Repdigits as Sums of Two Padovan Numbers

Ana Cecilia García Lomelí and Santos Hernández Hernández Unidad Académica de Matemáticas Universidad Autónoma de Zacatecas, Campus II Calzada Solidaridad entronque Paseo a la Bufa C.P. 98000 Zacatecas, Zac. Mexico

                aceciliagarcia.lomeli@gmail.com
                        shh@uaz.edu.mx

Abstract In this paper, we determine all repdigits in base 10 that can be written as a sum

of two Padovan numbers.

1 Introduction

We recall that a repdigit is a positive integer r that has only one distinct digit when written in base 10. That is, r is of the form d (\frac{10^{ℓ}−1}{9}) for some positive integers d, ℓ with 1 ≤ d ≤ 9.

Recently, the problem of looking for repdigits in a linear recurrence sequence has been studied. For example, Luca [10] and Marques [12] prove that 55 and 44 are the largest repdigits in the Fibonacci and Tribonacci sequences, respectively. More generally, let k ≥ 2 be an integer. Let F^{(k)} := (F_{n}^{(k)})_{n≥0} be the k-Fibonacci sequence given by

F_{0}^{(k)} = ⋯ = F_{k−2}^{(k)} = 0, F_{k−1}^{(k)} = 1

and the recurrence formula

F_{n+k}^{(k)} = F_{n+k−1}^{(k)} + F_{n+k−2}^{(k)} + ⋯ + F_{n}^{(k)} for all n ≥ 0.

Observe that the cases k = 2, 3 are the Fibonacci and Tribonacci sequences, respectively. Bravo and Luca [3] extended the results mentioned above by showing that the only repdigits with at least two digits in a k-Fibonacci sequence are, for k = 2, 3 the aforementioned ones, while for k > 3 there are no such repdigits belonging to F^{(k)}. Actually, Bravo and Luca [4] considered and solved the more general Diophantine equation

F_{n}^{(k)} + F_{m}^{(k)} = d (\frac{10^{ℓ} − 1}{9}) ,  (1)

in positive integers k ≥ 2, n ≥ m, ℓ ≥ 2, and 1 ≤ d ≤ 9. From their result we see that, for all k there are at most 4 solutions and the largest one is 777, which is found in F^{(4)}. Motivated by these results, in this note we study the similar problem with the Padovan sequence.

The Padovan sequence (P_{n})_{n≥0}, named after the architect R. Padovan, is the ternary recurrence sequence given by P_{0} = 0, P_{1} = P_{2} = 1 and the recurrence formula

P_{n+3} = P_{n+1} + P_{n}, for all n ≥ 0.  (2)

This is the sequence \underline{A000931} in Sloane’s Encyclopedia [14] and it was introduced by Stewart [15], where he relates it to the Fibonacci sequence. Its first few terms are

0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, …

In this note we study the problem of writing repdigits as sums of two Padovan numbers. More precisely, we solve the Diophantine equation

P_{n} + P_{m} = d (\frac{10^{ℓ} − 1}{9}) ,  (3)

in non-negative integers n, m, ℓ, d with n ≥ m, ℓ ≥ 2 and 1 ≤ d ≤ 9. To establish our result we make the following conventions. Since P_{1} = P_{2} = P_{3} = 1 and P_{4} = P_{5} = 2 we will assume that n, m ≠ 1, 2, 4. That is, whenever we think of 1 and 2 as members of the Padovan sequence, we think of them as being P_{3} and P_{5}, respectively. Thus, our result is

Theorem 1. All non-negative integer solutions (n, m, ℓ, d) of equation (3) with n ≥ m, ℓ ≥ 2 and 1 ≤ d ≤ 9 belong to the set

\{ \begin{matrix}(10, 5, 2, 1), & (9, 7, 2, 1), & (13, 3, 2, 2), & (14, 8, 2, 3) \\ (13, 11, 2, 3), & (15, 9, 2, 4), & (14, 12, 2, 4), & (17, 3, 2, 6) \\ (17, 11, 2, 7), & (16, 14, 2, 7), & (18, 5, 2, 8), & (31, 7, 4, 3) \end{matrix} \} .

