Product of Consecutive Tribonacci Numbers With Only One Distinct Digit

Eric F. Bravo and Carlos A. Gómez

Department of Mathematics Universidad del Valle Calle 13 No 100 – 00

Cali Colombia

eric.bravo@correounivalle.edu.co
carlos.a.gomez@correounivalle.edu.co

Florian Luca School of Mathematics

University of the Witwatersrand Johannesburg South Africa and

Research Group in Algebraic Structures and Applications

King Abdulaziz University

Jeddah Saudi Arabia and

Department of Mathematics University of Ostrava 30 Dubna 22, 701 03 Ostrava 1 Czech Republic

florian.luca@wits.ac.za

Abstract

Let (F_{n})_{n≥0} be the sequence of Fibonacci numbers. Marques and Togbé proved that if the product F_{n} ⋯ F_{n+ℓ−1} is a repdigit (i.e., a number with only distinct digit in its decimal expansion), with at least two digits, then (ℓ, n) = (1, 10). In this paper, we solve the same problem with Tribonacci numbers instead of Fibonacci numbers.

1 Introduction

A positive integer is called a repdigit if it has only one distinct digit in its decimal expansion. The sequence of numbers with repeated digits is included in Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS) [11] as sequence \underline{A010785}. In 2000, Luca [6] showed that the largest repdigit Fibonacci number is F_{10} = 55 and the largest repdigit Lucas number is L_{5} = 11.

Motivated by the results of Luca [6], several authors have explored repdigits in generalizations of Fibonacci numbers and Lucas numbers. For instance, Bravo and Luca [1] showed that the only repdigit in the k-generalized Fibonacci sequence^{1}, is F_{8}^{(3)} = 44. On the other hand, Bravo and Luca [2] showed that only repdigit in the k-generalized Lucas sequence^{2}, is L_{5}^{(4)} = 22.

Recently, this problem has been extended to study the product of consecutive Fibonacci or Lucas numbers which are repdigits. Marques and Togbé [9] proved that no product of more of two consecutive Fibonacci numbers can be a repdigit with at least two digits, while Irmak and Togbé [5] proved that 77 is the only repdigit with at least two digits appearing as the product of two or more consecutive Lucas numbers.

In this paper, we investigate the presence of repdigits in the product of consecutive Tribonacci numbers. More precisely, we prove the following result.

Theorem 1. The only solution of the Diophantine equation

T_{n} ⋯ T_{n+(ℓ−1)} = d (\frac{10^{m} − 1}{9}) , in positive integers n, ℓ, m, d,  (1)

with 1 ≤ d ≤ 9 and m ≥ 2 is (n, ℓ, m, d) = (8, 1, 2, 4); i.e., T_{8} = 44.

To solve equation (1), we use the 2-adic order of the Tribonacci numbers in order to bound the number of factors ℓ, then linear forms in logarithms to bound max{m, n} and finally a version of the Baker-Davenport Lemma to reduce such bounds to manageable values and finish off with a few calculations.

1The k-generalized Fibonacci sequence F^{(k)}, for an integer k ≥ 2, satisfies that its first k terms are 0, …, 0, 1 and each term afterwards is the sum of the preceding k terms. For k = 2, this reduces to the familiar Fibonacci numbers., while for k = 3 these are the Tribonacci numbers.

2The k-generalized Lucas sequence L^{(k)} satisfies that its first k terms are 0, …, 0, 2, 1 and each term afterwards is the sum of the preceding k terms. For k = 2, this reduces to the familiar Lucas numbers.

2 Preliminary results

2.1 The Tribonacci sequence

We consider T := (T_{n})_{n≥−1} given by T_{−1} = T_{0} = 0, T_{1} = 1 and T_{n+3} = T_{n+2} + T_{n+1} + T_{n}, for all n ≥ 0. Its characteristic equation, z^{3} − z^{2} − z − 1 = 0, has one real root α and two complex roots β and γ = β̄. In 1982, Spickerman [12] found the Binet formula for the Tribonacci numbers

T_{s} = aα^{s} + bβ^{s} + cγ^{s}, for all s ≥ 0,  (2)

where

a = \frac{1}{(α − β) (α − γ)}, b = \frac{1}{(β − α) (β − γ)}, c = \frac{1}{(γ − α) (γ − β)} = b̄.

It is easy to see that α ∈ (1.83, 1.84) , |β| = |γ| ∈ (0.73, 0.74) , a ∈ (0.18, 0.19) and |b| = |c| ∈ (0.35, 0.36). Since |β| = |γ| < 1, setting e_{s} := T_{s} − aα^{s}, we have

T_{s} = aα^{s} + e_{s},  with |e_{s}| < \frac{1}{α^{s/2}} for all s ≥ 1.  (3)

Further,

α^{s−2} ≤ T_{s} ≤ α^{s−1} for all s ≥ 1 (see [1]).  (4)

We recall a result of Marques and Lengyel [8] on the 2-adic order of a Tribonacci number. For a prime number p and a nonzero integer r, the p-adic order υ_{p}(r) is the exponent of the highest power of a prime p which divides r.

