Motivation

• Solving this will help find a "closed form" for double palindromes: A279092.

• Solving this, will help solve a similar Diophantine system whose solutions are "intersections" of: solutions to this problem for variable $b=b_0$ with solutions to this problem for variable $b=b_0-1$.

• Solving this problem, or solving the above linked similar Diophantine system using this problem's solution, will then help answer the question: Can a number be palindromic in more than 3 consecutive number bases?

Context

Let $n=(a_1,a_2,\dots,a_{l},a_{l+1},a_{l+2},\dots,a_{d-1},a_d)_b$ be digits of a $d=2l+1,l\in\mathbb N$ palindrome $n\in\mathbb N$ in some number base $b\in\mathbb N,b\gt 2$. "Palindrome" means: $a_i=a_{d-i+1},i=1,\dots,l+1$.

Solving the Diophantine system presented in the problem section, is equivalent to finding numbers that are simultaneously palindromic (palindromes) in two consecutive number bases $b,b-1$, and have exactly $d$ digits in both bases.

This can be generalized to $d_1,d_2$ digits in those two bases respectively, then $d=\max\{d_1,d_2\}$ is called the degree of palindrome $n$. We are observing the $d_1=d_2=d$ case, and the corresponding Diophantine system, in the problem section.

"Almost-all" numbers from A279092 are solutions to the below given Diophantine system. Specifically, all numbers from the linked OEIS sequence are either:

• a solution to the Diophantine system given in the problem section. (This system represents the $d_1=d_2$ case in the context of the mentioned generalization to $d$ as a degree.)
• a solution to the other part of the mentioned generalization. That is, the $d_1\ne d_2$ case.

We want to find "closed forms" for these simultaneous palindromes in two consecutive bases.

We formulated the following equivalent Diophantine system:

• Write the base $b$ palindrome $n$ in base $b-1$, using the binomial theorem on $a_i(b)^j=A_i((b+1)-1)^j,j=0,\dots,d-1$, to have base $b-1$ digits $A_i$ in terms of base $b$ digits $a_i$. Then introduce $o_i$ parameters defined to satisfy the inequalities on digits in base $b-1$, so we actually can have a valid base $b-1$ representation. Now solve for $A_{i}=A_{d-i+1},i=1,\dots,l+1$ to obtain palindromes, which is the given Diophantine system below (under conditions, so representations in both bases are valid):

**Problem**

Given $d=2l+1,l\in\mathbb N$, find all integers $a_1\in[1,b),a_2,\dots,a_{l+1}\in[0,b),b \gt 2$ such that:

$$ \sum_{s=1}^{i}\binom{d-s}{d-i}a_s + o_{i} - o_{i-1} (b-1)= \sum_{s=1}^{d-i+1}\binom{d-s}{i-1}a_s + o_{d-i+1} - o_{d-i+1} (b-1)$$

For $i=1,2,\dots,l+1$, where $o_1,\dots,o_{d-1}\in\mathbb Z$, $o_0=o_{d}=0$ are some integers, and such that for all $i\gt 1$, both LHS and RHS from all of the $l+1$ equalities are $\in[0,b-1)$, and for $i=1$, they are $\in[1,b-1)$. Notice that for $i=l+1$, equalitiy holds, and only RHS,LHS conditions need to be applied.

For every $x=(a_1,\dots,a_{l+1};b)$, there either exist unique $o_1,\dots,o_{d-1}$ such that (under which) $x$ is a solution to the system, or it is not a solution to the system.

The $d=3,5,7,\dots$ is called the degree, and $l=1,2,3,\dots$ the order, of this system.

Is this solvable for $d=2l+1$ in general, for all $x=(a_1,a_2,\dots,a_{l+1};b)$?

• So far, I found one family of solutions, that gives infinitely many solutions $x$, for every fixed $d$. But this is just a drop in the ocean of all solutions (families) that haven't been found.