The set of repdigits r that can be written as sums of two Padovan numbers is

{11, 22, 33, 44, 66, 77, 88, 3333},

and their respective representations are

11 = P_{10} + P_{5} = P_{9} + P_{7};
22 = P_{13} + P_{3};
33 = P_{14} + P_{8} = P_{13} + P_{11};
44 = P_{15} + P_{9} = P_{14} + P_{12};
66 = P_{17} + P_{3};
77 = P_{17} + P_{11} = P_{16} + P_{14};
88 = P_{18} + P_{5};
3333 = P_{31} + P_{7}.

In particular, we obtain that in the Padovan sequence there are no repdigits with more than one digit.

We also note that the problem of writing repdigits as sum of at most three Fibonacci numbers has been solved by Luca [11]. The problem of writing Fibonacci numbers as sum of two repdigits has been studied by Alvarado and Luca [6].

The method of proof of our result is the classical one with linear forms in logarithms and the reduction method of Baker-Davenport as used in [2, 4, 11], for example.

2 Tools

In this section, we gather the tools we need to prove Theorem 1. Let α be an algebraic number of degree d, let a > 0 be the leading coefficient of its minimal polynomial over ℤ and let α = α^{(1)}, …, α^{(d)} denote its conjugates. The logarithmic height of α is defined by

h(α) = \frac{1}{d} (log a + ∑_{i=1}^{d} log max \{|α^{(i)}|, 1\}) .

This height has the following basic properties. For α, β algebraic numbers and m ∈ ℤ we have

• h(α + β) ≤ h(α) + h(β) + log 2;

• h(αβ) ≤ h(α) + h(β);

• h(α^{m}) = |m|h(α).

Now let 𝕃 be a real number field of degree d_{𝕃}, α_{1}, …, α_{s} ∈ 𝕃 and b_{1}, …, b_{s} ∈ ℤ \ {0}. Let B ≥ max{|b_{1}|, …, |b_{s}|} and

Λ = α_{1}^{b_{1}} ⋯ α_{s}^{b_{s}} − 1.

Let A_{1}, …, A_{s} be real numbers with

A_{i} ≥ max{d_{𝕃} h(α_{i}), | log α_{i}|, 0.16}, i = 1, 2, …, s.

The first tool we need is the following result due to Matveev [13]; also see Bugeaud, Mignotte and Siksek [5, Theorem 9.4].

Theorem 2. Assume that Λ ≠ 0. Then

log |Λ| > −1.4 · 30^{s+3} · s^{4.5} · d_{𝕃}^{2} · (1 + log d_{𝕃}) · (1 + log B)A_{1} ⋯ A_{s}.

In this note, we always use s = 3. Further, 𝕃 = ℚ(γ), where γ is defined at the beginning of Section 3, has degree d_{𝕃} = 3. Thus, once and for all we fix the constant

C := 2.70444 × 10^{12} > 1.4 · 30^{3+3} · 3^{4.5} · 3^{2} · (1 + log 3).

Our second tool is a version of the reduction method of Baker and Davenport [1] based on their Lemma. We use the one given by Bravo, Gómez and Luca [2]. Also see Dujella and Pethő [7]. For a real number x, we write ∥x∥ for the distance from x to the nearest integer.

Lemma 3. Let M be a positive integer. Let τ, μ, A > 0, B > 1 be given real numbers. Assume that p/q is a convergent of τ such that q > 6M and ε := ∥μ q∥ − M∥τ q∥ > 0. If (n, m, w) is a positive solution to the inequality

0 < |n τ − m + μ| < \frac{A}{B^{w}}

with n ≤ M, then

w < \frac{log(Aq/ε)}{log B}.

Finally, the following result of Guzmán and Luca [8] will be very useful.

