EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 4, 2023, 2306-2322
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The Linear Algebra of the (r, β)-Stirling Matrices

Genevieve B. Engalan1,∗, Mary Joy R. Latayada1

1 Department of Mathematics, Caraga State University, Butuan City, Philippines

Abstract. This paper establishes the linear algebra of the (r, β)-Stirling matrix. Along the way,
this paper derives various identities, such as its factorization and relationship to the Pascal matrix
and the Stirling matrix of the second kind. Additionally, this paper develops a natural extension
of the Vandermonde matrix, which can be used to study and evaluate successive power sums of
arithmetic progressions.

2020 Mathematics Subject Classifications: 05, 11

Key Words and Phrases: (r, β)-Stirling numbers, (r, β)-Stirling matrix, Pascal matrix, Vander-
monde Matrix

1. Introduction

The (r, β)-Stirling numbers are a generalization of the classical Stirling numbers of the

second kind and r-Stirling numbers, and is denoted by,

〈
n
k

〉
β,r

. They were introduced by

Corcino in 1999 [5] by means of the following linear transformation:

tn =

n∑
k=0

〈
n
k

〉
β,r

(t− r)β,k (1)

where

(t− r)β,k =
k−1∏
i=0

(t− r − iβ). (2)

(t)β,k is called the generalized factorial of t with increment β, and as a convention (t)β,k = 0
if k ≤ 0. This numbers have applications in combinatorial and statistical problems.
Corcino and Aldema (2002) [6] further studied the (r, β)-Stirling numbers and derived some
combinatorial identities related to them. Corcino and Montero (2009) [7] also investigated
the (r, β)-Stirling numbers in the context of 0-1 tableaux, which are a tool used in algebraic
combinatorics.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4854

Email addresses: gequisto@carsu.edu.ph (G. Engalan), mrlatayada@carsu.edu.ph (M.J. Layatada)

https://www.ejpam.com 2306 © 2023 EJPAM All rights reserved.



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2307

In this paper, we introduce and study the (r, β)-Stirling matrix and we derive several
interesting identities about (r, β)-Stirling sequence. Two applications are given: we can
generalize the Vandermonde matrices and evaluate successive power sums of arithmetic
progressions.

2. Results

2.1. The (r, β)-Stirling Matrix

The key notions of the study are now defined.

Definition 1. The (r, β)-Stirling matrix is the n× n matrix defined by

S(β,r)(n) =

[〈
n
k

〉
β,r

]
0≤i,j≤n−1

Example 1. When n = 4, we have

S(β,r)(4) =


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β2 + 3βr + 3r2 3β + 3r 1

 . (3)

For the following propositions, we need the generalized n × n Pascal matrix and the
generalized n × n Stirling matrix of the second kind, defined by [2] and [3], respectively
as:

Pn[x] =

[
xi−j

(
i

j

)]
0≤i,j≤n−1

where Pn = Pn[1], and

S2(n)[x] =

[
xi−jS(i, j)

]
0≤i,j≤n−1

where S(i, j) is the Stirling numbers of the second kind.
The main technique used to prove the next propositions is the concept of the Riordan

group introduced by Shapiro et al. [10]. This, briefly, is a group of infinite lower triangular
arrays called Riordan matrices.

A pair of formal power series g and f in the ring C[[z]] define a Riordan matrix as
M = [mn,k]n,k≥0, where g(0) ̸= 0, f(0) = 0, f ′(0) ̸= 0, and [zn] is the coefficient extraction

operator. The matrixM is denoted by (g, f). Moreover, ifmn,k =
[
zn

n! ]g
fk

k! thenM is called
the exponential Riordan matrix or e-Riordan maxtrix, denoted by ⟨g, f⟩. For example,
the e-Riordan matrix representations of the three common e-Riordan matrices used in
this paper - the Pascal matrix, Stirling matrix of the second kind, and the (r, β)-Stirling
matrix:

Pn =
〈
ez, z

〉
, S2(n) =

〈
1, ez − 1

〉
, S(β,r)(n) =

〈
e(r)z,

eβz − 1

β

〉
.



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2308

The set of all e-Riordan matrices forms a group called e-Riordan group under the
Riordan multiplication defined by〈

g, f
〉
∗
〈
h, l
〉
=
〈
gh(f), l(f)

〉
.

