Repdigits as sums of three Padovan numbers

TL;DR

This paper classifies all base-10 repdigits that can be expressed as the sum of three Padovan numbers, providing a complete characterization of such representations.

Contribution

It determines all repdigits in base 10 that are sums of three Padovan numbers, a novel classification in the study of Padovan number representations.

Findings

Identifies all repdigits as sums of three Padovan numbers

Provides a complete list of such repdigits

Advances understanding of Padovan number representations

Abstract

Let $ \{P_{n}\}_{n\geq 0} $ be the sequence of Padovan numbers defined by $ P_0=0 $, $ P_1 =1=P_2$ and $ P_{n+3}= P_{n+1} +P_n$ for all $ n\geq 0 $. In this paper, we find all repdigits in base $ 10 $ which can be written as a sum of three Padovan numbers.

Equations

$$P_0=0,P_1=1,P_2=1\text{ and }{P}_{n+3}=P_{n+1}+P_n\text{ for all }n\ge 0.$$

$$\{P_n{\}}_{n\ge 0}=0,1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151,\dots .$$

$$N=d\left(\frac{10^{\ell }-1}{9}\right)$$

$$N=P_{n_1}+P_{n_2}+P_{n_3}=d\left(\frac{10^{\ell }-1}{9}\right),$$

$$N\in \{11,22,33,44,55,66,77,88,99,111,222,333,444,555,666,888,1111,3333,7777\}.$$

$$x^3-x-1=0$$

$$\alpha =\frac{r_1+r_2}{6},\beta =\frac{-(r_1+r_2)+\sqrt{-3}(r_1-r_2)}{12}$$

$$r_1=\sqrt[3]{108+12\sqrt{69}}\text{and}{r}_2=\sqrt[3]{108-12\sqrt{69}}.$$

$$P_n=a{\alpha }^n+b{\beta }^n+c{\gamma }^n\text{ for all}n\ge 0,$$

$$a=\frac{\alpha +1}{(\alpha -\beta )(\alpha -\gamma )},b=\frac{\beta +1}{(\beta -\alpha )(\beta -\gamma )},c=\frac{\gamma +1}{(\gamma -\alpha )(\gamma -\beta )}=\bar{b}.$$

$$1.32<\alpha <1.33$$

$$0.86<|\beta |=|\gamma |={\alpha }^{-\frac{1}{2}}<0.87$$

$$0.72<a<0.73$$

$$0.24<|b|=|c|<\mathrm{0.25.}$$

$$e(n):=P_n-a{\alpha }^n=b{\beta }^n+c{\gamma }^n\text{ then }|e(n)|<\frac{1}{{\alpha }^{n/2}}$$

$${\alpha }^{n-3}\le P_n\le {\alpha }^{n-1}\text{holds for all }n\ge 1.$$

$$a_0{x}^d+a_1{x}^{d-1}+\cdots +a_d=a_0\prod _{i=1}^D(x-{\eta }^{(i)}),$$

$$h(\eta ):=\frac{1}{D}\left(\log a_0+\sum _{i=1}^d\log \left(\max \{|{\eta }^{(i)}|,1\}\right)\right).$$

$$h({\eta }_1\pm {\eta }_2)$$

$$h({\eta }_1{\eta }_2^{\pm 1})$$

$$h({\eta }^s)$$

$$\Lambda :={\eta }_1^{b_1}\cdots {\eta }_t^{b_t}-1,$$

$$\log |\Lambda |>-1.4\times 30^{t+3}\times t^{4.5}\times D_{\mathbb{K}}^2(1+\log D_{\mathbb{K}})(1+\log B)A_1\cdots A_t,$$

$$B\ge \max \{|b_1|,\dots ,|b_t|\},$$

$$A_i\ge \max \{D_{\mathbb{K}}h({\eta }_i),|\log {\eta }_i|,0.16\},\text{for all}i=1,\dots ,t.$$