Lemma 4. If m ≥ 1, T > (4m^{2})^{m} and T > x/(log x)^{m}. Then

x < 2^{m}T (log T)^{m}.

3 Proof of Theorem 1

To start with, we recall some properties of the Padovan sequence. For a complex number z we write \overline{z} for its complex conjugate. Let ω ≠ 1 be a cubic root of 1. Let us consider the ℚ-irreducible polynomial X^{3} − X − 1. It is clear that its zeros are γ, δ, \overline{δ} where

γ := \sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 − \sqrt{69}}{18}}, δ := ω\sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \overline{ω}\sqrt[3]{\frac{9 − \sqrt{69}}{18}}.

It can be proved, by induction for example, the Binet formula

P_{n} = c_{1}γ^{n} + c_{2}δ^{n} + c_{3}\overline{δ}^{n} for all n ≥ 0,  (4)

where

c_{1} = \frac{γ(γ + 1)}{2γ + 3},  c_{2} = \frac{δ(δ + 1)}{2δ + 3},  c_{3} = \overline{c_{2}}.

Formula (4) follows from the general theorem on linear recurrence sequences since the above polynomial is the characteristic polynomial of the Padovan sequence. We note that

γ = 1.32471 …, |δ| = 0.86883 …, c_{1} = 0.54511 …, |c_{2}| = 0.28241 …

Further, the inequality

γ^{n−3} ≤ P_{n} ≤ γ^{n−1}  (5)

holds for all n ≥ 1. This can also be proved by induction.

Now we start the study of equation (3) in non-negative integers (n, m, ℓ, d) where, as we have said, n, m ≠ 1, 2, 4 and ℓ ≥ 2. Assume that n ≥ m. Since ℓ ≥ 2 we assume that n ≥ 1. From (5) and (3), we obtain

γ^{n−3} ≤ P_{n} ≤ P_{n} + P_{m} = d (\frac{10^{ℓ} − 1}{9}) ≤ 10^{ℓ},

and

γ^{n+2} ≥ 2P_{n} ≥ P_{n} + P_{m} = d (\frac{10^{ℓ} − 1}{9}) ≥ 10^{ℓ−1},

where we use γ^{3} > 2. Thus,

(n − 3)\frac{log γ}{log 10} ≤ ℓ and (n + 2)\frac{log γ}{log 10} ≥ ℓ − 1.  (6)

Since log γ/ log 10 = 0.122123 … we have, from (6), that if n ≤ 360 then ℓ ≤ 45. Running a Mathematica program in the range 0 ≤ m ≤ n ≤ 360, 1 ≤ d ≤ 9 and 2 ≤ ℓ ≤ 45 we obtain, with our conventions, all solutions listed in Theorem 1. We will prove that these are all of them.

From now on, we assume n > 360. From (6) we get that ℓ > 43 and also that n > ℓ. From the Binet formula (4), we rewrite our equation (3) and obtain

|c_{1}γ^{n} − \frac{d10^{ℓ}}{9} | < 2|c_{2}| + γ^{m−1} + 1 < γ^{2} + γ^{m−1} < γ^{m+5}.

Dividing through by d10^{ℓ}/9 we get

|\frac{9c_{1}}{d} γ^{n}10^{−ℓ} − 1| < \frac{9γ^{m+5}}{d10^{ℓ}} < \frac{1}{γ^{n−m−16}} ,  (7)

where we use γ^{n−3} ≤ 10^{ℓ} and 9 < γ^{8}. Let Λ be the expression inside the absolute value in the left-hand side of (7). Observe that Λ ≠ 0. To see this, we consider the ℚ-automorphism σ of the Galois extension ℚ(γ, δ) over ℚ given by σ(γ) := δ and σ(δ) := γ. Now, if Λ = 0, then σ(Λ) = 0. Thus,

10^{ℓ−1} ≤ \frac{d10^{ℓ}}{9} = |σ (c_{1}γ^{n})| = |c_{2}| |δ|^{n} < |c_{2}| < \frac{1}{3},

which is a contradiction since ℓ > 43. Hence, Λ ≠ 0 and we apply Matveev’s inequality to it. To do this we consider,

α_{1} = \frac{9c_{1}}{d} , α_{2} = γ, α_{3} = 10, b_{1} = 1, b_{2} = n, b_{3} = −ℓ.

