1. Introduction
On the set of natural numbers, the reverse number of a natural number
(with
the number of digits of a), denoted by
, is the natural number such that
. When a is equal to its reverse number (
), 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
yields
which is a palindrome. With
, we have
, and
. 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
be a natural number, where
denotes the number of digits of a and
a digit of a for all
. An iteration of a is the operation that consists of adding a to
. To express it conveniently, we define the unity function
which corresponds to its last digit, as follows:
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 |
|
1 |
1 |
2 |
4 |
4 |
4 |
4 |
5 |
5 |
6 |
|
0 |
3 |
5 |
8 |
0 |
0 |
7 |
1 |
0 |
7 |
The unity function
verifies some elementary properties:
Lemma 1. Let
be a digit of a natural number, and
.
1) For all
,
;
2) For all
, if
then
;
3) For all
, if
, then
.
Proof. The properties are straightforward. ,
Hence, if
is a digit of a natural number, then
and
.
Also, we consider the iteration function
, which is actually the one defined in [8], as follows:
Then, the first iteration of a will be denoted by
, the second iteration will be
and for all
,
(n factors). For example, in base 10, if
, then
.
Now let’s consider
. Then the nth iteration of a can be expressed as a function of a as
. As for the first iteration, its length is given by
, where
.
if and only if
or there exists
such that
and
for all
. For example, in base 10, if
, then we have
and
. Hence,
and
. In the second example, we consider the case where
. Then for
,
and
for all
. Hence,
and then
.
In order to successfully add a to its reverse number to obtain the first iteration of a, we define the vector parameter
(or
) which depends both on the length and the digits of a, as follows:
with
being calculated in the decreasing order.
Then the following are the expressions, depending on α, of the first iteration of a.
If
, then:
,
,
,
For every
,
.
If
, then:
a)
,
b)
,
c)
,
d) For every
,
.
For example, we consider the case where
, we have
,
. Therefore
. We then determine the epsilons in the decreasing order as follows:
,
because
,
because
and
because
. Hence, we can determine the digits of
as follows:
1)
,
2)
,
3)
,
4)
.
Thus,
.
In the more general way, if a and m are two natural numbers with
and
their respective lengths. Let’s say that our aim is to make the addition of the two numbers. Then the length of
is given by
(with
).
if and only if
or there exists
such that
and
for all j ≤i − 1. For example, if
and
, then we have
and
. Hence,
and
. The second example presents the case where
and
, then for
,
and
for all
. Hence,
and
.
In the above examples,
. If
, then we make the following adjustments: if
, then 0 is added
times to m and before the digit m1. For example, if
and
, then
. Hence 0 is added 2 times to m and before 9. m then becomes
with a greater length
and
. The same situation occurs with a if
.
We then define the vector parameter
which depends both on the digits of a and m as follows:
The
are also calculated in a decreasing order. The digits of
are then obtained as follows:
If
, then:
,
,
,
For every
,
.
If
, then:
1)
,
2)
,
3)
,
4) For every
,
.
3. A Probabilistic Characterization of Lychrel Numbers
Let
be a natural number. If a is a palindrome, then the couples of digits
of a most be chosen among the set of couples
, with
. Therefore the probability for a to be a palindrome is given by
This probability then depends on the length
of a. The more
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
and
;
The length of a can increase by 3. It is the case with most natural numbers a verifying
and
, or such that
and
;
The length of a can increase by 4, like in most of the cases where
and
;
The length of a can increase by 2. It happens with most natural numbers, especially when
.
3.1. Analyzing Some Particular Blocks of Consecutive Iterations
For the n first iterations of a (
and
), 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
, we have the numerical sequence a,
. The first block of 5 consecutive iterations is
and its starting point or origin is a and its destination is
. The second block of 5 consecutive iterations is
and its starting point or origin is
and its destination is
.
is the residual block with less than 5 iterations. The 2 blocks of 5 consecutive iterations are consecutive because
which is the destination of the first block, is also the starting point of the second block.
