INTERMATHS, VOL. 4, NO. 2 (2023), 38–53
https://doi.org/10.22481/intermaths.v4i2.13906
Article
cb licença creative commons
Peculiarities of smoothly undulating number
Eudes Antonio Costa
a
and Douglas Catulio dos Santos
b
a
Universidade Federal do Tocantins, Arraias - TO, Brasil;
b
Secretaria de Educação do Estado da Bahia, Barreiras - BA, Brasil
* Correspondence: catuliodouglas@outlook.com
Abstract: This note presents results related to divisibility or multiplicity between two numbers
in the class of integers called smoothly undulating numbers of the type
uz
[
n
]. The main result
is to characterize and display the types of divisors of some types of numbers
uz
[
n
], and we
show an algorithm to determine the greatest common divisor between two numbers uz [n].
Keywords: Divisibility; Undulating Numbers, Primality.
Classification MSC: 11A51; 11A67
1 Introduction
In this work, we present our study on properties related to divisibility criteria in
the class of integer numbers called undulating. It is observed that throughout the text,
the reference to number is directed to the elements of the set of non negative integers
(natural numbers). For example, if the digits of a number alternate more or less than
the digits adjacent to them, such as 3021 and 253612, then the number is called an
integer undulating. The term smoothly undulating, refers to numbers whose adjacent
digits alternate only between two digits, as in 7676767 or 858585. For example, see [
1
–
5
].
A formal definition is presented below:
Definition 1.1. Let
D
=
{
0
,
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
}
be the set of digits in the decimal
positional system:
(a) we say that a natural number N, with n ≥ 2 digits is undulating when,
N = a
1
a
2
· · · a
n−1
a
n
, with a
1
̸= 0 and a
i
∈ D, for i = 1, 2, ...n, (1)
and alternately
a
1
< a
2
,
a
2
> a
3
,
...
or
a
1
> a
2
,
a
2
< a
3
,
...
, that is, after the first digit
the next digits alternately increase and decrease or decrease and increase. However, the
absolute difference values between two adjacent digits can differ.
(b) we say that a natural number
N
, formed by
n >
2 digits is smoothly undulating
when,
N = aba · · · ab
| {z }
n even
or N = aba · · · ba
| {z }
n odd
, with a, b ∈ D, a ̸= b and a ̸= 0 . (2)
Submitted 06 November 2023; Accepted 20 November 2023; Available online: 30 December 2023.
ISSN 2675-8318 Copyright ©2023 INTERMATHS. Published by Edições UESB. This is an Open Access article under the CC BY 4.0 license.
Note that the integers 317, 4031, 523265 and 90634360 are undulating, while the
numbers 3535353 and 9494 are smoothly undulating. It is easy to see that in numbers
smoothly undulating the absolute value of the difference between two adjacent digits is
constant.
In the work by Costa and Costa [
6
, 2021], a study is presented on the primality of
numbers smoothly undulating and formed only by the digits 1 (one) and 0 (zero), in
which it is shown that among these numbers, only number 101 is prime. Already at
work, by Carvalho and Costa [
7
, 2021] presented properties related to base change,
divisibility, and primality of undulating numbers. They use list the prime, smoothly
undulating numbers smaller than 10
13
.
Here, we will show our study and the results obtained about the numbers smoothly
undulating. In this context, we consider divisibility criteria already well known in the
literature. The main objectives focus on establishing relations between the values
a, b
and
n
, according to Definition 1.1(b). We also present properties related to the change
of base and divisibility criteria, highlighting the difficulties in finding prime numbers in
the class of smoothly undulating numbers.
From now on, only the class of smoothly undulating numbers will be considered.
For simplicity (and convenience), we will denote by
AB
the set of numbers smoothly
undulating from Definition 1.1(b) and we will just say that N is a number (with n > 2
digits) smoothly undulating of type AB, if N ∈ AB.
Example 1.2. The numbers 10101
,
232323 and 5353535 belong to the set
AB
. Also,
the prime numbers 101
,
151
,
191
,
313
,
373
,
727
,
787
,
919
,
1212121, and 929292929, among
others, are of type AB.
We will use the notation
N
=
ab
[
n
], where
n
indicates the number of digits in the
number
N ∈ AB
, for
n >
2. For example, 10[5] = 10101 and 23[8] = 23232323. Thus,
according [7], the numbers smoothly undulating can be written in the form:
ab[n] =
a
P
n−1
2
i=0
10
2i
+ b
P
n−1
2
i=1
10
2i−1
, if n is odd,
a
P
n
2
i=1
10
2i−1
+ b
P
n
2
−1
i=0
10
2i
, if n is even .
(3)
This work is structured as follows. In the first two Sections, we make a literature
review collecting some results related to the theme, and then we present part of our
study involving numbers gently undulating. So, in Section 2 of this work, we present
some results already known about numbers of the type
ab
[
n
]. In Section 3 we specify
a
= 1 and
b
= 0, we list some results about numbers of the type
uz
[
n
], we present two
different proofs of the classic result of the non-existence of prime numbers in the class
uz
[
n
] for
n >
3. In Section 4 we present results related to divisibility or multiplicity
between two smoothly undulating numbers. In Section 5 we discuss the main results
(Theorems 5.2, 5.5, 5.8 and 5.13) of this work and we characterize and display types of
divisors of some types of numbers
uz
[
n
]. In Section 6 we show an algorithm to determine
the greatest common divisor between two numbers
uz
[
n
]. In Section 7, we used the fact
that no
uz
[
n
] is a perfect square, and furthermore we showed that no
uz
[
n
] is a perfect
cube.