$$\left|\tau -\frac{r}{s}\right|>\frac{1}{(a(M)+2)s^2},$$

$$0<|u\tau -v+\mu |<A{B}^{-w},$$

$$u\le M\text{and}w\ge \frac{\log (Aq/\epsilon )}{\log B}.$$

$$L<2^{r}H(\log H{)}^r.$$

$${\alpha }^{n_1-3}\le P_{n_1}\le P_{n_1}+P_{n_2}+P_{n_3}=d\left(\frac{10^{\ell }-1}{9}\right)\le 10^{\ell }$$

$$10^{\ell -1}\le d\left(\frac{10^{\ell }-1}{9}\right)=P_{n_1}+P_{n_2}+P_{n_3}\le 3P_{n_1}<{\alpha }^{n+3},$$

$$(n_1-3)\frac{\log \gamma }{\log 10}\le \ell \text{and}\ell -1\le (n_1+3)\frac{\log \gamma }{\log 10}.$$

$$(n_1-3)/5<\ell <(n_1+8)/5.$$

Full text

1 Introduction

Let $\{P_{n}\}_{n\geq 0}$ be the sequence of Padovan numbers given by

[TABLE]

The first few terms of this sequence are

[TABLE]

A repdigit is a positive integer $N$ that has only one distinct digit when written in base $10$. That is, $N$ is of the form

[TABLE]

for some positive integers $d,\ell$ with $1\leq d\leq 9$.

2 Main Result

In this paper, we study the problem of writing repdigits as sums of three Balancing numbers. More prcisely, we completely solve the Diophantine equation

[TABLE]

in non-negative integers $(N,n_{1},n_{2},n_{3},d,\ell)$ with $n_{1}\geq n_{2}\geq n_{3}\geq 0$, $\ell\geq 2$ and $1\leq d\leq 9$.

We discard the situations when $n_{1}=1$ and $n_{1}=2$ and just count the solutions for $n_{1}=3$ since $P_{1}=P_{2}=P_{3}=1$. For the same reasons, we discard the situation when $n_{1}=4$ and just count the solutions for $n_{1}=5$ since $P_{4}=P_{5}=2$. Thus, we always assume that $n_{1},n_{2},n_{3}\notin\{1,2,4\}$. Our main result is the following.

Theorem 2.1

All non-negative interger solutions $(N,n_{1},n_{2},n_{3},d,\ell)$ with $n_{1}\geq n_{2}\geq n_{3}\geq 0$, $\ell\geq 2$ and $1\leq d\leq 9$ to the Diophantine equation (2) arise from

[TABLE]

This paper serves as a continuation of the results in Ddamulira1 , Lomeli , Luca1 , Luca2 , Luca3 and Luca4 . The method of proof involves the application of Baker’s theory for linear forms in logarithms of algebraic numbers, and the Baker-Davenport reduction procedure. Computations are done using a simple computer program in Mathematica.

3 Preliminary results

3.1 The Padovan sequence

Here, we recall some important properties of the Padovan sequence $\{P_{n}\}_{n\geq 0}$. The characteristic equation

[TABLE]

has roots $\alpha,\beta,\gamma=\bar{\beta}$, where

[TABLE]

and

[TABLE]

Furthermore, the Binet formula is given by

[TABLE]

where

[TABLE]

Numerically, the following estimates hold:

[TABLE]

From (3), (4) and (3.1), it is easy to see that the contribution the complex conjugate roots $\beta$ and $\gamma$, to the right-hand side of (5), is very small. In particular, setting

[TABLE]

holds for all $n\geq 1$. Furthermore, by induction, we can prove that

[TABLE]

3.2 Linear forms in logarithms

Let $\eta$ be an algebraic number of degree $D$ with minimal primitive polynomial over the integers

[TABLE]

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

[TABLE]

In particular, if $\eta=p/q$ is a rational number with $\gcd(p,q)=1$ and $q>0$, then $h(\eta)=\log\max\{|p|,q\}$. The following are some of the properties of the logarithmic height function $h(\cdot)$, which will be used in the next sections of this paper without reference:

