Lucas Numbers Which Are Concatenations of Two Repdigits

mathematics

Article Lucas Numbers Which Are Concatenations of Two Repdigits

Yunyun Qu 1,2,* and Jiwen Zeng 1

1 School of Mathematical Sciences, Xiamen University, Xiamen 361005, Fujian, China; [email protected] 2 School of Mathematical Sciences, Guizhou Normal University, Guiyang 550001, Guizhou, China * Correspondence: [email protected]

Received: 21 July 2020; Accepted: 10 August 2020; Published: 13 August 2020

Abstract: In this paper, we ﬁnd all Lucas numbers written in the form c ··· cd ··· d, where c ··· cd ··· d is the concatenation of two repdigits in base 10 with c, d ∈ {0, 1, . . . , 9}, c 6= d and c > 0.

Keywords: Lucas numbers; concatenations of two repdigits; logarithmic height; continued fraction

1. Introduction Linear form in logarithms has many important applications in solving Diophantine equations [1–4]. In 2002, by applying linear form in logarithms, A. Dujella and B. Jadrijević[1] showed that the solutions to quartic Thue equations x4 − 4cx3y + (6c + 2)x2y2 + 4cxy3 + y4 = 1 are only (x, y) = (±1, 0) and (0, ±1) for an integer c ≥ 3. Suppose that {Fn}n≥0 is the Fibonacci sequence given by Fn+2 = Fn+1 + Fn, with initial values F0 = 0 and F1 = 1 and let {Ln}n≥0 be the Lucas sequence defined by Ln+2 = Ln+1 + Ln, where L0 = 2 and L1 = 1. In 2011, F. Luca and R. Oyono [2] concluded that there is no solution (m, n, s) to the Diophantine equation s s Fm + Fm+1 = Fn for integers m ≥ 2, n ≥ 1, s ≥ 3 by applying linear form in logarithms. There are many papers in the literature which solve Diophantine equations related to Fibonacci numbers and Lucas numbers [3–14]. In 2013, D. Marques and A. Togbé [3] found all solutions (n, a, b, c) to a b c a b c the Diophantine equation Fn = 2 + 3 + 5 and Ln = 2 + 3 + 5 for integers n, a, b, c with 0 ≤ max{a, b} ≤ c. In 2019, B. D. Bitim [4] investigated the solutions (n, m, a) to the Diophantine a equation Ln − Lm = 2 · 3 for nonnegative integers n, m, a with n > m. Let p be a prime number and max{a, b} ≥ 2, in 2009, F. Luca and P. Stanicˇ aˇ [5] concluded that there are only finitely many positive a b integer solutions (n, p, a, b) to the Diophantine equation Fn = p ± p . Assume that q ≥ 2 is an integer. A positive number n ∈ N is called a base q−repdigit if qt−1 n = c q−1 , for some t ≥ 1 and c ∈ {1, 2, ... , q − 1}. When q = 10, n is simply called a repdigit. We use B1 ··· Bt(q) to express an integer’s base−q representation which is the concatenation of the base−q representations of positive integers B1, ... , Bt. We ignore writing q if q = 10. Then we can denote the repdigit n by n = c ··· c and the concatenation of two repdigits in base 10 is c ··· c d ··· d, | {z } | {z } | {z } t times s times t times where c, d ∈ {0, 1, ... , 9}, c 6= d, c > 0, s ≥ 1 and t ≥ 1. There are many papers in the literature on investigating Diophantine equations related to repdigits [8,9,11–21]. In 2000, Luca [15] proved that if 10m−1 10m−1 Fn = a 9 and Ln = a 9 for some a ∈ {0, 1, ... , 9} and m ≥ 1, then n = 0, 1, 2, 3, 4, 5, 6, 10 and n = 0, 1, 2, 3, 4, 5 respectively. In 2012, all repdigits in base 10 expressible as sums of three Fibonacci numbers were found in [16]. In 2018, all repdigits in base 10 which are sums of four Fibonacci or Lucas numbers were determined in [17]. In 2019, all solutions to the Diophantine equation Fn = a ··· a b ··· b | {z } | {z } m times l times were found in [18], where a, b ∈ {0, 1, ... , 9} and a > 0. For the research of concatenations of

Mathematics 2020, 8, 1360; doi:10.3390/math8081360 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 1360 2 of 8 two repdigits in balancing numbers, Padovan numbers and Tribonacci numbers, please refer to the literature [19–21] respectively. In this paper, we ﬁnd all Lucas numbers which are concatenations of two repdigits. More precisely, we have the following result.

