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The wikipedia entry is reasonably coherent, and is pretty up-to-date. The twin prime conjecture itself remains open, but there has been recent remarkable progress on slightly weaker results. Namely, Y. Zhang proved that there is an infinite sequence of primes $p_n$ such that $p_{n+1} - p_n$ is uniformly bounded. (This result is referred to as bounded prime ...


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just to close my question. it's actually answered in the comments.


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Hmm, it is trivial that for any natural number $n \ge 1$ $$ \lfloor (3/2)^n \rfloor \lt (3/2)^n \lt \lceil (3/2)^n \rceil $$ and it is also easy to find that $$ (3/2)^n \lt {m \cdot 3^n-1\over m \cdot 2^n-1} \qquad \text{for $m \ge 1$ }$$ What you are looking for is a proof, that $$ {m \cdot 3^n-1\over m \cdot 2^n-1} \lt \lceil (3/2)^n \rceil \qquad ...


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I think Adam B has it. It's so common to look in problems for deeper links to more interesting structures (because those have often emerged in other proofs of open problems) it's easy to forget that - in a naive way - the problem is just those two points: 1) is there a loop? 2) is there a sequence that increases without bound? I'm going out on a limb ...


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Yes, we need more. Theorem. $\left\lfloor\left(\frac{3}{2}\right)^{n}\right\rfloor=\left\lfloor\left(\frac{3^{n}-1}{2^{n}-1}\right)\right\rfloor$ Proof We solve three inequalities for $n:1\rightarrow9$ and one larger $n$. \begin{equation} \left(\frac{3}{2}\right)^{n} < \frac{(2 m-1) 3^n-1}{(2 m-1) 2^n-1}\text{ (0.1) } \end{equation} ...



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