[TABLE]

Theorem 3.1

Let $\eta_{1},\ldots,\eta_{t}$ be positive real algebraic numbers in a real algebraic number field $\mathbb{K}\subset\mathbb{R}$ of degree $D_{\mathbb{K}}$, $b_{1},\ldots,b_{t}$ be nonzero integers, and assume that

[TABLE]

is nonzero. Then

[TABLE]

where

[TABLE]

and

[TABLE]

3.3 Reduction procedure

During the calculations, we get 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.

For the treatment of linear forms homogeneous in two integer variables, we use the well-known classical result in the theory of Diophantine approximation.

Lemma 1

Let $\tau$ be an irrational number, $\frac{p_{0}}{q_{0}},\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\ldots$ be all the convergents of the continued fraction of $\tau$ and $M$ be a positive integer. Let $N$ be a nonnegative integer such that $q_{N}>M$. Then putting $a(M):=\max\{a_{i}:i=0,1,2,\ldots,N\}$, the inequality

[TABLE]

holds for all pairs $(r,s)$ of positive integers with $0<s<M$.

For a nonhomogeneous linear form in two integer variables, we use a slight variation of a result due to Dujella and Pethő (see dujella98 , Lemma 5a). For a real number $X$, we write $||X||:=\min\{|X-n|:n\in\mathbb{Z}\}$ for the distance from $X$ to the nearest integer.

Lemma 2

Let $M$ be a positive integer, $\frac{p}{q}$ be a convergent of the continued fraction of the irrational number $\tau$ such that $q>6M$, and $A,B,\mu$ be some real numbers with $A>0$ and $B>1$. Let further $\varepsilon:=||\mu q||-M||\tau q||$. If $\varepsilon>0$, then there is no solution to the inequality

[TABLE]

in positive integers $u,v$ and $w$ with

[TABLE]

Finally, the following Lemma is also useful. It is Lemma 7 in guzmanluca .

Lemma 3

If $r\geqslant 1$, $H>(4r^{2})^{r}$ and $H>L/(\log L)^{r}$, then

[TABLE]

4 Bounding the variables

We assume that $n_{1}\geq n_{2}\geq n_{3}$. From (2) and (9), we have

[TABLE]

and

[TABLE]

where we use $\alpha^{4}>3$. Thus,

[TABLE]

Since $\log\gamma/\log 10=0.122123...<1/5$, we can conclude from the above that

[TABLE]

Running a Mathematica program in the range $0\leq n_{3}\leq n_{2}\leq n_{3}\leq 500$, $1\leq d\leq 9$ and $1\leq\ell\leq 100$ we obtain only the solutions listed in Theorem 2.1. From now onwards, we assume that $n_{1}>500$.

By using (8), equation (2) can be written as

[TABLE]

We then consider (13) in three different cases as follows.

4.1 Case 1

We have that

[TABLE]

This is equivalent to

[TABLE]

Thus, we have

[TABLE]

and so

[TABLE]

We divide through (14) by $a\alpha^{n_{1}}$ to get

[TABLE]

Thus, we have

[TABLE]

We put

[TABLE]

In order to apply Theorem 3.1 we need to check that $\Lambda_{1}\neq 0$. Suppose that $\Lambda_{1}=0$, then we have

[TABLE]

To see that this is not true, we consider the $\mathbb{Q}$-automorphism $\sigma$ of the Galois extension $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$ given by $\sigma(\alpha):=\beta$ and $\sigma(\beta):=\alpha$. Now, if $\Lambda_{1}=0$, then $\sigma(\Lambda_{1})=0$. Thus, conjugating the relation (16) under $\sigma$, and taking absolute values on both sides, we get

[TABLE]

which is false for $\ell\geq 2$ and $d\geq 1$. Therefore, $\Lambda_{1}\neq 0$.

