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I'm trying to prove (or disprove) the following "conjecture".
Given the following set of powers of two:

$$A = \{ x \mid x = 2^n \text{ and } 2^{n-1} < 3^m < x < 3^{m+1} < 2^{n+1}\}$$

(informally $2^n$ is "snapped" in an interval $[3^m,3^{m+1})$ that contains only one power of two)

Q1: Given an integer $k \geq 1$, is there a way to pick an infinite subset of $B \subseteq A$ (or possibly the whole $A$) such that the following holds:

For all integers $x,y,p,q \geq 1$, with $x,y \in B$, $x \neq y$ and all the factors of $p, q$ are smaller than $k$, does there always exist $n \geq 1$ such that:

$$x + np \in A,\quad y + nq \notin A\quad \text{?}$$

Informally whatever pair of elements of $B$ and deltas $p,q$ I choose, the two arithmetic progressions $x' = x + np$ and $y' = y +nq$ don't guarantee that $x', y'$ are always both inside or both outside of $A$.

Q2: And is the "conjecture" easier if the set is: $$A' = \{ x \mid 2^{n-1} < 3^m \leq x < 2^n < 3^{m+1} < 2^{n+1}$$

The only fuzzy idea for Q2 I have is that if $x,y$ are big enough and belong to the same $3^m \leq x<y < 2^n$ interval, then $p,q$ (that have factors smaller than $k$) don't give enough "resolution" (??) to keep $x'$ and $y'$ "aligned".

Sorry for the possibly inaccurate terminology but I'm not an expert.

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I suspect that irrationality of log3/log2 will yield that any subset B will work, and that Q2 is not essentially different from Q1 in this regard. You might try showing that having two such a.p.'s agreeing on A would imply the ratio above is rational, although I don't know yet how to do that. – The Masked Avenger Mar 11 '14 at 16:17
    
@TheMaskedAvenger: I agree, I think that it's enough to pick B=A ... but I have no idea on what tools are needed to prove it, too (and my math knowledges are poor :( ... perhaps something like a very basic version of Dirichlet's theorem (??) and picking $n = n'x$, $x' = x(1 + n'p)$, $y' = y + n'xq$ – Marzio De Biasi Mar 11 '14 at 16:42
    
First show that as an indicator function, np+x is nonperiodic for any x. That should get you the case q=p. If q is different from p, what would it mean if the two ap's did agree? – The Masked Avenger Mar 11 '14 at 16:55
    
@TheMaskedAvenger: I added another condition that can simplify the problem (the factors of $p, q$ can be made arbitrarily smaller than $x$ and $y$ ...) in other words $y$ and $xq$ or $x$ and $yp$ can be considered coprimes ... but I don't know if it can help. – Marzio De Biasi Mar 11 '14 at 17:07
    
The members of $A$ are tabulated at oeis.org/A237514 – Gerry Myerson Mar 11 '14 at 22:49

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