Lemma 2. For n ≥ 1, we have

υ_{2} (T_{n}) = \{ \begin{matrix}0, & if n ≡ 1, 2 (mod 4); \\ 1, & if n ≡ 3, 11 (mod 16); \\ 2, & if n ≡ 4, 8 (mod 16); \\ 3, & if n ≡ 7 (mod 16); \\ υ_{2}(n) − 1, & if n ≡ 0 (mod 16); \\ υ_{2}(n + 4) − 1, & if n ≡ 12 (mod 16); \\ υ_{2}((n + 1)(n + 17)) − 3, & if n ≡ 15 (mod 16). \end{matrix}

The Tribonacci sequence is included in the OEIS [11] as sequence \underline{A101292}.

2.2 Linear forms in logarithms

Let η be an algebraic number of degree d over ℚ with minimal primitive polynomial f(X) := a_{0} ∏_{i=1}^{d}(X − η^{(i)}) ∈ ℤ[X], where the leading coefficient a_{0} is positive. The logarithmic height of η is given by

h(η) := \frac{1}{d} (log a_{0} + ∑_{i=1}^{d} log max\{|η^{(i)}|, 1\}) .

In particular, if η = p/q ∈ ℚ with gcd(p, q) = 1 and q > 0, then h(η) = log max{|p|, q}. The following are some of the properties of the logarithmic height function h(·), which will be used in this paper: h(η ± γ) ≤ h(η) + h(γ) + log 2, h(ηγ^{±1}) ≤ h(η) + h(γ) and h(η^{s}) = |s|h(η) (s ∈ ℤ).

Many Diophantine problems can be solved by reducing them to an instance in which one can apply lower bounds for linear forms in logarithms of algebraic numbers. We will use the following theorem, which is a variation of a result of Matveev [10], proved by Bugeaud, Mignotte and Siksek [3, Theorem 9.1].

Theorem 3. Let 𝕃 be a number field of degree d_{𝕃} over ℚ, η_{1}, …, η_{t} nonzero elements of 𝕃, and b_{1}, …, b_{t} rational integers. Put Λ := η_{1}^{b_{1}} ⋯ η_{t}^{b_{t}} − 1 and B ≥ max{|b_{1}|, …, |b_{t}|}. Let A_{i} ≥ max{d_{𝕃}h(η_{i}), | log η_{i}|, 0.16} be real numbers, for i = 1, …, t. Then, assuming that Λ ≠ 0, we have

|Λ| > exp(−3 · 30^{t+4} · (t + 1)^{5.5} · d_{𝕃}^{2}(1 + log d_{𝕃})(1 + log tB)A_{1} ⋯ A_{t}).

3 Absolute bounds on the variables

3.1 On the number ℓ of factors

We claim that ℓ ≤ 6. Indeed, from Lemma 2 we have the following table for all possible values of n ≡ x (mod 16), x ∈ {0, 1, 2, …, 15}.

\begin{aligned}ℓ & x & υ_{2}(T_{n}T_{n+1} ⋯ T_{n+(ℓ−1)}) \\ 1 & 15 & ≥ 5 \\ 2 & 7, 11, 14 & ≥ 5, ≥ 4, ≥ 5 \\ 3 & 6, 10, 13 & ≥ 5, ≥ 4, ≥ 5 \\ 4 & 0, 4, 5, 9, 12 & ≥ 4, ≥ 5, ≥ 5, ≥ 4, ≥ 8 \\ 5 & 3, 8 & ≥ 6, ≥ 4 \\ 6 & 2 & ≥ 6 \\ 7 & 1 & ≥ 6 \end{aligned}

Table 1: 2-adic order of product of consecutive Tribonacci numbers

We analyze a pair of cases to illustrate the above table. The remaining cases are similar.

• n ≡ 1 (mod 16). Here, n + 2 ≡ 3 (mod 16), n + 3 ≡ 4 (mod 16) and n + 6 ≡ 7 (mod 16). So, by Lemma 2 we have υ_{2}(T_{n+2}) = 1, υ_{2}(T_{n+3}) = 2 and υ_{2}(T_{n+6}) = 3. Hence, with ℓ = 7, υ_{2}(T_{n}T_{n+1} ⋯ T_{n+6}) ≥ 6.

