Can a natural number be nontrivially palindromic in more than $3$ consecutive number bases?

Nontrivially means that I'm not counting one-digit palindromes.

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.