# $\lfloor \frac{3^n}{2^n} \rfloor = \lfloor \frac{3^n-1}{2^n-1} \rfloor$ for $n\geq2$

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 $$n\geq 2$$?

• This is an unsolved problem. See entry A002379 in the OEIS and references therein. – Nathaniel Johnston Nov 9 '19 at 18:26

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.