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Erdős conjectured that, except $1, 4$ and $256$, no power of $2$ is a sum of distinct powers of $3$.

A vice-versa conjecture may be: except $1$, $9$ and $81$, all powers of $3$ contains two consecutive powers of $2$ or/and two missed consecutive powers of $2$ in their binary expansions.

We can generate the powers $(3^n)$ as a cellular automaton where the $n$-th line represents the binary expansion of $3^n$ and we can observe that the conjecture is still verified as showed in the next image with 128 lines. 3 powers binary cellular automata

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    $\begingroup$ Every integer is a sum of powers of 3. Do you mean "distinct powers of 3"? $\endgroup$
    – Joël
    Commented Feb 14, 2017 at 15:25
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    $\begingroup$ Just on probabilistic grounds, if your conjecture is true for small $n$ it is quite likely true. Proving it is another matter. I would be surprised if it were tractable. $\endgroup$
    – Robert Israel
    Commented Feb 14, 2017 at 15:30
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    $\begingroup$ Yes, "sum of distinct powers of 3" $\endgroup$
    – Douzi Hassan
    Commented Feb 14, 2017 at 15:32
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    $\begingroup$ The first 640 powers of 3 in binary are tabulated at oeis.org/A004656/b004656.txt $\endgroup$
    – Gerry Myerson
    Commented Feb 14, 2017 at 23:04
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    $\begingroup$ I find better to visualize the powers of3 as an automata cellular image $\endgroup$
    – Douzi Hassan
    Commented Feb 15, 2017 at 9:39

2 Answers 2

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Your vice-versa conjecture is effectively a conjecture in the spirit of the one of Erdős, and as Robert Israel said in the comments, there is no hope to prove such conjectures.

Since the work of Narkiewicz, who studied the problem raised by Erdős, and whose main result remains very far from the conjecture of Erdős (Narkiewicz, W., A note on a paper of H. Gupta concerning powers of two and three, Univ. Beograd Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No 678 – No 715 (1980), 173–174; MSN), very little has been done.

Therefore, I don't pretend to answer your question by proving the conjecture, but simply by raising some further considerations.

Some considerations on your vice-versa conjecture. Quite natural conjectural hypotheses on the coefficients of binary expansions of powers of three, allow to extend the vice-versa conjecture in various ways. Here's a few examples:

Conjecture 1. Except for $1,3,9,27,81,729,2187$ and $6561$ all powers of $3$ contain three consecutive powers of $2$ in their unique expression as a sum of distinct powers of $2$.

Conjecture 2. For $n>25$ all $n$-th powers of $3$ contains four consecutive powers of $2$ in their unique expression as a sum of distinct powers of $2$.

and more generally:

Conjecture 3. For every $M$, there exists $N$ such that for $n>N$ all $n$-th powers of $3$ contains $M$ consecutive powers of $2$ in their unique expression as a sum of distinct powers of $2$.

Conjecture 4. Let us define a "word" as a finite sequence of 0's and 1's. For every word $W$, there exists $N$ such that for $n>N$ all $n$-th powers of three contains the word $W$ in the binary expansion of $3^n$.

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    $\begingroup$ Spelling note: Erdős is spelled thus, with an "o with double acute" ő, not an "o with diaeresis" ö. I have edited accordingly. I also removed "Yes" at the beginning of your post, since it looked like an answer to the question but isn't. I hope that was all right. $\endgroup$
    – LSpice
    Commented Oct 27, 2023 at 22:56
  • $\begingroup$ LSpice, Thank you. I appreciate your edit. I know how Erdős is spelled (I have Erdős number 1), but in my keybord there is noy key for that, so I have to copy paste from other sources. $\endgroup$
    – G. Melfi
    Commented Oct 28, 2023 at 10:53
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    $\begingroup$ Re, my apologies for assuming, and congratulations on your Erdős number! You can copy and paste the relevant character from, for example, the relevant Wiki page. $\endgroup$
    – LSpice
    Commented Oct 28, 2023 at 15:02
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After the latest edit, the question became much easier to answer: the only numbers whose binary expansion does not contain two consecutive powers of $2$ or two missed consecutive powers of $2$ are those whose binary expansion is $1010\dots$, i.e., number of the form $\lfloor2^n/3\rfloor$. Since $2^n$ or $2^n-2$, being even, cannot be a nontrivial power of $3$, this means the question is equivalent to: there are no integers $(n,m)$ such that $2^n-1=3^m$ other than $(1,0)$ or $(2,1)$. This special case of Mihăilescu’s theorem (Catalan’s conjecture) has an elementary proof due to Gersonides.

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  • $\begingroup$ Thank you for your answer, In fact there are 2 options "or" or "and". Indeed the “and” option is trivial. $\endgroup$
    – Douzi Hassan
    Commented Jan 3 at 21:16
  • $\begingroup$ If you want to postulate two conjectures, one with "or" and one with "and", you need to write it clearly in the question. The way it is written now, there is a single conjecture using "or/and", which on the face of it looks like a variant of the common expression "and/or", whose meaning is exactly the same as (non-exclusive) "or". $\endgroup$
    – Emil Jeřábek
    Commented Jan 4 at 9:34
  • $\begingroup$ OK, but it's not yet a theorem, it's just a conjecture. We have to leave space for imagination. Your comments are a proof of this. Isn't it? $\endgroup$
    – Douzi Hassan
    Commented Jan 5 at 10:49

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