On the Problem of Pillai with Tribonacci Numbers and Powers of 3

Mahadi Ddamulira Institute of Analysis and Number Theory Graz University of Technology Kopernikusgasse 24/II A-8010 Graz Austria mddamulira@tugraz.at mahadi@aims.edu.gh

Abstract Let (T_{n})_{n≥0} be the sequence of tribonacci numbers defined by T_{0} = 0, T_{1} = T_{2} = 1, and T_{n+3} = T_{n+2} +T_{n+1} +T_{n} for all n ≥ 0. In this note, we find all integers c admitting at least two representations as a difference between a tribonacci number and a power of 3.

1 Introduction

We consider the sequence (T_{n})_{n≥0} of tribonacci numbers defined by

T_{0} = 0, T_{1} = 1, T_{2} = 1, and T_{n+3} = T_{n+2} + T_{n+1} + T_{n} for all n ≥ 0.

The tribonacci sequence is sequence \underline{A000073} on the Online Encyclopedia of Integer Sequences (OEIS) [18]. The first few terms of the tribonacci sequence are

(T_{n})_{n≥0} = 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, ….

In this paper, we study the Diophantine equation

T_{n} − 3^{m} = c,  (1)

for a fixed integer c and variable integers n and m. In particular, we are interested in finding those integers c admitting at least two representations as a difference between a tribonacci number and a power of 3. This equation is a variation of the Pillai equation

a^{x} − b^{y} = c,  (2)

where x, y are non-negative integers and a, b, c are fixed positive integers.

In the 1930’s, Pillai [19, 20] conjectured that for any given integer c ≥ 1, the number of positive integer solutions (a, b, x, y), with x ≥ 2 and y ≥ 2 to the equation (2) is finite. This conjecture is still open for all c ≠ 1. The case c = 1 is the conjecture of Catalan which was proved by Mihăilescu [17]. The work of Pillai work was an extension of the work of Herschfeld [14, 15], who had already studied a particular case of the problem with (a, b) = (2, 3). Since then, different variations of the Pillai equation have been studied. Several recent results for the different variations of the Pillai problem involving Fibonacci numbers, tribonacci numbers, Pell numbers, k-generalized Fibonacci numbers, and other linearly recurrent sequences, with powers of 2, have been completely studied. For example, see [3, 4, 5, 6, 7, 8, 10, 11].

We discard the situation when n = 1 and just count the solutions for n = 2 since T_{1} = T_{2} = 1. The reason for the above convention is to avoid trivial parametric families such as 1 − 3^{m} = T_{1} − 3^{m} = T_{2} − 3^{m}. Thus, we always assume that n ≥ 2. The main aim of this paper is to prove the following result.

Theorem 1. The only integers c having at least two representations of the form T_{n} − 3^{m} with n ≥ 2 and m ≥ 0, are c ∈ {−2, 0, 1, 4}. Furthermore, all the representations of the above integers as T_{n} − 3^{m} with integers n ≥ 2 and m ≥ 0 are given by

−2 = T_{5} − 3^{2} = T_{2} − 3^{1},
0 = T_{9} − 3^{4} = T_{2} − 3^{0},  (3)
1 = T_{4} − 3^{1} = T_{3} − 3^{0},
4 = F_{6} − 3^{2} = T_{5} − 3^{1}.

2 Preliminary results

2.1 The tribonacci sequence

The characteristic polynomial of the tribonacci sequence (T_{n})_{n≥0} is given by

Ψ(x) := x^{3} − x^{2} − x − 1.

Ψ(x) is irreducible in ℚ[x], and has a positive real zero

α = \frac{1}{3} (1 + (19 + 3\sqrt{33})^{1/3} + (19 − 3\sqrt{33})^{1/3}) ,

lying strictly outside the unit circle and two complex conjugate zeros β and γ lying strictly inside the unit circle. Furthermore, |β| = |γ| = α^{−1/2}. According to Dresden and Zu [9], a Binet-like formula for the k-generalized Fibonacci sequences is established. For the tribonacci sequence, it states that

T_{n} = C_{α}α^{n−1} + C_{β}β^{n−1} + C_{γ}γ^{n−1} for all n ≥ 1,  (4)

where C_{X} = (X − 1)/(4X − 6). Dresden and Zu [9], also showed that the contribution of the complex conjugate zeros β and γ to the right-hand side of (4) is very small. More precisely,

|T_{n} − C_{α}α^{n−1}| < \frac{1}{2} for all n ≥ 1.  (5)

