda 1(1) = 3+
\surd
8 and \lambda 2(1) = 3 -
\surd
8 are
balancing constant and its conjugate for the balancing numbers.
In order to find the Binet formula for balancing polynomials, once again method of induction is
used to get
Bn(x) =
\lambda n
1 (x) - \lambda n
2 (x)
\lambda 1(x) - \lambda 2(x)
, (3.3)
where \lambda 1(x) = 3x+
\surd
9x2 - 1 and \lambda 2(x) = 3x -
\surd
9x2 - 1 with 9x2 - 1 \geq 0.
The following result demonstrates the limit of the quotient of two consecutive terms of balancing
polynomials.
Lemma 3.1. For x >
1
3
, \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty
Bn+1(x)
Bn(x)
= \lambda 1(x).
Proof. Using the Binet formula (3.3), we have
\mathrm{l}\mathrm{i}\mathrm{m}
n\rightarrow \infty
Bn+1(x)
Bn(x)
= \mathrm{l}\mathrm{i}\mathrm{m}
n\rightarrow \infty
\lambda n+1
1 (x) - \lambda n+1
2 (x)
\lambda n
1 (x) - \lambda n
2 (x)
=
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 557
= \mathrm{l}\mathrm{i}\mathrm{m}
n\rightarrow \infty
\lambda 1(x) -
\biggl(
\lambda 2(x)
\lambda 1(x)
\biggr) n
\lambda 2(x)
1 -
\biggl(
\lambda 2(x)
\lambda 1(x)
\biggr) n .
Since \lambda 2(x) < \lambda 1(x) for every x >
1
3
, \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty
\biggl(
\lambda 2(x)
\lambda 1(x)
\biggr) n
= 0 and the desired result is obtained.
Observation 3.1. It can be observed that, for x = - 1
3
or
1
3
, \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty
\bigm| \bigm| \bigm| \bigm| Bn+1
Bn
\bigm| \bigm| \bigm| \bigm| = 1 and for
x < - 1
3
, \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty
\bigm| \bigm| \bigm| \bigm| Bn+1
Bn
\bigm| \bigm| \bigm| \bigm| = \lambda 1(x). However, this limit does not exist for x lies between - 1
3
and
1
3
.
The following result straightforwardly deduce by iterating recurrence relation for balancing poly-
nomials.
Proposition 3.1. For an integer r lying between 1 and n - 1,
Bn+1(x) = Br(x)Bn - (r - 2)(x) - Br - 1(x)Bn - (r - 1)(x).
The following result is obtained on replacing n - r - 2 by n and r by m+ 1 in the expression
given in Proposition 3.1.
Proposition 3.2. Let m and n are any two integers, then
Bm+n(x) = Bm+1(x)Bn(x) - Bm(x)Bn - 1(x).
In particular, for m = n - 1 and m = n in the above identity respectively, an expression
for the polynomial of even degree B2n - 1(x) = B2
n(x) - B2
n - 1(x) and the expression B2n(x) =
= Bn(x)[Bn+1(x) - Bn - 1(x)] which is equivalently, B2n(x) =
B2
n+1(x) - B2
n - 1(x)
6x
are obtained.
The matrix corresponds to balancing polynomials is represented by a second order matrix N
whose entries are the first three balancing polynomials, that is N =
\biggl(
6x - 1
1 0
\biggr)
. It can be easily
verified by induction, that the matrix N when raised to the n th power is given by
Nn =
\biggl(
Bn+1(x) - Bn(x)
Bn(x) - Bn - 1(x)
\biggr)
,
for any integer n. In fact, N0 represent the identity matrix and N - n is the matrix inverse of N .
Since
\mathrm{d}\mathrm{e}\mathrm{t}N = 1, \mathrm{d}\mathrm{e}\mathrm{t}Nn = (\mathrm{d}\mathrm{e}\mathrm{t}N)n = 1.