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 39
2 Primality of smoothly undulating numbers
In this Section, our intention is to study the primality of the numbers in the set
AB
,
motivated and inspired by studies and topics already presented (or consecrated) in the
literature, related to prime numbers and primes repunidades [6–8].
The first results present a characterizations about smoothly undulating numbers,
occurs when b = 0 in ab[n], obtaining the following result:
Proposition 2.1. [
7
] No smoothly undulating number
ab
[
n
] is prime if
a >
1 and
b
= 0.
Proof. Just notice that,
a0[n] = a · 10[n] .
See that
a
0[
n
] is a composite number, divisible by
a
and also by 10[
n
]. Therefore,
a
0[
n
]
is not prime.
The next presents characterization about the numbers smoothly undulating when
n
is
even.
Proposition 2.2. [
2
,
3
] For any digits
a
and
b
, if
n >
3 is even, then
ab
[
n
] is never
prime.
Proof.
Considering even
n >
3, we have
n
= 2
k
, for some positive integer
k
. Now, just
see that, ab[n] = ab[2k] is of the form,
ab[2k] = ab
|{z}
ab
|{z}
· · · ab
|{z}
| {z }
k times
= ab · 10
2k−2
+ ab · 10
2k−4
+ · · · + ab · 10
2
+ ab
= ab · (10
2(k−1)
+ 10
2(k−2)
+ · · · + 10
2
+ 1)
= ab · 10[ 2k − 1] .
Therefore, ab[n] will not be a prime number if n > 3 is even.
A specific case, which we will use later:
Lemma 2.3. [7] If n = 6, then ab[6] = 3 · 7 · 13 · 37 · ab, for any a, b ∈ D.
Proof.
It follows from Proposition 2.2 that,
ab
[6] =
ab ·
(10[5]), as 10101 = 3
·
7
·
13
·
37,
it follows that,
ab[6] = 3 · 7 · 13 · 37 · ab .
From now on, we consider
ab
[
n
] to be numbers of type
AB
, with
n
odd. We present
other results we obtained about smoothly undulating numbers.
Proposition 2.4. [7] If a ∈ {2, 4, 5, 6, 8} and n ≥ 3 is odd, then ab[n] is not prime.
Proof.
Just note that if
a ∈ {
2
,
4
,
5
,
6
,
8
}
and
n
is odd, then the smoothly undulating
number ab[n] is even or a multiple of 5. It is not a prime number.
40 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
Before it is worth remembering an auxiliary result, the criterion of divisibility by 3,
which can be consulted in [9, 10]:
Lemma 2.5. An integer
n
is divisible by 3 if and only if the sum of its digits is a
multiple of 3.
Proposition 2.6. [
7
] If
a
equals 3 or 9 and
n >
3 is odd, with
n ≡
1
mod
3, then
ab[n] is not prime.
Proof.
Since
n
is odd, it is observed that the number of times the number
b
appears
is
n−1
2
. Since
n −
1 is even and
n −
1
≡
0
mod
3, it follows that
n −
1 = 6
t
for some
positive integer t, like this
n − 1
2
=
6t
2
= 3t ,
that is,
n−1
2
≡
0
mod
3. We also have that,
a ≡
0
mod
3, so the sum of the digits
of
ab
[
n
] is a multiple of 3. Therefore, by Lemma 2.5, it is concluded that
ab
[
n
] is not
prime.
Proposition 2.7. [
7
] If
b ∈ {
3
,
6
,
9
}
and
n >
3 is odd, with
n ≡
2
mod
3, then
ab
[
n
]
is not prime.
Proof.
If
n
is odd, the number of times the digit
a
appears is
n+1
2
. Since
n
+ 1 is even
and n + 1 ≡ 0 mod 3, it follows that n + 1 = 6t, for some t positive integer, like this
n + 1
2
=
6t
2
= 3t ,
that is,
a ·
n+1
2
≡
0
mod
3. We also have that,
b ≡
0
mod
3, so
ab
[
n
] is a multiple of 3.
Therefore, ab[n] is not prime (Lemma 2.5).
Proposition 2.8. [
7
] For
k ≥
1, if
n
= 3
k
is an odd integer and
a
+
b
+
a ≡
0
mod
3,
then ab[n] is not prime.
Proof.
The case where
k
= 1, like
a
+
b
+
a ≡
0
mod
3, follows that
ab
[3] is a multiple
of 3.
If n = 3k > 3 is odd, we have
ab[n] = ababab · 10
n−6
+ ababab · 10
n−12
+ · · · + ababab · 10
3
+ aba .
It follows from the Corollary 2.3 that, 3 divides the number
ababab
. Therefore,
ab
[
n
] is
a sum of multiples of 3. Therefore, it is not prime (Lemma 2.5).