The minimal polynomial of C_{α} over the integers is given by

44X^{3} − 44X^{2} + 12X − 1,

has zeros C_{α}, C_{β}, C_{γ} with |C_{α}|, |C_{β}|, |C_{γ}| < 1. Numerically,

1.83 < α < 1.84,
0.73 < |β| = |γ| = α^{−1/2} < 0.74,
0.61 < |C_{α}| < 0.62,
0.19 < |C_{β}| = |C_{γ}| < 0.20.

It is also a well known fact (see [3, 8]) that

α^{n−2} ≤ T_{n} ≤ α^{n−1} holds for all n ≥ 1.  (6)

Let 𝕂 := ℚ(α, β) be the splitting field of the polynomial Ψ over ℚ. Then, [𝕂, ℚ] = 6. Furthermore, [ℚ(α) : ℚ] = 3. The Galois group of 𝕂 over ℚ is given by

G := Gal(𝕂/ℚ) ≅ {(1), (αβ), (αγ), (βγ), (αβγ), (αγβ)} ≅ S_{3}.

Thus, we identify the automorphisms of G with the permutations of the zeros of the polynomial Ψ. For example, the permutation (αγ) corresponds to the automorphism σ : α → γ, γ → α, β → β.

2.2 Linear forms in logarithms

In order to prove Theorem 1, we need to use several times a Baker–type lower bound for a nonzero linear form in logarithms of algebraic numbers. There are many such bounds in the literature like that of Baker and Wüstholz [2]. We use the one of Matveev [16]. Matveev [16] proved the following theorem, which is one of our main tools in this paper.

Let γ be an algebraic number of degree d with minimal primitive polynomial over the integers

a_{0}x^{d} + a_{1}x^{d−1} + ⋯ + a_{d} = a_{0} ∏_{i=1}^{d}(x − γ^{(i)}),

where the leading coefficient a_{0} is positive and the η^{(i)}’s are the conjugates of γ. Then the logarithmic height of γ is given by

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

In particular, if γ = p/q is a rational number with gcd(p, q) = 1 and q > 0, then h(γ) = log max{|p|, q}. Some of the properties of the logarithmic height function h(·), that will be used in the next sections of this paper without reference are as follows:

h(η ± γ) ≤ h(η) + h(γ) + log 2,
h(ηγ^{±1}) ≤ h(η) + h(γ),  (7)
h(η^{s}) = |s|h(η) (s ∈ ℤ).

Theorem 2 (Matveev). Let γ_{1}, …, γ_{t} be positive real numbers in a number field 𝕂 ⊆ ℝ of degree D, and let b_{1}, …, b_{t} be nonzero integers. Assume that

Λ := γ_{1}^{b_{1}} ⋯ γ_{t}^{b_{t}} − 1,  (8)

is nonzero. Then,

log |Λ| > −1.4 × 30^{t+3} × t^{4.5} × D^{2}(1 + log D)(1 + log B)A_{1} ⋯ A_{t},

where

B ≥ max{|b_{1}|, …, |b_{t}|}

and

A_{i} ≥ max{Dh(γ_{i}), | log γ_{i}|, 0.16}, for all i = 1, …, t.

2.3 Baker-Davenport reduction procedure

During the course of our calculations, we get some upper bounds on our variables which are too large. Thus, we need to reduce them. To do so, we use some results from the theory of continued fractions. Specifically, for a nonhomogeneous linear form in two integer variables, we use a slight variation of a result due to Dujella and Pethő [12], which itself is a generalization of a result of Baker and Davenport [1].

For a real number X, we write ∥X∥ := min{|X − n| : n ∈ ℤ} for the distance from X to the nearest integer.

Lemma 3 (Dujella, Pethő). Let M be a positive integer, p/q be a convergent of the continued fraction of the irrational number τ such that q > 6M, and let A, B, and μ be some real numbers with A > 0 and B > 1. Let further ε := ∥μq∥ − M∥τq∥. If ε > 0, then there is no solution to the inequality

0 < |uτ − v + μ| < AB^{−w},

in positive integers u, v, and w with

u ≤ M and w ≥ \frac{log(Aq/ε)}{log B}.

