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I posed a conjecture which is a generalization of Conjecture on palindromic numbers

My question is to find a proof or a disproof of it. I have put it in the comments section of the page of the OEIS sequence A266577. Where OEIS is the online encyclopedia of integer sequences. Here is the link of the sequence: https://oeis.org/A266577

The conjecture is as follows: For integers $n$ and $m > 1$, let $b(n,m)=n^m+1$, $S(n,m) =$ sequence of numbers of the form $(b(n,m)^k+\cdots+b(n,m)^{(n-1)k}+1)/n$, where $k$ is any nonnegative integer. Then for each positive integer $s$, $(b(n,m)^s-1)\cdot$(product of any $m$ not necessarily distinct terms of $S(n,m)$) is palindromic in base $b(n,m)$.

Example: assume $n=5$ and $m=3$, then $b(5,3)=5^3+1=126$. Assume $k_1=1$ and $k_2=1$ and $k_3=2$ (they are three values since $m=3$). Assume $s=3$; then we have the calculation $$((126+126^2+126^3+126^4+1)/5)^2\cdot (126^2+126^4+126^6+126^8+1)/5\cdot (126^3-1)$$ which is equal to: $32807046133985032885720309126001$ and this number has the base-$126$ expansion $(1,3,7,12,19,25,31,34,37,37,37,34,31,25,19,12,7,3,1)_{126}$ which is a palindromic number in base $126$.

Professor Amiram Eldar wrote the following code in the PAR/GP language:

ispal(d) = Vecrev(d) == d ;
is(n, m, s) = {my(b = n^m + 1, c = (b^s - 1) * prod(k = 1, m, (1 + sum(i = 1, n-1, b^(i*k)))/n), d = digits(c,b)); ispal(d);}
for(n = 2, 30, for(m = 2, 30, for(s = 1, 30, if(!is(n, m, s), print(n, m, s); break))))

It runs successfully, without finding any counter-example.

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    $\begingroup$ This is not really readable. Can you please use MathJax to typeset it properly? $\endgroup$ Commented May 15 at 11:42
  • $\begingroup$ @Dave Benson Thank you for advice. I have some learning difficulties but I promise you to do my best. $\endgroup$ Commented May 15 at 11:48

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The conjecture is true and is proved similarly to my proof of the previous conjecture.

Let $n,m$ be fixed and $b:=b(n,m)=n^m+1$.

First, note that the numbers in question has the form: $$N:=(b^s-1)\frac{b^{nk_1}-1}{(b^{k_1}-1)n}\cdots \frac{b^{nk_m}-1}{(b^{k_m}-1)n} = \frac{b^s-1}{b-1}\frac{b^{nk_1}-1}{b^{k_1}-1}\cdots \frac{b^{nk_m}-1}{b^{k_m}-1}.$$

Second, note that $$M:=\frac{b^{nk_1}-1}{b^{k_1}-1}\cdots \frac{b^{nk_m}-1}{b^{k_m}-1} = \sum_{j=0}^{L} s_j b^j,$$ where $L:=(k_1+\dots+k_m)(n-1)$, is a base-$b$ palindromic number, whose base-$b$ digit sum equals $n^m=b-1$. Indeed, $s_j$ equals the number of solutions to $k_1x_1 + \dots + k_mx_m = j$ with $(x_1,\dots,x_m)\in \{0,1,\dots,n-1\}^m$. That is, each $s_j\leq n^m$ and their sum is $n^m$. By considering complements $x'_i:=n-1-x_i$, we can also see that $s_j = s_{L-j}$ for any $j$, i.e., $M$ is a base-$b$ palindromic number of length $L+1$ (starting and ending with digit $s_0=s_L=1$).

Third, the number $\frac{b^s-1}{b-1}$ is a base-$b$ repunit of length $s$.

Finally, note that in $$N=\frac{b^s-1}{b-1}M = \sum_{j=0}^L\frac{b^s-1}{b-1} s_j b^j$$ the summation produces no carries in base $b$ (since $\sum_j s_j = n^m < b$). Since $N$ can be viewed as the sum of base-$b$ palindromes of the form $\frac{b^s-1}{b-1}s_j(b^j+b^{L-j})$ of fixed length $L+s$ (padded with leading zeros as needed), the absence of carries implies that $N$ is a base-$b$ palindrome itself (with the digit sum equal $sn^m$).

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