Lychrel Numbers in Base 10: A Probabilistic Approach

Abstract

For decades, Lychrel numbers have been studied on many bases. Their existence has been proven in base 2, 11 or 17. This paper presents a probabilistic proof of the existence of Lychrel number in base 10 and provides some properties which enable a mathematical extraction of new Lychrel numbers from existing ones. This probabilistic approach has the advantage of being extendable to other bases. The results show that palindromes can also be Lychrel numbers.

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Kuitché, R. (2024) Lychrel Numbers in Base 10: A Probabilistic Approach. Advances in Pure Mathematics, 14, 667-694. doi: 10.4236/apm.2024.148037.

1. Introduction

On the set of natural numbers, the reverse number of a natural number a= a 1 a l( a ) (with l( a ) the number of digits of a), denoted by Rev( a ) , is the natural number such that Rev( a )= a l( a ) a l( a )1 a 1 . When a is equal to its reverse number ( a=Rev( a ) ), we say that a is a palindrome. It is the case with 11, 22, 33, 44, 55, 66, 77, 121 or 985,589. If a natural number a is not a palindrome, one can eventually obtain a palindrome after the iterative process of reversion and addition with its reverse number. The example with a=145 yields 145+Rev( 145 )=145+541=686 which is a palindrome. With a=159 , we have 159+Rev( 159 )=159+951=1110 , and 1110+Rev( 1110 )=1110+0111=1221 . Hence, 159 leads to a palindrome after two iterations. The natural number 89 leads to a palindrome after 24 iterations; 903,282,208 produces a palindrome after 33 iterative reversions and additions. With the same process, a palindrome is obtained after 45 iterative reversions and additions of 90,914,509. However, there are natural numbers that do not seem to form a palindrome after the computation of the iterative process of reversing and adding their reverse numbers. These numbers are usually called Lychrel Numbers. A more detailed definition of Palindromes and Lychrel numbers can be found in [1].

They are often used for the encryption and decryption of messages in cryptography, especially in cryptography on elliptic curves ([2]), and for the search of palindromes which are important in biotechnology, notably in DNA and RNA sequences, as they help to stabilize these types of molecules ([3]).

To the best of our knowledge, the problem of Lychrel numbers was first raised in [4] when the author mentioned the number of natural numbers for which no palindrome is reached after 100 iterations. Since then, the existence of Lychrel numbers has been proved in many bases, among which the bases 2, 11, 17 or 26, and presented in [1].

In base 10, some works have been done to verify whether 196 is a Lychrel number: in 1987, the author from [5] presented an algorithm to verify that 196 is a Lychrel number, and this algorithm ran for many years and stopped after 2416 million iterations, without reaching a palindromic number. In 1995, the author from [6] obtained a non-palindromic number with 2 million digits with a super computer. Jason Doucette took over the research in the years 2000 and reached a number with 12.5 million digits, which is still not a palindrome. In 2006, Wade VanLandingham continued with the computation and reached a non-palindromic number with more than 300 million digits. Romain Dolbeau took over the work and after more than 2.4 billion iterations, he obtained in 2015 a non-palindromic number with a billion digits. The summary of works done on 196 and other Lychrel candidates can be found in [7] and [5]. In 2017, the authors from [8] established that the existence of Lychrel numbers in base 10 is due to the symmetry breaking in the space of the natural numbers. However, their demonstration remains a global one because they were not able to provide proof for a particular natural number.

The probabilistic approach has been used in many domains, notably in the study of the reliability of cementless hip prostheses in the presence of mechanical uncertainties and its application to the investigation of the influence of bone-implant interface properties ([9]). Also, the implementation of non-recursive base conversion is presented in [10] with a probabilistic approach, notably the deterministic Markov process.

In this work, we use a simple probability calculation.

The rest of the paper is organized as follows: in section 2 we present some tools which enable the explanation of how to obtain the length and the expression of an iteration. In section 3 we discuss some blocs of consecutive iterations and present some properties of the iteration function which enable a probabilistic characterization of Lychrel numbers. Section 4 presents some properties for extracting new Lychrel numbers from others. Section 5 addresses the notion of Palindrome-Lychrel numbers and section 6 concludes the paper.

2. Length and Expression of an Iteration

In this section, we define the notion of iteration and provide some tools that help to obtain the length and the expression of an iteration.

Let a= a 1 a l( a ) be a natural number, where l( a ) denotes the number of digits of a and a i a digit of a for all i{ 1,,l( a ) } . An iteration of a is the operation that consists of adding a to Rev( a )= a l( a ) a 1 . To express it conveniently, we define the unity function U which corresponds to its last digit, as follows:

U:NN a a l( a )

Some examples in base 10 are contained in Table 1 below.

Table 1. The unity of some numbers.

a

0

3

25

1268

1000

7960

1997

40,961

20,000

178,997

l( a )

1

1

2

4

4

4

4

5

5

6

U( a )

0

3

5

8

0

0

7

1

0

7

The unity function U verifies some elementary properties:

Lemma 1. Let a0 be a digit of a natural number, and ρ{ 0,1 } .

1) For all x,yN , U( x+y )=U( U( x )+U( y ) ) ;

2) For all xN , if U( x ){ 0,,8 } then U( x+ρ )=U( x )+ρ ;

3) For all i{ 1,,10 } , if a+i10 , then U( a+i )=a10+i .

Proof. The properties are straightforward. ,

Hence, if a0 is a digit of a natural number, then U( a+10 )=a and U( a+101 )=a1 .

Also, we consider the iteration function ϕ , which is actually the one defined in [8], as follows:

ϕ:NN aa+Rev( a )

Then, the first iteration of a will be denoted by ϕ 1 ( a )=ϕ( a ) , the second iteration will be ϕ 2 ( a )=ϕ( ϕ 1 ( a ) )=ϕ( ϕ( a ) ) and for all nN ,

ϕ n ( a )=ϕ( ( ϕ( a ) ) ) (n factors). For example, in base 10, if a=75 , then ϕ 2 ( 75 )=ϕ( ϕ( 75 ) )=ϕ( 75 )+Rev( ϕ( 75 ) )=75+Rev( 75 )+Rev( 75+Rev( 75 ) ) =75+57+Rev( 75+57 )=132+Rev( 132 )=132+231=363 .

Now let’s consider ϕ( a )=a+Rev( a ) . Then the nth iteration of a can be expressed as a function of a as ϕ n ( a )=a+ i=1 n1 Rev( ϕ i ( a ) ) . As for the first iteration, its length is given by l( ϕ( a ) )=l( a )+α , where α{ 0,1 } . α=1 if and only if a 1 + a l( a ) 10 or there exists i{ 1,,l( a ) } such that a i + a l( a )i+1 10 and a j + a l( a )j+1 =9 for all ji1 . For example, in base 10, if a=4006 , then we have l( a )=4 and a 1 + a l( a ) =4+6=1010 . Hence, α=1 and l( ϕ( a ) )=l( a )+1=4+1=5 . In the second example, we consider the case where a=456745 . Then for j=3 , a 3 + a 4 =6+7=1310 and a j + a 7j =9 for all j2 . Hence, α=1 and then l( ϕ( a ) )=l( a )+1=6+1=7 .

In order to successfully add a to its reverse number to obtain the first iteration of a, we define the vector parameter ϵ( a ) (or ϵ 1 ( a ) ) which depends both on the length and the digits of a, as follows:

ϵ j ( a )={ 0 ifj=l( a )+α 1 ifj<l( a )+αand a j + a l( a )j+1 + ϵ j+1 ( a )10 0 else

with ϵ j ( a ) being calculated in the decreasing order.

Then the following are the expressions, depending on α, of the first iteration of a.

If α=1 , then:

  • l( ϕ( a ) )=l( a )+1 ,

  • ϕ ( a ) 1 =1 ,

  • ϕ ( a ) l( ϕ( a ) ) =U( a l( a ) + a 1 ) ,

  • For every j{ 2,,l( a ) } , ϕ ( a ) j =U( a j1 + a l( a )j+2 + ϵ j ( a ) ) .

If α=0 , then:

a) l( ϕ( a ) )=l( a ) ,

b) ϕ ( a ) 1 =U( a 1 + a l( a ) + ϵ 1 ( a ) ) ,

c) ϕ ( a ) l( ϕ( a ) ) =U( a l( a ) + a 1 ) ,

d) For every j{ 2,,l( a ) } , ϕ ( a ) j =U( a j + a l( a )j+1 + ϵ j ( a ) ) .

