The conjecture is as follows:
Let $n\in\mathbb{N}\setminus\{1\}$. Define $a(n)=2^n+1$ and the set: $$S(n) = \{ (a(n)^m+1)/2\ :\ m\in \mathbb{N}_0\}.$$ Then for all $c\in\mathbb{N}$, the number $(a(n)^c-1)s_1s_2\cdots s_n$, where $s_i\in S(n)$, is palindromic in base $a(n)$ .
I pose this as a conjecture, which I tested well. My question is to find a proof or a disproof for this conjecture .
Example: Let $n=5$, and so $a(n)=2^5+1=33$. Take 5 values of $m$ (since $n=5$), e.g., $m_1=m_2=2$, $m_3=m_4=3$, $m_5=5$, and $c=4$. We have : $(33^2+1)/2=545$, $(33^3+1)/2=17969$, $(33^5+1)/2=19567697$, and $(33^4-1)=1185920$. Multiplying these numbers (with repetitions), we get : $$545^2*17969^2*19567697*1185920=2225542694023717115676496000$$ $$=1+1*33+3*33^2+5*33^3+5*33^4+10*33^5+9*33^6+11*33^7+14*33^8+10*33^9+14*33^{10}+11*33^{11}+9*33^{12}+10*33^{13}+5*33^{14}+5*33^{15}+3*33^{16}+1*33^{17}+1*33^{18},$$ which is palindromic in base 33.
(Note:what is written above is the conjecture that Max Alekseyev proved in Apr 20 2018, and what is written below is a modifying of the question by adding another conjecture which I ask for a proof or a disproof of it). The another conjecture says that for all $n\in\mathbb{N}\setminus\{1\}$ , $S(n)$ is the maximally dense set in $a(n)$ base numeral system with the property described in the conjecture that has been proven earlier in the question , That is , if we find an infinite set of numbers (call it $T$) ,such that for all $c\in\mathbb{N}$, the number $(a(n)^c-1)t_1t_2\cdots t_n$ , where $t_i\in T$, is palindromic in base $a(n)$ ;then $T$ is equal to $S(n)$ or $T$ is a proper subset of $S(n)$ or $T$ is a finite set of numbers union a proper subset of $S(n)$ . Thanks goes to Max Alekseyev .
