Return to Question

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.

Was initially asked on MSE - but wasn't solved, and still isn't.

Smallest number $N$ which is nontrivially palindromic in $x$ consecutive number bases:

$$ \begin{array}{|c|} \hline x& N & \text{Palindrome} \\ \hline 1& 3 & 11_2 \\ 2& 10 & 101_3=22_4\\ 3& 178 & 454_6 =343_7 = 262_8\\ 4& ?& \\ \hline \end{array} $$

Solution for $x=4$ probably does not exist;

One idea I had is to rely on these observations (taken in short, from below):

• $(1)$ That $3$ digit numbers will never be palindromic in four consecutive bases , where by 3 digits I mean the 3 digits when written in the consecutive palindromic bases.

• $(*)$ all numbers palindromic in $3$ out of $4$ (including first and last in those three) consecutive bases (and not divisible by the other one) have $3$ digits in those bases, except the $2$ observed exceptions so far. (see last paragraph at the end of the post) - in other words, a solution for $x\ge4$ should have $3$ digits in its palindromic bases, if it exists.

But I do not know how to show (prove) these observations to be true.

If $(1)$ and $(*)$ are true, they contradict each other, thus $x\ge4$ does not have a solution.

$$\text{Looking at consecutive palindromes in three bases}$$

We can look at patterns for $3$ consecutive bases.

It was shown on MSE for the 3 digit pattern, those solutions won't extend to a fourth base, which isn't hard to see. If only we could also show that these are the only solutions (including the one exception), then $(1)$ is true.

The patterns for other digits so far seem to follow with similar patterns.

$(0)$ Lets start by excluding even length palindromes ($2p$ digits, $p\in\mathbb N$) since they can't be palindromic in consecutive number bases. This follows directly from the fact that a palindrome in base $b$ is divisible by $b+1$.

(To be more precise, I consider base $b$ for digits and bases $b+1,b+2,\dots$ for additional consecutive bases; Example of $10=101_3=22_4$ is contained as $3$ digit example for $x=2$ consecutive bases in those two bases.)

Following that, then we choose to observe odd digit length palindromes of $2d+1$ digits, $d\in\mathbb N$, which are palindromic in $3$ consecutive number bases $b\in\mathbb N \ge 2$, and we have so far ($x=3$):

Pick ($n=2k+3, k\in\mathbb N$), then we get a new example for every $n$, of form:

$$\frac{1}{2}(n^3 + 6n^2 + 14n + 11)$$

Which is palindromic in bases $n+1, n+2, n+3$.

Other than these, we have one more example: $300 = 606_7 = 454_8 = 363_9$

$(1)$ How to show (prove) that all 3 digit examples other than $300$ are in this pattern?

(No new exceptions or patterns exist for $b\lt2333$, so far. Verified here using python code)

You can see that the first pattern emerges at number base $b=6$, and no new patterns emerge for the next $2300$ number bases, which makes the existence of a second pattern (or more exceptions) very unlikely.

Pick ($n=4k+40, k\in\mathbb N$), then we get a new example for every $n$, of form:

$$\frac{1}{4}(3n^5 + 30n^4 + 125n^3 + 270n^2 + 307n + 148)$$

Which is palindromic in bases $n+1, n+2, n+3$; No other examples (exceptions) exist.

$(2)$ How to show (prove) that all 5 digit examples are in this pattern?

(No exceptions or new patterns exist for $b\le333$, so far. Verified here using python code)

Update: Based on computed examples and observations so far, multiple patterns and exceptions seem to exist for this case. Three patterns seem to exist:

Pick ($n=2k+79, k\in\mathbb N\cup \{-3,-2,-1\}$), there are infinitely many examples of form:

$$\frac{1}{2}(n^7+14n^6+88n^5 + 320n^4 + 718n^3 + 980n^2 + 748n + 247)$$

Pick ($n=6k+67, k\in\mathbb N\cup \{-2,-1,0\}$), there are infinitely many examples of form:

$$\frac{1}{6}(n^7+14n^6+88n^5 + 320n^4 + 718n^3 + 980n^2 + 748n + 245)$$

Pick ($n=12k+24, k\in\mathbb N$), there are infinitely many examples of form:

$$\frac{1}{12}(2 n^7 + 30 n^6 + 209 n^5 + 852 n^4 + 2117 n^3 + 3114 n^2 + 2474 n + 816)$$

All examples generated by the above are palindromic in number bases $n+1, n+2, n+3$.

Beside patterns, we have exceptions, (examples that do not fit into any patterns) :

9 3360633
13 19987816
15 43443858 
22 532083314 
26 1778140759 
28 2721194733 
28 11325719295 
36 47622367425 
40 97638433343 
42 224678540182 
43 265282702996 
48 561091062285 

These are palindromic in column given base $b$ and bases $b+1,b+2$.

$(3)$ How to show (prove) that all 7 digit examples other than these $12$ exceptions are in one of these three patterns (that these are the only patterns)?

(No new exceptions or patterns exist for $b\le111$, so far. Verified here using python code)

No examples have been found so far. I've checked number bases $b\le50$ so far, here.

Haven't searched for examples yet, as $d=4$ is already taking a long time per number base.

Some of these examples and patterns are also mentioned in an OEIS sequence.

How can one find these polynomial pattern expressions for some $d$ algebraically? Rather than needing to compute a lot of examples and then fitting them in a polynomial of degree $2d+1$?

Is there anything out there that can actually be used on this problem?

You can run the python code here and modify digit = 1 variable to check 2*digit+1 digit examples; and also modify variables under #overwrite: if you wish.

P.S. Can the python code I'm using be more optimized? (Is there a faster way to compute this?)

$$\text{Looking at almost four consecutive palindromes}$$

"Almost palindromic in four bases" - if it is palindromic in bases $b, b+3$ and in $b+1$ or $b+2$.
I checked how many of these are in the following digit groups up to some number base:

($b\le900$) For $3$ digits, there are $\gt1484$ examples.

($b\le150$) For $5$ digits, only two examples at $b=16$ and at $b=17$

($b\le50$) For $7$ digits, there are no examples.

($b\le30$) For $9$ digits, there are no examples.

Haven't checked $\ge11$ digit examples, yet.

(More digits get harder to compute/check)

Edit: Removed odd cases from above as $(*)$ was rewritten a bit (below).

The example output is here (including even digit cases), and the example code can be run here; where you can modify the bases (and digits) being checked.

Two observed exceptions so far, are $71240$ for $b=16$ and $1241507$ for $b=17$, both having $5$ digits. The second exception being a twin prime.

Then we can form the $(*)$ observation as:

All numbers that are not divisible by $b,b+1,b+2,b+3$ and are palindromic in $b,b+3$ and either $b+1,b+2$ number bases, must have $3$ digits in those bases or are in the finite set $E$ where $E$ so far is $E=\{71240, 1241507\}$.

But proving this seems equally hard, if not harder, than proving the initial question itself.

From this: If only 3 digit examples have a chance to be palindromic in $4$ consecutive bases, and $E$ has no more new exceptions (as observed here so far), then it is only needed to prove no new 3 digit solutions exist other than ones under $(1)$ above.

$2017:$ 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)

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.

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:

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.

$(3)$ 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:

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.