Theorem 1. If Ln = c ··· c d ··· d, (1) | {z } | {z } s times t times with c, d ∈ {0, 1, . . . , 9}, c 6= d, c > 0, s ≥ 1 and t ≥ 1, then

(n, Ln) ∈ {(6, 18), (7, 29), (8, 47), (9, 76), (11, 199), (12, 322)}.

2. Preliminaries Firstly, the Binet’s formula for Lucas sequence is

n n Ln = α + β , n ≥ 0, √ √ 1+ 5 1− 5 where α = 2 and β = 2 . For all positive integers n, we have the following inequality

n−1 n+1 α ≤ Ln ≤ α . (2)

Secondly, we recall the deﬁnition and properties for logarithmic height of an algebraic number. Let η be an algebraic number of degree m and suppose that the minimal primitive polynomial of η is m (i) f (X) := a0 ∏i=1(X − η ) ∈ Z[X] with a0 > 0. We give the logarithmic height of η by

m ! 1 (i) h(η) := log a0 + ∑ log max{|η |, 1} . m i=1

In this paper, for any two integers a and b, we denote the greatest common divisor of a and b by p gcd(a, b). Speciﬁcally, h(η) = log max{|p|, q} when η = q ∈ Q with gcd(p, q) = 1 and q > 0. We have the following properties of the logarithmic height h(·):

h(η ± γ) ≤ h(η) + h(γ) + log 2,

h(ηγ±1) ≤ h(η) + h(γ),

h(ηk) = |k|h(η)(k ∈ Z). We need the following lemma to prove our theorem.

Lemma 1. (see [22]) Let dL be the degree of an algebraic number field L over Q and L ⊆ R. Let γ1, γ2,..., γl ∈ L b1 bl be non-zero elements and let b1,..., bl be rational integers. If Γ := γ1 ··· γl − 1 6= 0, then

| | ≥ (− · l+3 4.5 2 ( + )( + ) ··· ) Γ exp 1.4 30 l dL 1 logdL 1 logB A1 Al , where Aj are real numbers such that

Aj ≥ max{dLh(γj), |logγj|, 0.16} for j = 1, . . . , l and B ≥ max{|b1|,..., |bl|, 3}.

Thirdly, we need the following Lemma2 and Lemma3 to reduce some large upper bounds on the variables in the course of our calculations. Mathematics 2020, 8, 1360 3 of 8

p Lemma 2. (see [23]) Let M be a positive integer and let q be a convergent of the continued fraction of the irrational number α such that q > 6M, and let A, B, τ be some real numbers with A > 0 and B > 1. Let e := kτqk − Mkαqk, where k · k denotes the distance from the nearest integer. If e > 0, then there exists no solution to the inequality 0 < |uα − v + τ| < AB−ω

log(Aq/e) in positive integers u, v, and ω with u ≤ M and w ≥ logB .

Lemma 3. (see [24]) Let τ be an irrational number, M be a positive integer and pk (k = 0, 1, 2, ...) be all qk the convergents of the continued fraction [a0, a1, ...] of τ. Let N be such that qN > M. Then putting aM := max{ai : i = 0, 1, . . . , N}, the inequality

1 |mτ − n| > (aM + 2)m holds for all pairs (n, m) of integers with 0 < m < M.