• n ≡ 12 (mod 16). Then, from Lemma 2 we have υ_{2}(T_{n}) = υ_{2}(n + 4) − 1 ≥ 3, since n + 4 ≡ 0 (mod 16). On other hand, n + 3 ≡ 15 (mod 16) and υ_{2}(T_{n+3}) = υ_{2}((n + 4)(n + 20)) − 3 ≥ 5, where we used Lemma 2 and the fact that n + 20 ≡ 0 (mod 16). Therefore, with ℓ = 4, υ_{2}(T_{n}T_{n+1}T_{n+2}T_{n+3}) ≥ 8.

Since υ_{2} (d (\frac{10^{m}−1}{9})) = υ_{2}(d) ≤ 3 for all 1 ≤ d ≤ 9, it then follows that ℓ ≤ 6.

3.2 An absolute bound for m and n

First, assume that n ≥ 20. Combining (1) and (4), we get

10^{m−1} < α^{ℓ(n−1)+ \frac{ℓ(ℓ−1)}{2}} .

Thus,

m < ℓn + ℓ(ℓ − 1)/2.  (5)

Now, by (3), we get that

T_{n} ⋯ T_{n+(ℓ−1)} = (aα^{n} + e_{n}) ⋯ (aα^{n+(ℓ−1)} + e_{n+ℓ−1})
= a^{ℓ}α^{ℓn+ℓ(ℓ−1)/2} + r(a, α, n, ℓ)

where r(a, α, n, ℓ) involves the part of the expansion of the previous line that contains the product of powers of a, α and the errors e_{i}, for i = n, … n + (ℓ − 1). Moreover, r(a, α, n, ℓ) is the sum of 63 terms with maximum absolute value a^{6}α^{(ℓ−1)n+ℓ(ℓ−1)/2}α^{−n/2}.

Combining the above equality with (1), we obtain

a^{ℓ}α^{ℓn+ℓ(ℓ−1)/2} − \frac{d}{9}10^{m} = −\frac{d}{9} − r(a, α, n, ℓ).

Dividing both sides of the above equality by a^{ℓ}α^{ℓn+ℓ(ℓ−1)/2} and taking the absolute value, we conclude that

|\frac{d}{9a^{ℓ}} α^{−(ℓn+ℓ(ℓ−1)/2)}10^{m} − 1| ≤ (\frac{d}{9} + |r(a, α, n, ℓ)|) · a^{−ℓ}α^{−(ℓn+ℓ(ℓ−1)/2)}  (6)
< (1 + 63a^{ℓ−1}α^{(ℓ−1)n+ℓ(ℓ−1)/2}α^{−n/2}) · a^{−ℓ}α^{−(ℓn+ℓ(ℓ−1)/2)}
≤ 64a^{−1}α^{−3n/2},

Below, we use Matveev’s theorem to find a lower bound for the left-hand side of (6), with the parameters

t := 3, (η_{1}, b_{1}) := ((d/9)a^{−ℓ}, 1), (η_{2}, b_{2}) := (α, −(ℓn+ℓ(ℓ−1)/2)) and (η_{3}, b_{3}) := (10, m).

The number field containing η_{1}, η_{2}, η_{3} is 𝕃 := ℚ(α, β), which has d_{𝕃} := 6. We claim that Λ := η_{1}^{b_{1}}η_{2}^{b_{2}}η_{3}^{b_{3}} − 1 ≠ 0. Otherwise, we get

a^{ℓ}α^{ℓn+ℓ(ℓ−1)/2} = d · 10^{m}/9.

Conjugating the above relation by the automorphism σ : α → β, β → α, γ → γ (here, we use the fact that the Galois group of 𝕃 over ℚ is isomorphic to S_{3}), and then taking absolute values on both sides of the resulting equality, we obtain

|b|^{ℓ}|β|^{ℓn+ℓ(ℓ−1)/2} = d · 10^{m}/9,

which is not possible because |b|^{ℓ}|β|^{ℓn+ℓ(ℓ−1)/2} < 1 and d · 10^{m}/9 > 10. Thus, Λ ≠ 0. Next, h(η_{1}) ≤ h(d)+h(9a^{ℓ}) ≤ log 9+h(9)+ℓh(a), h(η_{2}) = \frac{1}{3} log α and h(η_{3}) = log 10. Now we need to estimate h(a). For it, the minimal polynomial of a is 44X^{3} + 4X − 1. So, h(a) = \frac{1}{3} log 44 and h(η_{1}) ≤ 2 log 9 + 2 log 44. Thus, we can take A_{1} := 72, A_{2} := 2 and A_{3} := 14. According to (5), we take B := ℓn + ℓ(ℓ − 1)/2. Applying Matveev’s theorem we get a lower bound for |Λ|, which by comparing it to (6), leads to

exp (−2.72251 × 10^{19}(1 + log 3(ℓn + ℓ(ℓ − 1)/2))) < \frac{356}{α^{3n/2}} .