So we apply Theorem 3.1 with the data

[TABLE]

It is a well–known fact that

[TABLE]

the mimimal polynomial of $a$ is $23x^{3}-23x^{2}+6x-1$ and has roots $a,b,c$. Since $|b|=|c|<|a|=a<1$ (by (3.1)), then

[TABLE]

Since $\eta_{1},\eta_{2},\eta_{2}\in\mathbb{Q}(\alpha)$, we take the field $\mathbb{K}:=\mathbb{Q}(\alpha)$ with degree $D_{\mathbb{K}}:=3$. Since $\max\{1,\ell,n_{1}\}\leq n_{1}$, we take $B:=n_{1}$. Further, the minimal polynomial of $\alpha$ over $\mathbb{Z}$ is $x^{3}-x-1$ has roots $\alpha,~{}\beta,~{}\gamma$ with $1.32<\alpha<1.33$ and $|\beta|=|\gamma|<1$. Thus, we can take $h(\alpha)=\frac{1}{3}\log\alpha$. Similarly, $h(10)=\log 10$ . Also,

[TABLE]

Thus, we can take $A_{1}:=3\log 10$, $A_{2}:=\log\alpha$ and $A_{3}:=15\log 3$. So, Theorem 3.1 tells us that the left-hand side of (15) is bounded below by

[TABLE]

By comparing the above inequality with the right-hand side of (15) we get that

[TABLE]

4.2 Case 2

We have that

[TABLE]

This is equivalent to

[TABLE]

Thus, we have

[TABLE]

and so

[TABLE]

We divide through (14) by $a(\alpha^{n_{1}}+\alpha^{n_{2}})$ to get

[TABLE]

Thus, we have

[TABLE]

We put

[TABLE]

As before, in order to apply Theorem 3.1 we need to check that $\Lambda_{2}\neq 0$. Suppose that $\Lambda_{2}=0$, then we have

[TABLE]

To see that this is not true, we again consider the $\mathbb{Q}$-automorphism $\sigma$ of the Galois extension $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$ given by $\sigma(\alpha):=\beta$ and $\sigma(\beta):=\alpha$. Now, if $\Lambda_{2}=0$, then $\sigma(\Lambda_{2})=0$. Thus, conjugating the relation (20) under $\sigma$, and taking absolute values on both sides, we get

[TABLE]

which is false for $\ell\geq 2$ and $d\geq 1$. Therefore, $\Lambda_{2}\neq 0$.

So we apply Theorem 3.1 with the data

[TABLE]

Since $\eta_{1},\eta_{2},\eta_{2}\in\mathbb{Q}(\alpha)$, we take the field $\mathbb{K}:=\mathbb{Q}(\alpha)$ with degree $D_{\mathbb{K}}:=3$. Since $\max\{1,\ell,n_{2}\}\leq n_{1}$, we take $B:=n_{1}$. Further,

[TABLE]

Thus, we can take $A_{1}:=3\log 10$, $A_{2}:=\log\alpha$ and $A_{3}:=5.31\times 10^{14}\log n_{1}$. So, Theorem 3.1 tells us that the left-hand side of (19) is bounded below by

[TABLE]

By comparing the above inequality with the right-hand side of (19) we get that

[TABLE]

4.3 Case 3

We have that

[TABLE]

This is equivalent to

[TABLE]

Thus, we have

[TABLE]

and so

[TABLE]

We divide through (14) by $a(\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}})$ to get

[TABLE]

Thus, we have

[TABLE]

We put

[TABLE]

As before, in order to apply Theorem 3.1 we need to check that $\Lambda_{3}\neq 0$. Suppose that $\Lambda_{3}=0$, then we have

[TABLE]

To see that this is not true, we again consider the $\mathbb{Q}$-automorphism $\sigma$ of the Galois extension $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$ given by $\sigma(\alpha):=\beta$ and $\sigma(\beta):=\alpha$. Now, if $\Lambda_{3}=0$, then $\sigma(\Lambda_{3})=0$. Thus, conjugating the relation (24) under $\sigma$, and taking absolute values on both sides, we get

[TABLE]

which is false for $\ell\geq 2$ and $d\geq 1$. Therefore, $\Lambda_{2}\neq 0$.