Finally, the following lemma is also useful. It is Lemma 7 in [13].

Lemma 4 (Gúzman, Luca). Let m ⩾ 1, Y > (4m^{2})^{m}, and Y > x/(log x)^{m}. Then

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

3 Proof of Theorem 1

Let n, m, n_{1}, and m_{1} be non-negative integers such that (n, m) ≠ (n_{1}, m_{1}) and

T_{n} − 3^{m} = T_{n_{1}} − 3^{m_{1}}.

Without loss of generality, we assume that m ≥ m_{1}. If m = m_{1}, then T_{n} = T_{n_{1}}, so (n, m) = (n_{1}, m_{1}), which gives a contradiction to our assumption. Thus, m > m_{1}. Since

T_{n} − T_{n_{1}} = 3^{m} − 3^{m_{1}},  (9)

and the right-hand side is positive, we get that the left-hand side is also positive and so n > n_{1}. Thus, n ≥ 3 and n_{1} ≥ 2.

Using the equation (9) and the inequality (6), we get

α^{n−4} ≤ T_{n−2} ≤ T_{n} − T_{n_{1}} = 3^{m} − 3^{m_{1}} < 3^{m},  (10)
α^{n−1} ≥ T_{n} ≥ T_{n} − T_{n_{1}} = 3^{m} − 3^{m_{1}} ≥ 3^{m−1},  (11)

from which we get that

1 + (\frac{log 3}{log α}) (m − 1) < n < (\frac{log 3}{log α}) m + 4.  (12)

If n ≤ 300, then m ≤ 200. We ran a Mathematica program for 2 ≤ n_{1} < n ≤ 300 and 0 ≤ m_{1} < m ≤ 200 and found only the solutions from the list (3). From now, we assume that n > 300. Note that the inequality (12) implies that m < 0.6n + 0.4. Therefore, to solve the Diophatine equation (1), it suffices to find an upper bound for n.

3.1 Bounding n

From (4) and (5), we get

|C_{α}α^{n−1} − 3^{m}| = |(C_{α}α^{n−1} − T_{n}) + (T_{n_{1}} − 3^{m_{1}})|

= |(C_{α}α^{n−1} − T_{n}) + (T_{n_{1}} − C_{α}α^{n_{1}−1}) + (C_{α}α^{n_{1}−1} − 3^{m_{1}})|

< 1 + \frac{7}{10}α^{n_{1}−1} + 3^{m_{1}}

< α^{n_{1}} + 3^{m_{1}}

< 2 max{α^{n_{1}}, 3^{m_{1}}}.

In the above we have used the fact that |C_{α}| < 0.62 < 0.7. Multiplying through by 3^{−m}, using the relation (10) and using the fact that α < 3, we get

|C_{α}α^{n−1}3^{−m} − 1| < 2 max \{\frac{α^{n_{1}}}{3^{m}} , 3^{m_{1}−m}\} < max\{α^{n_{1}−n+6}, 3^{m_{1}−m+1}\}.  (13)

For the left-hand side, we apply the result of Matveev, Theorem 2 with the following data:

t := 3, γ_{1} := C_{α}, γ_{2} := α, γ_{3} := 3, b_{1} := 1, b_{2} := n − 1, and b_{3} := −m.

Through out we work with the field 𝕂 := ℚ(α) with D = 3. Since max{1, n − 1, m} ≤ n, we take B := n. The minimal polynomial of C_{α} over the integers is given by

44x^{3} − 44x^{2} + 12x − 1.

Since |C_{α}|, |C_{β}|, |C_{γ}| < 1, we get that h(C_{α}) = \frac{1}{3} log 44. So we can take A_{1} := 3h(γ_{1}) = log 44. We can also take A_{2} := 3h(γ_{2}) = log α and A_{3} := 3h(γ_{3}) = 3 log 3. We put

Λ := C_{α}α^{n−1}3^{−m} − 1.

First we check that Λ ≠ 0, if it were, then C_{α}α^{n−1} = 3^{m} ∈ ℤ. Conjugating this relation by the automorphism (αβ), we obtain that C_{β}β^{n−1} = 3^{m}, which is a contradiction because |C_{β}β^{n−1}| < 1 while 3^{m} ≥ 3 for all m ≥ 1. Thus, Λ ≠ 0. Hence, by Theorem 2, the left-hand side of (13) is bounded as follows:

log |Λ| > −1.4 · 30^{6} · 3^{4.5} · 3^{2}(1 + log 3)(1 + log n)(log 44)(log α)(3 log 3).