For example, we consider the case where a=974 , we have l( a )=3 , a 1 + a l( a ) = a 1 + a 3 =9+4=1310 . Therefore l( ϕ( a ) )=l( a )+1=3+1=4 . We then determine the epsilons in the decreasing order as follows: ϵ 4 ( a )=0 , ϵ 3 ( a )=1 because a 3 + a 1 10 , ϵ 2 ( a )=1 because a 2 + a 2 + ϵ 3 ( a )10 and ϵ 1 ( a )=1 because a 1 + a 3 + ϵ 2 ( a )10 . Hence, we can determine the digits of ϕ( a ) as follows:

1) ϕ ( a ) 1 =1 ,

2) ϕ ( a ) l( ϕ( a ) ) =ϕ ( a ) 4 =U( a l( a ) + a 1 )=U( a 3 + a 1 )=U( 4+9 )=U( 13 )=3 ,

3) ϕ ( a ) 3 =U( a 2 + a l( a )1 + ϵ 3 ( a ) )=U( 7+7+ ϵ 3 ( a ) )=U( 14+ ϵ 3 ( a ) ) =U( 14+1 )=U( 15 )=5 ,

4) ϕ ( a ) 2 =U( a 1 + a l( a ) + ϵ 2 ( a ) )=U( 9+4+ ϵ 2 ( a ) )=U( 13+ ϵ 2 ( a ) ) =U( 13+1 )=U( 14 )=4 .

Thus, ϕ( a )=ϕ ( a ) 1 ϕ ( a ) 2 ϕ ( a ) 3 ϕ ( a ) 4 =1453 .

In the more general way, if a and m are two natural numbers with l( a ) and l( m ) their respective lengths. Let’s say that our aim is to make the addition of the two numbers. Then the length of a+m is given by max( l( a ),l( m ) )+α (with α{ 0,1 } ). α=1 if and only if a 1 + b 1 10 or there exists i{ 1,,max( l( a ),l( m ) ) } such that a i + m i 10 and a j + m j =9 for all ji − 1. For example, if a=900 and m=105 , then we have l( a )=l( m )=3 and a 1 + m 1 =9+1=10 . Hence, α=1 and l( a+m )=max( l( a ),l( m ) )+1=l( a )+ 1=3+1=4 . The second example presents the case where a=4545 and m=5456 , then for j=4 , a 4 + m 4 =1110 and a j + m j =9 for all j3 . Hence, α=1 and l( a+m )=max( l( a ),l( m ) )+1=l( a )+1=4+1=5 .

In the above examples, l( a )=l( m ) . If l( a )l( m ) , then we make the following adjustments: if l( m )<l( a ) , then 0 is added l( a )l( m ) times to m and before the digit m1. For example, if a=105 and m=9 , then l( a )l( m )=2 . Hence 0 is added 2 times to m and before 9. m then becomes d=009 with a greater length l( d )=3=l( a ) and a+m=a+d . The same situation occurs with a if l( a )<l( m ) .

We then define the vector parameter σ( a,m ) which depends both on the digits of a and m as follows:

σ j ( a,m )={ 0 ifj=max( l( a ),l( m ) )+α 1 ifj<max( l( a ),l( m ) )+αand a j + m j + σ j+1 ( a,m )10 0 else

The σ j ( a,m ) are also calculated in a decreasing order. The digits of a+m are then obtained as follows:

If α=1 , then:

  • l( a+m )=l( a+d )=l( d )+1 ,

  • ( a+m ) 1 = ( a+d ) 1 =1 ,

  • ( a+m ) l( a+d ) = ( a+d ) l( a+d ) =U( a l( a ) + d l( d ) ) ,

  • For every j{ 2,,l( d )+1 } , ( a+m ) j = ( a+d ) j =U( a j1 + d j1 + σ j ( a,m ) ) .

If α=0 , then:

1) l( a+m )=l( a+d )=l( d ) ,

2) ( a+m ) 1 =U( a 1 + d 1 + σ 1 ( a,m ) ) ,

3) ( a+m ) l( a+m ) = ( a+d ) l( a+d ) =U( a l( a ) + d l( d ) ) ,

4) For every j{ 2,,l( d ) } , ( a+m ) j = ( a+d ) j =U( a j + d j + σ j ( a,m ) ) .

3. A Probabilistic Characterization of Lychrel Numbers

Let a= a 1 a 2 a l( a ) be a natural number. If a is a palindrome, then the couples of digits ( a i , a l( a )i+1 ) of a most be chosen among the set of couples { ( 0,0 ),( 1,1 ),,( 9,9 ) } , with ( a 1 , a l( a ) )( 0,0 ) . Therefore the probability for a to be a palindrome is given by

p( a )={ C 9 1 10 l( a ) 2 1 C 90 1 100 l( a ) 2 1 = ( 1 10 ) l( a ) 2 ifl( a )iseven C 9 1 C 10 1 10 l( a )1 2 1 C 90 1 C 10 1 100 l( a )1 2 1 = ( 1 10 ) l( a )1 2 else

This probability then depends on the length l( a ) of a. The more l( a ) increases, the more this probability decreases.

If a is not a palindrome, then a palindrome can eventually be reached after several consecutive iterations of a. We suppose that 5 consecutive iterations are computed, then several situations can be noticed:

  • The length of a cannot increase by 0 or by 5;

  • The length of a can increase by 1. It is the case with most natural numbers such that a 1 =1 and a l( a ) =0 ;

  • The length of a can increase by 3. It is the case with most natural numbers a verifying ( a 1 , a l( a ) ){ ( 9,8 ),( 8,8 ) } and a 2 + a l( a )1 0 , or such that a 1 = a l( a ) =9 and a 2 + a l( a )1 =3 ;

  • The length of a can increase by 4, like in most of the cases where a 1 = a l( a ) =9 and a 2 = a l( a )1 =0 ;

  • The length of a can increase by 2. It happens with most natural numbers, especially when { a 1 , a l( a ) }{ 8,7 } .

3.1. Analyzing Some Particular Blocks of Consecutive Iterations

For the n first iterations of a ( nN and n5 ), its numerical sequence can be decomposed in several blocks of 5 consecutive iterations. For each block of iterations, we name the starting point or origin of the block, the element (out of the block) that generates the 5 iterations; and the destination, the last iteration of the block. Two blocks of consecutive iterations are said to be consecutive when the starting point of one of them is the destination of the other. For example, if n=13 , we have the numerical sequence a, ϕ 1 ( a ),, ϕ 13 ( a ) . The first block of 5 consecutive iterations is ϕ 1 ( a ), ϕ 2 ( a ), ϕ 3 ( a ), ϕ 4 ( a ), ϕ 5 ( a ) and its starting point or origin is a and its destination is ϕ 5 ( a ) . The second block of 5 consecutive iterations is ϕ 6 ( a ), ϕ 7 ( a ), ϕ 8 ( a ), ϕ 9 ( a ), ϕ 10 ( a ) and its starting point or origin is ϕ 5 ( a ) and its destination is ϕ 10 ( a ) . ϕ 11 ( a ), ϕ 12 ( a ), ϕ 13 ( a ) is the residual block with less than 5 iterations. The 2 blocks of 5 consecutive iterations are consecutive because ϕ 5 ( a ) which is the destination of the first block, is also the starting point of the second block.

Now assume that n ( n5 ) iterations are computed from a. We split the n iterations into blocks of 5 consecutive iterations. We set A n ( a ) the set of blocks for which the length of the starting point increases by 4; B n ( a ) the set of blocks for which the length of the origin increases by 3; C n ( a ) the set of blocks for which the length of the starting point increases by 2 and D n ( a ) the set of blocks for which the length of the origin increases by 1. Then we have B n ( a )= B 1n ( a ) B 2n ( a ) B 3n ( a ) , where B 3n ( a ) is the set of blocks from B n ( a ) such that their destinations generate their next blocks of 5 consecutive iterations also from B n ( a ) , B 2n ( a ) is the set of blocks from B n ( a ) such that their destination produces their next blocks of 5 consecutive iterations from C n ( a ) and B 1n ( a ) is the set of blocks from B n ( a ) such that their destination generate their next blocks of 5 consecutive iterations from D n ( a ) .