So we apply Theorem 3.1 with the data

[TABLE]

Since $\eta_{1},\eta_{2},\eta_{2}\in\mathbb{Q}(\alpha)$, we take the field $\mathbb{K}:=\mathbb{Q}(\alpha)$ with degree $D_{\mathbb{K}}:=3$. Since $\max\{1,\ell,n_{3}\}\leq n_{1}$, we take $B:=n_{1}$. Further,

[TABLE]

Thus, we can take $A_{1}:=3\log 10$, $A_{2}:=\log\alpha$ and $A_{3}:=5.16\times 10^{28}(\log n_{1})^{2}$. So, Theorem 3.1 tells us that the left-hand side of (23) is bounded below by

[TABLE]

By comparing the above inequality with the right-hand side of (19) we get that

[TABLE]

Now, we apply Lemma 3 on the above inequality (25) with the data: $r:=3,~{}H:=1.94\times 10^{42},1~{}L:=n_{1}$. We obtain that $n_{1}<2.7\times 10^{48}$. We record what we have proved

Lemma 4

Let $(N,n_{1},n_{2},n_{3},d,\ell)$ be the nonnegative integer solutions to the Diophantine equation (2) with $n_{1}\geq n_{2}\geq n_{3}\geq 0$, $1\leq d\leq 9$ and $\ell\geq 2$. Then we have

[TABLE]

5 Reducing the bounds

The bounds ontained in Lemma 4 are too large to carry out meaningful computations on the computer. Thus, we need to reduce these bounds. To do so, we return to (15), (19) and (23) and apply Lemma 2 via the following procedure.

First, we put

[TABLE]

For technical reasons, we assume that $n_{1}-n_{2}\geq 20$ and go to (15). Note that $e^{\Gamma_{1}}-1=\Lambda_{1}\neq 0$. Thus, $\Gamma_{1}\neq 0$. If $\Gamma_{1}<0$ then

[TABLE]

If $\Gamma_{1}>0$ then we have that $|e^{\Gamma_{1}}-1|<1/2$. Hence $e^{\Gamma_{1}}<2$. Thus, we get that

[TABLE]

Therefore, in both cases, we have that

[TABLE]

Dividing through the above inequality by $\log\alpha$, we get

[TABLE]

If we put

[TABLE]

we can rewrite (26) as

[TABLE]

We now apply Lemma 2 on (27). We put $M:=3\times 10^{48}$. A quick computer search in Mathematica reveals that the convergent

[TABLE]

of $\tau$ is such that $q_{106}>6M$ and $\varepsilon_{d}\geq 0.0129487>0$. Therefore, with $A:=36$ and $B:=\gamma$ we calculated each value of $\log(36q_{106}/\varepsilon_{d})/\log\alpha$ and found that all of them are at most $432$. Thus, we have that $n_{1}-n_{2}\leq 432.$

Next, we put

[TABLE]

For technical reasons, as before we assume that $n_{2}-n_{3}\geq 20$ and go to (19). Note that $e^{\Gamma_{2}}-1=\Lambda_{2}\neq 0$. Thus, $\Gamma_{2}\neq 0$. If $\Gamma_{2}<0$ then

[TABLE]

If $\Gamma_{2}>0$ then we have that $|e^{\Gamma_{2}}-1|<1/2$. Hence $e^{\Gamma_{2}}<2$. Thus, we get that

[TABLE]

Therefore, in both cases, we have that

[TABLE]

Dividing through the above inequality by $\log\alpha$, we get

[TABLE]

We put

[TABLE]

where $k:=n_{1}-n_{2}$. We can rewrite (28) as

[TABLE]