By comparing with (13), we get

min{(n − n_{1} − 5) log α, (m − m_{1} − 1) log 3} < 2.06 × 10^{13}(1 + log n),

which gives

min{(n − n_{1}) log α, (m − m_{1}) log 3} < 2.12 × 10^{13}(1 + log n).

Now we split the argument into two cases.

Case 1. min{(n − n_{1}) log α, (m − m_{1}) log 3} = (n − n_{1}) log α.

In this case, we rewrite (9) as

|C_{α}α^{n−1} − C_{α}α^{n_{1}−1} − 3^{m}| = |C_{α}α^{n−1} − T_{n}) + (T_{n_{1}} − C_{α}α^{n_{1}−1}) − 3^{m_{1}} |
< 1 + 3^{m_{1}} ≤ 3^{m_{1}+1},

which implies

|C_{α}(α^{n−n_{1}} − 1)α^{n_{1}−1}3^{−m} − 1| < 3^{m_{1}−m+1}.  (14)

We put

Λ_{1} := C_{α}(α^{n−n_{1}} − 1)α^{n_{1}−1}3^{−m} − 1.

As before, we take 𝕂 := ℚ(α), so we have D = 3. We have Λ_{1} ≠ 0, for if Λ_{1} = 0, then

C_{α}(α^{n−n_{1}} − 1)α^{n_{1}−1} = 3^{m}.

By conjugating the above relation by the Galois automorphism (αβ), we get that

C_{β}(β^{n−n_{1}} − 1)β^{n_{1}−1} = 3^{m}.

The absolute value of the left-hand side is at most |C_{β}(β^{n−n_{1}} − 1)β^{n_{1}−1}| ≤ |C_{β}β^{n−1}| + |C_{β}β^{n_{1}−1}| < 2, while the absolute value of the right-hand side is at least 3^{m} ≥ 3 for all m ≥ 1, which is a contradiction.

We apply Theorem 2 on the left-hand side of (14) with the following data:

t := 3, γ_{1} := C_{α}(α^{n−n_{1}} − 1), γ_{2} := α, γ_{3} := 3, b_{1} := 1, b_{2} := n_{1} − 1, and b_{3} := −m.

Since

h(γ_{1}) ≤ h(C_{α}) + h(α^{n−n_{1}} − 1)
< \frac{1}{3} log 44 + \frac{1}{3}(n − n_{1}) log α + log 2
< \frac{1}{3}(log 11 + log 32) + \frac{1}{3} × 2.12 × 10^{13}(1 + log n)
< \frac{1}{3} × 2.50 × 10^{13}(1 + log n).  (15)

So, we can take A_{1} := 2.50 × 10^{13}(1 + log n). Furthermore, as before, we take A_{2} := log α and A_{3} := 3 log 3. Finally, since max{1, n_{1} − 1, m} ≤ n, we can take B := n. Then we get

log |Λ_{1}| > −1.4 · 30^{6} · 3^{4.5} · 3^{2}(1 + log 3)(1 + log n)(2.50 × 10^{13}(1 + log n))(log α)(3 log 3).

Then,

log |Λ_{1}| > −1.36 × 10^{25}(1 + log n)^{2}.

By comparing the above relation with (14), we get that

(m − m_{1}) log 3 < 1.40 × 10^{26}(1 + log n)^{2}.  (16)

Case 2. min{(n − n_{1}) log α, (m − m_{1}) log 3} = (m − m_{1}) log 3.

In this case, we rewrite (9) as

|C_{α}α^{n} − (3^{m−m_{1}} − 1) · 3^{m_{1}} | = |(C_{α}α^{n−1} − T_{n}) + (T_{n_{1}} − C_{α}α^{n_{1}−1}) + C_{α}α^{n_{1}−1}|

< 1 + \frac{7}{10}α^{n−1} < α^{n−1} (beacause n ≥ 3),

which implies that

|C_{α}(3^{m−m_{1}} − 1)^{−1}α^{n−1}3^{−m_{1}} − 1| < \frac{α^{n_{1}}}{3^{m} − 3^{m_{1}}} ≤ \frac{3α^{n_{1}}}{3^{m}}
< 3α^{n_{1}−n+4} < α^{n_{1}−n+6}.  (17)

We put

Λ_{2} := C_{α}(3^{m−m_{1}} − 1)^{−1}α^{n−1}3^{−m_{1}} − 1.