Thus, B = n. We have that h(α_{2}) = log γ/3 and h(α_{3}) = log 10. For α_{1} we use the height properties to conclude that

h(α_{1}) ≤ log γ + 13 log 2.

So, we choose A_{1} = 27.9, A_{2} = 0.3 and A_{3} = 7. From Matveev’s inequality we obtain

log |Λ| > −C(1 + log n) · 27.9 · 0.3 · 7,

which, combined with (7) gives us

(n − m) log γ < 1.58454 × 10^{14}(1 + log n).  (8)

Now we find an upper bound on n. To do this, from the Binet formula (4) we rewrite our equation (3) as

|c_{1} (γ^{n−m} + 1) γ^{m} − \frac{d10^{ℓ}}{9} | ≤ 4|c_{2}| + 1 < γ^{3}.

Dividing through by d10^{ℓ}/9 we obtain

|\frac{9c_{1}(γ^{n−m} + 1)}{d}γ^{m}10^{−ℓ} − 1| < \frac{9γ^{3}}{d10^{ℓ}} < \frac{1}{γ^{n−14}} .  (9)

Let Λ_{1} be the expression inside the absolute value in the left-hand side of (9). Note that with a similar argument as the above one, it can be proved that Λ_{1} ≠ 0. So, we apply Matveev’s inequality to it. We consider

α_{1} = \frac{9c_{1}(γ^{n−m} + 1)}{d}, α_{2} = γ, α_{3} = 10, b_{1} = 1, b_{2} = m, b_{3} = −ℓ.

Thus, B = n. The heights of α_{2} and α_{3} have already been calculated. From the properties of the heights we get that

h(α_{1}) ≤ log γ + 14 log 2 + (n − m)\frac{log γ}{3} < \frac{1.58455 × 10^{14}(1 + log n)}{3},

where we have used (8). We choose A_{1} = 1.58455 × 10^{14}(1 + log n), A_{2} and A_{3} as above. Therefore, from Theorem 2 we obtain

log |Λ| > −C(1 + log n) · (1.58455 × 10^{14}(1 + log n)) · 0.3 · 7,

which, combined with (9) gives us n < 1.28011 × 10^{28}(log n)^{2}. Now, from Lemma 4 we get

n < 2.14474 × 10^{32}.  (10)

Now we will reduce the above bound on n. To do this, we first consider

Γ = ℓ log 10 − n log γ + log (\frac{d}{9c_{1}} ) , 1 ≤ d ≤ 9.

Assume n − m ≥ 20 and go to (7). Note that e^{−Γ} − 1 = Λ ≠ 0. Thus, Γ ≠ 0. If Γ < 0 then

0 < |Γ| < e^{|Γ|} − 1 = |Λ| < \frac{1}{γ^{n−m−16}} .

If Γ > 0 we have that 1 − e^{−Γ} = |e^{−Γ} − 1| < 1/2. Hence, e^{Γ} < 2. Thus, we get

0 < Γ < e^{Γ} − 1 = e^{Γ}|Λ| < \frac{2}{γ^{n−m−16}} .

So, in both cases we have

0 < |Γ| < \frac{2}{γ^{n−m−16}} .

Dividing through by log γ we get

|ℓ τ − n + μ_{d}| < \frac{640}{γ^{n−m}} ,  (11)

where

τ := \frac{log 10}{log γ} and μ_{d} := \frac{log(d/9c_{1})}{log γ}, 1 ≤ d ≤ 9.