We now apply Lemma 2 on (29). We put $M:=3\times 10^{48}$. A quick computer search in Mathematica reveals that the $106$-th convergent of $\tau$ is such that $q_{106}>6M$ and $\varepsilon_{d,k}\geq 0.000134829>0$ for all $1\leq d\leq 9$ and $1\leq k\leq 432$ except for the case $\varepsilon_{9,11}$, which is always negative. Therefore, with $A:=22$ and $B:=\alpha$ we calculated each value of $\log(22q_{106}/\varepsilon_{d,k})/\log\alpha$ and found that all of them are at most $446$. Thus, we have that $n_{2}-n_{3}\leq 446.$

The problem in the case of $\varepsilon_{9,11}$ is due to the fact that

[TABLE]

Thus, if we consider the identity (30), the inequality (28) becomes

[TABLE]

In this case we apply the classical result from Diophantine approximation given in Lemma 1. We assume that $n_{2}-n_{3}$ is so large that the right-hand side of the inequality in (31) is smaller than $1/(2\ell^{2})$. This certainly holds if

[TABLE]

Since $\ell<n_{1}<3\times 10^{48}$, it follows that the last inequality (32) holds provided that $n_{2}-n_{3}\geq 415$, which we now assume. In this case $r/s:=(n_{2}+9)/\ell$ is a convergent of the continued fraction of $\tau:=\log 10/\log\alpha$ and $\ell<3\times 10^{48}$. We are now set to apply Lemma 1.

We write $\tau:=[a_{0};a_{1},a_{2},a_{3},\ldots]=[8;5,3,3,1,5,1,8,4,6,1,4,1,1,1,9,1,4,4,9,\ldots]$ for the continued fraction of $\tau$ and $p_{k}/q_{k}$ for the $k-$th convergent. We get that $r/s=p_{j}/q_{j}$ for some $j\leq 106$. Furthermore, putting $a(M):=\max\{a_{j}:j=0,1,\ldots,106\}$, we get $a(M):=564$. By Lemma 1, we get

[TABLE]

giving

[TABLE]

leading to $n_{2}-n_{3}\leq 435$. Thus, in both cases we have that $n_{2}-n_{3}\leq 446$.

Lastly, we put

[TABLE]

For technical reasons, as before we assume that $n_{1}\geq 20$ and go to (23). Note that $e^{\Gamma_{3}}-1=\Lambda_{3}\neq 0$. Thus, $\Gamma_{3}\neq 0$. If $\Gamma_{3}<0$ then

[TABLE]

If $\Gamma_{3}>0$ then we have that $|e^{\Gamma_{3}}-1|<1/2$. Hence $e^{\Gamma_{3}}<2$. Thus, we get that

[TABLE]

Therefore, in both cases, we have that

[TABLE]

Dividing through the above inequality by $\log\alpha$, we get

[TABLE]

We put

[TABLE]

where $1\leq k:=n_{1}-n_{3}=(n_{1}-n_{2})+(n_{2}-n_{3})\leq 878$ and $1\leq s:=n_{2}-n_{3}\leq 446$. We can rewrite (33) as

[TABLE]

We now apply Lemma 2 on (34). We put $M:=3\times 10^{48}$. A quick computer search in Mathematica reveals that the $106$-th convergent of $\tau$ is such that $q_{106}>6M$ and $\varepsilon_{d,k,s}\geq 0.000125>0$. Therefore, with $A:=36$ and $B:=\alpha$ we calculated each value of $\log(36q_{106}/\varepsilon_{d,k,s})/\log\alpha$ and found that all of them are at most $485$. Thus, we have that $n_{1}\leq 485.$ This contradicts our assumption that $n_{1}>500$. Hence, Theorem 2.1 holds. ∎

Acknowledgements.

The author was supported by the Austrian Science Fund (FWF) grants: F5510-N26 – Part of the special research program (SFB), “Quasi Monte Carlo Metods: Theory and Applications” and W1230 –“Doctoral Program Discrete Mathematics”.