3. Proof of Theorem1

3.1. Bounding n According to (1), we get Ln = c ··· c d ··· d | {z } | {z } s times t times = c ··· c · 10t + d ··· d | {z } | {z } (3) s times t times 1 = (c10s+t − (c − d)10t − d). 9 n−1 s+t Suppose that n > 1000. From inequality (2), we can get α ≤ Ln < 10 and s+t−1 n+1 10 ≤ Ln ≤ α , which implies that

(s + t)log10 − log10 − logα ≤ nlogα < (s + t)log10 + logα. (4)

Thus, we can get 4.78(s + t) − 5.8 < n < 4.79(s + t) + 1. (5)

n−1 From (5), we get s + t > 4.79 > 208 and n > s + t. According to (3) and Binet’s formulae for Lucas sequences, we get

|9αn − c10s+t| = | − 9βn − ((c − d)10t + d)| ≤ 9α−n + 9 · 10t + 9 < 27 · 10t, (6) which implies that

9 − − 27 αn10 s t − 1 < . (7) c 10s

s+t Let Γ := 9 αn10−s−t − 1, then Γ 6= 0. If Γ = 0, then αn = 10 c ∈ , 1 c √ √ 1 1 9 Q 10s+tc (1+ 5)n f +g 5 thus we have 9 = 2n = 2n , where f , g ∈ Z, f > 0, g > 0, this implies √ 10s+tc2n 9 − f that 5 = g ∈ Q, which is impossible. According to Lemma 1, we take l = 3, 9 γ1 = c , γ2 = α, γ3 = 10 and b1 = 1, b2 = n, b3 = −s − t. Thus, we have L = Q(α), dL = [L : Q] = 2. 9 1 Note that h(γ1) = h( c ) ≤ log9, h(γ2) = 2 logα, h(γ3) = log10. Thus, we can take A1 = 2log9, A2 = 0.5, A3 = 4.8. Note that B = max{|b1|, |b2|, |b3|, 3} = max{1, n, s + t, 3} = n. Hence, we get

| Γ1 |> exp(−C1(1 + logn)), (8) Mathematics 2020, 8, 1360 4 of 8

13 where C1 = 1.025 × 10 . Thus from (7) and (8) , we can get

slog10 < C1(1 + logn) + log27. (9)

We rewrite Equation (3), then we get

s c10 − (c − d) d − αn − · 10t = βn + ≤ α n + 1 < 2. (10) 9 9

It follows that s c10 − (c − d) − 2 · α n · 10t − 1 < . (11) 9 αn

s s c10 −(c−d) −n t n c10 −(c−d) t Let Γ2 := 9 · α · 10 − 1, then Γ2 6= 0. If Γ2 = 0, then α = 9 · 10 ∈ Q, which is c10s−(c−d) false. According to Lemma 1, we take l = 3,γ1 = 9 , γ2 = α, γ3 = 10 and b1 = 1, b2 = −n, b3 = t. Thus, we have L = Q(α), dL = [L : Q] = 2. From (9), we can get

s h(γ1) ≤ h(c10 − (c − d)) + h(9) ≤ 3log9 + slog10 + log2 (12) ≤ C1(1 + logn) + log27 + 3log9 + log2 ≤ 1.03 · 1013 · (1 + logn),

1 13 and we have h(γ2) = 2 logα, h(γ3) = log10. Thus, we can take A1 = 2.06 · 10 · (1 + logn), A2 = 0.5, A3 = 4.8. Note that B = max{|b1|, |b2|, |b3|, 3} = max{1, n, t, 3} = n. Hence, we get

2 | Γ2 |> exp(−C2(1 + logn) ), (13)

25 where C2 = 4.8 × 10 . Thus from (11) and (13) , we can get

2 nlogα < C2(1 + logn) + log2, (14) this implies that n < 4.8 × 1029. Hence we can conclude that

n + 5.8 s + t < < 1.01 · 1029. 4.78 To sum up, we have the lemma as follows.

