1. Introduction
We start by recalling that the Fibonacci sequence
is defined by the recurrence
with initial values
and
(see, e.g., [
1,
2,
3]). This sequence admits many generalizations and one of the most known is its higher order version. The Fibonacci sequence is a binary (or second order) recurrence and then, for any integer
, the sequence of the
k-generalized Fibonacci numbers (or
k-bonacci numbers) is defined by the
kth order recurrence
with initial values
and
. Clearly, for
, we have the Fibonacci numbers and for
, we have Tribonacci numbers (which is one of the most well-studied generalizations of Fibonacci numbers).
On the other hand, a repdigit (short for “repeated digit”) is a number of the form
where
and
(here, as usual, for integers
, we denote
), that is, a number with only one distinct digit (in this case
a) in its decimal expansion.
We point out that many authors have been interested in solving Diophantine equations involving repdigits (their sums, products, concatenations, etc.) and some special forms of linear recurrences (like their product, sums, etc.). For some works in this direction, we refer the reader to [
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19,
20,
21,
22,
23,
24,
25,
26,
27,
28,
29,
30,
31,
32,
33] and the references therein.
Luca [
34], in 2000, and Marques [
35], in 2012, proved that the largest repdigits in the Fibonacci and Tribonacci sequence are
and
, respectively. Recently, Bednařík and Trojovská [
36] and Trojovský [
37] found all repdigits of the form
and
, respectively.
The aim of this paper is to continue this program and generalize the main result of [
36]. More precisely, we search for repdigits which are the product of the
nth
k-bonacci number by the
nth
-bonacci number, for
. Our main result is the following:
Theorem 1. The Diophantine equationdoes not have a solution in positive integers , with , and . The main tools in the proof is the transcendental method (lower bounds for linear logarithm of real algebraic numbers) together with the theory of continued fractions (reduction method). It is important to stress that the method can be implemented for any given range of values for m. However, we chose in order to avoid too much time of computation (by using Mathematica software).
2. Auxiliary Results
The results of this section can be found in the classical literature about this kind of Diophantine equation (see, for example, [
20] and the references therein). For this reason, we shall present here these tools as succinctly as possible.
The first useful result is due to Dresden and Du [
38] (Theorem 1) who proved that
with
, where
is the dominant root of the polynomial
. Moreover, we have the notation
. Furthermore, Bravo and Luca [
39] showed that
Another very useful ingredient is the following result à la Baker:
Lemma 1. Let be algebraic numbers and let be nonzero integer numbers. Let D be the degree of the number field extension and let be any constant such that In addition, choose a constant B for which If , then This result is a version of a Matveev theorem [
40] due to Bugeaud et al. (see its proof in [
41]). In the previous statement,
denotes the logarithmic height of an
-degree algebraic number
. This function satisfies the following properties (the proof of these facts can be found in [
42]):
Lemma 2. We have
- i.
;
- ii.
;
- iii.
, for all .
Finally, the last tool was proved by Dujella and Pethő [
43] (Lemma 5(a)):
Lemma 3. Let M be a positive integer and let be a convergent of the continued fraction expansion of the irrational number γ such that . Let and define , where μ is a given real number. If , then there is no solution to the Diophantine inequalityin positive integers and satisfying In the previous statement, we used the notation .
Now, we are ready to deal with the proof of the theorem.
3. The Proof of Theorem 1
3.1. Upper Bounds for n and ℓ in Terms of k
Let
and
be the dominant roots of the sequences
and
, respectively. Moreover, set
,
,
and
. Thus, by using (
4) in Equation (
3), we obtain
After some computations, we get
where we used that
and
. Now, we divide the inequality by
, to derive
where we applied the inequality
(see [
44]). Let us define
where
(for
). Then, Equation (
8) can be rewritten as
First, we claim that
(for
). On the contrary, we would have
. Now, we can apply to the previous equality, a nontrivial automorphism
(
and
) of the Galois group of
to obtain (after applying absolute values):
where
and
(both are smaller than 2). The previous inequality implies
and so
yielding ta contradiction. Therefore, we have that
as desired.
In order to use Lemma 1, we take
,
and
For this choice, we have
, for
. Note that
, and by Lemma 2
Moreover, in [
39] (p. 73), an estimate for
was given. More precisely, it was proved that
. Thus, by using again Lemma 2, we have
where we used that
, for all
.
Note that
. However, by (
3), we deduce that
yielding
and so we can take
.
Now, we are in a position to use Lemma 1 which provides us (after some manipulations)
By combining (
10) and (
11), we obtain
Since, for
, the function
is increasing, then it is a simple exercise to show that
Thus, by using (
13) in (
12) for
and
, we get
where
for all
.
In conclusion, we arrive at
Now, the proof splits into two cases, as follows.
3.2. The Case
By using (
14), we have that, if
, then
Now, we desire to apply the reduction method (based on Lemma 3) to make the bounds much smaller. The further arguments work for
and
in a similar way. So, to avoid unnecessary repetitions, we shall consider only the case when
(and then
). Thus, by (
8), we have
We divide by
to get
where
and
.