We find the probability for a natural number m to be the starting point of the blocks from A n ( a ) , B 2n ( a ) and B 3n ( a ) . To get there, we consider each natural number m= m 1 m 2 m l( m ) as a set of l( m ) 2 couples of digits ( m i , m l( m )i+1 ) if l( m ) is even, and as a set of l( m )1 2 couples of digits ( m i , m l( m )i+1 ) and a middle digit in { 0,1,,9 } if l( m ) is odd. The total number of possible couples of digits is 100 because each digits belongs to { 0,1,,9 } . Moreover, for each number, the first couple ( m 1 , m l( m ) ) must not belong to the set { ( 0,0 ),( 0,1 ),( 0,2 ),( 0,3 ),( 0,4 ),( 0,5 ),( 0,6 ),( 0,7 ),( 0,8 ),( 0,9 ) } . Hence, the first couple must be chosen among the total of 90 cases instead of 100. If l( m ) is odd, the digit in the middle m l( m )1 2 +1 must be chosen among { 0,1,,9 } , leading us to the total number of 10 possible cases. Now we study the different situations.

Case of A n ( a )

For m to be the origin of a block from A n ( a ) , it is necessary in the very large majority of cases that its digits verify m 1 = m l( m ) =9 and m 2 = m l( m )1 =0 . Hence, ϕ( m )=1U( 8+ ϵ 1 1 ( m ) ) ϵ 2 1 ( m )BU( m 3 + m l( m )2 )18 where B is the rest of digits of ϕ( m ) .

It implies that ϕ( m )=18 ϵ 2 1 ( m )BU( m 3 + m l( m )2 )18 because ϵ 1 1 ( m )=0 .

ϕ 2 ( m )=U( 9+ ϵ 1 2 ( m ) )U( 9+ ϵ 2 2 ( m ) )U( ϵ 2 1 ( m )+U( m 3 + m l( m )2 )+ ϵ 3 2 ( m ) ) CU( 1+U( m 3 + m l( m )2 )+ ϵ l( m )2 2 ( m ) )99

where C is the rest of digits of ϕ 2 ( m ) . We impose that ϵ 1 2 ( m )=0 . That is the case if ϵ 2 2 ( m )=0 , which implies that ϵ 2 1 ( m )+U( m 3 + m l( m )2 )+ ϵ 3 2 ( m )9 . For that to be possible, it requires at least that U( m 3 + m l( m )2 )7 or U( m 3 + m l( m )2 )=8 and ( ϵ 3 2 ( m ), ϵ 3 2 ( m ) ){ ( 0,1 ),( 1,0 ) } or U( m 3 + a l( m )2 )=9 and ( ϵ 3 2 ( m ), ϵ 3 2 ( m ) )=( 0,0 ) . The last two cases are minor cases and depend on others digits m 4 , m 5 , of m. Hence, for simplification, we consider only the case U( m 3 + m l( m )2 )7 . That leads us to 0 m 3 + m l( m )2 7 or 10 m 3 + m l( m )2 17 . It gives us 80 possible couples ( m 3 , m l( m )2 ) . The probability of m to be the starting point in a block of iterations from A n ( a ) is then given by

P( A n ( a ) )( m ) C 1 1 C 1 1 C 80 1 100 l( m ) 2 3 C 90 1 100 l( m ) 2 1 = 80 900000 = 8 90000 if l( m ) is even and P( A n ( a ) )( m ) C 1 1 C 1 1 C 80 1 C 10 1 100 l( m )1 2 4 C 90 1 C 10 1 100 l( m )1 2 2 = 80 900000 = 8 90000 if l( m ) is odd.

Case of B 3n ( a )

For m to be the starting point of a block from B 3n ( a ) , it is necessary in the very large majority of cases that its digits verify m 1 = m l( m ) =9 and m 2 + m l( m )1 =3 and U( m 3 + m l( m )2 )0 .

Hence, the length of m increases by 3 after 3 iterations and also by 3 after 5 iterations. Since m 2 + m l( m )1 =3 and U( m 3 + m l( m )2 )0 , the first and last digits of the 7th iteration will also be equal to 9. Hence, from the 6th to the 10th iterations, the length also increases by 3. Therefore we obtain that for m to be the starting point of the blocks in B 3n , it is necessary at least that ( m 1 , m l( m ) )=( 9,9 ) and m 2 + m l( m )1 =3 . The number of possible couples ( m 2 , m l( m )1 ) such that m 2 + m l( m )1 =3 is 4. The probability of m to be the starting point in a block of iterations from B 3n ( a ) is then given by

P( B 3n ( a ) )( m ) C 1 1 C 4 1 100 l( m ) 2 2 C 90 1 100 l( m ) 2 1 = 4 9000 if l( m ) is even and

P( B 3n ( a ) )( m ) C 1 1 C 4 1 C 10 1 100 l( m )1 2 3 C 90 1 C 10 1 100 l( m )1 2 2 = 4 9000 if l( m ) is odd.

Case of B 2n ( a )

For m to be the starting point of a block of iterations from B 2n ( a ) , it requires at least that ( m 1 , m l( m ) ){ ( 9,8 ),( 8,8 ) } . Moreover, If ( m 1 , m l( m ) )=( 9,8 ) , then in the very large majority of cases, ( m 2 , m l( m )1 ) has to verify

m 2 + m l( m )1 N( [ 3,9 ][ 12,18 ] ) , which gives us a total of 67 possible cases for couples ( m 2 , m l( m )1 ) . If ( m 1 , m l( m ) )=( 8,8 ) , then in the very large majority of cases, ( m 2 , m l( m )1 ) has to verify m 2 + m l( m )1 N( [ 4,9 ][ 13,18 ] ) , which gives us a total of 56 possible cases for couples ( m 2 , m l( m )1 ) . The probability of m to be the starting point of a block of iterations from B 2n ( a ) is then givenby

P( B 2n ( a ) )( m ) C 1 1 C 67 1 100 l( m ) 2 2 C 90 1 100 l( m ) 2 1 + C 1 1 C 56 1 100 l( m ) 2 2 C 90 1 100 l( m ) 2 1 = 123 9000 if l( m ) is even and P( B 2n ( a ) )( m ) C 1 1 C 67 1 C 10 1 100 l( m )1 2 3 C 90 1 C 10 1 100 l( m )1 2 2 + C 1 1 C 56 1 C 10 1 100 l( m )1 2 3 C 90 1 C 10 1 100 l( m )1 2 2 = 123 9000 if l( m ) is odd.

3.2. Some Properties of the Iteration Function

The blocks of consecutive iterations studied in the above subsection lead us to some properties of the iteration function. These properties relate the progression of the length of iterations to the number of iterations, and present the link with the detection of Lychrel numbers.

Proposition 2. Let a= a 1 a 2 a l( a )1 a l( a ) N be a natural number in base 10 such that l( a )3 . For all n1 , we denote by ϕ n ( a ) the nth iteration of a

lim n+ l( ϕ n ( a ) ) n =κ0.414

Proof. Let consider a= a 1 a 2 a l( a )1 a l( a ) N . Then after the first iteration, we have l( ϕ( a ) )=l( a )+ α 1 ( a ) , where α 1 ( a ){ 0,1 } ( α 1 ( a )=1 if the length of a increases and α 1 ( a )=0 if not). After the second iteration, l( ϕ 2 ( a ) )= l( ϕ( a ) )+ α 2 ( a ) (with α 2 ( a ){ 0,1 } ), which implies that l( ϕ 2 ( a ) )=l( a )+ α 1 ( a )+ α 2 ( a ) . Hence, after n iterations, l( ϕ n ( a ) )=l( a )+ i=1 n α i ( a ) , where α i ( a ){ 0,1 } for all i{ 1,,n } .

Notice that after 5 consecutive iterations, we have 1 i=1 5 α i ( a ) 4 . Hence we split the n iterations into blocks of 5 consecutive iterations. Then we will have E( n 5 ) blocks of 5 consecutive iterations, where E( n ) stands for the floor of n. Setting

A n ( a )={ k{ 1,,E( n 5 ) }: i k =1 5 α i k ( a ) =4 } ;

B n ( a )={ k{ 1,,E( n 5 ) }: i k =1 5 α i k ( a ) =3 } ;

C n ( a )={ k{ 1,,E( n 5 ) }: i k =1 5 α i k ( a ) =2 } ;

D n ( a )={ k{ 1,,E( n 5 ) }: i k =1 5 α i k ( a ) =1 } .