It follows that
B2
n(x) - Bn - 1(x)Bn+1(x) = 1,
which we call as Cassini formula for balancing polynomials. It is also observed that\biggl(
B - n+1(x) - B - n(x)
B - n(x) - B - n - 1(x)
\biggr)
= N - k = (Nk) - 1 =
\biggl(
- Bn - 1(x) Bn(x)
- Bn(x) Bn+1(x)
\biggr)
.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
558 P. K. RAY
Equating (2.1) entry from above matrices, we get B - n(x) = - Bn(x), that shows balancing polyno-
mials are also extended negatively. Further, as NmNn = Nm+n for all m,n \in \BbbZ , the left expression
is
NmNn =
\biggl(
Bm+1(x)Bn+1(x) - Bm(x)Bn(x) Bm(x)Bn - 1(x) - Bm+1(x)Bn(x)
Bm(x)Bn+1(x) - Bm - 1(x)Bn(x) Bm - 1(x)Bn - 1(x) - Bm(x)Bn(x)
\biggr)
,
while the right-hand side expression reduces to
Nm+n =
\biggl(
Bm+n+1(x) - Bm+n(x)
Bm+n(x) - Bm+n - 1(x)
\biggr)
.
Equating (2.1) entry from both the matrices, we obtain the identity given in Proposition 3.2.
Proposition 3.3 (Catalan identity). For integers n, r with n > r, Bn - r(x)Bn+r(x) - B2
n(x) =
= - B2
r (x).
Proof. Using Binet formula (3.3), the left-hand side expression reduces to
-
\biggl(
\lambda 2r
1 (x) + \lambda 2r
2 (x) - 2
[\lambda 1(x) - \lambda 2(x)]2
\biggr)
,
and we obtain the desired result.
It can be seen that, for r = 1, the Catalan identity reduces to the Cassini formula for balancing
polynomials. Further, replacing n by 4n and r = 2n, and then n = 2n + r in the Catalan identity,
we get the following corollary.
Corollary 3.1. B2n(x)[B2n(x)+B6n(x)] = B2
4n(x) and B2n(x)B2n+2r(x)+B2
r (x) = B2
2n+r(x).
Proposition 3.4 (General bilinear index reduction formula). For integers n, r, a, b, c, d with
a+ b = c+ d,
Ba(x)Bb(x) - Bc(x)Bd(x) = Ba - r(x)Bb - r(x) - Bc - r(x)Bd - r(x).
Proof. Using Binet formula (3.3), for all integers n, h and k, the following identity is easily
verified:
Bn+h(x)Bn+k(x) - Bn(x)Bn+h+k(x) = Bh(x)Bk(x).
On setting n = c, h = a - c and k = b - c in the above identity gives
Ba(x)Bb(x) - Bc(x)Ba+b - c(x) = Ba - c(x)Bb - c(x).
Again putting n = c - r, h = a - c and k = b - c in the same identity, we obtain
Ba - r(x)Bb - r(x) - Bc - r(x)Ba+b - c - r(x) = Ba - c(x)Bb - c(x).
Comparing these two identities and using a+ b - c = d follows the proof of the formula.
Setting a = n + 1, b = m, c = n, d = m + 1 and r = n - 1 in the general bilinear index
reduction formula, we have the following result.
Corollary 3.2. For n,m integers with n \leq m, Bm - n(x) = Bn+1(x)Bm(x) - Bn(x)Bm+1(x).
We end this section with the derivation of a product formula for balancing polynomials.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 559
Theorem 3.1. If Bn(x) be the n th balancing polynomial, then
Bn(x) =
\prod
1\leq l\leq n - 1
\biggl[
6x - 2 \mathrm{c}\mathrm{o}\mathrm{s}
l\pi
n
\biggr]
.