3 Smoothly undulating numbers of the type UZ
Specifying
a
= 1 and
b
= 0 in the Definition 1.1 we have say that a natural number
uz
[
k
] is a smoothly undulating of the type
UZ
if it is formed only by the digits 0 and 1
and is an alternating sequence of the digits 0 and 1, starting with 1. For any natural
m ∈ UZ
, we will use the notation
m
=
uz
[
n
] (general) or
m
= 10[
n
] (specific) where
n ≥
2 indicates the number of digits in the number smoothly undulating
m
. It follows
from Proposition 2.1 and Proposition 2.2 that some smoothly undulating numbers of
type
ab
[
n
] are multiples of
uz
[
n
], so in the rest of the text we will concentrate our efforts
on this class of numbers.
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 41
Example 3.1. The numbers 10[2] = 10, 10[ 3] = 101, 10[4] = 1010, and successively. In
particular the number 10[ 2023] =
1010 . . . 0101
| {z }
2023
with 1012 digits 1 alternating with 1011
digits 0; and more, it follows from the Equation Eq. (3) that
10[2023] = 1 · 10
2022
+ 1 · 10
2020
+ · · · + 1 · 10
2
+ 1 .
Remark 3.2. [6, 11] Note that:
1. if n is even then uz[n] ends in 0, otherwise uz[n] ends in 1.
2.
given the natural numbers
n, k
1
and
k
2
such that
k
1
+
k
2
=
n
, let
uz
[
n
] be a
smoothly undulating number with
k
1
digits 1 alternated by
k
2
digits 0 and we write
uz
(
n
) = 1
k
1
0
k
2
. And more,
n
= 2
k
is even so the amount, or number of digits, of
1s and 0s in
uz
[
n
] are equal and
uz
[
n
] = 1
k
0
k
. Whereas if
n
= 2
k
+ 1 is odd then,
in
uz
[
n
], the quantity of 1s is
k
+ 1 and that of 0s is
k
, that is, in
uz
[
n
] = 1
k+1
0
k
.
Example 3.3. [
6
,
11
] By direct inspection of the divisors of the first six numbers
smoothly undulating U Z we obtain that:
1. the number uz[2] = 10 = 2 · 5 is not prime;
2. the number uz[3] = 101 is prime;
3. the number uz[4] = 1010 = 10 · 101 is not prime;
4. the number uz[5] = 10101 = 111 · 91 is not prime;
5. the number uz[6] = 101010 = 10 · 10101 is not prime;
6. the number uz[7] = 1010101 = 101 · (10
4
+ 1) is not prime .
In Example 3.3 we list only one factorization, by way of example, when the number
uz
[
n
] is composite. In addition to the Example 3.3, we also present Table 1, with the
smoothly undulating numbers uz[n] and some factors for odd n, with 5 ≤ n ≤ 13.
Table 1. Factors prime
n uz[n] Factors
5 10101 3 · 7 · 13 · 37
7 1010101 73 · 101 · 137
9 101010101 41 · 271 · 9091
11 10101010101 3 · 7 · 13 · 37 · 101 · 9901
13 1010101010101 239 · 4649 · 909091
An interesting result about the smoothly undulating numbers uz[n] is the following:
Theorem 3.4. [
6
, Theorem 5] Except 101, no smoothly undulating number
uz
[
n
] is
prime.
42 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
Our interest in the primality of these numbers is motivated by reading Ribenboin [
8
].
Here we will present forward two alternative income statements, different from the one
presented in [6].
The next result presents the Binet expression for the smoothly undulating numbers
Theorem 3.5. For every integers k ≥ 2 we have uz[n] =
10
2k
− 1
99
, where n = 2k − 1.
Proof. By induction on k, for k = 2 we have
10[3] =
10
4
− 1
99
=
(10
2
− 1)(10
2
+ 1)
(10 − 1)(10 + 1)
=
10
2
− 1
10
2
− 1
· (10
2
+ 1) = 10
2
+ 1 ,
it follows from the Equation Eq. (3) that 10[3] = 1
·
10
2
+ 1. Assume that for some
k > 2 we have,
10
2k
− 1
99
= 1 · 10
2·k
+ 1 · 10
2(k−1)
+ · · · + 1 · 10
2
+ 1 , (4)
let’s show it for the successor of k. Let’s see
10
2(k+1)
− 1
99
=
10
2k+2
− 1
99
=
10
2k+2
− 10
2
+ 10
2
− 1
99
=
10
2
(10
2k
− 1)
99
+
10
2
− 1
99
Eq. (4)
= 10
2
· (1 · 10
2·k
+ 1 · 10
2(k−1)
+ · · · + 1 · 10
2
+ 1) + 1
= 1 · 10
2·(k+1)
+ 1 · 10
2k
+ · · · + 1 · 10
2
+ 1 .
Again, it follows from the Equation Eq. (3) that
uz
[
n
+ 2] =
10
2(k+1)
− 1
99
, as we
wanted.
In the previous proposition, we have that
n
is odd, in the case where the number of
digits is even, it follows directly that
Corollary 3.6. For every even integers
n ≥
2 we have
uz
[
n
] =
10
2k
− 1
99
·
10, where
n = 2k.
We will make use of the following auxiliary results, and three results establish
conditions on a polynomial factorization, and will be used throughout the work.