Then, we have

| A n ( a ) |+| B n ( a ) |+| C n ( a ) |+| D n ( a ) |=E( n 5 ) (1)

Moreover, any block of 5 iterations from B n ( a ) produces the next block of 5 iterations either from B n ( a ) , or from C n ( a ) or D n ( a ) . Then we have | B n ( a ) |=| B 1n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) | where B 3n ( a ) is the set of blocks that yield blocks from B n ( a ) , B 2n ( a ) the set of blocks that yield blocks from C n ( a ) and B 1n ( a ) the set of blocks that produce blocks from D n ( a ) . Also, any block from A n ( a ) produces the next block of 5 iterations from D n ( a ) . Hence,

| D n ( a ) |=| A n ( a ) |+| B 1n ( a ) | (2)

moving further, we have

i=1 n α i ( a ) =4| A n ( a ) |+3| B n ( a ) |+2| C n ( a ) |+| D n ( a ) |+ β n ( a ), (3)

with β n ( a ) the cardinality of a possible block of less than 5 iterations β n ( a )4 .

From (1), we get | C n ( a ) |=E( n 5 )| A n ( a ) || B n ( a ) || D n ( a ) | . Hence,

l( ϕ n ( a ) )=l( a )+ i=1 n α i ( a ) =l( a )+4| A n ( a ) |+3| B n ( a ) |+2| C n ( a ) |+| D n ( a ) |+ β n ( a ) =l( a )+4| A n ( a ) |+3| B n ( a ) |+2( E( n 5 )| A n ( a ) || B n ( a ) || D n ( a ) | ) +| D n ( a ) |+ β n ( a )

=2E( n 5 )+l( a )+2| A n ( a ) |+| B n ( a ) || D n ( a ) |+ β n ( a ) =2E( n 5 )+l( a )+2| A n ( a ) |+| B n ( a ) |( | A n ( a ) |+| B 1 n( a ) | )+ β n ( a ) =2E( n 5 )+l( a )+| A n ( a ) |+| B n ( a ) |+| B 1n ( a ) |+ β n ( a ) =2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |+ β n ( a ) 2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |+4.

In the same way, we have

l( ϕ n ( a ) )=l( a )+ i=1 n α i ( a ) =l( a )+4| A n ( a ) |+3| B n ( a ) |+2| C n ( a ) |+| D n ( a ) |+ β n =l( a )+4| A n ( a ) |+3| B n ( a ) |+2( E( n 5 )| A n ( a ) || B n ( a ) || D n ( a ) | ) +| D n ( a ) |+ β n ( a ) =2E( n 5 )+l( a )+2| A n ( a ) |+| B n ( a ) || D n ( a ) |+ β n ( a ) =2E( n 5 )+l( a )+2| A n ( a ) |+| B n ( a ) |( | A n ( a ) |+| B 1n ( a ) | )+ β n ( a ) =2E( n 5 )+l( a )+| A n ( a ) |+| B n ( a ) || B 1n ( a ) |+ β n ( a ) =2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |+ β n ( a ) 2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |since β n ( a )0.

Hence,

2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |l( ϕ n ( a ) ) 2E( n 5 )+l( a )+| A n ( a ) |+| B 2n ( a ) |+| B 3n ( a ) |+4

Which implies that

2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n l( ϕ n ( a ) ) n 2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n + 4 n

Hence, from relation (1), the numerical sequences | A n ( a ) | n , | B 2n ( a ) | n and | B 3n ( a ) | n are convergent. Therefore,

2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n and

2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n + 4 n are convergent with the same limit and then from the gendarme theorem, we have

lim n+ l( ϕ n ( a ) ) n = 2 5 + lim n+ ( | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n )=κ

We approximate lim | A n ( a ) | n by the probability of a given natural number m to be the starting point of a block of iterations from A n ( a ) , which is 8 90000 We approximate lim | B 3n ( a ) | n by the probability of a given natural number m to be the starting point of a block of iterations from B 3n ( a ) , which is 4 9000 . Also, lim | B 2n ( a ) | n is approximated by the probability of a given natural number m to be the starting point of a block of iterations from B 2n ( a ) , which is 123 9000 .

Hence,

κ= 2 5 +lim | A n ( a ) | n +lim | B 2n ( a ) | n +lim | B 3n ( a ) | n 2 5 + 8 90000 + 123 9000 + 4 9000 0.414 .

,

We now provide a link between the above result and Lychrel numbers.

Let a be an element of N such that l( a )3 . Then we consider the map defined as follow:

ω a : N * N n{ l( ϕ n ( a ) ) n if ϕ i ( a )2 ϕ i1 ( a )foralli{ 1,2,,n } 0else

Assume that a is a Lychrel number. Then ω a ( n )= l( ϕ n ( a ) ) n for all n N * .

Hence,

lim n+ ω a ( n )= lim n+ l( ϕ n ( a ) ) n =κ0.414

Conversely, suppose that

lim n+ ω a ( n )=κ0.414

Then for all ϵ , there exists n ϵ N such that | ω a ( n )κ |<ϵ for all n> n ϵ . For 0< ϵ 0 <κ , there exists n ϵ 0 N such that | ω a ( n )κ |< ϵ 0 for all n> n ϵ 0 . Assume there is n 0 > n ϵ 0 such that ϕ n 0 ( a )=2 ϕ n 0 1 ( a ) . Then n 0 +1> n 0 > n ϵ 0 and then ω a ( n 0 +1 )=0 , which implies that

κ=| 0κ |=| ω a ( n 0 +1 )κ |< ϵ 0 , which is absurd because ϵ 0 <κ . Hence, ϕ n ( a )2 ϕ n1 ( a ) for all n> n ϵ 0 . Moreover, ω a ( n ϵ 0 +1 )= l( ϕ n ϵ 0 +1 ( a ) ) n ϵ 0 +1 , which implies that for all p< n ϵ 0 +1 , ϕ p ( a )2 ϕ p1 ( a ) , and then that for all p n ϵ 0 , ϕ p ( a )2 ϕ p1 ( a ) . It thus comes that ϕ n ( a )2 ϕ n1 ( a ) for all n N * .

That leads us to the following result:

Theorem 3. Let a be a natural number such that l( a )3 . Then the following assertions are equivalent:

1) a is a Lychrel number

2) lim n+ ω a ( n )=κ0.414

3.3. A Probabilistic Detection of Lychrel Numbers

The last result of the above subsection does not enable us to point a specific natural number as a Lychrel number, since the evaluation of ω a ( n ) any given aN remains difficult. In order to solve this particular problem, we propose a probabilistic definition of Lychrel numbers and associate to this result a probability that enables the detection of Lychrel numbers.

Hence, a natural number a will be qualified to be a Lychrel number when its probability P( a ) to reach the first palindrome after several iterations is less than ( 1 10 ) 330 .

The choice of ( 1 10 ) 330 is explained by the principle that for a to be a lychrel number, P( a ) should be equal to 0. But computationally, ( 1 10 ) 330 is equivalent

to 0 (in many softwares like Excel, R, Anaconda,…). Another reason is that this probability is small enough to give the assurance that ω a ( n ) is close to κ0.414 , which gives us sufficient power to consider a as a Lychrel number

according to theorem 3.

It means that for a natural number a to be a Lychrel number, the minimum number of iterations n 0 to be computed must be as enormous as possible for ω a ( n 0 ) to be closed to κ , let say | ω a ( n 0 )κ |< ϵ 0 = 10 5 .

Having established that

2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n l( ϕ n ( a ) ) n 2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n + 4 n for all nN , let κ 0 be the limit of | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n . Then there exists r 0 N such that for all n> r 0 , | | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n κ 0 |< ϵ 0 . It implies that for all n> r 0 , κ 0 ϵ 0 < | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n . We then conclude that 2 5 E( n 5 ) n 5 + l( a ) n + κ 0 ϵ 0 < 2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n for all n> r 0 .

Also, there exists r 1 N such that for all n> r 1 , | l( ϕ n ( a ) ) n κ |< ϵ 0 , which implies that l( ϕ n ( a ) ) n <κ+ ϵ 0 for all n> r 1 .

Setting r 2 =max{ r 0 , r 1 } , we have

2 5 E( n 5 ) n 5 + l( a ) n + κ 0 ϵ 0 < 2 5 E( n 5 ) n 5 + l( a ) n + | A n ( a ) | n + | B 2n ( a ) | n + | B 3n ( a ) | n l( ϕ n ( a ) ) n <κ+ ϵ 0

For all n> r 2 , which implies that

2 5 E( n 5 ) n 5 + l( a ) n + κ 0 ϵ 0 <κ+ ϵ 0 for all n> r 2 .