Taking logarithms in the above inequality, we get

\frac{3n}{2} log α − log 356 < 2.72251 × 10^{19}(1 + log 3(6n + 15))
< 5.44503 × 10^{19} log(6n + 15),

where we used the fact that 1 + log 3h < 2 log h, for all h ≥ 9. Hence,

n < 6.1 × 10^{19} log(6n + 15).

Therefore, we obtain n < 3.1 × 10^{21} and record what we have proved so far.

Lemma 4. If (n, ℓ, m, d) is a positive integer solution of (1) with n ≥ 20, m ≥ 2 and 1 ≤ d ≤ 9, then 1 ≤ ℓ ≤ 6 and

max{m, n} < 3.1 × 10^{21}.

4 Reducing max{m, n}

To lower the bound of n, we will use the following result of diophantine approximation, see [4].

Lemma 5. Let κ be an irrational number, M be a positive integer, and p/q be a convergent of the continued fraction of κ such that q > 6M. Let A, B, μ be some real numbers with A > 0 and B > 1. Let ϵ := ∥μq∥ − M ∥κq∥, where ∥·∥ denotes the distance from the nearest integer. If ϵ > 0, then there is no solution of the inequality

0 < |rκ − s + μ| < AB^{−w}

in positive integers r, s and w with r ≤ M and w ≥ log(Aq/ϵ)/ log B.

Let Γ := m log 10−(ℓn+ℓ(ℓ−1)/2) log α+log(d/9a^{ℓ}). Therefore, (6) can be rewritten as |e^{Γ} −1| < 356/α^{3n/2}. Since |e^{Γ} −1| < 1/2 for all n ≥ 20 (because 365/α^{3n/2} < 1/2), it follows that e^{|Γ|} < 2 and so 0 < |Γ| ≤ e^{|Γ|} − 1 = e^{|Γ|}|e^{Γ} − 1| < 712/α^{3n/2}. Hence, 0 < |Γ| < 712/α^{3n/2} holds for n ≥ 20.

Replacing Γ in the above inequality and dividing both sides of the resulting inequality by log α, we obtain

0 < |mκ − (ℓn + ℓ(ℓ − 1)/2) + μ| < 1180(α^{3/2})^{−n},  (7)

with κ := log 10/ log α and μ := log(d/9a^{ℓ})/ log α. Here, we took M := 1.9 × 10^{22}, which is an upper bound on m by Lemma 4, and we applied Lemma 5 to inequality (7) for each 1 ≤ ℓ ≤ 6 and 1 ≤ d ≤ 9. With the help of the computer algebra system Mathematica, we found that

q_{47} = 938425170962281070635339 > 6M

is a denominator of a convergent of the continued fraction of κ such that the minimum value of ϵ := ∥μq_{47}∥−M ∥κq_{47}∥ is greater than 0.00437116. The conditions of Lemma 5 are fulfilled for A := 1180 and B := α^{3/2}. Then, there are no solutions of (1) on the interval

[⌊\frac{log(1180q_{47}/ϵ)}{log B}⌋ + 1, M) = [75, 1.9 × 10^{22}).

So, m ≤ 459. Now, we start again the entire process using this much smaller bound of m. In this application of Lemma 5 we found that the assumption n ≥ 20 implies n ≤ 26. Thus, n ≤ 26 holds. Hence, it remains to check equation (1) for 1 ≤ n ≤ 26, 1 ≤ ℓ ≤ 6, 2 ≤ m ≤ 171 and 1 ≤ d ≤ 9. For this, we use Mathematica and conclude that the quadruple (n, ℓ, m, d) = (8, 1, 2, 4) is the only solution of the diophantine equation (1). This completes the proof of Theorem 1.

5 Acknowledgments

The authors are grateful to the referee for the useful comments to improve of this paper. E. F. B thanks Colciencias for support during his Ph.D. studies. F. L. was supported in part by NRF (South Africa) Grant CPRR160325161141, RTNUM18 from CoEMaSS at Wits and by CGA (Czech Republic) Grant 17-02804S.

References

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[8] D. Marques and T. Lengyel, The 2-adic order of the Tribonacci numbers and the equation T_{n} = m!, J. Integer Sequences 17 (2014), Article 14.10.1.

[9] D. Marques and A. Togbé, On repdigits as product of consecutive Fibonacci numbers, Rend. Istit. Mat. Univ. Trieste. 44 (2012), 393–397.

[10] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers II, Izv. Ross. Akad. Nauk Ser. Mat. 64 (2000), 125–180. In Russian. English translation in Izv. Math. 64 (2000), 1217–1269.

[11] N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2019. Available

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11J86. Keywords: Tribonacci number, lower bound, linear form in logarithms of algebraic numbers, repdigit.

Received January 10 2019; revised versions received January 11 2019; August 14 2019. Published in Journal of Integer Sequences, August 24 2019.

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