Now assume that n (
) iterations are computed from a. We split the n iterations into blocks of 5 consecutive iterations. We set
the set of blocks for which the length of the starting point increases by 4;
the set of blocks for which the length of the origin increases by 3;
the set of blocks for which the length of the starting point increases by 2 and
the set of blocks for which the length of the origin increases by 1. Then we have
, where
is the set of blocks from
such that their destinations generate their next blocks of 5 consecutive iterations also from
,
is the set of blocks from
such that their destination produces their next blocks of 5 consecutive iterations from
and
is the set of blocks from
such that their destination generate their next blocks of 5 consecutive iterations from
.
We find the probability for a natural number m to be the starting point of the blocks from
,
and
. To get there, we consider each natural number
as a set of
couples of digits
if
is even, and as a set of
couples of digits
and a middle digit in
if
is odd. The total number of possible couples of digits is 100 because each digits belongs to
. Moreover, for each number, the first couple
must not belong to the set
. Hence, the first couple must be chosen among the total of 90 cases instead of 100. If
is odd, the digit in the middle
must be chosen among
, leading us to the total number of 10 possible cases. Now we study the different situations.
Case of
For m to be the origin of a block from
, it is necessary in the very large majority of cases that its digits verify
and
. Hence,
where B is the rest of digits of
.
It implies that
because
.
where C is the rest of digits of
. We impose that
. That is the case if
, which implies that
. For that to be possible, it requires at least that
or
and
or
and
. The last two cases are minor cases and depend on others digits
of m. Hence, for simplification, we consider only the case
. That leads us to
or
. It gives us 80 possible couples
. The probability of m to be the starting point in a block of iterations from
is then given by
if
is even and
if
is odd.
Case of
For m to be the starting point of a block from
, it is necessary in the very large majority of cases that its digits verify
and
and
.
Hence, the length of m increases by 3 after 3 iterations and also by 3 after 5 iterations. Since
and
, 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
, it is necessary at least that
and
. The number of possible couples
such that
is 4. The probability of m to be the starting point in a block of iterations from
is then given by
if
is even and
if
is odd.
Case of
For m to be the starting point of a block of iterations from
, it requires at least that
. Moreover, If
, then in the very large majority of cases,
has to verify
, which gives us a total of 67 possible cases for couples
. If
, then in the very large majority of cases,
has to verify
, which gives us a total of 56 possible cases for couples
. The probability of m to be the starting point of a block of iterations from
is then givenby
if
is even and
if
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
be a natural number in base 10 such that
. For all
, we denote by
the nth iteration of a
Proof. Let consider
. Then after the first iteration, we have
, where
(
if the length of a increases and
if not). After the second iteration,
(with
), which implies that
. Hence, after n iterations,
, where
for all
.
Notice that after 5 consecutive iterations, we have
. Hence we split the n iterations into blocks of 5 consecutive iterations. Then we will have
blocks of 5 consecutive iterations, where
stands for the floor of n. Setting
;
;
;
.
Then, we have
(1)
Moreover, any block of 5 iterations from
produces the next block of 5 iterations either from
, or from
or
. Then we have
where
is the set of blocks that yield blocks from
,
the set of blocks that yield blocks from
and
the set of blocks that produce blocks from
. Also, any block from
produces the next block of 5 iterations from
. Hence,
(2)
moving further, we have
(3)
with
the cardinality of a possible block of less than 5 iterations
.
From (1), we get
. Hence,
In the same way, we have
Hence,
Which implies that
Hence, from relation (1), the numerical sequences
,
and
are convergent. Therefore,
and
are convergent with the same limit and then from the gendarme theorem, we have
We approximate
by the probability of a given natural number m to be the starting point of a block of iterations from
, which is
We approximate
by the probability of a given natural number m to be the starting point of a block of iterations from
, which is
. Also,
is approximated by the probability of a given natural number m to be the starting point of a block of iterations from
, which is
.
Hence,
.
,
We now provide a link between the above result and Lychrel numbers.
Let a be an element of N such that
. Then we consider the map defined as follow:
Assume that a is a Lychrel number. Then
for all
.
Hence,
Conversely, suppose that
Then for all
, there exists
such that
for all
. For
, there exists
such that
for all
. Assume there is
such that
. Then
and then
, which implies that
, which is absurd because
. Hence,
for all
. Moreover,
, which implies that for all
,
, and then that for all
,
. It thus comes that
for all
.