Proposition 1. Let Pn be the n× n generalized Pascal matrix, then

S(β,r)(n) = PnS
(β,r−1)(n)

Proof. Consider the e-Riordan matrix representations,

Pn =
〈
ez, z

〉
and S(β,r−1)(n) =

〈
e(r−1)z, e

βz−1
β

〉
.

By using the e-Riordan matrix multiplication, we have

PnS
(β,r−1)(n) =

〈
ez, z

〉
∗
〈
e(r−1)z,

eβz − 1

β

〉
=
〈
eze(r−1)z,

eβz − 1

β

〉
=
〈
ez+(r−1)z,

eβz − 1

β

〉
=
〈
ez+rz−z,

eβz − 1

β

〉
=
〈
erz,

eβz − 1

β

〉
= S(β,r)(n).

Example 2. Let n = 4. Then

P4S
(β,r−1)(4) =


1 0 0 0
1 1 0 0
1 2 1 0
1 3 3 1




1 0 0 0
r − 1 1 0 0

(r − 1)2 β + 2(r − 1) 1 0
(r − 1)3 β2 + 3β(r − 1) + 3(r − 1)2 3β + 3r 1



=


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β2 + 3βr + 3r2 3β + 3r 1


= S(β,r)(4)

Proposition 2. Let Pn[r − s] be the n× n Pascal matrix defined by

Pn[r − s] =
[
(r − s)i−j

(n
k

)]
0≤i,j≤n−1

.



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2309

Then,
S(β,r)(n) = Pn[r − s]S(β,s)(n),

provided that r ≥ s.

Proof. Consider Pn[r − s] = ⟨e(r−s)z, z⟩ and S(β,s)(n) = ⟨erz, eβz−1
β ⟩. Then,

Pn[r − s]S(β,s)(n) =
〈
e(r−s)z, z

〉
∗
〈
esz,

eβz − 1

β

〉
=
〈
e(r−s)zesz,

eβz − 1

β

〉
=
〈
e(r−s)z+sz,

eβz − 1

β

〉
=
〈
erz−sz+sz,

eβz − 1

β

〉
=
〈
erz,

eβz − 1

β

〉
= S(β,r)(n).

Example 3. Let n = 4. Then

P4[r − s]S(β,s)(4) =


1 0 0 0

r − s 1 0 0
(r − s)2 2(r − s) 1 0
(r − s)3 3(r − s)2 3(r − s) 1



1 0 0 0
s 1 0 0
s2 β + 2s 1 0
s3 β2 + 3βs+ 3s2 3β + 3s 1



=


1 0 0 0
r 1 0 0

−1 + 2r + (r − 1)2 b+ 2r 1 0

−2 + 3r + 3 (r − 1)2 + (r − 1)3 b2 + 3b+ 6r − 3 + 3b (r − 1) + 3 (r − 1)2 3b+ 3r 1



=


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β2 + 3βr + 3r2 3β + 3r 1


= S(β,r)(4)

Proposition 3. Let Pn[r] and S2(n)[β] be the n×n Pascal matrix and Stirling matrix of
the second kind. Then,

S(β,r)(n) = Pn[r]S2(n)[β],

where (S2(n)[β]i,j = βi−jS(i, j) and S(i, j) is a Stirling number of the second kind and
0 ≤ i, j ≤ n− 1.



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2310

Proof. It was previously shown that the e-Riordan matrix representations of Pn[r] and
S2(n)[β] are ⟨erz, z⟩ and ⟨1, eβz − 1⟩, respectively.
Using the e-Riordan multiplication, we have

Pn[r]S2(n)[β] =
〈
erz, z

〉
∗
〈
1,

eβz − 1

β

〉
=
〈
erz(1),

eβz − 1

β

〉
=
〈
erz,

eβz − 1

β

〉
= S(β,r)(n).

Example 4. Let n = 4. Then

P4[r]S2(4)[β] =


1 0 0 0
r 1 0 0
r2 2r 1 0
r3 3r2 3r 1



1 0 0 0
0 1 0 0
0 β 1 0
0 β2 3β 1



=


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β2 + 3βr + 3r2 3β + 3r 1


= S(β,r)(4)

Proposition 4. LetPn[r − rβ] and S(1,r)(n)[β] be n× n matrices, where

(S(1,r)(n)[β])ij = βi−j

〈
n
k

〉
1,r

.

