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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 .

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    $\begingroup$ I've edited your question to make it readable. $\endgroup$ – Max Alekseyev Apr 20 '18 at 1:27
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    $\begingroup$ If you have a new question, Ahmad, I think it's best to post it as a new question (with links between it and this question), rather than to bury it as an edit here. $\endgroup$ – Gerry Myerson Apr 27 '18 at 21:21
  • $\begingroup$ @Gerry Myerson ,I was unlucky because I saw your advise after starting the question as a bounty , but I hope the question will take attention , thank you very much . $\endgroup$ – Ahmad Jamil Ahmad Masad Apr 27 '18 at 21:24
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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 $c2^n$).

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