Or, how can we go about solving this, and obtaining solutions, for arbitrary fixed $d$?

• I've solved it for $d=3,5$. For fixed $d=5$ already, the "closed form" for all of the solutions seems messy, as you will see by the end of this post.

**My progress on families of solutions across all $d$**

I don't know how to solve for all solutions in general. But I did find one family of solutions, giving infinitely many solutions for every $d$ (Thanks to @Peter). - This result is given in the context of double palindromes in the linked Peter's claim. This claim (result) is now proven.

That is, we have the following family of solutions;

$$x=\left(\left\{a_i=\begin{cases}b-\binom{2l_0}{2l_0-i},& i\text{ is odd}\\\binom{2l_0}{2l_0-i}-1, &i \text{ is even}\end{cases},i=1,\dots,l_0\right\};b\ge \binom{2l_0}{l_0}\right)$$

...is a solution to the Diophantine system for every $d=2l_0-1,l_0\in\mathbb N$ and $b\ge \binom{2l_0}{l_0}$. Since $d=1$ is not considered in the problem statement: let $l_0\gt 1$.

That is, substituting the above $x$ into the Diophantine system, will result in $"b-2=b-2","0=0"$ for "LHS=RHS" equations, for odd,even $i$ respectively,for all $d=2l+1,l=l_0+1$, for corresponding $o_i$ parameters.

For example, for $l=1,2,3,4,\dots$ we have $(o_i,i=1,\dots,2l)$ equal to: $$(2,1),(4,6,6,2),(6,15,24,21,12,3),(8,28,62,85,80,49,20,4),\dots$$ These are easy to determine since we know expected "LHS==RHS" for this family. That is, a closed form is possible for these $o_i$, but it is irrelevant since we know all $a_i,i=1,\dots,l+1$ and $b$ explicitly, for this family.

Question $1$. How can we generalize this $x=(a_1,\dots,a_{l+1};b)$, to find similar families, to encompass more solutions across more different $o_i$ sets of parameters, for every $d$?

**My progress on solving for all solutions for a fixed $d$**

I've also made computational progress, in cases of first few fixed values of $d$.

I have solved it computationally for smallest case, $d=3$, finding all solutions $(a_1,a_2;b)$.

For the next case, $d=5$, I needed to make some workarounds. That is, solve the system under fixed $o_i$ parameters. I individually look at sets of $o_i$ parameters under which the system has solutions, after eliminating sets of $o_i$ parameters under which the system can't have solutions, computationally, to be able to now solve for all $(a_1,a_2,a_3;b)$ computationally. Like this, I also managed to solve the $d=5$ case, for all soltuions.

But for $d\ge 7$, even when trying to solve under individual fixed $o_i$ parameters, some sets of such parameters can't be solved (with my implementation). I have some families of solutions for $d=7$, but I have not solved this case completely (for all families of solutions), using my computational implementation.

For $d\ge 9$, my implementation can't solve for entire families. I can only computationally solve for individual solutions, under fixed $(d,b)$ parameters. - This gets on average, exponentially solver in regards to increasing $d$ needed to be solved.

Even If I could solve for them, there does not seem to be a "nice closed form" to represent the solutions, when working with fixed cases of $d$.

Question $2$. Is it possible to make further progress on this problem?

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**More details about my progress on solving fixed $d$**

I tried using a Computer-Algebra-System, namely Mathemtica, to try to solve this for small $d$.

First case, $d=3$, can be solved using Reduce[], after implementing the system in Mathematica.

$(d=3)$ That is, we have the equalities $1,\dots,l$ (that is, one equality in this case): $$ a_1+o_1=2 a_1+a_2-o_2(b-1) $$ With conditions on LHS,RHS for $i=1,\dots,l+1$ as: $$ a_1+o_1\in[1,b-1)\\ 2 a_1+a_2-o_2(b-1)\in[1,b-1)\\ 2 a_1+a_2-o_2(b-1)\in[0,b-1)\\ $$ Where the problem conditions are $o_1,o_2\ge 0,a_1\in[1,b),a_2\in[0,b),b\gt 2$.