Then,
S(β,r)(n) = Pn[r − rβ]S(1,r)(n)[β]

Proof. Consider the e-Riordan matrix representations

Pn[r − rβ] =
〈
e(r−rβ)z, z

〉
and

S(1,r)(n)[β] =

[〈
erz,

ez − 1

1

〉]β
=
〈
erβz,

eβz − 1

β

〉



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2311

Now,

Pn[r − rβ]S(1,r)(n)[β] =
〈
e(r−rβ)z, z

〉
∗
〈
erβz,

eβz − 1

β

〉
=
〈
e(r−rβ)z+rβz,

eβz − 1

β

〉
=
〈
erz−rβz+rβz,

eβz − 1

β

〉
=
〈
erz,

eβz − 1

β

〉
= S(β,r)(n)

Example 5. Let n = 4. Then

P4[r − rβ]S(1,r)(4)[β] =


1 0 0 0

(r − rβ) 1 0 0
(r − rβ)2 2(r − rβ) 1 0
(r − rβ)3 3(r − rβ)2 3(r − rβ) 1



·


1 0 0 0
βr 1 0 0

(βr)2 β + 2βr 1 0
(βr)3 β2 + 3β2r + 3(βr)2 3β + 3βr 1



=


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β2 + 3βr + 3r2 3β + 3r 1


= S(β,r)(4).

2.2. Factorization of the (r, β)-Stirling Matrix

To factor the (r, β)-Stirling matrix, we need the following matrices defined by Zhang
in [13]:

Sn[x] =
[
xi−j

]
0≤j,i≤n−1

For example, when n = 4,

S4[x] =


1 0 0 0
x 1 0 0
x2 x 1 0
x3 x2 x 1

 .

Zhang also define the n× n matrix Gk[x] as

Gk[x] = In−k ⊕ Sk[x], (1 ≤ k ≤ n− 1),



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2312

where Gn[k] = Sn[k] and ⊕ denotes the matrix direct sum.

Proposition 5. For any integer n,m ≥ 1 and r ≥ 0, we have

S(β,r)(n) = Gn[r]Gn−1[r] · · ·G1[r]P̄n−1[β] · · · P̄1[β]

Proof. By proposition 3, S(β,r) = Pn[r]S2(n)[β].
Note that by Theorem 1 of [13],

Pn[r] = Gn[r]Gn−1[r] · · ·G1[r].

Also, based on one of the results of Cheon and Kim in [3],

S2(n)[β] = P̄n−1[β] · · · P̄1[β].

This follows that,

S(β,r)(n) = Gn[r]Gn−1[r] · · ·G1[r]P̄n−1[β] · · · P̄1[β].

2.3. Relationship between the (r, β)-Stirling matrix and a Generalized
Vandermonde Matrix

In this section, we introduce a generalization of the Vandermonde matrix which will
be useful in the study of successive power sums of arithmetic progression. To do that, we
define the following matrices.

Definition 2. Let S(β,r)(n) be the (r, β)-Stirling matrix. The matrix factorial of the
(r, β)-Stirling matrix, denoted by S̃(β,r)(n) is defined by

S̃(β,r)(n) := S(β,r)(n) · diag(0!, 1!, . . . , n!). (4)

Example 6. Let n = 4. Then

S̃(β,r)(4) = S(β,r)(4) · diag(0!, 1!, 2!, 3!)

=


0! 0 0 0
r 1! 0 0
r2 β + 2r 2! 0
r3 β2 + 3βr + 3r2 (3β + 3r)2! 3!



=


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β2 + 3βr + 3r2 6β + 6r 6

 .



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2313

Theorem 1. Let Vβ,r
n (t) be the n× n generalized Vandermonde matrix defined by

Vβ,r
n (t) : = Vβ,r

n (βt+ r, βt+ β + r, βt+ 2β + r, . . . , βt+ (n− 1)β + r)

=


1 1 1 · · · 1

βt+ r βt+ β + r βt+ 2β + r · · · βt+ (n− 1)β + r
(βt+ r)2 (βt+ β + r)2 (βt+ 2β + r)2 · · · (βt+ (n− 1)β + r)2

...
...

...
. . .