Lemma 3.7. [
9
, Proposition 3.6] Let
a
and
b
be integers and
n
natural, then
a − b
divides a
n
− b
n
.
Lemma 3.8. [
9
, Proposition 3.7] Let
a
and
b
be integers and
n
natural, then
a
+
b
divides a
2n
− b
2n
.
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 43
Lemma 3.9. [
9
, Proposition 3.8] Let
a
and
b
be integers and
n
natural, then
a
+
b
divides a
2n+1
+ b
2n+1
.
As a direct application of the Lemma 3.7, the next result will display a smoothly
undulating divisor of a smoothly undulating number, in the case where n is composite.
Proposition 3.10. For any
m
,
n ∈ N
, if
m
is a multiple of
n
then
uz
[2
m −
1] is a
multiple of uz[2n − 1].
Proof.
If
m
is a multiple of
n
then
m
=
n · k
for some integer
k
. See that, according
Theorem 3.5
uz[2m − 1] =
10
2m
− 1
99
=
10
2nk
− 1
99
=
10
2nk
− 1
10
2n
− 1
·
10
2n
− 1
99
=
10
2nk
− 1
10
2n
− 1
· uz[2n − 1] .
By Lemma 3.7 we have that 10
2n
−
1 divides 10
2nk
−
1 = (10
2n
)
k
−
1
k
and thus the
smoothly undulating uz[2n − 1] divides uz[2m − 1].
3.1 First proof of Theorem 3.4
It follows from Corollary 3.6 that the number smoothly undulating
uz
(2
k
) is a multiple
of 10 (even and multiple of 5) and therefore is not prime, justifying the following
property.
Proposition 3.11. No smoothly undulating number uz[2k] is prime.
Proposition 3.12. No smoothly undulating number
uz
[2
k −
1] is prime, for
k >
2
integer.
Proof. According Theorem 3.5 that uz[2k − 1] =
10
2k
− 1
99
, as
uz[2k − 1] =
10
2k
− 1
99
=
(10
k
− 1)(10
k
+ 1)
(10 − 1)(10 + 1)
.
We have
k >
2, if
k
is odd, then by the Lemma 3.7 we hat that
10
k
− 1
10 − 1
is the integer,
and by the Lemma 3.9 we hat that
10
k
+ 1
10 + 1
the same fact.
Now if we have
k
even and
k
= 2
t
for some
t >
1 then by the Lemma 3.7 and
Lemma 3.8 we hat that
10
k
− 1
(10 − 1)(10 + 1)
is the integer. Finally the case where
k
is even
and
k ̸
= 2
t
. For some
k >
1 then
k
=
t
2
· t
1
, where
t
1
, t
2
and
t
3
are integers, such that
t
1
= 2
t
3
is even and t
1
is odd, so
uz[2k − 1] =
10
2k
− 1
99
=
(10
k
− 1)(10
k
+ 1)
(10 − 1)(10 + 1)
=
(10
t
1
− 1)(10
t
1
+ 1)
(10 − 1)(10 + 1)
· (10
2t
1
+ 1) . . . (10
t
2
·t
1
+ 1).
44 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
Now follows from the first case that
(10
t
1
− 1)(10
t
1
+ 1)
(10 − 1)(10 + 1)
is integer.
The Theorem 3.4 is a direct consequence of Propositions 3.11 and 3.12.
3.2 Second proof of Theorem 3.4
The numbers repunit are written, in the decimal system, as the repetition of the unit,
represented by the set
R
n
=
{
1
,
11
,
111
, . . . , R
n
, . . .}
. According to [
8
,
12
–
14
] the
equation
R
n
=
10
n
− 1
9
presents the Binet expression for the repunit numbers. Soo the
Theorem 3.5 presents the Binet expression for the smoothly undulating numbers.
Proposition 3.13. If k is odd and n = 2k − 1, then uz[n] is a multiple of R
k
.
Proof. Just note that
uz[n] =
10
k
− 1
9
·
10
k
+ 1
11
=R
k
·
10
k
+ 1
11
,
like 11 | 10
k
+ 1 if k odd (Lemma 3.9), then the result follows.
Proposition 3.14. If k is even and n = 2k − 1, then uz[n] is a multiple of 10[3].
Proof. Just notice that
uz[n] =
10
4k
− 1
99
=
10
4
− 1
99
·
10
4k
− 1
10
4
− 1
=10[3] ·
10
4k
− 1
10
4
− 1
.
The result follows from the fact that
10
4k
− 1
10
4
− 1
=
(10
4
)
k
− 1
10
4
− 1
is an integer, according to
Lemma 3.7.
Combining Propositions 3.13 and 3.14 it turns out that except 101, no smoothly
undulating number is prime.
4 Divisibility in UZ
In this Section we will approach some results related to divisibility or multiplicity
between two smoothly undulating numbers.
From now on we only consider smoothly undulating numbers of the type
UZ
and ending
in 1, that is, smoothly undulating numbers
uz
[
n
] where
n
is odd, that is,
n
= 2
k −
1, for
some integer k. As uz[n] ends in 1, this is enough to justify the following fact:
Proposition 4.1. If n is odd then neither 2 nor 5 divides uz[n].
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 45
According to Proposition 4.1 and by definition, if
n
is odd no
uz
[
n
] ends with the
digits 0, 2, 4, 6, 8 or 5; that is, no smoothly undulating numbers is a multiple of 2 or 5.
