Double palindrome:
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...is a number nontrivially palindromic in two consecutive bases $b,b\pm1$
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Let $d_1,d_2$ be numbers of digits in the two bases: nontrivially means $d_1,d_2\gt 1$.
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Let $d=\max\{d_1,d_2\}$ be called the degree of a double palindrome.
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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 $d$.
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Example: $(b^{2l}-1)/(b+1)$ is palindromic in $(b,b+1)$ for all $b\gt \binom{2l}{l}$, with $d=2l-1$.
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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:
$$(1,0)_{10}=(1,0,1)_3=(2,2)_4$$
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For $d=5$, the following should be all of the solutions: $$ 130=(1, 1, 2, 1, 1)_{3}=(2, 0, 0, 2)_{4}\\ 651=(1, 0, 1, 0, 1)_{5}=(3, 0, 0, 3)_{6}\\ 2997=(1, 1, 5, 1, 1)_{7}=(5, 6, 6, 5)_{8}\\ 6886=(1, 0, 4, 0, 1)_{9}=(6, 8, 8, 6)_{10} $$
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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?
from Hot Weekly Questions - Mathematics Stack Exchange
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