...
(βt+ r)n−1 (βt+ β + r)n−1 (βt+ 2β + r)n−1 · · · (βt+ (n− 1)β + r)n−1


and

Cβ
n (t) =

[
βi

(
t+ j
i

)]
0≤i,j≤n−1

.

Then we can factor Vβ,r
n (t) as

Vβ,r
n (t) = S̃(β,r)(n)Cβ

n (t). (5)

Proof. Consider the Equation (1),

tn =
n∑

k=0

〈n
k

〉
β,r

(t− r)β,k.

Note that we can write this as

tn =
n∑

k=0

〈n
k

〉
β,r

( t−r
β

k

)
βkk!.

Replacing t by βt+ r, we have

(βt+ r)n =
n∑

k=0

〈n
k

〉
β,r

(
(βt+r)−r

β

k

)
βkk!

(βt+ r)n =
n∑

k=0

〈n
k

〉
β,r

(
t
k

)
βkk!. (6)

This equation (6) can be represented by the following system of matrix equation for each
n = 0, 1, 2, . . .

v(t) = S̃r,β(n)cn(t), (7)

where
v(t) = [1, βt+ r, (βt+ r)2, (βt+ r)3, . . . , (βt+ r)n−1]

and

cn(t) =
[(t

0

)
, β

(
t
1

)
, β2

(
t
2

)
, . . . , βn−1

(
t

n− 1

)]



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2314

which is the first column of the Vβ,r
n (t). That is,

1
βt+ r

(βt+ r)2

(βt+ r)3

...
(βt+ r)n−1


=



1 0 0 0 0
r 1 0 0 0
r2 β + 2r 2 0 0
...

...
...

. . .
...

rn−1

〈
n− 1
2

〉
β,r

〈
n− 1
3

〉
β,r

2! · · · (n− 1)!





(
t
0

)
β
(
t
1

)
β2
(
t
2

)
...

βn−1
(

t
n−1

)

 .

Thus, by equation (5) we can generalize that,

Vβ,r
n (t) = S̃(β,r)(n)Cβ

n (t).

Example 7. Let n = 4. Then we have matrices

S̃(β,r)(4) =


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β2 + 3βr + 3r2 6β + 6r 6


and

Cβ
4 (t) =


1 1 1 1
βt β(t+ 1) β(t+ 2) β(t+ 3)

β2t(t−1)
2

β2(t+1)t
2

β2(t+2)(t+1)
2

β2(t+3)(t+2)
2

β3t(t−1)(t−2)
6

β3(t+1)t(t−1)
6

β3(t+2)(t+1)(t)
6

β3(t+3)(t+2)(t+1)
6

 .

Now,

S̃(β,r)(4)Cβ
4 (t) =


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β2 + 3βr + 3r2 6β + 6r 6



·


1 1 1 1
βt β(t+ 1) β(t+ 2) β(t+ 3)

β2t(t−1)
2

β2(t+1)t
2

β2(t+2)(t+1)
2

β2(t+3)(t+2)
2

β3t(t−1)(t−2)
6

β3(t+1)t(t−1)
6

β3(t+2)(t+1)(t)
6

β3(t+3)(t+2)(t+1)
6



=


1 1 1 1

βt+ r βt+ β + r βt+ 2β + r βt+ 3β + r
(βt+ r)2 (βt+ β + r)2 (βt+ 2β + r)2 (βt+ 3β + r)2

(βt+ r)3 (βt+ β + r)3 (βt+ 2β + r)3 (βt+ 3β + r)3


= Vβ,r

4 (t)



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2315

Corollary 1. For any real number t, we have

Vβ,r
n (t) = S̃(β,r)(n)△n(t)P

T
n

where [△n(t)]ij = βi

(
t

i− j

)
and P T

n is the transpose of the Pascal Matrix Pn

Proof. From Vandermonde’s convolution identity,

(
m+ n

r

)
=

(mi )(
n

r−i),∑
i=0

we can obtain the LU factorization of Cβ
n (t). That is,

Cβ
n (t) =

[
βi

(
t+ j
i

)]
0≤i,j≤n−1

=

[
βi

i∑
k=0

(
j
i

)(
t

i− k

)]
0≤i,j≤n−1

=

[ i∑
k=0

βi

(
j
i

)(
t

i− k

)]
0≤i,j≤n−1

Let [△n(t)]ij = βi

(
t

i− j

)
. Note the

[(
j
i

)]
ij

is the transpose of the Pascal matrix

Pn =

((
i
j

))
ij

. Then we can write Cβ
n (t) as

Cβ
n (t) = △n(t)P

T
n .