Clearly, Λ_{2} ≠ 0, for if Λ_{2} = 0, then C_{α} = (α^{−1})^{n−1}(3^{m}−3^{m_{1}}) implying that C_{α} is an algebraic integer, a contradiction. We again apply Theorem 2 with the following data:

t := 3, γ_{1} := C_{α}(3^{m−m_{1}} − 1)^{−1}, γ_{2} := α, γ_{3} := α, b_{1} := 1, b_{2} := n, and b_{3} := −m_{1}.

We note that

h(γ_{1}) = h(C_{α}(3^{m−m_{1}} − 1)^{−1}) ≤ h(C_{α}) + h(3^{m−m_{1}} − 1)

= \frac{1}{3} log 44 + h(3^{m−m_{1}} − 1) < log(3^{m−m_{1}+2})

= (m − m_{1} + 2) log 3 < 2.50 × 10^{13}(1 + log n).

So, we can take A_{1} := 7.5 × 10^{13}(1 + log n). Further, as in the previous applications, we take A_{2} := log α and A_{3} := 3 log 3. Finally, since max{1, n − 1, m_{1}} ≤ n, we can take B := n. Then, we get

log |Λ_{2}| > −1.4 · 30^{6} · 3^{4.5} · 3^{2}(1 + log 3)(1 + log n)(7.5 × 10^{13}(1 + log n))(log α)(3 log 3).

Thus,

log |A_{2}| > −4.08 × 10^{26}(1 + log n)^{2}.

Now, by comparing with (17), we get that

(n − n_{1}) log α < 4.10 × 10^{26}(1 + log n)^{2}.  (18)

Therefore, in both Case 1 and Case 2, we have

min{(n − n_{1}) log α, (m − m_{1}) log 3} < 2.12 × 10^{13}(1 + log n),
max{(n − n_{1}) log α, (m − m_{1}) log 3} < 4.10 × 10^{26}(1 + log n)^{2}.  (19)

Finally, we rewrite the equation (9) as

|C_{α}α^{n−1} − C_{α}α^{n_{1}−1} − 3^{m} + 3^{m_{1}} | = |(C_{α}α^{n−1} − T_{n}) + (T_{n_{1}} − C_{α}α^{n_{1}−1})| < 1.

Dividing through by 3^{m} − 3^{m_{1}}, we get

|\frac{C_{α}(α^{n−n_{1}} − 1)}{3^{m−m_{1}} − 1}α^{n_{1}−1}3^{−m_{1}} − 1| < \frac{1}{3^{m} − 3^{m_{1}}} ≤ \frac{3}{3^{m}}
≤ 3α^{−(n−4)} ≤ α^{6−n},  (20)

since 3 < α ≤ α^{n_{1}}. We again apply Theorem 2 on the left-hand side of (20) with the following data:

t := 3, γ_{1} := \frac{C_{α}(α^{n−n_{1}} − 1)}{3^{m−m_{1}} − 1}, γ_{2} := α, γ_{3} := 3, b_{1} := 1, b_{2} := n_{1} − 1, and b_{3} := −m_{1}.

By using the algebraic properties of the logarithmic height function, we get

3h(γ_{1}) = 3h (\frac{C_{α}(α^{n−n_{1}} − 1)}{3^{m−m_{1}} − 1}) ≤ h (C_{α}(α^{n−n_{1}} − 1)) + h(3^{m−m_{1}} − 1)
< log 352 + (n − n_{1}) log α + 3(m − m_{1}) log 3
< 6.80 × 10^{26}(1 + log n)^{2},

where in the above inequalities, we used the argument from (15) as well as the bounds (19). Thus, we can take A_{1} := 6.80 × 10^{26}(1 + log n), and again as before A_{2} := log α and A_{3} := 3 log 3. If we put

Λ_{3} := \frac{C_{α}(α^{n−n_{1}} − 1)}{3^{m−m_{1}} − 1}α^{n_{1}−1}3^{−m_{1}} − 1,

we need to show that Λ_{3} ≠ 0. If not, Λ_{3} = 0 leads to

C_{α}(α^{n−1} − α^{n_{1}−1}) = 3^{m} − 3^{m_{1}}.