However, from the Proposition 4.3 we deduce that for any integer
n
multiple of 3, or
multiple of power of 3, there will be an infinite smoothly undulating numbers multiples
of
n
. Furthermore, below we will characterize some divisors of smoothly undulating
numbers.
Proposition 4.2. [7] If n is odd, with n ≡ 2 mod 3, then uz[n] is multiple of 3.
Proof.
If
n
is odd then
n
+ 1 is even and
n
+ 1
≡
0
mod
3, it follows that
n
+ 1 = 6
t
,
for some t positive integer, like this
n + 1
2
=
6t
2
= 3t ,
that is,
n+1
2
≡
0
mod
3. According Lemma 2.5, we have that, 1
·
n + 1
2
≡
0
mod
3, so
uz[n] is a multiple of 3.
Proposition 4.3. If
q
= 3
k
with
k >
0 integer, then
q
divides
uz
[
n
], where
n
= 2
q −
1.
Proof. Just note that, according Theorem 3.5, Lemma 3.7 and Lemma 2.5,
uz[n] =
10
2q
− 1
99
=
10
2·3
k
− 1
99
=
(10
2
)
3
k
− 1
3
k
10
2
− 1
=10
2(3
k
−1)
+ 10
2(3
k
−2)
+ · · · + 10
4
+ 10
2
+ 10
0
≡ 1 + 1 + 1 + · · · + 1 + 1
| {z }
3
k
times
≡ 0 mod 3
k
.
Now it is worth remembering an auxiliary result, the criterion of divisibility by 11,
which can be consulted in [9, 10]:
Lemma 4.4. A number is divisible by 11 if the alternating sum of its digits is divisible
by 11.
Example 4.5. See that R
2
and R
11
divides uz[21]. In fact, according Theorem 3.5,
uz[21] =
10
22
− 1
10
2
− 1
=
10
11
− 1
10 − 1
·
10
11
+ 1
10 + 1
=R
11
·
10
11
+ 1
10 + 1
.
By the Lemma 3.9 we hat that
10
11
+ 1
10 + 1
is the integer, and
10
11
+ 1
10 + 1
=10
10
− 10
9
+ · · · + 10
2
− 10 + 1
=10
9
(10 − 1) + 10
7
(10 − 1) + · · · + 10
3
(10 − 1) + 10(10 − 1) + 1
=9090909091
Now follows from Lemma 4.4 that 9090909091 is a multiple of 11. Soon
R
2
= 11 and
R
11
divides uz[21].
46 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
Proposition 4.6. For all integer
k >
0, if
n
= 11
· k −
1 then
R
2
and
R
11
divide
uz
[
n
].
Proof. According Theorem 3.5,
uz[n] =
10
2(11k)
− 1
99
=
10
11k
− 1
9
·
10
11k
+ 1
11
=
10
11
− 1
9
·
(10
11
)
k
− 1
10
11
− 1
!
·
(10
11
)
k
+ 1
10
11
+ 1
·
10
11
+ 1
11
!
=R
11
·
10
11
+ 1
11
·
(10
11
)
k
− 1
10
11
− 1
·
(10
11
)
k
+ 1
10
11
+ 1
!
.
In the face of the Lemmas 3.7-3.9, each factor is an integer and follows from the
Example 4.5 that
10
11
+ 1
10 + 1
is a multiple of 11. Soon
R
2
= 11 and
R
11
divides
uz
[
n
].
There is a close relationship between the repunit and smoothly undulating numbers,
in the Proposition 3.13 we saw that a smoothly undulating number is a multiple of a
repunit, now a symmetrical situation
Proposition 4.7. For every integers
k >
1, then
R
2k
is a multiple of
uz
[
n
], were
n = 2k − 1.
Proof. Just note that
R
2k
=11 · 10
2k−2
+ 11 · 10
2k−4
+ · · · + 11 · 10
2
+ 11 · 10
0
=11 · (10
2k−2
+ 10
2k−4
+ · · · + 10
2
+ 10
0
)
=11 · uz[2k − 1] .
Proposition 4.8. For every integers
k ≥
1 we have
uz
[5] divides
uz
[
n
], where
n
= 6
k−
1.
Proof. By induction on k, for k = 1 it’s trivial.
Assume that for some k ≥ 1 we have that uz[5] divides uz[6k − 1], that is,
uz[6k − 1] = uz[5] · q , (5)
for some integer q. let’s show it for the successor of k. Let’s see
uz[6(k + 1) − 1] = uz[6k + 5]
= 10
6k+4
+ 10
6k+2
+ 10
6k
+ 10
6k−2
+ · + 10
2
+ 10
0
= 10
6k
· (10
4
+ 10
2
+ 10
0
) + uz[6k − 1]
= uz[5] · (10
6k
+ q)
In the second addend we use the induction hypothesis, Equation Eq. (5), And we get
the result
Corollary 4.9. For every integers
k ≥
1 we have that
uz
[
n
] is multiple of 3, 7, 13 and
37; where n = 6k − 1.
Proof.
This follows from Proposition 4.8 and that
uz
[5] = 3
·
7
·
13
·
37, looking at Table
1.