By Theorem 1,
Vβ,r

n (t) = S̃(β,r)(n)△n(t)P
T
n .

Example 8. Let t = 1, and n = 4. Then,

S̃(β,r)(4) =


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β3 + 3βr + r2 6β + 6r 6

 ,

△4(t) =


1 0 0 0
β beta 0 0
0 β2 β2 0
0 0 β3 β3 0





G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2316

and

P T
4 =


1 1 1 1
0 1 2 3
0 0 1 3
0 0 0 1

 .

Now,

S̃(β,r)(n)△n(t)P
T
n =


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β3 + 3βr + r2 6β + 6r 6



×


1 0 0 0
β β 0 0
0 β2 β2 0
0 0 β3 β3



1 1 1 1
0 1 2 3
0 0 1 3
0 0 0 1



=


1 0 0 0

β + r β 0 0
(β + r)2 β(3β + 2r) 2β2 0
(β + r)3 β(7β2 + 9βr + 3r2) 6β2(2β + r) 6β3



1 1 1 1
0 1 2 3
0 0 1 3
0 0 0 1



=


1 1 1 1

β + r 2β + r 3β + r 4β + r
(β + r)2 (2β + r)2 (3β + r)2 (4β + r)2

(β + r)3 (2β + r)3 (3β + r)3 (4β + r)3


= Vβ,r

4 (t).

Corollary 2. For any real number t,

det(Vβ,r
n (t)) =

n−1∏
k=0

k!βk.

Proof. Let Vβ
n(t) be the generalized Vandermonde matrix defined by,

Vβ,r
n (t) = Vβ,r

n (βt+ r, βt+ β + r, βt+ 2β + r, . . . , βt+ (n− 1)β + r).

By Kalman [11], the formula for getting the determinant of a Vandermonde matris is

det((V β
n (t)) =

n−1∏
k=0

(ti − tj).



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2317

Now,

det((V β
n (t)) = [(βt+ (n− 1)β + r)− (βt+ r)] · · · [(βt+ (n− 1)β + r)− (βt+ (n− 2)β + r)]

× [(βt+ (n− 2)β + r)− (βt+ r)] · · · [(βt+ (n− 2)β + r)− (βt+ (n− 3)β + r)]

× [(βt+ (n− 3)β + r)− (βt+ r)] · · · [(βt+ (n− 3)β + r)− (βt+ (n− 4)β + r)]

...

× [(βt+ β + r)− (βt+ r)]

= (n− 1)β · (n− 2)β · (n− 3)β · · ·β × (n− 2)β · (n− 3)β · (n− 4)β · · ·β
× (n− 3)β · (n− 4)β · (n− 5)β · · ·β
...

× β

=

n−1∏
k=0

k!βk.

Example 9. Let n = 4. Then we have,

Vβ,r
4 (t) =


1 1 1 1

β + r 2β + r 3β + r 4β + r
(β + r)2 (2β + r)2 (3β + r)2 (4β + r)2

(β + r)3 (2β + r)3 (3β + r)3 (4β + r)3


Now,

det(Vβ,r
4 (t)) = (βt+ 3β + r − (βt+ r))(βt+ 3β + r − (βt+ β + r))(βt+ 3β + r − (βt+ 2β + r))

× (βt+ 2β + r − (βt+ r))(βt+ 2β + r − (βt+ β + r))

× (βt+ β + r − (βt+ r))

= (3β · 2β · β)(2β · β)(β)
= (3 · 2 · 1β3)(2 · 1β2)(β)

= (3!β3)(2!β2)(1!β)

=
3∏

k=0

k!βk.

Lemma 1. Let Ln[β] be an n× n matrix defined by

[Ln[β]]1≤i,j≤n =

(
j

i− 1

)
βi−1(i− 1)!.

For the n× n (r, β)-Stirling matrix S(β,r)(n),

Vβ,r
n (1) = S(β,r)(n)Ln[β]

T .



