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I stumbled upon the following claim online: $\lfloor \frac{3^n}{2^n} \rfloor = \lfloor \frac{3^n-1}{2^n-1} \rfloor$ for all integers $n\in \mathbb{N}$, $n\geq2$. Checking with the computer, the claim seems to be true at least for integers $n$ such that $2 \leq n \leq 10000$. How could I prove this claim is true for all integers such that $n\geq 2$?

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    $\begingroup$ This is an unsolved problem. See entry A002379 in the OEIS and references therein. $\endgroup$ – Nathaniel Johnston Nov 9 '19 at 18:26
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Mahler proved in 1957 (see here) that if $q$ is a positive rational number which is not an integer, then the distance of $q^n$ to the nearest integer is $(1-o(1))^n$. In particular, taking $q=3/2$, we have for $n$ sufficiently large that $$\lfloor q^n\rfloor+1-q^n>(q/2)^n>\lfloor q^n\rfloor/2^n.$$ Rearranging the two sides, we get for $n$ sufficiently large that $$\left\lfloor\frac{3^n}{2^n}\right\rfloor>\frac{3^n-1}{2^n-1}-1,$$ hence also that $$\left\lfloor\frac{3^n}{2^n}\right\rfloor=\left\lfloor\frac{3^n-1}{2^n-1}\right\rfloor.\tag{$\ast$}$$ Mahler's proof is ineffective (i.e. it does not produce a lower bound for $n$), because it relies on Roth's approximation theorem, which is ineffective. As far as I know, we still don't have an effective lower bound for $n$ beyond which $(\ast)$ is guaranteed to hold.

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