Now we apply Lemma 3. Put M := 2.14474×10^{32}. Since ℓ < n, from (10) we have that M is the upper bound on ℓ. A quick computation with Mathematica reveals that the convergent

\frac{p_{81}}{q_{81}} = \frac{7680712497306530625416042950269134710}{937994904644870281019845134758679913}

of τ is such that q_{81} > 6M and ε_{d} ≥ 0.0440913 > 0. Thus, with A := 640, B := γ we calculated each value log(q_{81}640/ε_{d})/ log γ and found that all of them are at most 328. Thus we have that

n − m ≤ 328.

Next, we consider

Γ_{1} = ℓ log 10 − m log γ + log (\frac{d}{9c_{1}(γ^{n−m} + 1)}) , 1 ≤ d ≤ 9,

and go to inequality (9). Note that e^{−Γ_{1}} − 1 = Λ_{1} ≠ 0. Thus, Γ_{1} ≠ 0. Since n > 360, with an argument similar to the above one we get

0 < |Γ_{1}| < \frac{2}{γ^{n−14}} .

Dividing through by log γ the we obtain

|ℓ τ − m + μ_{d}| < \frac{365}{γ^{n}},  (12)

where τ is as above, and

μ_{d} := \frac{log(d/9c_{1}(γ^{n−m} + 1))}{log γ} for 1 ≤ d ≤ 9.

Consider

μ_{d,k} = \frac{log(d/9c_{1}(γ^{k} + 1))}{log γ},  d = 1, …, 9,  k = 1, …, 328.

Note that m ≠ 0, since if it is not the case we would have n ≤ 328, which contradicts n > 360. Thus, we apply Lemma 3 again. A quick computation with Mathematica gives that the 81-th convergent of τ again well works, that is, it is such that q_{81} > 6M and ε_{d,k} ≥ 0.0000152967 > 0 for all d = 1, …, 9, k = 1, …, 328 except for the case ε_{9,11}, which is always negative. Thus, for all of these values we calculated log(q_{81} 365/ε_{d,k}) and we find that all of them are at most 354. Thus, n ≤ 354 for all of these values.

The problem in case of the pair (9, 11) is that we have the identity

\frac{2γ + 3}{γ(γ + 1)(γ^{11} + 1)} = \frac{1}{γ^{9}} .

Thus, inequality (12) is

|ℓ τ − (m + 9)| < \frac{365}{γ^{n}}

and we use the theory of continued fractions to study it. Recall, for example, that if

|α − \frac{p}{q}| < \frac{1}{2q^{2}}

then p/q is a convergent of α. See [9, Thm. 184, p. 153]. This result is known as Legendre’s theorem. Since n > 360 and M > ℓ, we have 730ℓ < γ^{n}. From Legendre’s theorem, this implies that (m + 9)/ℓ is a i-th convergent of τ. Let [a_{0}, a_{1}, a_{2}, …] = [8, 5, 3, …] be the continued fraction expression of τ. A quick computation with Mathematica gives that

q_{71} ≤ M < q_{72} and b := max{a_{j} : j = 1, 2, …, 72} = 49.

In particular, we have that b ≥ a_{i+1}. Then, from the properties of the convergents of a continued fraction we obtain

\frac{1}{(b + 2)ℓ} < |ℓ τ − (m + 9)| < \frac{365}{γ^{n}},

which yields

γ^{n} < 365 · 51 · M.

Thus n ≤ 299. So, together with the results above, this shows that in all cases we have n ≤ 354, which contradicts what we assumed about n. This finishes the proof of Theorem 1.

4 Acknowledgments

We wish to express our gratitude to the anonymous referee for careful reading and for the numerous suggestions and remarks that improved the presentation of this work. The first author was supported by a CONACyT Doctoral Fellowship and partly supported by Fundación Kovalevskaia de la Sociedad Matemática Mexicana.

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2010 Mathematics Subject Classification: Primary 11J86; Secondary 11D61, 11D45. Keywords: repdigit, Padovan sequence, linear form in logarithms, reduction method.

(Concerned with sequence \underline{A000931}.)

Received September 28 2018; revised version received October 8 2018; January 11 2019. Published in Journal of Integer Sequences, February 23 2019.

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