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2318

Proof. Let Ln[β] be an n× n matrix defined by

[Ln[β]]1≤i,j≤n =

(
j

i− 1

)
βi−1(i− 1)!.

Then,

[Ln[β]
T ]1≤i,j≤n =

(
i− 1
j

)
βjj!.

Now,

[S(β,r)(n)Ln[β]
T ]ij =

i−1∑
k=0

〈i− 1
k

〉
βkjk

(
i− 1
k

)
.

By equation (5),

[S(β,r)(n)Ln[β]
T ]ij = (βj + r)i−1

= (β + (j − 1)β + r)i−1

= [Vβ,r
n (1)]ij .

Theorem 2. For any real number t, and the generalized Pascal matrix,

Vβ,r
n (t) = Pn[β(t− 1)]S(β,r)(n)Ln[β]

T .

Proof. From Lemma 1,

Pn[β(t− 1)]S(β,r)(n)Ln[β]
T = Pn[β(t− 1)]Vβ,r

n (1)

=

i−1∑
k=0

(
i− 1
k

)
(β(t− 1))i−1−k(βj + r)k

= (β(t− 1) + βj + r)i−1

= (βt+ β(j − 1) + r)i−1

= [Vβ,r
n (x)]ij .

Example 10. Let n = 4. Then,

P4[β(t− 1)] =


1 0 0 0

β(t− 1) 1 0 0
(β(t− 1))2 2β(t− 1) 1 0
(β(t− 1))3 3(β(t− 1))2 3β(t− 1) 1

 .



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2319

S(β,r)(4) =


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β3 + 3βr + r2 3β + 3r 1

 ,

and

L4[β]
T =


1 1 1 1
β 2β 3β 4β
0 2β2 6β2 12β2

0 0 6β3 24β3

 .

Now,

P4[β(t− 1)]S(β,r)(4)L4[β]
T =


1 0 0 0

β(t− 1) 1 0 0
(β(t− 1))2 2β(t− 1) 1 0
(β(t− 1))3 3(β(t− 1))2 3β(t− 1) 1



×


1 0 0 0
r 1 0 0
r2 β + 2r 1 0
r3 β3 + 3βr + r2 3β + 3r 1



1 1 1 1
β 2β 3β 4β
0 2β2 6β2 12β2

0 0 6β3 24β3



=


1 0 0 0

β(t− 1) 1 0 0
(β(t− 1))2 2β(t− 1) 1 0
(β(t− 1))3 3(β(t− 1))2 3β(t− 1) 1



×


1 1 1 1

β + r 2β + r 0 0
(β + r)2 (2β + r)2 (3β + r)2 (4β + r)2

(β + r)3 (2β + r)3 (3β + r)3 (4β + r)3



=


1 1 1 1

βt+ r βt+ β + r βt+ 2β + r βt+ 3β + r
(βt+ r)2 (βt+ β + r)2 (βt+ 2β + r)2 (βt+ 3β + r)2

(βt+ r)3 (βt+ β + r)3 (βt+ 2β + r)3 (βt+ 3β + r)3


= Vβ,r

4 (t).

2.4. Successive Sum of Powers of Arithmetic Progressions

We will show that matrix S̃(β,r)(n) can be used to derive a summation formula for
arithmetic progressions. The following equations defined by Bazsó and Pintér in [1], and
Mezo and Ramı́rez in [12] will be utilized for the proof of the following theorem.



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2320

Definition 3. [1] For k = 1, 2, . . . , n, the numbers Zk
1,β,r(l), Z

k
2,β,r(l) are defined by the

recursive formulas:

Zk
1,β,r(l) = rk + (β + r)k + (2β + r)k + · · ·+ ((l − 1)β + r)k =

l−1∑
j=0

(jβ + r)k, (8)

Zk
p,β,r(l) =

n∑
j=1

Zk
p−1,β,r(l). (p = 2, 3, 4, . . .). (9)

Note that if p = β = r = 1, we obtain the sum of powers of the first n positive integers,
that is

Zk
1,1,1(l) = 1k + 2k + 3k + 4k + · · ·+ lk

Definition 4. [12] For each i = 1, 2, . . . , n, and for p ≥ 0,

ti(p) =

[(
p+ i− 2
p− 1

)
,

(
p+ i− 2

p

)
, . . . ,

(
p+ i− 2
p+ k − 2

)]T
(10)

zβ,ri (p) =

[(
p+ i− 2
p− 1

)
, Z1

p,β,r(i), . . . , Z
k−1
p−1,β,r(i)