A contradiction is reached upon a conjugation by the automorphism (αβ) in 𝕂 and by taking absolute values on both sides. Thus, Λ_{3} ≠ 0. Applying Theorem 2 gives

log |Λ_{3}| > −1.4 · 30^{6} · 3^{4.5} · 3^{2}(1 + log 3)(1 + log n)(6.80 × 10^{26}(1 + log n)^{2})(log α)(3 log 3), a comparison with (20) gives

(n − 6) < 3.70 × 10^{39}(1 + log n)^{3},

or

n < 3.8 × 10^{39}(1 + log n)^{3}.  (21)

Now, by applying Lemma 4 on (21) with the data m := 3, Y := 3.8 × 10^{39}, and x := n, leads to n < 3 × 10^{46}.

3.2 Reducing the bound for n

We need to reduce the above bound for n and to do so we make use of Lemma 3 several times. To begin, we return to (13) and put

Γ := (n − 1) log α − m log 3 + log C_{α}.

For technical reasons we assume that min{n − n_{1}, m − m_{1}} ≥ 20. We go back to the inequalities for Λ, Λ_{1}, and Λ_{2}, Since we assume that min{n − n_{1}, m − m_{1}} ≥ 20 we get |e^{Γ} − 1| = |Λ| < \frac{1}{4}. Hence, |Λ| < \frac{1}{2} and since the inequality |y| < 2|e^{y} − 1| holds for all y ∈ (−\frac{1}{2}, \frac{1}{2}), we get

0 < |Γ| < 2 max{α^{n_{1}−n+6}, 3^{m_{1}−m+1}} ≤ max{α^{n_{1}−n+8}, 3^{m_{1}−m+2}}.

By substituting for Γ in the above inequality and dividing through by log 3, we get the inequality

0 < |(n − 1) (\frac{log α}{log 3}) − m + \frac{log C_{α}}{log 3} | < max \{\frac{α^{8}}{(log 3)α^{n−n_{1}}} , \frac{9}{(log 3)3^{m−m_{1}}} \} .

We apply Lemma 3 with the following data

τ := \frac{log α}{log 3}, μ := \frac{log C_{α}}{log 3} , and (A, B) := (\frac{α^{8}}{log 3}, α) or (\frac{9}{log 3}, 3) .

Let τ = [a_{0}; a_{1}, a_{2}, …] = [0; 1, 1, 4, 13, 1, 6, 1, 4, 1, 10, 7, 1, 24, 3, 3, 2, 12, 4, 4, …] be the continued fraction expansion of τ. We choose M := 3 × 10^{46} which is the upper bound on n. With the help of Mathematica, we find out that the convergent

\frac{p}{q} = \frac{p_{88}}{q_{88}} = \frac{383979914200993729068715782793592146551951600940}{692255294546383107303758900444711151890883197059}

is such that q = q_{88} > 6M. Furthermore, it yields ε > 0.0428119, and therefore either

n − n_{1} ≤ \frac{log ((α^{8}/ log 3)q/ε)}{log α} < 193, or m − m_{1} ≤ \frac{log ((9/ log 3)q/ε)}{log 3} < 105.

Thus, we have that either n − n_{1} ≤ 193 or m − m_{1} ≤ 105.

Now we distinguish between the cases n − n_{1} ≤ 193 and m − m_{1} ≤ 105. First, we assume that n − n_{1} ≤ 193. In this case we consider the inequality for Λ_{1}, (14) and also assume that m − m_{1} ≥ 20. We put

Γ_{1} := (n_{1} − 1) log α − m log 3 + log (C_{α}(α^{n−n_{1}} − 1)) .

Then, inequality (14) implies that

|Γ_{1}| < \frac{6}{3^{m−m_{1}}} .

If we substitute for Γ_{1} in the above inequality and divide through by log 3, we then get

0 < |(n_{1} − 1) (\frac{log α}{log 3}) − m + \frac{log(C_{α}(α^{n−n_{1}} − 1))}{log 3}| < \frac{6}{(log 3)3^{m−m_{1}}} .

