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 .



Conjecture on palindromic numbers
The conjecture is true.


Let $b=a(n)=2^n+1$.


First, notice that the number in question is
$$N=(b^c-1)\frac{b^{m\_1}+1}2\cdots \frac{b^{m\_n}+1}2 = \frac{b^c-1}{b-1}(b^{m\_1}+1)\cdots (b^{m\_n}+1).$$


Second, notice that 
$$M=(b^{m\_1}+1)\cdots (b^{m\_n}+1) = \sum\_{k=0}^{L} s\_k b^k,$$
where $L=m\_1+\dots+m\_n$, is a base-$b$ palindromic number, whose base-$b$ digit sum equals $2^n$. Indeed, $s\_k$ equals the number of subsets of $\{ m\_1, \dots, m\_n \}$ with the sum $k$. Since the total number of subsets is $2^n$, each $s\_k\leq 2^n$ and their sum is $2^n$. By considering subsets complements, we can also see that $s\_k = s\_{L-k}$, 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 $\frac{b^c-1}{b-1}$ is a base-$b$ repunit of length $c$.


Finally, notice that in 
$$N=\frac{b^c-1}{b-1}M = \sum\_k\frac{b^c-1}{b-1} s\_k b^k$$
the summation produces no carry-on's in base $b$ (since $\sum\_k s\_k = 2^n < b$).
Since $N$ can be viewed as the sum of base-$b$ palindromes of the form $\frac{b^c-1}{b-1}s\_k(b^k+b^{L-k})$ of fixed length $L+c$ (padded with leading zeros if needed), the absence of carry-on's implies that $N$ is a base-$b$ palindrome itself (with the digit sum equal $c2^n$). 