]T
(11)

Theorem 3. For each p = 1, 2, 3, . . . , n, we have

S̃β,r(k)
[(

n+p−1
p

)
, β
(
n+p−1
p+1

)
, . . . , βk−1

(
n+p−1
p+k−1

)]T
=
[(

n+p−1
p

)
, Z1

p,β,r(n), . . . , Z
k−
p,β,r(n)

]T (12)

Proof. Let n and k be positive integers such that n ≥ k and p = 1, 2, 3, . . . , n. Now,
we will prove equation (3.7) by induction on n+ p.

Note that the sum of the entries in the second row of Vβ,r
n (0) is

r + (β + r) + (2β + r) + · · ·+ ((n− 1)β + r) = r1 + (β + r)1 + (2β + r)1 + · · ·+ ((n− 1)β + r)1

= Zk
1,β,r(n).

Now, substituting t = 0 to equation (12), we have

Vβ,r
n (0) = S̃β,rCβ

n (0)

= S̃β,r
[(

n
1

)
, β
(
n
2

)
, . . . , βk−1

(
n−1
k−1

)]T
=
[(

n−1
1

)
, Z1

1,β,r(n), . . . , Z
k−1
1,β,r(n)

]T
.

Thus, equation (12) is true for p = 1.
Consider p ≥ 2, and supposed the result is true for all i ≤ n+ p. Using the identity(

n+ 1

k + 1

)
=

n∑
l−0

(
l

k

)
,



G. Engalan, M.R. Latayada / Eur. J. Pure Appl. Math, 16 (4) (2023), 2306-2322 2321

and equations (10) and (11), by induction we have,

S̃β,r(k)ti(p+ 1) = S̃β,r(k)(t1(p) + t2(p) + · · ·+ tn(p))

= Zβ,r
1 (p) + Zβ,r

2 (p) + · · ·+ Zβ,r
n (p)

= Zβ,r
n (p+ 1).

Thus, equation (12) follows.

Example 11. Equation (12) in Theorem 3 yields nice formulas to sums of powers of
integers.
For example, if p = 1, and k = 3, we obtain

1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β3 + 3βr + 3r2 6β + 6r 6



(
n
1

)
β
(
n
2

)
β2
(
n
3

)
β3
(
n
4

)


=


n

1
2n(βn− β + 2r

1
6n(2β

2n2 − 3β2n+ β2 + 6βnr − 6βr + 6r2)
1
4n(βn− β + 2r)(β2n2 − β2n+ 2βnr − 2βr + 2r2)


= [n,Z1

1,β,r(n), Z
2
1,β,r(n), Z

3
1,β,r(n)]

T

If p = 2, and k = 3, we obtain


1 0 0 0
r 1 0 0
r2 β + 2r 2 0
r3 β3 + 3βr + 3r2 6β + 6r 6




(n+1
2

)
β
(n+1

3

)
β2

(n+1
4

)
β3

(n+1
5

)


=


1
2
n(n+ 1)

1
6
n(n+ 1)(βn− β + 3r)

1
12

n(n+ 1)(β2n2 − β2n+ 4βnr + 6r2)
1
60

n(n+ 1)(3β3n3 − 3β3n2 − 2β3n+ 2β3 + 15β2n2r − 15β2nr + 30βnr2 − 30βr2 + 30r3)


= [∗, Z1

2,β,r(n), Z
2
2,β,r(n), Z

3
2,β,r(n)]

T .

Note that Z1
2,m,r(n), k = 1, 2, 3, expresses

rk+(rk+(β+r)k)+· · ·+(rk+(β+r)k+(2β+r)k)+· · ·+((n−1)β+r)k) =
n∑

l=1

l−1∑
j=0

(jβ+r)k.

Corollary 3. For each p = 1, 2, . . . , n, we have

np∑
np−1=1

np−1∑
np−2=1

· · ·
n1−1∑
i=0

(iβ + r)k =

k∑
i=0

i!
〈k
i

〉
β,r

(
n+ p− 1

p+ i

)
= Zk

p,β,r(n),

where np = n.



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