13
$\begingroup$

Double palindrome:

  • ...is a number nontrivially palindromic in two consecutive bases $b,b\pm1$

  • Let $d_1,d_2$ be numbers of digits in the two bases: nontrivially means $d_1,d_2\gt 1$.

  • Let $d=\max\{d_1,d_2\}$ be called the degree of a double palindrome.

  • Example: $10$ is palindromic in bases $(b,b-1)=(4,3)$ with $(d_1,d_2)=(2,3)$ digits: $$10=(1,0)_{10} =(2,2)_4=(1,0,1)_3$$

Theorem 1. If $d$ is even, there are no examples.

  • Even length (amount of digits) palindromes in base $b$ are divisible by $b+1$.
  • Thus, such palindrome will end in $0$ in the other base, and can't be a double palindrome.

From now on, assume we have an odd degree $d=2l+1,l\in \mathbb N$.

Theorem 2. If $d_1=d_2$, there are infinitely many double palindromes for every fixed $d$.

  • Example: $(b^{2l}-1)/(b+1)$ is palindromic in $(b,b+1)$ for all $b\gt \binom{2l}{l}$, with $d=2l-1$.

  • the above result was discussed and proven in my other question.


Conjecture. If $d_1\ne d_2$, there are finitely many double palindromes, for every fixed $d$.

Question. Is there any hope in proving this conjecture?


Results on small cases of $d$ via brute force search:

  • If $d=3$, it can be shown the only solution is $10$ in bases $3,4$, as:

$$(1,0)_{10}=(1,0,1)_3=(2,2)_4$$

  • For $d=5$, the following should be all of the solutions: $$ 130=(1, 1, 2, 1, 1)_{3}=(2, 0, 0, 2)_{4}\\ 651=(1, 0, 1, 0, 1)_{5}=(3, 0, 0, 3)_{6}\\ 2997=(1, 1, 5, 1, 1)_{7}=(5, 6, 6, 5)_{8}\\ 6886=(1, 0, 4, 0, 1)_{9}=(6, 8, 8, 6)_{10} $$

  • For $d=7$, the following should be all of the solutions: $$ 9222=(2, 1, 0, 0, 0, 1, 2)_{4}=(2, 4, 3, 3, 4, 2)_{5}\\ 26691=(1, 3, 2, 3, 2, 3, 1)_{5}=(3, 2, 3, 3, 2, 3)_{6}\\ 27741=(1, 3, 4, 1, 4, 3, 1)_{5}=(3, 3, 2, 2, 3, 3)_{6}\\ 626626=(1, 1, 5, 4, 5, 1, 1)_{9}=(6, 2, 6, 6, 2, 6)_{10}\\ 1798303=(1, 0, 1, 9, 1, 0, 1)_{11}=(7, 2, 8, 8, 2, 7)_{12}\\ 1817179=(1, 0, 3, 1, 3, 0, 1)_{11}=(7, 3, 7, 7, 3, 7)_{12} $$

And so on. For every $d$, solutions seem to only exits in relatively small bases.

For a general fixed $d=2l+1,l\in\mathbb N$, is it possible to set upper bounds on base $b$, after which solutions can't exits? - to prove the conjecture?


That is, how to show that double palindromes can't exist in (arbitrarily large) number bases $(b,b\pm1)$, when $b\gt b_0$, for some value $b_0:=b_0(d)$, if degree $d$ is fixed, and $d_1\ne d_2$?

Given $d=2l+1$ digits and bases $b,b+1$, then:

I have following data: digits [number of terms] (last b with terms / last b checked) {terms}

3  [1]    (3/100)  {10} 
5  [4]    (9/100)  {130, 651, 2997, 6886} 
7  [6]    (11/100) {9222, 26691, 27741, 626626, 1798303, 1817179} 
9  [12?]  (17/50)  {11950, 76449, 193662, 704396, 723296, 755846, 883407, 4252232, 10453500, 65666656, 2829757066, 7064428133} 
11 [14?]  (21/30)  {175850, 2347506, 2593206, 48616624, 160258808, 630832428, 5435665345, 8901111098, 9565335659, 37180386607, 131349132446, 746852269786, 7056481741113, 17985535104496} 
13 [>32?] (25/25)  {6643, 749470, 1181729, 17099841, 17402241, 25651017, 32317933, 295979506, 297379006, 402327541, 9689802280, 54453459798, 54464523606, 55027793502, 827362263728, 2909278729092, 2926072706292, 4036309890977, 7448647872250, 8013269088838, 17901027912530, 34577567573550, 34811609537160, 35194041720930, 54489277730565, 54768340178775, 55150772362545, 142077571662616, 682765460591464, 683230317449824, 733909097713709, 59777562308125626, ...} 
15 [>19]  (15/15)  {11435450, 203509031168, 204191148800, 231773764784, 321015775216, 3741580511478, 19404342621340, 41275222257214, 42143900934124, 218053292350812, 218210353012812, 218254595452812, 251569181965152, 259799383997952, 3338546970154550, 3617178283518590, 23044579418585216, 26926823266016368, 38322172687372936, ...}
17 [>21]  (12/12)  {16516113567, 16619231967, 198522549056, 204185363456, 240971251611, 246467321391, 303520083621, 330347455102, 341225573632, 4102350269485, 12262956787888, 13267882222408, 68995850733945, 1366179755723700, 1767662936108630, 4782537117352874, 5987078778707895, 140538057123815013, 388816019726293166, 396289206590671310, 411924791551509530, ...}
19 [>15]  (9/9)    {916821671, 956613659, 1136307905, 155784877126, 4262839618051, 126532386891655, 6615812399178042, 6622944330543930, 6641481107049786, 10688365729164780, 81877825421774500, 120168724989001390, 190076027720670091, 194216405504612491, 547906983389609745, ...}
21 [>9]   (6/6)    {1422773, 2806999337418, 3101308506654, 275956595195822, 451853066660344, 6116904274791985, 6875219172190387, 10229280954883514, 10231408608585002, ...}
23 [>8]   (5/5)    {5415589, 46746179770, 77887660577, 37004798195346, 47470618709562, 48517516968462, 3099677168429681, 9779924118261554, ...}
25 [>2]   (4/4)    {635913760790, 383478037564629, ...}
27 [>1]   (4/4)    {5892002867556037, ...}
...