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 47
5 Prime factor of smoothly undulating number
To demonstrate the following Theorem 5.2, a well-known result in mathematics will
be used, namely the little Theorem of Fermat, see [9, 10]:
Lemma 5.1. If a, p ∈ Z, with p prime and (a, p) = 1, then a
p−1
≡ 1 mod p.
Where (
a, b
) is the greatest common divisor of numbers
a
and
b
. As a consequence of
the Lemma we have.
Theorem 5.2. If p = 7 or p > 11 is a prime number, then p divides uz[2p − 3].
Proof.
According to Fermat’s little theorem, 10
p−1
≡
1
mod p
, or equivalent (10
p−1
)
2
≡
1
2
= 1
mod p
, with
p
prime. In particular, 10
2(p−1)
−
1
≡
0
mod p
. Therefore,
p
divides 10
2(p−1)
−
1 = 99
· uz
[2
p −
3], as
p
= 7 or
p >
11 so
mdc
(
p,
99) = 1 it can be
concluded that p necessarily divides uz[2p − 3].
Example 5.3. For the prime
p
= 7 we have that 7 divide
uz
[11], because
n
= 2
·
7
−
3 = 11.
And more, since 11
≡
2
mod
3, according Proposition 4.2, we have 3 divide
uz
[11]. In
fact uz[11] = 10101010101 = 3 · 7 · 13 · 37 · 101 · 9901.
For all natural numbers, let
φ
(
m
) be the number of natural numbers less than or equal
to
m
, that is, (
a, m
) = 1 for
a ≤ m
. The function
φ
(
m
) is known as Euler’s function.
We will make use of the following result, and the proof can be consulted in [9, 10].
Lemma 5.4. (Euler-Fermat theorem) Let
a, m
be natural numbers. If (
a, m
) = 1, then
a
φ(m)
≡ 1 mod m.
Now, we consider an integer
m
that is not a multiple of 3 or 11, with (2
, m
) = 1
and (5
, m
) = 1, the next result displays a smoothly undulating numbers multiple of
m
,
formally we have:
Theorem 5.5. For any integers
k
1
, k
2
≥
0 and
q, m >
1, if (2
, m
) = (5
, m
) = 1 and
m ̸= 3
k
1
· 11
k
1
· q, then m divides uz
h
2φ(m) − 1
i
, where φ(m) is the Euler function.
Proof.
Let
m
be a natural
m ̸
= 3
k
1
·
11
k
1
· q
, as (2
, m
) = (5
, m
) = 1 then (10
2
, m
) =
(10
, m
) = 1, since (2
,
5) = 1. It follows from the Lemma 5.4 that 10
φ(m)
≡
1
mod m
,
or equivalent (10
φ(m)
)
2
≡ 1
2
mod m . Like this
99 · uz
h
2φ(m) − 1
i
= (10
φ(m)
)
2
− 1 ≡ 0 mod m .
If m ̸= 3
k
1
· 11
k
1
· q then (m, 99) = 1, that is, m divides uz
h
2φ(m) − 1
i
.
Let
m
be a natural number and
a
any integer such that (
a, m
) = 1. Let
h >
1 be the
smallest natural number such that
a
h
≡
1
mod m
. In this case, we say that the order
of
a
module
m
is denoted by
ord
m
(
a
) =
h
. We will also need the following auxiliary
results, additional details and proof can be found at [9, 10].
Lemma 5.6. If
ord
m
(
a
) =
h
, the positive integers
k
such that
a
k
≡
1
mod m
are
precisely those for which h | k.
48 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
Lemma 5.7. If (a, m) = 1 and ord
m
(a) = h then h divides φ(m).
Theorem 5.8. Let
p, q
be distinct prime numbers with
p >
2. If
q
is a divisor of
uz
[
n
],
then q is of the form 2px + 1, for all natural x, with n = 2p − 1.
Proof.
For all
p >
2 prime, we have
uz
[
n
] =
10
2p
− 1
99
composite and let
q
be a prime
divisor of
uz
[
p
], see that
q ̸
= 5, according Proposition 4.1. Since
q
is a divisor of
uz
[
n
]
then there is an integer
x
1
>
0 such that
uz
[
n
] =
q · x
1
, hence 99
·
(10
2p
−
1) = 99
· q · x
1
.
So q | (10
2p
− 1) where
10
2p
≡ 1 mod q . (6)
Since (10
, q
) = 1, we have 10
φ(q)
≡
1
mod q
. Let
h
=
ord
q
(10) and according to
Lemma 5.7, we get
h | φ
(
q
) =
q −
1. It follows from Equation Eq. (6) and Lemma 5.6
that
h
divides
p
2
resulting in
h
divides the prime
p
, then
h
=
p
and
q
=
py
+ 1 for some
natural
y
, like
q
is odd entails that
y
= 2
x
is even, that is,
q
= 2
px
+ 1 for some natural
x.
Example 5.9. For the prime
p
= 7 we have
n
= 2
·
7
−
1 = 13 and
uz
[13] =
1010101010101 = 239
·
4649
·
909091. See yet that 239 = 2
·
7
·
17 + 1, 4649 = 2
·
7
·
332 + 1
and 909091 = 2 · 7 · 64935 + 1
In Theorem 5.8 it is a characterisation is displayed for a prime factor of a type of
smoothly undulating number. In addition to the previously presented result, we provide
a characterisation for a prime factor of a composite of smoothly undulating number. To
prove the Theorem 5.13 that will be presented later, we will make use of well-established
results about quadratic residues, which can be found in [9, 10, 15].