Proof. Clearly, the degree of the balancing polynomial Bn(x) is n - 1 for n \geq 1. To find out
a formula for the zeros of Bn(x), we express Bn(x) in terms of hyperbolic functions using Binet
formula. Setting 3x = \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h} z, we have
\lambda 1(x) = \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h} z + \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} z = ez
and
\lambda 2(x) = \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h} z - \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} z = e - z.
It follows that
Bn(x) =
enz - e - nz
ez - e - z
=
\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}(nz)
\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} z
.
Let z = u + iv where the imaginary number i =
\surd
- 1. As \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} z \not = 0, Bn(x) will be zero only
when \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}(nz) = 0, that is e2nz = 1 which is equivalent to e2nu(\mathrm{c}\mathrm{o}\mathrm{s} 2nv+ i \mathrm{s}\mathrm{i}\mathrm{n} 2nv) = 1. Equating
the real part, u = 0 and therefore \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}(inv) = i \mathrm{s}\mathrm{i}\mathrm{n}nv = 0. It follows that, v =
l\pi
n
for any integer
l, and hence z = i
l\pi
n
. Consequently, for any integer l, we have
3x = \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
i
l\pi
n
\biggr)
= \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
l\pi
n
\biggr)
.
As a result, the theorem follows when 1 \leq l \leq n - 1.
Observation 3.2. It is observed that, the number of zeros of the balancing polynomial Bn(x)
is the range of l, i.e., the number of roots of the balancing polynomial Bn(x) are in between 1 to
(n - 1). For instance, if the balancing polynomial B4(x) = 216x3 - 12x is considered, then it can
be observed that the number of roots of B4(x) are 3 and they are Indeed, x =
1
3
\surd
2
,
1
3
, - 1
3
\surd
2
.
Observation 3.3. It is also observed that, for x = 1, the product formula mentioned in Theorem
3.1 gives an alternating expression for Bn . For example
B4 =
\prod
1\leq k\leq 3
\biggl(
6 - 2 \mathrm{c}\mathrm{o}\mathrm{s}
k\pi
4
\biggr)
=
\biggl(
6 - 2 \cdot 1\surd
2
\biggr)
(6 - 2 \cdot 0)
\biggl(
6 + 2 \cdot 1\surd
2
\biggr)
= 204.
4. Derivative sequences of balancing polynomials. In this section, the sequences that are ob-
tained by differentiating the balancing polynomials are studied. Several properties of these sequences
and some relations between the balancing polynomials and their derivatives are also shown.
The identity (3.2) can be rewritten as
Bn(x) =
\lfloor n - 1
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - 1 - j
j
\biggr)
(6x)n - 1 - 2j . (4.1)
Differentiation of (4.1) with respect to x yields
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
560 P. K. RAY
B\prime
n(x) =
\lfloor n - 2
2
\rfloor \sum
j=0
( - 1)j(n - 1 - 2i)
\biggl(
n - 1 - j
j
\biggr)
(6x)n - 2j - 2, (4.2)
with B\prime
0(x) = 0 and B\prime
1(x) = 0. Few first derivative sequences of balancing polynomials with
respect to x are given below
B\prime
1(x) = 0, B\prime
2(x) = 6,
B\prime
3(x) = 72x, B\prime
4(x) = 648x2 - 12,
B\prime
5(x) = 5184x3 - 216x,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Different types of numerical sequences can be generated on substitution of integers in the variable x
of the balancing polynomials such as
\{ B\prime
n(1)\} = \{ 0, 6, 72, 636, 4968, . . .\} ,
\{ B\prime
n(2)\} = \{ 0, 6, 144, 2580, 41040, . . .\} ,
\{ B\prime
n(3)\} = \{ 0, 6, 216, 5820, 139320, . . .\} ,
\{ B\prime
n(4)\} = \{ 0, 6, 288, 10356, 323612, . . .\} .
The following results are some interesting connections between balancing polynomials with their
first derivatives.
