A palindromic number (also known as a numeral palindrome or a numeric palindrome) is a number that remains the same when its digits are reversed.
Consider a number $n>0$ in base $b\geq2$, 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, $0\leq a_{i}<b$ for all $i$ and $a_{k}\neq0$. Then $n$ is palindromic if and only if $a_{i}=a_{k-i}$ for all $i$. Zero is written $0$ in any base and is also palindromic by definition.
In this context, a Lychrel number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. It is conjectured that $196$ and other numbers that have not yet yielded a palindrome are Lychrel numbers, but no number in base 10 has yet been proven to be Lychrel.
I was curious about the sequence of "candidate Lychrel" numbers, which can be found at https://oeis.org/A023108. I tried to find some kind of pattern among them, and I found the following: every number of the form $99k-2$ is a candidate Lychrel for $1<k<9$. This “curious” fact led me to check if some similar property ocurred for higher candidate Lychrel numbers, and I have checked that every number of the form $999k-1$ is a candidate Lychrel for $1<k<10$ excepting $k=5$, which is a palindromic number; and every number of the form $9999k$ is a candidate Lychrel for $1<k<10$. I have not checked further, only that $99999*2+1$ is also a candidate Lychrel.
Clearly, it can be made the following rough conjecture:
Conjecture. Let it be some number in base $10$ of the form $n=999…9k-m$. Let us denote with $j$ the number of "9" composing the form of $n$. Then, if $1<{k}<9$, $j\geq2$ and $m=4-j$, $n$ is either a palindromic number or a Lychrel number.
Any guess about the relationship between this clear pattern, and the dynamic of the add-and-reverse-digits process? Any counterexample to the conjecture?
Thanks in advance!
Best Answer
There is a counterexample.
Take $(j,k)=(6,2)$ for which we have $$n=(10^j-1)k-(4-j)=2000000$$
This is not a palindromic number.
Also, this is not a Lychrel number since under the "add reverse" operation, we have $$2000000+2=2000002$$ which is a palindromic number.
Added :
Let us call a number good if it is either a palindromic number or a Lychrel number.
According to A023108, we have the followings :
For $j=2$, $n=(10^2-1)k-2$ is good for $k=2,3,\cdots, 8$.
For $j=3$, $n=(10^3-1)k-1$ is good for $k=2,3,\cdots, 8,\color{red}{9}$.
For $j=4$, $n=(10^4-1)k-0$ is good for $k=2,3,\cdots, 8,\color{red}{9,10,11}$.
For $j=5$, $n=(10^5-1)k+1$ is good for $k=2$.
Also, we can see the followings :
For $j=6$ and $k=2$, $n=2000000$ is not good.
For $j=7$ and $k=2$, $n=20000001$ is not good.
For $j=7$ and $k=3$, $n=30000000$ is not good.
For $j=8$ and $k=3$, $n=300000001$ is not good.
For $j=8$ and $k=4$, $n=400000000$ is not good.
So, it seems that there are functions $F(j),G(j)$ such that $n=(10^j-1)k-(4-j)$ is good for $j\ge 2$ and $F(j)\le k\le G(j)$.
Something like $$F(j)=2+(j-5)\left\lfloor\frac j6\right\rfloor,\qquad G(j)=j+6+\left\lfloor\frac j4\right\rfloor$$