Definition 5.10. Let
p >
2 be a prime number and
a
any integer. We indicate the
Legendre symbol by:
a
p
!
=
1, if p ∤ a e a is residue quadratic modulo p;
−1, se p | a;
0, otherwise otherwise .
Lemma 5.11. [Euler’s criterion] [
9
,
15
] Let
p >
2 be a prime and
a
be a any integer.
So
a
p
!
= a
p−1
2
mod p .
Lemma 5.12. For every prime
p >
5 we have
10
p
!
=
(−1)
p−1
2
, that is, 10 is quadratic
residue modulo p if, and only if, p ≡ ±1 mod 6.
Theorem 5.13. If
p >
2 is a prime and
n
= 2
p −
1. Then each prime divisor
q >
5 of
uz[n] is of the form 6z ± 1, for some natural z.
Proof. It follows from Proposition 5.8 that q = 2px + 1, with natural x. See that
10
q−1
2
= 10
2px+1−1
2
= (10
p
)
x
≡ 1 mod q .
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 49
By Euler’s criterion, we have
10
q
!
= 1 and 10 is quadratic residue modulo
q
. Now, it
follows from Lemma 5.12 that q = 6z ± 1, for some natural z .
Such a situation can be described or specified in the following example.
Example 5.14. Again for the prime
p
= 7 we have
n
= 2
·
7
−
1 = 13 and
uz
[13] =
1010101010101 = 239
·
4649
·
909091. From which we notice that 239 = 6
×
40
−
1 and
4649 = 6 × 775 − 1, and 909091 = 6 × 151515 + 1.
6 Greatest common divisor in UZ
Our next result is known as Euclid’s Algorithm, and it establishes a method for
calculating the greatest common divisor between two numbers, so given the integers
a
and
b
we indicate the greatest divisor between them by
gcd
(
a, b
), additional details and
proof can be found at [9, 10].
Lemma 6.1. [Euclid’s Algorithm] Given the integers
a, b
, with
b
=
aq
+
r
for
q
and 0
≤ r < |b|
integers. The greatest common divisor between
a
and
b
is given by
gcd(a, b) = gcd(a, r).
As an application of Lemma 6.1 we present the following result:
Proposition 6.2. For any natural m ≥ n, with m, n ≥ 1,
gcd(uz[m], uz[n]) = uz[gcd(m, n)] .
Proof.
In fact, since
m ≥ n
there are integers
q, r
1
from the Euclidean division of
m
by
n
, that is,
m
=
nq
+
r
1
with 0
≤ r
1
< n
. Analogously, let
r
2
, r
3
, . . . , r
s
, r
s+1
be the
partial remainders in Euclid’s algorithm. By Lemma 6.1, we have
r
s
= (
m, n
). On the
other hand, see that uz[m] = 10
2r
1
uz[n] + uz[r
1
], it follows from Lemma 6.1 that
gcd(uz[m], uz[n]) = gcd(uz[n], uz[r
1
]) = . . . = gcd(uz[r
s
], uz[r
s+1
])
= gcd(uz[r
s
], 0) = uz[gcd(m, n)] .
Example 6.3. Note that
gcd
uz
[100]
, uz
[60]
=
uz
[20], since
gcd
(100
,
60) = 20. As
well as gcd
uz[2023], uz[34]
= uz[17], since gcd(2023, 34) = 17.
Let
a
and
b
be any two integers, we say that
a
is rarely prime with
b
, which are
co-primes, when (a, b) = 1.
Corollary 6.4. Two consecutive smoothly undulating numbers are co-prime.
Proof. Just note that gcd(n + 1, n) = 1 for all natural n.
We can get the Proposition 3.10 as a direct application of Proposition 6.2, see
Corollary 6.5 (Proposition 3.10). Let
m, n
be natural, if
n
is a multiple of
m
then
uz[m] divides uz[n].
Proof.
As by assumption
n
is a multiple of
m
, then there exists an integer
q
such that
n
=
mq
, as well as
gcd
(
m, n
) =
gcd
(
m, mq
) =
m
the result follows from Proposition 6.2.
50 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
7 Smoothly undulating numbers and powers
In this Section we show some results that relate the smoothly undulating numbers and
powers with a natural exponent.
Proposition 7.1. The difference between two consecutive smoothly undulating numbers
is a perfect square.
Proof. According to Theorem 3.5 that
uz[n + 1] − uz[n] =
10
2n+2
− 1
99
−
10
2n
− 1
99
=
10
2n+2
− 10
2n
99
=
10
2n
(10
2
− 1)
99
= (10
n
)
2
.
Corollary 7.2. If
ab
is a perfect square then the difference
ab
[2
n
]
− ab
[2(
n −
1)] is also
a perfect square.
However, it is known that
Proposition 7.3. [11] Except of uz[1], no other uz[n] is a perfect square.
As consequence we have to
Proposition 7.4. For n ≥ 2 no uz[n] is an even power.
Proof.