Proposition 4.1. Let B\prime
n(x) denote the first derivative of Bn(x). Then
B\prime
n(x) =
3nBn+1(x) - 18xBn(x) - 3nBn - 1(x)
2(9x2 - 1)
for x \not = \pm 1
3
. (4.3)
Proof. Recall that \lambda 1(x) = 3x+
\surd
9x2 - 1 and \lambda 2(x) = 3x -
\surd
9x2 - 1 with \lambda 1(x)\lambda 2(x) = 1.
Their first derivatives are respectively
\lambda \prime
1(x) =
6\lambda 1(x)
\bigtriangleup
and \lambda \prime
2(x) = - 6\lambda 2(x)
\bigtriangleup
,
where \bigtriangleup = 2
\surd
9x2 - 1. Therefore, use of Binet formula (3.3) yields
B\prime
n(x) =
d
dx
\biggl[
\lambda n
1 (x) - \lambda n
2 (x)
\bigtriangleup
\biggr]
=
=
6n\bigtriangleup
\biggl[
\lambda n
1 (x) + \lambda n
2 (x)
\bigtriangleup
\biggr]
- 36x
\biggl[
\lambda n
1 (x) - \lambda n
2 (x)
\bigtriangleup
\biggr]
\bigtriangleup 2
=
=
6n
\biggl[
\lambda n
1 (x)\{ \lambda 1(x) - \lambda 2(x)\} + \lambda n
2 (x)\{ \lambda 1(x) - \lambda 2(x)\}
\bigtriangleup
\biggr]
- 36xBn(x)
\bigtriangleup 2
=
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 561
=
6n
\biggl[
\lambda n+1
1 (x) - \lambda n+1
2 (x)
\bigtriangleup
- \lambda n - 1
1 (x) - \lambda n - 1
2 (x)
\bigtriangleup
\biggr]
- 36xBn(x)
\bigtriangleup 2
=
=
3nBn+1(x) - 18xBn(x) - 3nBn - 1(x)
2(9x2 - 1)
,
which completes the proof.
In particular, for x = 1, the identity (4.3) reduces to
B\prime
n(1) =
3n(Bn+1 - Bn - 1) - 18Bn
16
.
Using the well known identity Bn+1 - Bn - 1 = 2Cn [9], the above identity further reduces to
B\prime
n(1) =
3nCn - 9Bn
8
.
Using the recurrence relation for balancing polynomials, (4.3) may also be rewritten as
B\prime
n(x) =
9(n - 1)xBn(x) - 3nBn - 1(x)
9x2 - 1
for x \not = \pm 1
3
. (4.4)
The derivative of balancing polynomials may be obtained by the self-convolution of balancing
polynomials. The following proposition establishes the fact.
Proposition 4.2 (the derivative of balancing polynomials and the convolved balancing polyno-
mials).
B\prime
1(x) = 0 and B\prime
n(x) =
n - 1\sum
j=1
6Bj(x)Bn - j(x) for n > 1. (4.5)
Proof. The proof of this result is based on induction on n. Clearly the result holds for n = 2.
As an inductive hypothesis, let the formula is true for every polynomial B\prime
m(x) with m \leq n. For
inductive step, differentiate the recurrence relation for balancing polynomials w.r.t. x and use the
inductive hypothesis to get
B\prime
n+1(x) = 6Bn(x) + 6xB\prime
n(x) - B\prime
n - 1(x) =
= 6Bn(x) + 6x
n - 1\sum
j=1
6Bj(x)Bn - j(x) -
n - 2\sum
j=1
6Bj(x)Bn - 1 - j(x) =
= 6Bn(x) + 36xBn - 1(x)B1(x) + 6x
n - 2\sum
j=1
6Bj(x)Bn - j(x) -
n - 2\sum
j=1
6Bj(x)Bn - 1 - j(x) =
= 6Bn(x) + 36xBn - 1(x) +
n - 2\sum
j=1
6Bj(x)[6Bn - j(x) - Bn - 1 - j(x)] =
= 6Bn(x) + 36xBn - 1(x) +
n - 2\sum
j=1
6Bj(x)Bn+1 - j(x) =
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
562 P. K. RAY
= 6Bn(x)B1(x) + 6Bn - 1(x)B2(x) +
n - 2\sum
j=1
6Bj(x)Bn+1 - j(x) =
=
n\sum
j=1
Bj(x)Bn+1 - j(x),
which ends the proof.