Lemma 4. If (n, c, d, s, t) is a solution in non-negative integers of Equation (1), with c, d ∈ {0, 1, . . . , 9}, c 6= d and c > 0, then s + t < n < 4.8 · 1029, s + t < 1.01 · 1029.

3.2. Reducing the Bound on n We use the Lemmas2 and3 to reduce the bound for n. Let 9 Λ := (s + t)log10 − nlogα − log . 1 c

From (7), we conclude that 27 e−Λ1 − 1 < . (15) 10s Mathematics 2020, 8, 1360 5 of 8

−Λ1 27 1 1 −Λ1 3 If s ≥ 2, then |e − 1| < 10s < 2 , which implies that 2 < e < 2 . If Λ1 > 0, Λ1 Λ1 −Λ1 54 |Λ1| −Λ1 27 then 0 < Λ1 < e − 1 = e (1 − e ) < 10s . If Λ1 < 0, then 0 < |Λ1| < e − 1 = e − 1 < 10s . 54 In any case, it is always holds true 0 < |Λ1| < 10s , which implies

54 log10 log 9 < ( + ) − − c < logα 0 s t n s . (16) logα logα 10

The continued fraction of log10 is [a , a , a , a , a , ... ] = [4, 1, 3, 1, 1, 1, 6, ... ], and let pk be its kth logα 0 1 2 3 4 qk 29 log10 convergent. Note that s + t < 1.01 · 10 by Lemma 4. It is easy to see that logα is irrational. In fact, log10 p p q if logα = q (p, q ∈ Z and p > 0, q > 0, gcd(p, q) = 1), then α = 10 ∈ Q, which is an absurdity. 29 For all c ∈ {1, ... , 8}, according to (16) and Lemma 2, we take M = 1.01 · 10 and q60 > 6M, hence we get the minimum value of e is 0.061483 . . . and s < 34. If c = 9, from (16), we get

54 log10 logα 0 < (s + t) − n < . (17) logα 10s

29 According to Lemma 3, we take M = 1.01 · 10 and q60 > M, hence we get aM := max{ai : i = 0, 1, . . . , 60} = 106 and we have

log10 1 1 ( + ) − > > s t n 29 . (18) logα (aM + 2)(s + t) 108 · 1.01 · 10

Thus, from (17) and (18), we get

54 1 logα < , 108 · 1.01 · 1029 10s this leads to s < 34. So we always have s < 34. Let c10s − (c − d) Λ := tlog10 − nlogα + log . 2 9 From (11) and n > 1000, we conclude that

2 1 eΛ2 − 1 < < , (19) αn 2

1 Λ2 3 Λ2 2 which implies that 2 < e < 2 . If Λ2 > 0, then 0 < Λ2 < e − 1 < αn . If Λ2 < 0, −Λ2 −Λ2 Λ2 4 4 then 0 < |Λ2| < e − 1 = e (1 − e ) < αn . In any case, since 0 < |Λ2| < αn , thus we have

c10s−(c−d) 4 log10 log < − + 9 < logα 0 t n n , (20) logα logα α where s ≤ 33, c ∈ {1, ... , 9} and d ∈ {0, 1, ... , 9}. For inequality (20), we consider the following two 29 cases: if (s, c, d) 6= (1, 1, 0), according to (20) and Lemma 2, we take M = 1.01 × 10 and q60 > 6M, hence we obtain 25 negative values of e , the minimum value in the values of positive e is 0.00004477 ... and n < 171. For the values of (s, c, d) corresponding to the 25 negative values of e , we take q63 > 6M, according to (20) and Lemma 2, we get the minimum value in the values of e is 0.005613 ... and n < 168. If (s, c, d) = (1, 1, 0), from (20), we get

4 log10 logα 0 < t − n < . (21) logα αn Mathematics 2020, 8, 1360 6 of 8

29 According to Lemma 3, we take M = 1.01 · 10 and q60 > M, hence we get aM := max{ai : i = 0, 1, . . . , 60} = 106 and we have

log10 1 1 − > > t n 29 . (22) logα (aM + 2)t 108 · 1.01 · 10

Thus, from (21) and (22), we get

4 1 logα < , 108 · 1.01 · 1029 αn which leads to n < 153. In summary, we have n < 171. This contradicts the assumption n > 1000. Finally, we search for the solutions to (1) in the range n ≤ 1000 by applying a program written in Mathematica and we obtain the solutions (n, Ln) ∈ {(6, 18), (7, 29), (8, 47), (9, 76), (11, 199), (12, 322)}. We complete the proof.