Again we apply Lemma 3 with the same τ as in the case of Γ. We use the 88-th convergent p/q = p_{88}/q_{88} of τ as before. But in this case we choose (A, B) := (\frac{6}{log 3}, 3) and use

μ_{ℓ} := \frac{log(C_{α}(α^{ℓ} − 1))}{log 3},

instead of μ for each possible value of ℓ := n − n_{1} ∈ [1, 2, …, 193]. For all values of ℓ, we get ε > 0.0000420218. Hence, by Lemma 3, we get

m − m_{1} < \frac{log ((6/ log 3)q/ε)}{log 3} < 110.

Thus, n − n_{1} ≤ 193 implies that m − m_{1} ≤ 110.

Now let us turn to the case m − m_{1} ≤ 105 and we consider the inequlity for Λ_{2}, (17). We put

Γ_{2} := (n − 1) log α − m_{1} log 3 + log (\frac{C_{α}}{3^{m−m_{1}} − 1}) ,

and we also assume that n − n_{1} ≥ 20. We then have

|Γ_{2}| < \frac{α^{8}}{α^{n−n_{1}}} .

If we substitute for Γ_{2} in the above inequality and divide through by log 3, we then get

0 < |(n − 1) (\frac{log α}{log 3}) − m_{1} + \frac{log(C_{α}/(3^{m−m_{1}} − 1))}{log 3}| < \frac{α^{8}}{(log 3)α^{n−n_{1}}} .

We apply again Lemma 3 with the same τ, q, M, (A, B) := (\frac{α^{8}}{log 3}, α), and

μ_{ℓ} := \frac{log(C_{α}/(3^{ℓ} − 1))}{log 3} for ℓ = 1, 2, …, 105.

We get ε > 0.00218297, therefore

n − n_{1} < \frac{log ((α^{8}/ log 3)q/ε)}{log α} < 198.

To conclude, we first get that either n − n_{1} ≤ 193 or m − m_{1} ≤ 105. If n − n_{1} ≤ 193, then m − m_{1} ≤ 110, and if m − m_{1} ≤ 105, then n − n_{1} ≤ 198. Thus, we conclude that we always have n − n_{1} ≤ 198 and m − m_{1} ≤ 110.

Finally, we go to the inequality of Λ_{3}, (20). We put

Γ_{3} := (n_{1} − 1) log α − m_{1} log 3 + log (\frac{C_{α}(α^{n−n_{1}} − 1)}{3^{m−m_{1}} − 1}) .

Since n > 300, the inequality (20) implies that

|Γ_{3}| < \frac{2}{α^{n−6}} = \frac{α^{8}}{α^{n}} .

Substituting for Γ_{3} in the above inequality and dividing through by log 3, we get

0 < |(n_{1} − 1) (\frac{log α}{log 3}) − m_{1} + \frac{log(C_{α}(α^{k} − 1)/(3^{ℓ} − 1))}{log 3}| < \frac{α^{8}}{(log 3)α^{n}} ,

where (k, ℓ) := (n−n_{1}, m−m_{1}). We again apply Lemma 3 with the same τ, q, M, (A, B) := (\frac{α^{8}}{log 3}, α), and

μ_{k,l} := \frac{log(C_{α}(α^{k} − 1)/(3^{ℓ} − 1))}{log 3} for 1 ≤ k ≤ 198 and 1 ≤ ℓ ≤ 110.

For the cases, we get ε > 0.0000115272, so we obtain

n ≤ \frac{log ((α^{8}/ log 3)q/ε)}{log α} < 207.

Hence, n ≤ 207. However, this contradicts our working assumption that n > 300. This completes the proof of Theorem 1.

4 Acknowledgments

The author would like to thank the referee for the careful reading of the manuscript, the useful comments, and suggestions that greatly improved on the presentation of this paper. The author was supported by the Austrian Science Fund (FWF) grants: F5510-N26 – Part of the special research program (SFB), “Quasi-Monte Carlo Methods: Theory and Applications”, P26114-N26 –“Diophantine Problems: Analytic, geometric and computational aspects”, and W1230 –“Doctoral Program Discrete Mathematics”.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11J86. Keywords: tribonacci sequence, linear form in logarithms, Baker’s method.

(Concerned with sequence \underline{A000073}.)

Received June 1 2019; revised versions received July 31 2019; August 21 2019. Published in Journal of Integer Sequences, August 23 2019.

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