So, there's this equation that I've been pondering upon:(to slove for n) $\left\lfloor N\alpha ^{n} \right\rfloor = \left\lfloor N\alpha ^{n+1}\right\rfloor+1$
I can't remove the floors for approximation purposes because that would distort the answer too much. But, I thought of using the definition of floors.That is if
$\left\lfloor k\right\rfloor = m$
$m\leq k < m+1$.
Thus, here we can say that we've
$\left\lfloor N \alpha ^{n+1}\right\rfloor+1 \leq N\alpha ^{n} < \left\lfloor N \alpha ^{n+1}\right\rfloor+2$
But that would only complicate the matters more.
Thus, I'm stuck down here. Any help appreciated.


Hint: you are on the right track with thinking at the definitions. Write it as:

$$N \alpha^{n+1} \lt \lfloor N \alpha^n \rfloor \le N \alpha^n \lt \lfloor N \alpha^n \rfloor + 1 = \lfloor N \alpha^{n+1} \rfloor + 2 \le N \alpha^{n+1} + 2$$

$$ \implies \quad N \alpha^{n+1} \lt N \alpha^n \lt N \alpha^{n+1} + 2 \quad \iff \quad \alpha \lt 1 \lt \alpha + \frac{2}{N \alpha^n} $$

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    $\begingroup$ I am not said downvoter, but just to check, I think that should be a $2$ on the right hand side? Or a $\leq$ $\endgroup$ – Badam Baplan Mar 1 '17 at 4:32
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    $\begingroup$ @BadamBaplan Thank you for the correction, edited and fixed. $\endgroup$ – dxiv Mar 1 '17 at 4:44
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    $\begingroup$ Then finding the value of $n$ using the inequality $ 1<\alpha+\frac{2}{N\alpha ^{n}}$ ? @dxiv $\endgroup$ – Mooncrater Mar 1 '17 at 4:54
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    $\begingroup$ @Mooncrater Right, and $\,\alpha^n \lt \frac{2}{N(1-\alpha)}\,$ will give a lower bound for $n\,$. $\endgroup$ – dxiv Mar 1 '17 at 4:58
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    $\begingroup$ @Mooncrater No, that inequality actually gives a lower bound since $\alpha \lt 1$ and once the inequality is satisfied for $n_0$ it will automatically be satisfied for $\forall n \ge n_0\,$. But I realize that my hint doesn't address the upper bound. Is there any additional known relation between $N$ and $\alpha$ in that context? $\endgroup$ – dxiv Mar 1 '17 at 5:07

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