Taking into consideration the fact that E( n 5 ) n 5 E( n 5 )+1 and κ= κ 0 + 2 5 , we conclude that l( a )2 2 ϵ 0 <n .

Hence, l( a )2 2 ϵ 0 <n and then for l( ϕ n 0 ( a ) ) n 0 to be near κ , at least n 0 iterations has to be produced without reaching a palindrome.

For all nN , let P n ( a ) be the probability of reaching the first palindrome at the nth iteration.

From an iteration ϕ i1 ( a ) ( 2i n 0 +1 ) to ϕ i ( a ) , there are two possibilities: ϕ i ( a ) is a palindrome or ϕ i ( a ) is not a palindrome. We set A i the event which makes ϕ i ( a ) a palindrome and B i the event for which ϕ i ( a ) is not a palindrome. Also, we set q i ( a )=p( A i ) the probability for ϕ i ( a ) to be a palindrome and p i ( a )=p( B i )=1 q i ( a ) the probability for ϕ i ( a ) not to be a palindrome.

Then P n 0 +1 ( a )=p( A n 0 +1 B 1 B 2 B 3 B 4 B n 0 )p( A n 0 +1 ) . Hence, 0 P n 0 +1 ( a )p( A n 0 +1 ) .

However

p( A n 0 +1 )={ ( 1 10 ) l( ϕ n 0 +1 ( a ) ) 2 ifl( ϕ n 0 +1 ( a ) )iseven ( 1 10 ) l( ϕ n 0 +1 ( a ) )1 2 else

For, u= l( a )2 2 ϵ 0 , l( ϕ u ( a ) )=E( κu )E( 0.414u )=E( 0.414 l( a )2 2 ϵ 0 ) .

Hence,

if l( ϕ u ( a ) ) is even, then

P u ( a ) ( 1 10 ) E( 0.414 l( a )2 2 ϵ 0 ) 2 ( 1 10 ) E( 0.414 l( a )2 2 ϵ 0 )1 2

if l( ϕ u ( a ) ) is odd, then

P u ( a ) ( 1 10 ) E( 0.414 l( a )2 2 ϵ 0 )1 2

Therefore,

P( a ) P n 0 +1 ( a ) P u ( a ) ( 1 10 ) E( 0.414 l( a )2 2 ϵ 0 )1 2 ,

We then have the following result:

Theorem 4. Let a= a 1 a 2 a l( a )1 a l( a ) N be a natural number such that l( a )3 , n 0 the minimum number of iterations to be computed without the reach of a palindrome and such that | ω a ( n 0 )κ |< ϵ 0 ( κ0.414 , 0< ϵ 0 <1 ), P( a ) the probability of reaching de first palindrome and for all nN , P n ( a ) the probability to reach the first palindrome at the nth iteration of a.

Then

1) l( a )2 2 ϵ 0 < n 0 ;

2) P( a ) P n 0 +1 ( a ) P u ( a ) ( 1 10 ) E( 0.414u )1 2 , with u= l( a )2 2 ϵ 0 .

Now we test the above properties on 196, 879, 1997 and 7059.

For the test to be carried out successfully, we have written a code in Anaconda Navigator software, which takes a natural number a, executes the iterative process of reversing and adding the reverse number, returns the last valueW of l( ϕ n ( a ) ) n , C of | l( ϕ n ( a ) ) n 0.414 | and P of p( A n+1 ) in both cases where a is not a Lychrel number and where a is a Lychrel number. The code can be seen in Figure 1 below:

Figure 1. A programme for testing Lychrel numbers.

The code is globally conceived in a R.A.I (Reverse, Add and Iterate) process, and is therefore described in three mean steps: firstly, a natural input number a is taken and reversed. Secondly, the reverse number obtained (reverse-num) is added to the input and the first iteration (nu) is obtained. These two steps are described from line 1 to line 20 of the above code. Thirdly, the process described in the first and the second steps is iterated. At the end, P, C and W are returned.

If P is less than ( 1 10 ) 330 , then the number is concluded to be a Lychrel number. The results of the test are presented in the following Table 2.

Table 2. Lychrel number test on some natural numbers.

a

196

879

1997

7059

16,909,736,969,870,700,090,800

l( a )

3

3

4

4

23

ω a ( n )

0.41409

0.41429

0.41464

0.41429

0

| ω a ( n )0.414 |

9.69 × 10−05

0.00029

0.00064

0.00029

0.414

p( A n+1 )

0

0

0

0

2.51 × 1028

The 0s in the last line of the table above represent the values of p( A n+1 ) which are inferior to ( 1 10 ) 330 . Hence, From the result in this table, p( A n+1 ) is less than ( 1 10 ) 330 for 196, 879, 1997 and 7059. Therefore, P( a )< P n+1 ( a )< P u ( a )<p( A n+1 )< ( 1 10 ) 330 . We conclude that 196, 879, 1997 and 7059 are Lychrel numbers according to the probabilistic definition and to theorem 4. However 16,909,736,969,870,700,090,800, which comes from [11], is not a Lychrel number since 2.51× 10 28 ( 1 10 ) 330 .

4. Extracting New Lychrel Numbers from Others

Having presented a (probabilistic) characterization of Lychrel numbers, we now answer the question of how to extract new Lychrel numbers from others. In this section, we especially extend our results to bases b such that ( 3b10 ).

Let l be a natural number and N l the set of all natural numbers of length l in base b ( 3b10 ). Then we define on N l the binary relation R l as follows: ( a,m ) R l if and only if for all i{ 1,,l } , a i + a li+1 = m i + m li+1 . Then R l is reflexive, symmetric and transitive and is then an equivalence relation on the set N l . In other words, ( a,m ) R l if and only if ϕ( a )=ϕ( m ) . The following properties help us to search, for a given number a, the number m verifying ϕ( a )=ϕ( m ) under some conditions.

4.1. Some Properties of Construction of New Lychrel Numbers

The following results hold:

Proposition 5. Let a= a 1 a 2 a l( a )1 a l( a ) N be a natural number in base b such that 3b10 and l( a )2 .

1) If 1 a 1 b2 and a l( a ) >0 , then ϕ( a )=ϕ( a+ 10 l( a )1 1 ) ;

2) If l( a ) is even, 1 a l( a ) 2 b2 and a l( a ) 2 +1 >0 , then

ϕ( a )=ϕ( a+( b1 ) 10 l( a ) 2 1 ) ;

3) If l( a ) is even, ( a 1 , a l( a ) )( b1,b1 ) , a 1 + a l( a ) 10 ,

a l( a ) 2 1 + a l( a ) 2 +2 10 , 1 a l( a ) 2 b2 and a l( a ) 2 +1 =0 , then ϕ 2 ( a )= ϕ 2 ( a+( b1 ) 10 l( a ) 2 1 ) ;

4) If l( a )5 , l( a ) is odd, a 1 b1 , a 2 b1 and a l( a )1 0 , then ϕ( a )=ϕ( a+ k=1 l( a )3 ( b1 ) 10 k ) .

Proof. Let consider such number a= a 1 a 2 a l( a )1 a l( a ) N .

1) We set x=a+( 10 l( a )1 1 ) . Then l( x )=l( a ) because a 1 b2 and l( 10 l( a )1 1 )=l( a )1 . Moreover, setting m= 10 l( a )1 1 , l( m )=l( a )1 . We then consider d the number obtained by adding 0 before the first digit of m. Then l( a )=l( d ) and a+m=a+d . Hence, σ l( x ) ( a,d )=0 by definition and since in base b, all the digits of d are equal to b1 except the first one and a l( a ) >0 , we have σ i ( a,d )= 1 { a i+1 +101+ σ i+1 ( a,d )10 } =1 for all i{ l( a )1,,1 } (the σ i ( a,d ) are calculated in a decreasing order). Hence, the digits of x are then given by x 1 =U( a 1 +0+ σ 1 ( a,d ) )=U( a 1 +1 )= a 1 +1 (since a 1 b2 ), x l( x ) = x l( a ) =U( a l( a ) +101 )= a l( a ) 1 (by lemma 1), and for all i{ 2,,l( a )1 } , x i =U( a i +101+ σ i ( a,d ) )=U( a i +101+1 )=U( a i +10 ) = a i (also by lemma 1). Hence, whether l( ϕ( x ) )=l( ϕ( a ) )=l( a )=l( x ) or l( ϕ( x ) )=l( ϕ( a ) )=l( a )+1=l( x )+1 , we have ϕ ( x ) i =ϕ ( a ) i for all i{ 1,,l( x ) } .