That leads us to the following result:
Theorem 3. Let a be a natural number such that
. Then the following assertions are equivalent:
1) a is a Lychrel number
2)
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
any given
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
to reach the first palindrome after several iterations is less than
.
The choice of
is explained by the principle that for a to be a lychrel number,
should be equal to 0. But computationally,
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
is close to
, 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
to be computed must be as enormous as possible for
to be closed to
, let say
.
Having established that
for all
, let
be the limit of
. Then there exists
such that for all
,
. It implies that for all
,
. We then conclude that
for all
.
Also, there exists
such that for all
,
, which implies that
for all
.
Setting
, we have
For all
, which implies that
for all
.
Taking into consideration the fact that
and
, we conclude that
.
Hence,
and then for
to be near
, at least
iterations has to be produced without reaching a palindrome.
For all
, let
be the probability of reaching the first palindrome at the nth iteration.
From an iteration
(
) to
, there are two possibilities:
is a palindrome or
is not a palindrome. We set
the event which makes
a palindrome and
the event for which
is not a palindrome. Also, we set
the probability for
to be a palindrome and
the probability for
not to be a palindrome.
Then
. Hence,
.
However
For,
,
.
Hence,
if
is even, then
if
is odd, then
Therefore,
,
We then have the following result:
Theorem 4. Let
be a natural number such that
,
the minimum number of iterations to be computed without the reach of a palindrome and such that
(
,
),
the probability of reaching de first palindrome and for all
,
the probability to reach the first palindrome at the nth iteration of a.
Then
1)
;
2)
, with
.
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
, C of
and P of
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
, 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 |
|
3 |
3 |
4 |
4 |
23 |
|
0.41409 |
0.41429 |
0.41464 |
0.41429 |
0 |
|
9.69 × 10−05 |
0.00029 |
0.00064 |
0.00029 |
0.414 |
|
0 |
0 |
0 |
0 |
2.51 × 10−28 |
The 0s in the last line of the table above represent the values of
which are inferior to
. Hence, From the result in this table,
is less than
for 196, 879, 1997 and 7059. Therefore,
. 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
.
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 (
).
Let l be a natural number and
the set of all natural numbers of length l in base b (
). Then we define on
the binary relation
as follows:
if and only if for all
,
. Then
is reflexive, symmetric and transitive and is then an equivalence relation on the set
. In other words,
if and only if
. The following properties help us to search, for a given number a, the number m verifying
under some conditions.
4.1. Some Properties of Construction of New Lychrel Numbers
The following results hold:
Proposition 5. Let
be a natural number in base b such that
and
.
1) If
and
, then
;
2) If
is even,
and
, then
;
3) If
is even,
,
,
,
and
, then
;
4) If
,
is odd,
,
and
, then
.
Proof. Let consider such number
.
1) We set
. Then
because
and
. Moreover, setting
,
. We then consider d the number obtained by adding 0 before the first digit of m. Then
and
. Hence,
by definition and since in base b, all the digits of d are equal to
except the first one and
, we have
for all
(the
are calculated in a decreasing order). Hence, the digits of x are then given by
(since
),
(by lemma 1), and for all
,
(also by lemma 1). Hence, whether
or
, we have
for all
.
2) Let consider such number
. We set
. Then
because
and
. Moreover, setting
,
since
is even. We then consider d the number obtained by adding 0 to m
times and just before the first digit of m. Then
and
. The digits of d are all equal to 0, except the one in the position
which is equal to
. Hence, we have
,
for all
(Note that the
are calculated in a decreasing order) and since
,
. Also, for all
,
, since
. Hence, the digits of x are then expressed as the function of the digits of a as follows: for all
,
.
because
.
(because of lemma 1)
(also by lemma 1)
. for all
,
. its comes that
,
. We then conclude that
, and that
for all
.