$(d=3)$ solutions are possible only if $(o_1,o_2)\in\{(1,1),(2,1)\}$. Each gives one family:

$$\begin{array}{} (o_1,o_2) & a_1 & a_2 & b \\ (1,1) & x+1 & y+4 & a_1+a_2 \\ (2,1) & x+2 & 5 & a_1+4 \end{array}$$

Where $x,y\in\mathbb N=\{0,1,2,\dots\}$. Here is the Mathematica code.

But for $d\ge 5$, the Reduce[] halts - keeps running forever, and can't solve it for all $((a_i);b)$.

For $d=5$, it is still possible to extract and solve all fixed $o_i$ parameter sets with solutions, if handled individually (after eliminating family of sets that do not have solutions).

$(d=5)$ That is, we have the equalities $1,\dots,l=2$, in this case: $$\begin{align} a_1 + o_1 &= 2 a_1 + 2 a_2 + a_3 - o_4 (b-1) \\ 4 a_1 + a_2 - o_1 (b-1) + o_2 &= 4 a_1 + 4 a_2 + 2 a_3 - o_3(b-1)+o_4 \end{align}$$ With conditions on LHS,RHS for $i=1,\dots,l+1$ as: $$\begin{align} a_1 + o_1&\in[1,b-1)\\ 2 a_1 + 2 a_2 + a_3 - o_4(b-1)&\in[1,b-1)\\ 4 a_1 + a_2 - o_1(b-1) + o_2&\in[0,b-1)\\ 4 a_1 + 4 a_2 + 2 a_3 - o_3(b-1) + o_4&\in[0,b-1)\\ 6 a_1 + 3 a_2 + a_3 - o_2(b-1) + o_3&\in[0,b-1) \end{align}$$ Where the problem conditions are $o_1,o_2,o_3,o_4\ge 0,a_1\in[1,b),a_2,a_3\in[0,b),b\gt 2$.

I have solved this case computationally to obtain all the solutions:

$(d=5)$ There are $12$ sets $(o_1,o_2,o_3,o_4)$ under which solutions can be obtained:

$$\begin{array}{} (o_1,o_2,o_3,o_4) & a_1 & a_2 & a_3 & b \\ (2,4,3,1) & 2 & \in\{2,3\} & a_1-a_2+1 & 2a_1+a_2 \\ (2,4,3,1) & \in\{3,4\}& \in\{1,2\} & a_1-a_2+1 & 2a_1+a_2 \\ (2,4,3,1) &\in[4,8]&0 & a_1-a_2+1 & 2a_1+a_2 \\ (2,4,3,1) &\in\{5,6\} & 1 & a_1-a_2+1 & 2a_1+a_2 \\ (2,5,5,2) &2 &\in\{3,4\} &3a_1 & 2a_1+a_2 \\ (2,5,5,2) &1 &4 &3a_1 & 2a_1+a_2 \\ (2,4,5,2) &\in\{1,7\} &\in\{7,8\} &3a_1-2 & 2a_1+a_2-1 \\ (2,4,5,2) &\in\{2,3,6\} &\in\{6,7,8\} &3a_1-2 & 2a_1+a_2-1 \\ (2,4,5,2) &\in\{4,5\} &\in[5,8] &3a_1-2 & 2a_1+a_2-1 \\ (2,4,5,2) &8 &8 &3a_1-2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+4 &4 &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+5 &5 &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+6 &\in\{3,6\} &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+7 &7 &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+8 &2 &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+10 &1 &a_1-a_2 & 2a_1+a_2-1 \\ (2,3,3,1) &x+12 &0 &a_1-a_2 & 2a_1+a_2-1 \\ (4,8,8,3) &8 &13 &9 &14 \\ (4,8,8,3) &9 &13 &11 &15 \\ (4,8,8,3) &10 &13 &13 &16 \\ (4,8,8,3) &11 &13 &15 &17 \\ (4,8,8,3) &12 &13 &17 &18 \\ (4,6,6,2) &2x+14 &14 &a_1-14 & 2x+20 \\ (4,6,6,2) &2x+15 &14 &a_1-14 & 2x+21 \\ (1,3,4,2) &x+1 &x+y+9 &3a_1-1 &2a_1+a_2 \\ (1,3,2,1) &x+y+3 &y+4 &a_1-a_2+1 &2a_1+a_2+1 \\ (1,4,4,2) &x+1 &x+\{5,6\} &3a_1+1 &a_1+a_2+1 \\ (1,4,4,2) &x+\{2,3,4\} &x+5 &3a_1+1 &a_1+a_2+1 \\ (3,6,5,2) &2x+2y+12 &2x+12 &a_1-a_2+1 &3x+2y+18 \\ (3,6,5,2) &2x+2y+11 &2x+12 &a_1-a_2+1 &3x+2x+17 \\ (3,6,7,3) &2x+2y+20 &4x+2y+37 &2x+3y+20 &4x+3y+38 \\ (3,6,7,3) &2x+2y+21 &4x+2y+39 &2x+3y+21 &4x+3y+40 \\ (3,7,7,3) &2 (4+x) &2 (x+\{6,7\}) &4 (4+x) - (x+\{6,7\}) &2 (4+x) + (x+\{6,7\}) \\ (3,7,7,3) &2 (5+x) &2 (x+\{6,9\}) &4 (5+x) - (x+\{6,9\}) &2 (5+x) + (x+\{6,9\}) \\ (3,7,7,3) &2 (6+x) &2 (x+11) &4 (6+x) - (x+11) &2 (6+x) + (x+11) \\ (3,7,7,3) &2 (7+x) &2 (x+13) &4 (7+x) - (x+13) &2 (7+x) + (x+13) \\ (3,7,7,3) &2 (8+x) &2 (x+15) &4 (8+x) - (x+15) &2 (8+x) + (x+15) \\ (3,7,7,3) &2 (x+[3,5]) + 1 &2 (x+6) +1 &4 (x+[3,5]) - (x+6) +2 &2 (x+[3,5]) + (x+6) +1 \\ (3,7,7,3) &2 (x+4) + 1 &2 (x+8) +1 &4 (x+4) - (x+8) +2 &2 (x+4)+ (x+8) +1 \\ (3,7,7,3) &2 (x+5) + 1 &2 (x+10) +1 &4 (x+5) - (x+10) +2 &2 (x+5) + (x+10) +1 \\ (3,7,7,3) &2 (x+6) + 1 &2 (x+12) +1 &4 (x+6) - (x+12) +2 &2 (x+6) + (x+12) +1 \\ (3,7,7,3) &2 (x+7) + 1 &2 (x+14) +1 &4 (x+7) - (x+14) +2 &2 (x+7) + (x+14) +1 \\ \end{array}$$

Where $x,y\in\mathbb N=\{0,1,2,\dots\}$. Here is the raw solution output.

$(d=7) \text{ Partial solution.}$ We can similarly obtain some solution families for some $o_i$ parameters for the $d=7$ case, but my implementation couldn't solve it in general. This can be seen by the end of the following answer - which also has $d=5$ written out in a different format of expressions, separating finite and infinite families.

$(d\ge 9) \text{ Unsolved.}$ I couldn't solve for entire families of solutions with my implementation, for $d\ge 9$ cases of the Diophantine system. Solutions for fixed $(d,b)$ cases can be obtained using the Mathematica code from the end of the following answer that solves a similar system in the context of double and triple palindromes.

I also forgot to mention, that it is sufficient to observe $o_i\in\mathbb N=\{0,1,2,\dots\}$ instead in $\mathbb Z$, to obtain all solutions for some $d$. (Look at the equalities when $o_i\le 0$.)