Suppose that for some
n ≥
1 we have
uz
[
n
] =
a
2k
with
a
and
k >
0 integers,
this implies that
uz
[
n
] would be a perfect square, considering that
uz
[
n
] = (
a
k
)
2
, which
contradicts the Proposition 7.3.
Remark 7.5. An open question is whether the smoothly undulating numbers are a sum
of two powers of even index. Something that doesn’t happen in repunit numbers.
Proposition 7.6. For n ≥ 2 no uz[n] is a perfect cube.
The result will follow with the help of the lemma ahead.
Lemma 7.7. [
16
, Theorem 1] Let
x, y, m, n
be natural, with
x >
1
, y >
1
, m >
2, n > 1, the equation
x
n
− 1
x − 1
= y
m
,
no has a solution (
x, y, m, n
) satisfying
gcd
(
xφ
(
x
)
, m
) = 1, where
φ
(
x
) é the function
Euler’s action of x.
Now we display a proof for the Proposition 7.6
Proof. According Theorem 3.5, for n ≥ 2, that
uz[n] =
10
2n
− 1
99
= 1 + 10
2
+ 10
4
+ 10
6
+ . . . + 10
2(n−1)
.
Now, applying the Lemma 7.7, we must show that the Diophantine equation
1 + 10
2
+ 10
4
+ 10
6
+ . . . + 10
2(n−1)
= y
3
;
does not have integer solutions for any
n
and
y
, since
mdc
(
xφ
(
x
)
, m
) = 1 , for
x
= 10
2
,
φ(10
2
) = 40 and m = 3.
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 51
8 Considerations
Here we present some results about smoothly undulating numbers, specifically those
formed alternately by the digits 1 and 0, such a set being denoted by
UZ
. Furthermore,
we show that no element of
UZ
can be expressed as a perfect cube or as a power of the
even index. Moreover, we approach some properties related to the divisibility between
two elements of
UZ
; in particular, we highlight the relationship between the elements
of
UZ
and the repunit numbers. We hope that the presented results can inspire and
motivate further studies on this class of numbers.
Acknowledgments. This work was partially supported by the PROPESQ-UFT.
Disclosure statement. The authors declare no conflict of interest in the writing of the
manuscript, or in the decision to publish the results.
ORCID
Eudes Antonio Costa https://orcid.org/0000-0001-6684-9961
Douglas Catulio dos Santos https://orcid.org/0000-0002-5221-6087
References
1.
C. A. Pickover . “Is There a Double Smoothly Undulating Integer?”, Journal of Recreational
Mathematics , v.22, n.1, p. 52-53, 1990.
2. C. A. Pickover. Keys to Infinity (Chapter 20). New York, 1995.
3.
C. A. Pickover. Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning .
(Chapter 52 and 88). Oxford University Press, 2003.
4.
D. F. Robinson. “There are no double smoothly undulating integers in both decimal and
binary representation”. Journal of Recreational Mathematics , v. 26, n. 2, p. 102-103, 1994.
5.
K. Shirriff. “Comments on Double Smoothly Undulating Integers”. Journal of Recreational
Mathematics , v. 26, n. 2, p. 103-104, 1994.
6.
E. A. Costa; G. A. Costa, G. A. “Existem números primos na forma 101... 01”. Revista do
Professor de Matemática , n. 103, p. 21-22, 2021.
7.
F. S. Carvalho, ; E.A. Costa. “Um passeio pelos números ondulantes”. REMAT: Revista
Eletrônica da Matemática, v. 8, n. 2, p. e3001-e3001, 2022. https://doi.org/10.35819/re
mat2022v8i2id6043
8. P. Ribenboim. The little book of bigger primes. 2nd ed. New York: Springer, 2004.
9. A. Hefez. Aritmética . SBM-Coleção PROFMAT, 2a. ed. Rio de Janeiro-RJ. SBM, 2016.
10.
I. Niven; H.S. Zuckerman; H. L. Montegomery. An introduction to the theory of numbers .
John Wiley and Sons. 1991.
11.
E. A. Costa; A. B. Souza. “Números ondulantes na forma 101...01”. Gazeta Matemática ,
SPM, 2024 (a aparecer).
12.
A. H. Beiler. Recreations in the theory of numbers: the queen of mathematics entertains.
2nd ed. New York: Dover, 1966.
13.
E. A. Costa; D. C. Santos. “Algumas propriedades sobre os números Monodígitos e
Repunidades”. Revista de Matemática , v. 2, p. 47-58, 2022.
14.
S. Yates. Repunits and repetends. Star Publishing Co., Inc. Boynton Beach, Florida, 1992.
52 | https://doi.org/10.22481/intermaths.v4i2.13906 E. A. Costa, D. C. Santos
15.
C. G. T. de Moreira; F. E. B. Martinez; N. C. Saldanha. Tópicos de teoria dos números.
Rio de Janeiro: SBM, 2012.
16.
L. Maohua. “A note on perfect powers of the form
x
m−1
+
. . .
+
x
+ 1”. Acta Arithmetica,
v. 69, n. 1, p. 91-98, 1995.
© INTERMATHS
CC BY 4.0
E. A. Costa, D. C. Santos INTERMATHS, 4(2), 38–53, December 2023 | 53