2) Let consider such number a= a 1 a 2 a l( a )1 a l( a ) N . We set

x=a+( b1 ) 10 l( a ) 2 1 . Then l( x )=l( a ) because a l( a ) 2 b2 and l( ( b1 ) 10 l( a ) 2 1 )= l( a ) 2 . Moreover, setting m=( b1 ) 10 l( a ) 2 1 ,

l( m )= l( a ) 2 N since l( a ) is even. We then consider d the number obtained by adding 0 to m l( a ) 2 times and just before the first digit of m. Then l( a )=l( d ) and a+m=a+d . The digits of d are all equal to 0, except the one in the position l( a ) 2 +1 which is equal to b1 . Hence, we have σ l( x ) ( a,d )=0 , σ i ( a,d )= 1 { a i+1 +0+ σ i+1 ( a,d )10 } =0 for all i{ l( a )1,, l( a ) 2 +1 } (Note that the σ i ( a,d ) are calculated in a decreasing order) and since a l( a ) 2 +1 >0 , σ l( a ) 2 ( a,d )=1 . Also, for all i{ l( a ) 2 1,,1 } , σ i ( a,d )= 1 { a i+1 +0+ σ i+1 ( a,d )10 } =0 , since a l( a ) 2 b2 . Hence, the digits of x are then expressed as the function of the digits of a as follows: for all i{ 1,, l( a ) 2 1 } , x i =U( a i +0+ σ i ( a,d ) )=U( a i +0 )= a i .

x l( a ) 2 =U( a l( a ) 2 +0+ σ l( a ) 2 ( a,d ) )=U( a l( a ) 2 +1 )= a l( a ) 2 +1 because 1 a l( a ) 2 b2 . x l( a ) 2 +1 =U( a l( a ) 2 +1 +101+ σ l( a ) 2 +1 ( a,d ) )=U( a l( a ) 2 +1 +101 )+ σ l( a ) 2 +1 ( a,d ) (because of lemma 1) = a l( a ) 2 +1 1+ σ l( a ) 2 +1 ( a,d ) (also by lemma 1) = a l( a ) 2 +1 1 . for all i{ l( a ) 2 +2,,l( a ) } ,

x i =U( a i +0+ σ i ( a,d ) )=( a i +0 )= a i . its comes that i{ 1,,l( x ) } , x i + x l( x )i+1 = a i + a l( a )i+1 . We then conclude that l( ϕ( x ) )=l( ϕ( a ) ) , and that ϕ ( x ) i =ϕ ( a ) i for all i{ 1,,l( a ) } .

3) Let consider such number a= a 1 a 2 a l( a ) N . We set x=a+( b1 ) 10 l( a ) 2 1 . Then l( x )=l( a ) because 1 a l( a ) 2 b2 and l( ( b1 ) 10 l( a ) 2 1 )= l( a ) 2 . Moreover, setting m=( b1 ) 10 l( a ) 2 1 , l( m )= l( a ) 2 N since l( a ) is even. We then consider d the number obtained by adding 0 to m l( a ) 2 times and just before the first digit of m. Then l( a )=l( d ) and a+m=a+d . The digits of d are all equal to 0, except the one in position l( a ) 2 +1 which is equal to b1 . Hence, we have σ l( x ) ( a,d )=0 and σ i ( a,d )=0 for all i{ l( a )1,,1 }\{ l( a ) 2 } (Note that the σ i ( a,d ) are calculated in a decreasing order) and since a l( a ) 2 +1 =0 , we also have σ l( a ) 2 ( a,d )=0 . We then conclude that for all i{ 1,,l( x ) }\{ l( a ) 2 +1 } , x i =U( a i +0+ σ i ( a,d ) )=U( a i +0 )= a i . Moreover,

x l( a ) 2 +1 =U( a l( a ) 2 +1 +101+ σ l( a ) 2 +1 ( a,d ) )=U( 0+101+0 )=101 . Now we search for the digits of ϕ( x ) . In fact, ϕ( x )=ϕ ( x ) 1 ϕ ( x ) 2 ϕ ( x ) l( ϕ( x ) ) . Since x 1 + x l( x ) = a 1 + a l( a ) 10 , then l( ϕ( x ) )=l( x )+1=l( a )+1 . Hence,

ϕ ( x ) 1 =1=ϕ ( a ) 1 and ϕ ( x ) l( ϕ( x ) ) =U( x l( x ) + x 1 )=U( a l( a ) + a 1 )=ϕ ( a ) l( ϕ( a ) ) . Also, for all i{ 2,, l( a ) 2 1 } , ϕ ( x ) i =ϕ ( a ) i , ϕ ( x ) l( x ) 2 =ϕ ( a ) l( a ) 2 +1 , (using lemma 1) ϕ ( x ) l( x ) 2 +1 =ϕ ( a ) l( a ) 2 +1 , ϕ ( x ) l( x ) 2 +2 =ϕ ( a ) l( a ) 2 +2 1 (also because of lemma 1) and for all i{ l( a ) 2 +3,,l( a )+1 } , ϕ ( x ) i =ϕ ( a ) i . After finding the digits of ϕ( x ) , it is quite clear that ϕ( x )ϕ( a ) . As for the digits of ϕ 2 ( x ) , it can be seen that ϕ 2 ( x ) i = ϕ 2 ( a ) i for all i{ 1,,l( x )+1 } .

4) Let suppose l( a ) is odd. We consider the natural number

m= k=1 l( a )3 ( b1 ) 10 k . For example, if b=10 and l( a )=5 ,

m= k=1 53 9 10 k = k=1 2 9 10 k =9 10 1 +9 10 2 =990 . Then by construction, l( m )=l( a )2 and all de digits of c are all equal to 101 , except the last digit which is equal to 0. Moreover, considering d the number obtained by adding 0 to m twice and just before the first digit of m, we have l( a )=l( d ) and a+m=a+d . The digits of d are all equal to b1 , except the first, the second and the last which are equal to 0. We suppose that a l( a )1 0 . Then setting x=a+d , l( x )=l( a ) if and only if a 1 b1 and a 2 b1 . Hence, we have σ l( a ) ( a,d )=0 by definition of σ , and σ l( a )1 ( a,d )=0 because a l( a ) +0b1 . Also, σ 1 ( a,d )=0 because a 2 b1 , and for all i{ 2,,l( a )2 } , σ i ( a,d )=1 since a i+1 +101+ σ i+1 ( a,d )10 (we remind that the the σ i ( a,d ) are determined in the decreasing order). From the σ i ( a,d ) , the digits of x are now given by x 1 =U( a 1 + σ 1 ( a,d ) )= a 1 since a 1 b1 and σ 1 ( a,d )=0 . x 2 =U( a 2 + σ 2 ( a,d ) )=U( a 2 +1 )= a 2 +1 since a 2 b1 . Moreover, for all i{ 3,,l( a )2 } , x i =U( a i +b1+ σ i ( a,d ) )=U( a i +b1+1 )=U( a i +b )=U( a i +10 )= a i (by lemma 1). Also,

x l( a )1 =U( a l( a )1 +b1+ σ l( a )1 ( a,d ) )=U( a l( a )1 +b1 )=U( a l( a )1 +101 )= a l( a )1 1 (also by lemma 1) and x l( a ) =U( a l( a ) +0+ σ l( a ) ( a,d ) )=U( a l( a ) +0+0 )= a l( a ) . Hence, whether l( ϕ( x ) )=l( ϕ( a ) )=l( a ) or l( ϕ( x ) )=l( ϕ( a ) )=l( a )+1 , ϕ ( x ) i =ϕ ( a ) i for all i{ 1,,l( ϕ( a ) ) } . ,

Corollary 1. Let a= a 1 a 2 a l( a )1 a l( a ) N be a Lychrel number in base b such that 3b10 and l( a )2 .

i) If 1 a 1 b2 and a l( a ) >0 , then a+ 10 l( a )1 1 is a Lychrel number;

ii) If l( a ) is even, 1 a l( a ) 2 b2 and a l( a ) 2 +1 >0 , then a+( b1 ) 10 l( a ) 2 1 is a Lychrel number;

iii) If l( a ) is even, ( a 1 , a l( a ) )( b1,b1 ) , a 1 + a l( a ) 10 , a l( a ) 2 1 + a l( a ) 2 +2 10 , 1 a l( a ) 2 b2 and a l( a ) 2 +1 =0 , then a+( b1 ) 10 l( a ) 2 1 is a Lychrel number;

iv) If l( a )5 , l( a ) is odd, a 1 b1 , a 2 b1 and a l( a )1 0 , then a+ k=1 l( a )3 ( b1 ) 10 k is a Lychrel number.