4. Conclusions and Future Research

(k) For a ﬁxed integer k ≥ 2, let {Fn }n≥2−k be the k−generalized Fibonacci sequence deﬁned by (k) = (k) + (k) + ··· + (k) (k) = (k) = ··· = (k) = (k) = Fn Fn−1 Fn−2 Fn−k with the initial values F−(k−2) F−(k−3) F0 0, F1 1 (k) (k) (k) (k) (k) and {Ln }n≥2−k be the k−generalized Lucas sequence given by Ln = Ln−1 + Ln−2 + ··· + Ln−k (k) = (k) = ··· = (k) = (k) = (k) = with the initial values L−(k−2) L−(k−3) L−1 0, L0 2, L1 1. Suppose that c, d ∈ {0, 1, . . . , 9}, c 6= d, c > 0, s ≥ 1 and t ≥ 1, our aim is to solve the two Diophantine equations

(k) Fn = c ··· c d ··· d (23) | {z } | {z } s times t times and (k) Ln = c ··· c d ··· d. (24) | {z } | {z } s times t times For k = 2 and k = 3, the Diophantine Equation (23) has been solved in [18] and [21], respectively. In this paper, we solve the Diophantine Equation (24) for the case of k = 2. Our future research work is to solve the Diophantine Equations (23) and (24) completely for the case of k ≥ 3. In addition, for the main Mathematica programs used in this paper, readers can refer to AppendixA.

Author Contributions: Writing—original draft preparation, Y.Q. and J.Z.; Writing—review and editing, Y.Q. and J.Z.; Y.Q. and J.Z. have equally contributed to this work. All authors have read and agreed to the published version of the manuscript. Funding: The research was supported by Guizhou Provincial Science and Technology Foundation(Grant No. QIANKEHEJICHU[2019]1221) and the National Natural Science Foundation of China(Grant No. 11261060). Acknowledgments: The authors would like to thank the anonymous reviewers for their very useful and detailed suggestions. Conﬂicts of Interest: The authors declare no conﬂict of interest.

Appendix A. Mathematica Programs We give the main Mathematica programs used in this paper as follows : √ = 1+ 5 = log[10] • α 2 ; γ log[α] ; • Generates a list of the ﬁrst n terms in γ0s continued fraction representation:

ContinuedFraction[γ, n] Mathematics 2020, 8, 1360 7 of 8

• The denominator of the nth(n = 0, 1, 2, . . .) convergent of γ0s continued fraction:

log[10] q[n ] := Module[{γ = √ }, Last[Denominator[Convergents[γ, n + 1]]]]; 1+ 5 log[ 2 ]

• The function kxk which denotes the distance from x to the nearest integer:

cldist[x , jd ] := Module[{}, Abs[N[Round[x] − x, jd]]];

• The number e := kτqk − Mkαqk in Lemma2:

epsilon[τ , q , M , α , jd ] := Module[{}, cldist[τ ∗ q, jd] − M ∗ cldist[α ∗ q, jd]];

9 [ 9 ] = − log c [ ] = − log c • The number τ : logα in (16): τ c : log[α] ; s−( − ) s−( − ) log c10 c d log[ c10 c d ] = 9 [ ] = 9 • The number τ : logα in (20): τ s , c , d : log[α] ; • The nth term of Lucas sequence Ln: √ √ 1 + 5 1 − 5 Lucas[n ] := Module[{α = , β = }, Simpli f y[Expand[αn + βn]]]; 2 2

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