Observation 4.1. It is observed that, the identities (4.4) and (4.5) together yields the following
formula:
n - 1\sum
j=1
Bj(x)Bn - j(x) =
3(n - 1)xBn(x) - nBn - 1(x)
2(9x2 - 1)
for n > 1 and x \not = \pm 1
3
.
Setting x = 1, the corresponding formula for balancing numbers will obtain.
Proposition 4.3. For n \geq 1, B\prime
n+1(x) =
\sum \lfloor n - 1
2
\rfloor
j=0
6(n - 2j)Bn - 2j .
Proof. Method of induction is used to prove this result. For n = 1, the result is trivial, since
B\prime
2(x) =
0\sum
j=0
6(1 - j)B1 - j = 6.
We assume that the formula is true for k \leq n. Then
B\prime
n(x) =
\lfloor n - 2
2
\rfloor \sum
j=0
6(n - 1 - 2j)Bn - 1 - 2j
and
B\prime
n+1(x) =
\lfloor n - 1
2
\rfloor \sum
j=0
6(n - 2j)Bn - 2j .
Differentiating the recurrence relation for balancing polynomials Bn+2(x) = 6xBn+1(x) - Bn(x),
using the hypothesis and after some algebra, the result is easily verified for n+ 1.
Proposition 4.4. For n \geq 1, Bn(x) =
1
6n
\Bigl[
B\prime
n+1(x) - B\prime
n - 1(x)
\Bigr]
.
Proof. Using identity (4.2), the right-hand side expression reduces to
\lfloor n
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - j
j
\biggr)
(6x)n - 2j -
\lfloor n - 2
2
\rfloor \sum
j=0
( - 1)j
\biggl(
n - 2 - j
j
\biggr)
(6x)n - 2 - 2j .
Further simplification yields
(6x)n +
\lfloor n
2
\rfloor \sum
i=1
( - 1)j
\biggl[ \biggl(
n - j
j
\biggr)
+
\biggl(
n - 1 - j
j - 1
\biggr) \biggr]
(6x)n - 2j .
Therefore, using usual properties for combinations, we obtain
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
BALANCING POLYNOMIALS AND THEIR DERIVATIVES 563
Bn+1(x) - Bn - 1(x) = (6x)n + n
\lfloor n
2
\rfloor \sum
j=1
( - 1)j
\biggl(
n - 1 - j
j - 1
\biggr)
1
j
(6x)n - 2j .
Differentiating the above equation with respect to x, we get
B\prime
n+1(x) - B\prime
n - 1(x) = 6n(6x)n - 1 + 6n
\lfloor n
2
\rfloor \sum
j=1
( - 1)j
\biggl(
n - 1 - j
j - 1
\biggr)
n - 2i
j
(6x)n - 1 - 2j .
It follows that
B\prime
n+1(x) - B\prime
n - 1(x)
6n
=
\lfloor n - 1
2
\rfloor \sum
j=0
( - 1)j - 1
\biggl(
n - 1 - j
j - 1
\biggr)
(6x)n - 1 - 2j = Bn(x),
and the result follows.
Explicit formulas for any derivative can be obtained similarly. For example, differentiating (4.2)
with respect to x, the second derivative of balancing polynomials becomes
B\prime \prime
n(x) =
\lfloor n - 3
2
\rfloor \sum
j=0
( - 1)j6n - 1 - 2j(n - 1 - 2j)(n - 2j - 2)
\biggl(
n - 1 - j
j
\biggr)
xn - 2j - 3, (4.6)
with B\prime \prime
1 (x) = 0, B\prime \prime
2 (x) = 0 and for n \geq 2.