That is, the conjecture is: How to prove that each row in this table will be finite?

$\endgroup$
6
  • $\begingroup$ Do you require the digits to be less than 10, even in bases $d_1,d_2$? $\endgroup$
    – almagest
    Commented Sep 24, 2019 at 12:49
  • $\begingroup$ @almagest (I assume you meant bases $b_1,b_2=b,b\pm 1$) The digits in base $b$ are as usual $\in[0,b-1]$. $\endgroup$
    – Vepir
    Commented Sep 24, 2019 at 12:51
  • 1
    $\begingroup$ Ah, yes. Sorry, you require the bases to be consecutive. Otherwise it would be too easy and we would be drowning in double palindromes! $\endgroup$
    – almagest
    Commented Sep 24, 2019 at 12:56
  • $\begingroup$ I suggest to add these numbers to the OEIS as a new sequence. $\endgroup$ Commented Sep 26, 2019 at 8:07
  • 1
    $\begingroup$ I suspect that the conjecture may not hold. Almost a counterexample with half-integer digits: $$(b^{10}+1) + \tfrac{17}{2}(b^9+b) + 38(b^8+b^2) + (\tfrac{b}2 + 104)(b^7+b^3) + (\tfrac{b}2 + 181)(b^6+b^4) + \tfrac{433}2 b^5$$ $$= (b - \tfrac{1}2)(c^9+1) + \tfrac{13}2(c^8+c) + (\tfrac{b}2 - 17) (c^7+c^2) + 23(c^6+c^3) + \tfrac{b-23}2(c^5+c^4),$$ which holds fot any $b$ and $c:=b+1$. $\endgroup$ Commented Sep 30, 2019 at 15:32

1 Answer 1

4
+50
$\begingroup$

The almost-counterexample I gave in the comments has factor 2 in the denominators, and this is not without a reason. In fact, this factor prevents existence of an infinite series of solutions of a fixed length. Here is a proof.

First notice that in an infinite series of solutions, values of $b$ cannot be bounded. This immediately proves the case $|d_1 - d_2|>1$ as one palindrome in this case is asymptotically at least factor $b$ times larger than the other. Hence, it remains to consider the case $|d_1-d_2|=1$.

Let $d=2l+1$ be the length of one palindrome and $d-1=2l$ be the length of the other. If $b$ is the base of the first palindrome, then the second must be in base $b+1$ (not $b-1$ as this palindrome divisible by the base plus 1). Then we need to solve $$\sum_{i=0}^{l-1} a_i (b^i + b^{2l-i}) + a_l b^l = \sum_{i=0}^{l-1} c_i ((b+1)^i + (b+1)^{2l-1-i})$$ in integers $a_0\in[1,b-1]$, $c_0\in[1,b]$, $a_i\in [0,b-1]$ and $c_i\in[0,b]$ for $i\in\{1,2,\dots,l\}$.

Linearizing this equation as explained in my MO answer to a related question and expressing $a_0$, $a_1$, and $c_0$, we get $$\begin{cases} a_0 = -k_d,\\ a_1 = -\frac{d}2 k_0 b + k_1 b - k_0 - \frac{d}2 k_d,\\ c_0 = a_1 - k_d b + k_{d-1}, \end{cases} $$ where we have $k_0,k_1,k_{d-1},k_d$ are some integers whose lower and upper bounds depend on $d$ but not on $b$.

(Argument below is simplified.)

To keep $a_1\in[0,b-1]$ and $c_0\in[1,b]$ for large $b$, the coefficients of $b$ in $a_1$ and $c_0$ must be between $0$ and $1$. Together with $a_0\geq 1$ (i.e. $k_d\leq -1$) this implies that $k_d=-1$ and the coefficient of $b$ in $a_1$ and $c_0$ equal $1$ and $0$, respectively. Then, however, $a_1$ is a half-integer, which is impossible. Thus, an infinite series of solutions does not exist. QED

$\endgroup$
2
  • $\begingroup$ This looks satisfying, I'll have to take a closer look once I'm able. - Perhaps it is now possible to find some bound on maximal $b$ that can have solutions, and start to more easily, and fully, classify solutions. (Per case of $d$). $\endgroup$
    – Vepir
    Commented Oct 3, 2019 at 17:04
  • $\begingroup$ @Vepir: Yes, the bound for $b$ would follow from the bounds on $k_{d-1}$ and $k_d$, which should be possible to estimate analytically. Asymptotically I believe $b=O(2^d)$. $\endgroup$ Commented Oct 4, 2019 at 12:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .