$2017:$ Was initially asked on MSE - but wasn't solved or updated there since.
Update $2019$: I've returned to this problem, made some progress and updated the post here.
(I've basically rewritten this entire post here)
Introduction and problem
An $k$-palindrome is a number palindromic in $k$ consecutive number bases. Here, I'm wondering about the existence of $k=4$. Note that if it does not exits, $k\gt 4$ can't exist by definition.
Also note that if a number is even length (number of digits is even) palindrome in base $b$, then it is divisible by $b+1$ and thus can't be palindromic in that number base. So we can look for odd length cases of digits in the first (last, depending on how you define it) base only.
Can a natural number be nontrivially palindromic in more than $3$ consecutive number bases?
Nontrivially means that I'm not counting one-digit palindromes.
Smallest number $N$ which is nontrivially palindromic in $k$ consecutive number bases:
$$ \begin{array}{|c|} \hline k& N & \text{Palindrome} \\ \hline 1& 3 & 11_2 \\ 2& 10 & 101_3=22_4\\ 3& 178 & 454_6 =343_7 = 262_8\\ 4& ?& \\ \hline \end{array} $$
Where the index denotes the number base representation. For example $11_2$ is three in binary.
Note that $k=1$ is not special as those are just palindromes and easily can be constructed.
It is not hard to see $k=2,3$ have infinitely many examples. But finding all of them is hard.
I conjecture solution for $k=4$ does not exist. That is, there are no $4$-palindromes.
My question here is:
Can you provide arguments (help) or direction (on proving) why (why not) would this be true?
My ideas are presented below:
First approach to proving this
One idea to prove this is proving $(1)$ and $(2)$ which would imply this. The $(1)$ is now proven. The $(2)$ will be much harder to prove. The details are included below. Here are the claims:
$(1)$ A $3$ digit (when written in palindromic bases) number can't be a $4$-palindrome.
$(2)$ To be a $4$-palindrome, a number must have $3$ digits (when written in palindromic bases).
The second claim is based on the fact that almost-all "almost $4$-palindromes" have $3$ digits. That is, numbers palindromic in three out of four consecutive bases (that is, for example in bases $b,b+3$ and either $b+1$ or $b+2$).
Almost-all since the only two known "almost $4$-palindromes" that do not have $3$ digits are $71240, 1241507$ which have $5$ digits (when talking about digits, they are counted in the "pivot" number base among the consecutive bases in which the number is palindromic - "pivot" being either the smallest or largest base).
Proving $(2)$ is equally hard as proving $(3)$ from alternate approach below, since all cases of $d$ must be resolved. The $(1)$ specifically asked to resolve $d=3$ which was accomplished.
Second approach to proving this
$(3)$ I actually conjecture I have found all $3$-palindromes. If this is true, then a $4$-palindrome does not exist since neither of my $k=3$ solutions can be extended to a fourth consecuitve number base.
More details are linked at the end of the next section:
"$k$-problem system" results and conjectures
What I was able to prove both computationally and by hand:
Expanding $(1):$ In short, I have proven I have all $3$ digit $3$-palindromes, and nontrivial $d\le 3$ solutions don't exist. Now it is easy to see now that a $3$ digit $3$-palindrome can't be palindromic in fourth consecutive base, which means $4$-palindromes with $d\le 3$ digits do not exist. (Digits counted when written in palindromic number bases).
More details about this are linked at the end of this section, as well.
Expanding $(3):$ What I was able to conjecture is presented below:
In short, we can write the problem of finding $k$-palindromes in form of solving linear diophantine equations whose order grows when solving for longer palindromes of length $d$ (number of digits). Lets call that system needed to be solved "$k$-problem system".
There is the "general", and there is the "non-general" case of this problem system.
The non-general $k$-problem system is related to finding $d=2l+1$ (WLOG $d$ is odd) digit solutions, where all representations in consecutive bases have exactly $d$ digits.
The general $k$-problem system allows different lengths of digits in those $k$ consecutive number bases, where "pivot" number base among these $k$ consecutive number bases, usually smallest or largest, is taken to have $d$ digits.
The $k=2$ case has infinitely many solutions for every case of $d$. It is hard to find them all.
The $k=3$ is what will be the case of the "problem-system" from now on:
Expanded on $(3)$ in this context:
For the non-general case of the "problem-system", I have computationally solved (proved) I have all solutions for two smallest cases, $d=3,5$. I also strongly conjecture same is true for $d=7$. I also conjecture based only on computation, that a $3$-palindrome solution for $d\ge 9$ does not exist. That is, I conjecture I have all solutions.
For the general case of the "problem-system", I have computationally solved (proved) I have all solutions for smallest case, $d=3$. I also conjecture I have all solutions for $d\ge 5$ since another conjecture I strongly believe is that the general case is equivalent to the non-general case, except for having an extra finite set of solutions for sufficiently small bases $b$ which I have collected and computed all (conjectured).
Summing this all up: the important thing is I have all $3$ digit $3$-palindromes and that they can't be $4$-palindromes. I have all conjectured solutions for $3$-palindromes, but unable to prove the fact that large $d$ can't have solutions for $k=3$.
Note that I was trying to find counterexamples to these conjectures for a long time now but couldn't.
The more details as promised: A post mainly focusing on non-general "problem-system" can be read on MSE if you are interested in more details, or if you want to see the "all $3$-palindromes" solutions for the non-general case (the extra solutions from general case are easily re-computable if my conjectures are true).
I'll also include all infinite families (the non-sporadic solutions) for $k=3$ given there, here:
Given $k\in\mathbb N_0 = \mathbb N \cup \{0\}$, "infinite families" giving $3$-palindromes:
$$ \begin{array}{l,l,l,l} d & (a_i)&(c_i)k & b\\ d=3 & (2,6)&(1,1)k & 2k+8\\ d=5 & (31,32,0)&(3,2,1)k & 4k+47 \\ d=7 & (34,50,10,74)&(1,1,1,1)k & 2k+76 \\ d=7 & (8,33,0,41)&(1,3,1,3)k & 6k+58 \\ d=7 & (112,15,0,36)&(4,0,1,0)k & 6k+175 \\ d=7 & (227,160,187,200)&(5,3,5,3)k & 6k+280 \\ d=7 & (5,23,6,14)&(2,6,5,0)k & 12k+39 \\ d=7 & (93,78,30,50)&(10,6,7,0)k & 12k+119 \\ d=7 & (47,150,249,26)&(2,6,11,0)k & 12k+291 \\ \end{array} $$
Where these give digits of $3$-palindromes in base $b$, which are also palindromic in $b-1,b-2$.
That is, for example, observing the second row (family), the only $d=5$ case infinite family, means $(31+3k,32+2k,0+1k,32+2k,31+3k)$ are digits of a $3$-palindrome in base $4k+47$, for every $k$. We can convert this to a decimal value easily.
$(3):$ If we can prove these are all of the solutions for $k=3$ (and that sporadic solutions exist only for sufficiently small bases as I conjectured), we have solved the $k=4$ problem, the $4$-palindrome problem: There does not exist a number palindromic in four consecutive number bases.
The main thing needing proving is that $d\ge 9$ can't have solutions. Or can you find a counterexample, and actual solution? That would make this much more interesting.
The $d=9$ specifically for example will be very surprising if it had a solution (I found the period of solutions if they exist, must be larger than $500$ in this case, which is unlikely given $d=3,5$ have periods $2,4$ and $d=7$ has periods $2,6,12$). Period being families producing a solution every period amount of bases.