Some second derivative sequences of balancing polynomials with respect to x are given below
B\prime \prime
1 (x) = 0, B\prime \prime
2 (x) = 0,
B\prime \prime
3 (x) = 72, B\prime \prime
4 (x) = 1296x,
B\prime \prime
5 (x) = 15552x2 - 216,
. . . . . . . . . . . . . . . . . . . . . . . . . . .
For the second derivative of balancing polynomials, the numerical sequences by replacing x with an
integer are given by
\{ B\prime \prime
n(1)\} = \{ 0, 0, 72, 1296, 15336, . . .\} ,
\{ B\prime \prime
n(2)\} = \{ 0, 0, 72, 2592, 61992, . . .\} ,
\{ B\prime \prime
n(3)\} = \{ 0, 0, 72, 3888, 139752, . . .\} ,
\{ B\prime \prime
n(4)\} = \{ 0, 0, 72, 5184, 248832, . . .\} .
Differentiating r times the identity (4.6) yields the recurrence relation for derivative sequences
as follows:
B
(r)
n+1(x) =
\left\{
0, if n < r,
6rr!, if n = r,
1
n - r
\Bigl[
6nxB
(r)
n (x) - (n+ r)B
(r)
n - 1(x)
\Bigr]
, if n > r.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
564 P. K. RAY
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Received 13.12.14
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 4
|
| id | umjimathkievua-article-1715 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:11:14Z |
| publishDate | 2017 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/5b/b11c0286891416a2607e8503510e375b.pdf |
| spelling | umjimathkievua-article-17152019-12-05T09:24:35Z Balancing polynomials and their derivatives Балансуючi полiноми та їх похiднi Ray, P. K. Раі, П. К. We study the generalization of balancing numbers with a new sequence of numbers called $k$-balancing numbers. Moreover, by using the Binet formula for $k$-balancing numbers, we obtain the identities including the generating function of these numbers. In addition, the properties of divisibility of these numbers are investigated. Further, balancing polynomials that are natural extensions of the $k$-balancing numbers are introduced and some relations for the derivatives of these polynomials in the form of convolution are also proved. Вивчається узагальнення балансуючих чисел з новою послiдовнiстю чисел, що називаються $k$-балансуючими числами. Бiльш того, за допомогою формули Бiне для $k$-балансуючих чисел отримано тотожностi, що включають породжуючу функцiю для цих чисел. Вивчено властивостi подiльностi цих чисел. Крiм того, ми вводимо балансуючi полiноми, що є природним узагальненням $k$-балансуючих чисел, i доводимо деякi спiввiдношення для похiдних цих полiномiв у формi згорток. Institute of Mathematics, NAS of Ukraine 2017-04-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1715 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 4 (2017); 550-564 Український математичний журнал; Том 69 № 4 (2017); 550-564 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1715/697 Copyright (c) 2017 Ray P. K. |
| spellingShingle | Ray, P. K. Раі, П. К. Balancing polynomials and their derivatives |
| title | Balancing polynomials and their derivatives |
| title_alt | Балансуючi полiноми та їх похiднi |
| title_full | Balancing polynomials and their derivatives |
| title_fullStr | Balancing polynomials and their derivatives |
| title_full_unstemmed | Balancing polynomials and their derivatives |
| title_short | Balancing polynomials and their derivatives |
| title_sort | balancing polynomials and their derivatives |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1715 |
| work_keys_str_mv | AT raypk balancingpolynomialsandtheirderivatives AT raípk balancingpolynomialsandtheirderivatives AT raypk balansuûčipolinomitaíhpohidni AT raípk balansuûčipolinomitaíhpohidni |