3) Let consider such number
. We set
. Then
because
and
. Moreover, setting
,
since
is even. We then consider d the number obtained by adding 0 to m
times and just before the first digit of m. Then
and
. The digits of d are all equal to 0, except the one in position
which is equal to
. Hence, we have
and
for all
(Note that the
are calculated in a decreasing order) and since
, we also have
. We then conclude that for all
,
. Moreover,
. Now we search for the digits of
. In fact,
. Since
, then
. Hence,
and
. Also, for all
,
,
, (using lemma 1)
,
(also because of lemma 1) and for all
,
. After finding the digits of
, it is quite clear that
. As for the digits of
, it can be seen that
for all
.
4) Let suppose
is odd. We consider the natural number
. For example, if
and
,
. Then by construction,
and all de digits of c are all equal to
, 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
and
. The digits of d are all equal to
, except the first, the second and the last which are equal to 0. We suppose that
. Then setting
,
if and only if
and
. Hence, we have
by definition of
, and
because
. Also,
because
, and for all
,
since
(we remind that the the
are determined in the decreasing order). From the
, the digits of x are now given by
since
and
.
since
. Moreover, for all
,
(by lemma 1). Also,
(also by lemma 1) and
. Hence, whether
or
,
for all
. ,
Corollary 1. Let
be a Lychrel number in base b such that
and
.
i) If
and
, then
is a Lychrel number;
ii) If
is even,
and
, then
is a Lychrel number;
iii) If
is even,
,
,
,
and
, then
is a Lychrel number;
iv) If
,
is odd,
,
and
, then
is a Lychrel number.
Proof. Let
be such Lychrel number. i) Since
and
, then
, from i) of proposition 5. Hence, for all
,
. Since a is a Lychrel number, for all
,
, which implies that
for all
. Hence,
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,
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
.
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
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
. 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
. 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
.
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
. 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
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
.
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 |
|
4 |
4 |
5 |
5 |
5 |
5 |
|
0.41464 |
0.41429 |
0.41466 |
0.41495 |
0.41468 |
0.41495 |
|
0.00064 |
0.00029 |
0.00066 |
0.00095 |
0.00068 |
0.00095 |
|
0 |
0 |
0 |
0 |
0 |
0 |
The 0s in the last line of the above table represent the values of
which are less than
. Hence, for each of the chosen numbers,
is less than
. Therefore, the probability
of reaching the first palindrome is such that
. 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
. If a Palindrome-Lychrel number
can be obtained from a, then
is both a lychrel number and a palindrome. But
is obtained from a by the iterative application of 1) of proposition 5 to a. By so doing,
always remains the same for all the iterative numbers obtained, including
. Hence,
and since
is a palindrome,
is also a palindrome and then
is a palindrome too. Moreover, when 1) of proposition 5 is applied to a number a to obtain a number m, one always has
. We then have
and since
is a palindrome,
and then
is even. It then comes that
is a couple of digits with the same parity.
Conversely, we suppose that
is a Lychrel number with
been a palindrome and
a couple of digits with the same parity. We set
. Then after iteratively applying 1) of proposition 5 to a n times, the number
is obtained and is a palindrome. Since
is a Lychrel number, we conclude that
is a Palindrome-Lychrel number. We can now present the following result:
Proposition 6. Let
be a Lychrel number. Then the following assertions are equivalent:
1) A Palindrome-Lychrel number can be obtained from a;
2)
is a palindrome and
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
is not a Lychrel number, by iteratively applying 1) proposition 5 on a, one can still obtain a palindrome if
is a palindrome and
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
, we can conclude that after iteratively applying 1) of proposition 5 to 26,568 3 times, we will reach a palindrome as follows:
,
and
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 (
), 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;
: natural numbers;
: the reverse number of
;
: the length of a, meaning the number of digits of
;
: the set of natural numbers of length l;
: the ith digit of
;
: the ith iteration of
;
: the vector parameter used for the calculation of
;
: the jth element of the vector parameter used for the calculation of
;
: the ith digit of
;
: the vector parameter used for the calculation of
;
: the jth element of the vector parameter used for the calculation of
;
: the number of k-combinations in a set containing m elements;
: the event which makes
a palindrome;
: the event for which
is not a palindrome;
: the probability for
to be a palindrome;
: the probability of the event A;
: the probability for m to be a starting point of blocs of iterations from G;
: the probability of reaching the first palindrome after several iterations of
;
the probability to reach the first palindrome at the nth iteration of
.