Proof. Let a= a 1 a 2 a l( a )1 a l( a ) N be such Lychrel number. i) Since a 1 b2 and a l( a ) >0 , then ϕ( a )=ϕ( a+ 10 l( a )1 1 ) , from i) of proposition 5. Hence, for all nN , ϕ n ( a )= ϕ n ( a+ 10 l( a )1 1 ) . Since a is a Lychrel number, for all nN , ϕ n+1 ( a )2 ϕ n ( a ) , which implies that

ϕ n+1 ( a+ 10 l( a )1 1 )2 ϕ n ( a+ 10 l( a )1 1 ) for all nN . Hence, a+ 10 l( a )1 1 is also a Lychrel number.

ii), iii) and iv) are proven the same way like i). ,

4.2. An Application to the Classification of Lychrel Numbers

In base 10, when we take the first Lychrel number 196 and iteratively apply 1) of proposition 5 on it, we obtain the other elements of the first column of the table bellow, from 295 to 790 and we conclude from i) of corollary 1 that they too are Lychrel numbers. Also, considering the first iteration of 196 which is 887 and iteratively applying the same result, we obtain the numbers 689, 788 and 986 and conclude from the same corollary that they too are Lychrel numbers. The same situation occurs for 879 and we obtain 978. Hence, with 196, ϕ( 196 )=887 and 879, we obtain all the other elements of Table 3 below.

Table 3. Equivalence classes of 196, 879 and 887.

196

689

295

788

394

887

493

986

592

879

691

978

790

Now we take the second iteration of 196, which is 1675, and we iteratively apply 1) of proposition 5 on it to obtain the elements of the third column of Table 4 below, which are 2674, 3673, 4672, 5671 and 6670. We then conclude from i) of corollary 1 that they are also Lychrel numbers. On each element of this third column, we iteratively apply 2) of proposition 5 to obtain from each of them all the other elements of the corresponding line. We then conclude from ii) of corollary 1 that they too are Lychrel numbers. Hence, all the 35 other elements of the table below are obtained just from a single element which is 1675 and belong to an equivalence class of 4 .

Table 4. Equivalence class of 1675.

1495

1585

1675

1765

1855

1945

2494

2584

2674

2764

2854

2944

3493

3583

3673

3763

3853

3943

4492

4582

4672

4762

4852

4942

5491

5581

5671

5761

5851

5941

6490

6580

6670

6760

6850

6940

The same process is applied when we consider the Lychrel number 1857 which is the first iteration of 879, and 1997 as we can see in the following Table 5 and Table 6.

Table 5. Equivalence class of 1857.

1497

1587

1677

1767

1857

1947

2496

2586

2676

2766

2856

2946

3495

3585

3675

3765

3855

3945

4494

4584

4674

4764

4854

4944

5493

5583

5673

5763

5853

5943

6492

6582

6672

6762

6852

6942

7491

7581

7671

7761

7851

7941

8490

8580

8670

8760

8850

8940

Table 6. Equivalence class of 1997.

1997

2996

3995

4994

5993

6992

7991

8990

The elements of Table 6 above are iteratively obtained from 1997 by applying 1) of proposition 5. In this table, it can be seen that one of the obtained element, 4994 is a Palindrome. We called it Palindrome-Lychrel number.

Considering the third iteration of 196 which is 7436, when we iteratively apply 1) of proposition 5 to it we obtain the elements of the fifth column of Table 7 below. The elements of this column are then Lychrel numbers, from i) of corollary 1. Moreover when we iteratively apply 2) of proposition 5 to each element of this column we obtain the elements from column 1 to column 8 (except column 5). The 48 elements of the first 8 columns of the table belong to the same equivalence class of the relation 4 having 7436 as the representative, and are Lychrel numbers from ii) of corollary 1. Now we consider column 8 of the table and iteratively apply 3) of proposition 5 to each element of this column. We then obtain column 9 of the table and again, we iteratively apply 2) of proposition 5 to each element of this column to obtain the elements in columns 10 and 11 of the table. All the 18 elements in columns 9 to 11 of the table also belong to the same equivalence class which is different from the one containing the elements in columns 1 to 8 and are Lychrel numbers from iii) of corollary 1. Hence, all these 66 elements of the table are Lychrel numbers.

Table 7. Equivalence classes of 7436 and 4799.

4079

4169

4259

4349

4439

4529

4619

4709

4799

4889

4979

5078

5168

5258

5348

5438

5528

5618

5708

5798

5888

5978

6077

6167

6257

6347

6437

6527

6617

6707

9797

6887

6977

7076

7166

7256

7346

7436

7526

7616

7706

7796

7886

7976

8075

8165

8255

8345

8435

8525

8615

8705

8795

8885

8975

9074

9164

9254

9344

9434

9524

9614

9704

9794

9884

9974

The same situation occurs with Table 8 below where starting with the Lychrel number 9438 which is the second iteration of 879, and following the same steps as in the previous case, we obain all the 22 elements of the table. The first 8 columns of the table also constitute an equivalence class of the relation 4 . 3) of proposition 5 is also iteratively applied to each element of column 8 to obtain the elements of column 9, and 2) of proposition 5 is iteratively applied to each element of this column to obtain the other elements in columns 10 and 11 of the table. Also, all the 22 elements are Lychrel numbers.

Table 8. Equivalence classes of 9438 and 8799.

8079

8169

8259

8349

8439

8529

8619

8709

8799

8889

8979

9078

9168

9258

9348

9438

9528

9618

9708

9798

9888

9978

The following Table 9 is obtained in the same way. We start with the Lychrel number 9988 which is the first iteration of 1997 and iteratively apply 1) of proposition 5 to it to obtain the element of its column. After that, 2) of proposition 5 is applied to obtain the column 10 of the table. To end, 2) of proposition 5 enables the deduction of column 9 from column 10 and again, 2) of proposition 5 is applied to the elements of column 9 to obtain the other elements of the table.

Table 9. Equivalence classes of 9988 and 8089.

8089

8179

8269

8359

8449

8539

8629

8719

8809

8899

8989

9088

9178

9268

9358

9448

9538

9628

9718

9808

9898

9988

With Table 10 below, the smallest element is the Lychrel number 7059. Hence, 1) of proposition 5 is applied to 7059 to obtain the elements of its column, which are also Lychrels from i) of corollary 1. After, 2) of proposition 5 is applied to each element of this column to obtain the elements in columns 2 to 6 of the table, which are Lychrel numbers too, by ii) of corollary 1. Also, 3) of proposition 5 is applied to each element of column 6 to obtain the elements of column 7 and at the end, 2) of proposition 5 is iteratively applied to each element of column 7 to obtain the elements in columns 8 to 11 of the table. Note that columns 1 to 6 and columns 7 to 11 are different equivalent classes of 4 . All the elements of the table are Lychrel numbers.

Table 10. Equivalence classes of 7059 and 7599.

7059

7149

7239

7329

7419

7509

7599

7689

7779

7869

7959

8058

8148

8238

8328

8418

8508

8598

8688

8778

8968

8958

9057

9147

9237

9327

9417

9507

9597

9687

9777

9867

9957

Hence, it can then be observed that with only four Lychrel numbers 196, 879, 1997 and 7059, one can obtain all the 248 Lychrel numbers base 10 less than 10000, by using the iteration function on them and by iteratively applying proposition 5.

Proposition 5 also enables us to describe Lychrel numbers above 10,000 as it is shown in Table 11 below in which from 13,783 which is the fifth iteration of 196, and applying 1) of proposition 5, the numbers in the second column of the table are obtained and are Lychrel numbers too, by i) of corollary 1. After that, we apply 4) of proposition 5 on each element of this column to obtain all the elements of the other columns which are also Lychrel numbers. Note that this table is also an equivalence class of the relation 5 .

Table 11. Equivalence class of 13,783.

127,93

13,783

14,773

15,763

16,753

17,743

18,733

19,723

22,792

23,782

24,772

25,762

26,752

27,742

28,732

29,722

32,791

33,781

34,771

35,761

36,751

37,741

38,731

39,721

42,790

43,780

44,770

45,760

46,750

47,740

48,730

49,720

Also, considering the Lychrel numbers 17,787 and 18,887 which are respectively the third iteration of 879 and the second iteration of 1997, and applying 1) of proposition 5 and 4) of proposition 5 with the same process as in the previous case, we obtain the following Table 12 and Table 13 of new Lychrel numbers.