We claim that
is irrational for any integer
(In fact, it suffices to notice that
is irrational for all integer
). Let
be the denominator of the
-th convergent of the continued fraction of
. After taking
, we use Mathematica to obtain
Define
, for
and
. Then,
Observe that the conditions of Lemma 3 are fulfilled for
and
and hence there is no solution to inequality (
15) (and then no solution to the Diophantine Equation (
3)) for
n and
satisfying
Therefore, the possible solutions of (
3) are in the range
(since
was already studied) and
. By using Mathematica, we can explicit
, for
and
. Now, we search for 10 digits repdigits among these remainders
and none a solution was found. This completes the proof.
3.3. The Case
3.3.1. Bound for k in Terms of n
This allows us to apply the Bravo and Luca method (see [
39]). The main idea of their approach is to approximate
and
to 2 (in an explicit form). We shall omit the details since it is very well explained in [
39]. Therefore, we can write
and
where
,
,
and
. Moreover, they proved that
Now, we can use all these information to derive (after many straightforward manipulations) that
where
, for
. Thus
where we used the inequality (
7). We divide the previous inequality by
to have
Clearly, the left-hand side above is nonzero (otherwise 5 would divide
). We shall apply Lemma 1 again for
,
and
Therefore,
and
(
), and by Lemma 2,
. Thus, we can choose
and
and Lemma 1 gives
By combining (
16) and (
17), we get
3.3.2. Explicit Bounds for and k and the Reduction Method
By (
14) and (
18), we have
Since the left-hand side is
, while the right-hand one is
, then the previous inequality is valid only for finitely many values of
k (since
as
). In order to make this explicitly, after some calculations, the previous inequality becomes
Now, by using Mathematica software, we infer that
and
. Therefore, we need to make these bounds smaller. For that, we shall use Lemma 3. Take
. By (
16),
Again, we can suppose that
(otherwise, use that
, if
) to write
Dividing through by
, we obtain
where
and
, for
.
Clearly, is irrational and Let be the denominator of the -th convergent of its continued fraction. Taking , we use Mathematica again to obtain that .
Define
, for
. Then, we obtain
Thus, all conditions of Lemma 3 are satisfied for
and
and hence there is no solution to inequality (
19) for
ℓ and
k satisfying
Since , we have that . Thus . By repeating this process again for the new (we use ), we obtain and . To conclude, we apply one more time Lemma 3 for the new choice of (for ) and hence .
This contradicts our assumption that
. In conclusion, there is no solution to the Diophantine Equation (
3) for
. □
4. Other Similar Equations: The Elementary Method
It is important to notice that an elementary method does not provide a reasonable approach to deal with the Equation (
1). The possible reasons can be because the product (in the left-hand side) possesses only two terms,
n can be much larger than
as well as the very limited knowledge about arithmetic properties of repdigits (from that equation we infer only that
ℓ must be a composite number).
However, we shall provide here some similar Diophantine equations which can be solved by using basic tools.
The first one is when the order is larger than the index. More precisely
Proposition 1. The Diophantine equationdoes not have a solution in positive integers , with , and . The proof follows because , for all and so , for all . Since none repdigit is a multiple of 16, then . However, which contradicts .
Another possible problem is a symmetric equation (between k and n) which forces the previous case. For example:
Proposition 2. The Diophantine equationdoes not have a solution in positive integers , with , and . The proof is similar to the previous one by using that either or .
In the previous propositions, we used the fact that
is a power of 2, when
. However, we can use another approach (since
is never a power of two, for
, see [
39]) for equations related to the product of “many” consecutive
k-bonacci numbers. More precisely:
Proposition 3. The Diophantine equationdoes not have a solution in positive integers , with , and . For proving this, we notice that the recurrence of
yields
Thus,
, i.e.,
is a
-periodic sequence
. Thus, the sequence
contains infinitely many even numbers (for example,
, for all
). For this reason, at least one among the numbers
,
is even. The same happens for the lists
,
and
. Hence, the product
is a multiple of 16 and so it can not be a repdigit.
5. Conclusions
For any integer , the sequence of the k-generalized Fibonacci numbers (or k-bonacci numbers) is defined by the k initial values and and such that each term afterwards is the sum of the k preceding ones. In this paper, we search for solutions of the Diophantine equation for positive integers and a, with , and . In particular, the only repdigits, which can be written as a product of nth terms of two generalized Fibonacci sequences with consecutive orders, has only one digit. Our approach to proving this fact is to combine the Baker’s theory (on lower bounds for linear forms in the logarithms) with a reduction method from the theory of continued fractions (due to Dujella and Pethő). In the concluding section, we present some similar problems which can be solved by using only elementary tools.
Author Contributions
P.T. conceived the presented idea on the conceptualization, methodology, and investigation. Writing—review and editing and preparation of program procedures in Mathematica were done by P.C. All authors have read and agreed to the published version of the manuscript.
Funding
The authors was supported by the Project of Specific Research PrF UHK no. 2101/2021, University of Hradec Králové, Czech Republic.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Informed consent was obtained from all subjects involved in the study.
Conflicts of Interest
The authors declare no conflict of interest.
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