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Consider a number $n > 0$ in base $b ≥ 2$, where it is written in standard notation with $k+1$ digits $a_i$ as:

$${\displaystyle n=\sum _{i=0}^{k}a_{i}b^{i}}$$

with, as usual, $0 ≤ a_i < b$ for all $i$ and $a_k ≠ 0$. Then n is palindromic if and only if $a_i = a_{k-i}$ for all $i$. 

And defined $P_l$ as

$k+1$ is number of digits in $n$

$$ (\sum_{i=0}^{k}|a_{i}-a_{k-i}|)/2=l$$ Then

$$ n\in P_l$$

It means $P_0 =$ palindromic number

Example

● $121 \in P_0$

● $357 \in P_4$

● $9 \in P_0$

Let's defined function $\lambda_{b,n} (j)$ as

$\lambda_{b,n} (j)$ is legal if natural number $n \in P_r$ with base $b$ converted in $n \in P_j$ through the iterative process of repeatedly reversing its digits and adding the resulting numbers (This process is sometimes called the 196-algorithm). so $r$ is depend on $b,n$

Note If n is Lychrel number then $\lambda_{b,n} (0)$ is not legal

Example

Let's given $j=16$, $b=10$, $n=9519$

To check $\lambda_{10,9519} (16)$ is legal or not

$9519\in P_4$ so $ 9519+9159=18678$ again $18678+87681=106359$ where $106359 \in P_{16}$

Hence $\lambda_{10,9519} (16)$ is legal

Question

<p>1) Can we prove <span class="math-container">$\lambda_{10,n} (1)$</span> is legal for all <span class="math-container">$ n\in \mathbb{N} $</span> </p>

<p>2) Can we prove there exist legal <span class="math-container">$\lambda_{b,n} (j)$</span> for all <span class="math-container">$ n,j\in \mathbb{N} $</span> for some <span class="math-container">$b$</span></p>
Pruthviraj
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  • Some parts of your question are not clear to me. For example, you write that if $k+1$ is even, then $\sum_{i=0}^{k-1/2}|a_{i}-a_{k-i}| \in P_i$. First, I assume $k-1/2$ is supposed to be $(k-1)/2$. Next, you're using $i$ both for the summation and for $P_i$. With $P_i$, what connection, if any, does it have with $k$? With the in member set operator, does it apply to each value being summed, or the overall summation? – John Omielan Jul 10 '19 at 23:56
  • Yes you are right my apologies now I have edited my mistake please sure your confusion – Pruthviraj Jul 11 '19 at 02:09
  • 9151 should be 9159, @JohnOmielan I think it's supposed to represent the sum in the case of $i$ in the $P_i$, it seems to apply that way to all their examples anyways. –  Jul 11 '19 at 02:28
  • @Pruthviraj So given a number base $b$, we write $n$ in that base, and then start reversing and adding its digits until we reach a number that is $\in P_j$? If we call elements of $P_1$ "almost-palindromes", then you first question is asking if the 196-algorithm will always terminate for "almost-palindromes" for $b=10$. - That is, there are no "almost-Lychrel-Numbers" for base ten? For how many $n$ have you tested it? (Why do you think there are no counterexamples, that we can prove this is true for all $n$?). Edit: I think this will be as hard as Lychrel Numbers which is an open problem. – Vepir Aug 21 '19 at 10:04

1 Answers1

1

This is a partial answer too long for a comment:

  • I will explain why your first question is probably false and why we can't prove/disprove it.

  • I will prove your other question is trivial for all $n$ such that: ($j=0,1$) or if ($j$ divides $n\gt j^2$).
    Maybe someone else can work out the full proof If I don't get back to this.


Regarding your $1.$ question: "I don't think it is true and I do not think we can."

Your first question will be as hard as the Lychrel Numbers which is an open problem. We do not know for example if $196$ is truly a Lychrel Number. The sequence is just conjectured based on extensive computations, as we do not know how to solve these kind of problems (for base ten at least).

Because we do not know how to prove given $n$, that $\lambda_{10,n}(0)$ is "not legal".

And you are asking about $\lambda_{10,n}(1)$ which I don't see why would be any easier. I have ran a computation, and $n=101$ seems to be the smallest "not legal" number which would be a counter example to what you want proven. - But as I said, this is probably as hard to prove as $196$ being smallest "not legal" for $\lambda_{10,n}(0)$.

The $n=101$ does not turn into an element of $P_1$ for $\gt 3000$ reversal steps, I haven't computed more.

The number of steps numbers $1,\dots, 100$ need to become $\in P_1$, in comparison:

7, 6, 2, 5, 1, 1, 3, 4, 12, 0, 5, 0, 4, 2, 2, 3, 3, 11, 1, 5, 0, 4, 0, 2, 3, 3, 11, 1, 4, 3, 4, 0, 2, 0, 3, 11, 1, 4, 1, 4, 2, 2, 0, 3, 0, 1, 4, 1, 3, 2, 2, 3, 3, 0, 1, 0, 1, 3, 2, 2, 3, 3, 11, 1, 0, 1, 0, 2, 2, 3, 3, 11, 1, 4, 1, 0, 2, 0, 2, 3, 11, 1, 4, 1, 3, 2, 0, 2, 0, 11, 1, 4, 1, 3, 2, 2, 2, 0, 10, 0


Proving your $2.$ question for trivial cases:

You are asking if every number $n\in\mathbb N$ will turn into element of $P_j$ for every $j$, for some number base $b$?

I will denote digits of $n$ in base $b$ as $(n_1,n_2,\dots)_b$.

If $r=j$, then we are done, since $n\in P_j$ before we even start reversing digits.

I will find simplest of $b$ such that $r=j$ to prove the statement for given $j,n$ for trivial cases:

  • This is true for $j=0$ for all $n$. Take $b\gt n$ so $n=(n)_b$.

  • This is true for $j=1$ for all $n\gt 1$. Take $b=n$ so $n=(1,0)_b$.

  • For $j=1$, it works for $b=10$ for $n=1$. Same for all $n\lt 101$ as computed in $1.)$ question.

  • This is true if $n\gt j^2$ and is divisible by $j$, we have $n=(j,0)_b$ for $b=n/j$.

Given this, what is left to prove:

We are left to consider $n\le j^2$ and divisible by $j$, and $n\in\mathbb N$ not divisible by $j$, for $j\ge 2$ in both cases.

I will update the answer if I continue the proof. - I also encourage anyone reading this to try to continue the proof for non-trivial cases and post their own answer if they succeed.

Vepir
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  • thank you for your solution yes you are right my fist question is hard as 196 I'm not used computer for this observation. you motivate me for further study and work on palindromic number. it is not at all clear of your last two note (at bottom) please can you can explain it again. – Pruthviraj Aug 23 '19 at 06:57
  • @Pruthviraj Your $2.$ is easily true if $n\gt j^2$ and $n$ is divisible by $j$. This is because if we pick $b=n/j$ we have $n = j\cdot n/j + 0 = (j,0)_{n/j}$. That is, we have a two digit representation, and $j$ will be the first digit in that base, and $0$ the second. We have $r=j$ and $n\in P_j$ right away. Now, we haven't prove the $2.$ question claim otherwise yet, that is, if $n\le j^2$ and divisible by $j$, or if $n$ is not divisible by $j$. To continue this, one will need to start considering the reversal sums to find $b$ that works. – Vepir Aug 23 '19 at 07:36