Table 12. Equivalence class of 17,787.

16,797

17,787

18,777

19,767

26,796

27,786

28,776

29,766

36,795

37,785

38,775

39,765

46,794

47,784

48,774

49,764

56,793

57,783

58,773

59,763

66,792

67,782

68,772

69,762

76,791

77,781

78,771

79,761

86,790

87,780

88,770

89,760

Table 13. Equivalence class of 18,887.

17,897

18,887

19,877

27,896

28,886

29,876

37,895

38,885

39,875

47,894

48,884

49,874

59,893

58,883

59,873

69,892

68,882

69,872

79,891

78,881

79,871

89,790

88,880

89,870

One can observe that the number 18,887 yields another Palindrome-Lychrel number in the last table above, which is 48,884. Note that the two tables above are equivalent classes of 5 . It is also the case with Table 14 from the first iteration of 7059, which is 16,566.

Table 14. Equivalence class of 16,566.

13,596

14,856

15,576

16,566

17,556

18,546

19,536

23,595

24,585

25,575

26,565

27,555

28,545

29,535

33,594

34,584

35,574

36,564

37,554

38,544

39,534

43,593

44,583

45,573

46,563

47,553

48,543

49,533

53,592

54,582

55,572

56,562

57,552

58,542

59,532

63,591

64,581

65,571

66,561

67,551

68,541

69,531

73,590

74,580

75,570

76,560

77,550

78,540

79,530

Having shown that 196, 879, 1997 and 7059 are Lychrel numbers, and presented the list of all the Lychrel numbers related to them and which are below 10,000, we now focus on some numbers above 10,000 in the last four previous tables. The choices are 19,723, 89,760, 58,883 and 43,593, because they belong to different equivalence classes of 5 and for each table, there is no need testing all the elements. We also include 4994 and 8778 because they are palindromes, and also belong to different equivalence classes of 4 .

The results of the tests of the six chosen numbers using the algorithm presented in the previous section are presented in Table 15 below:

Table 15. Lychrel number test on some chosen natural numbers superior to 10,000.

a

4994

8778

19,723

89,760

43,593

58,883

l( a )

4

4

5

5

5

5

ω a ( n )

0.41464

0.41429

0.41466

0.41495

0.41468

0.41495

| ω a ( n )0.414 |

0.00064

0.00029

0.00066

0.00095

0.00068

0.00095

p( A n+1 )

0

0

0

0

0

0

The 0s in the last line of the above table represent the values of p( A n+1 ) which are less than ( 1 10 ) 330 . Hence, for each of the chosen numbers, p( A n+1 ) is less than ( 1 10 ) 330 . Therefore, the probability P( a ) of reaching the first palindrome is such that P( a )< P n+1 ( a )< P u ( a )<p( A n+1 )< ( 1 10 ) 330 . We then conclude that 19,723, 89,760, 58,883 and 43,593 are effectively Lychrel numbers. Also, 4884 and 8778 are palindrome-Lychrel numbers according to the probabilistic definition and to theorem 4.

5. Palindrome-Lychrel Numbers

The present section addresses the notion of Palindrome-Lychrel numbers and explains how to obtain them.

Let consider a Lychrel number a= a 1 a 2 a l( a )1 a l( a ) . If a Palindrome-Lychrel number k( a ) can be obtained from a, then k( a ) is both a lychrel number and a palindrome. But k( a ) is obtained from a by the iterative application of 1) of proposition 5 to a. By so doing, a 2 a l( a )1 always remains the same for all the iterative numbers obtained, including k( a ) . Hence, k ( a ) 2 k ( a ) l( a )1 = a 2 a l( a )1 and since k( a ) is a palindrome, k ( a ) 2 k ( a ) l( a )1 is also a palindrome and then a 2 a l( a )1 is a palindrome too. Moreover, when 1) of proposition 5 is applied to a number a to obtain a number m, one always has a 1 + a l( a ) = m 1 + m l( m ) . We then have a 1 + a l( a ) =k ( a ) 1 +k ( a ) l( k( a ) ) and since k( a ) is a palindrome, k ( a ) 1 =k ( a ) l( k( a ) ) and then k ( a ) 1 +k ( a ) l( k( a ) ) is even. It then comes that { a 1 , a l( a ) } is a couple of digits with the same parity.

Conversely, we suppose that a= a 1 a 2 a l( a )1 a l( a ) is a Lychrel number with a 2 a l( a )1 been a palindrome and { a 1 , a l( a ) } a couple of digits with the same parity. We set n=max( a 1 , a l( a ) ) a 1 + a l( a ) 2 . Then after iteratively applying 1) of proposition 5 to a n times, the number k( a )= a 1 a 2 a l( a )1 a l( a ) + a l( a ) a 1 | a l( a ) a 1 | n( 10 l( a )1 1 ) is obtained and is a palindrome. Since a= a 1 a 2 a l( a )1 a l( a ) is a Lychrel number, we conclude that k( a ) is a Palindrome-Lychrel number. We can now present the following result:

Proposition 6. Let a= a 1 a 2 a l( a )1 a l( a ) be a Lychrel number. Then the following assertions are equivalent:

1) A Palindrome-Lychrel number can be obtained from a;

2) a 2 a l( a )1 is a palindrome and { a 1 , a l( a ) } a couple of digits with the same parity.

The above result explains why Lychrel numbers 1997, 7779 or 18,887 generate Palindrome-Lychrel numbers in their respective equivalence classes.

However, even if a= a 1 a 2 a l( a )1 a l( a ) is not a Lychrel number, by iteratively applying 1) proposition 5 on a, one can still obtain a palindrome if a 2 a l( a )1 is a palindrome and { a 1 , a l( a ) } a couple of digits with the same parity. For example, when we consider the natural number 26,568, we can see that the digits 2 and 8 have the same parity and 656 is a palindrome.

Hence setting n=max( 2,8 ) 2+8 2 =3 , we can conclude that after iteratively applying 1) of proposition 5 to 26,568 3 times, we will reach a palindrome as follows: 26568+9999=36567 , 36567+9999=46566 and 46566+9999=56565 which is a palindrome.

The summary of the work is presented in de flowchart in Figure 2.

Figure 2. The Flowchart explaining the methodology of the work.

6. Conclusion

In this paper, we have defined some tools that permit the obtention of an iteration of a natural number and the proposal of a probabilistic characterization of Lychrel numbers. Moreover, we have provided some properties for the mathematical construction of new Lychrel numbers from others, without the use of a computer programme. These properties led us to the discovery of Lychrel numbers greater than 10,000 and to the discovery of Palindrome-Lychrel numbers. The characterization proposed in the paper can be extended to other bases b ( 3b9 ), provided the blocks of consecutive iterations of natural numbers in these bases are accurately studied. However one of the limitations of the work is that the characterization of Lychrel numbers here remains probabilistic. Our future work is then to find an algebraic Characterization of Lychrel numbers and Palindrome-Lychrel numbers, in order to make their handling more easy.

Acknowledgements

Sincere thanks to the reviewers for their comments and suggestions, and to the members of JAMP for their professionalism.

Notation

N: the set of natural numbers;

a,m,d,l,i,j,k : natural numbers;

Rev( a ) : the reverse number of a ;

l( a ) : the length of a, meaning the number of digits of a ;

N l : the set of natural numbers of length l;

a i : the ith digit of a ( 1il( a ) ) ;

ϕ i ( a ) : the ith iteration of a ;

ϵ i ( a ) : the vector parameter used for the calculation of ϕ i ( a ) ;

ϵ j i ( a ) : the jth element of the vector parameter used for the calculation of ϕ i ( a ) ;

ϕ ( a ) i : the ith digit of ϕ( a ) ;

σ( a,m ) : the vector parameter used for the calculation of a+m ;

σ j ( a,m ) : the jth element of the vector parameter used for the calculation of a+m ;

C m k : the number of k-combinations in a set containing m elements;

A i : the event which makes ϕ i ( a ) a palindrome;

B i : the event for which ϕ i ( a ) is not a palindrome;

p( a ) : the probability for a to be a palindrome;

p( A ) : the probability of the event A;

P( G )( m ) : the probability for m to be a starting point of blocs of iterations from G;

P( a ) : the probability of reaching the first palindrome after several iterations of a ;

P n ( a ) the probability to reach the first palindrome at the nth iteration of a .

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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