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My son, who is 16, is doing some independent research. A lower bound depending on $n$ for $\left\{ \left( \frac{4}{3} \right)^n \right\}=\left( \frac{4}{3} \right)^n-\left\lfloor \left(\frac{4}{3} \right)^n \right\rfloor$, the fractional part of $\left( \frac{4}{3} \right)^n$, might help him improve his results. Note that $\frac{1}{3^n}$ is an obvious bound. Is there a better bound known?

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    $\begingroup$ You can tart with articles by Arturas Dubickas like On the distance from a rational power to the nearest integer and Arithmetical properties of powers of algebraic numbers. $\endgroup$ Mar 25, 2018 at 10:20
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    $\begingroup$ You actually have $7/3^n$ as a lower bound since $4^n\not\equiv 1\pmod{3^n}$, else you would have $3^n$ dividing $4^n-1=(2^n-1)(2^n+1)$, as a result of which $3^n$ would have to divide one of $2^n-1$ or $2^n+1$; moreover, $4^n\not\equiv 2\pmod{3^n}$ in view of $4^n\not\equiv 2\pmod 3$, and there are equaly simple reasons why $4^n\notin\{3,4,5,6\}\pmod{3^n}$. $\endgroup$
    – Seva
    Mar 25, 2018 at 10:26
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    $\begingroup$ There may be more literature on the $(3/2)^n$ problem. As a survey the paper impan.pl/en/publishing-house/journals-and-series/… by Flatto, Lagarias,; Pollington On the range of fractional parts {ξ(p/q)n. Acta Arith. 70 (1995), no. 2, 125–147, mathscinet.ams.org/mathscinet-getitem?mr=1322557 looks good, including the more recent items that refer to this paper in mathscinet $\endgroup$ Mar 25, 2018 at 11:16
  • $\begingroup$ I probably miss something, but wouldn't it be insightful to use the Newton binomial for $ (1+1/3)^{n} $ and consider $ \binom{n}{k}\mod 3^k $? $\endgroup$ Mar 28, 2018 at 21:38
  • $\begingroup$ Indeed, I am thinking about that. Unfortunately, for k larger than n/2 we get significant contributions, and it is looking like the terms are less than 1 only when k gets above 0.7n (or about a little less than (log_3 2 ) times n). Also, the contribution of the larger terms may be unfortunate enough that it will cancel much of the gain incurred from looking at the smaller terms. Still thinking on it though. Gerhard "Don't Give Up Just Yet" Paseman, 2018.03.28. $\endgroup$ Mar 28, 2018 at 21:45

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A non-trivial lower bound can be obtained using linear forms in $p$-adic logarithms. Suppose that $\left\{\left(\frac{4}{3}\right)^n\right\}$ is small. Clearly it is a rational number with denominator $3^n$, so assume it is $\frac{a}{3^n}$. Then $3^n|4^n-a$, that is, $\nu_3(4^n-a)$ is exceptionally large.

In general, the theory of linear forms gives lower bounds for the (archimedean or non-archimidean) distance of an expression of the form $\alpha_1^{b_1}\alpha_2^{b_2}\cdots\alpha_k^{b_k}$ from 1. In this case we have $k=2$, and there are better results known then for $k\geq 3$. Using a result by Bugeaud and Laurent (J. Number Theory 61 (1996), 311-342, Corollary 2) one can get $a>\frac{n}{12000}$, that is, $\frac{a}{3^n}>\frac{n}{12000\cdot 3^n}$.

Unless there is some elementary trick, I doubt that something significantly better can be proven.

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    $\begingroup$ Has anyone looked at using the binomial theorem on (1 + 1/3)^n? Maybe the tail terms yield a nice provable lower bound of 1/3^{n - something}. Maybe I will try it. Gerhard "Don't Spoil It For Us" Paseman, 2018.03.28. $\endgroup$ Mar 28, 2018 at 21:20
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It may be interesting to note that, subject to the ABC conjecture, you have the fantastically good estimate $$ \left\{ \left( \frac43 \right)^n\right\} \gg_\delta \delta^n,\quad \delta\in(0,1). $$

The proof goes as follows.

Let $$ 4^n = 3^nk+r,\quad 0<r<3^n. $$ The greatest common divisor of $k$ and $r$ divides $4^n$, and we write $$ (k,r)=2^d,\ k=2^dk_0,\ \text{and}\ r=2^dr_0, $$ so that $2^{-d}4^n=3^nk_0+r_0$. Assuming the ABC, for any $\varepsilon>0$ we have then \begin{align*} 2^{-d}4^n &\ll_\varepsilon \left( {\rm rad}(2^{-d}4^n\cdot3^nk_0\cdot r_0) \right)^{1+\varepsilon} \\ &\le\ \ \ (6k_0r_0)^{1+\varepsilon} \\ &\ll_\varepsilon (kr)^{1+\varepsilon}\cdot 2^{-d}. \end{align*} In view of $k<(4/3)^n$, this implies $$ 4^n \ll_{\varepsilon} \left(\frac43\right)^{(1+\varepsilon)n} r^{1+\varepsilon} $$ and, as a result, $$ \left\{ \left( \frac43 \right)^n \right\} = \frac r{3^n} \gg_{\varepsilon} 4^{-\varepsilon n/(1+\varepsilon)}. $$

The assertion follows by choosing $\varepsilon$ to satisfy $4^{-\varepsilon/(1+\varepsilon)}=\delta$.

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    $\begingroup$ Probably Baker's explicit abc-conjecture will give explicit constant $C(\delta)$ in $\ll_\delta$. $\endgroup$ Mar 30, 2018 at 13:33
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I'll give a refined version of the idea expressed in my comments as an answer as it seems to capture the real thing.

Let $ r_{k}(n) $ be the representant of $ \binom{n}{k} $ in $ \mathbb{Z}/3^{k}\mathbb{Z} $ whose absolute value is minimal. Then the fractional part of $ (4/3)^n $ is $ \sum_{k=1}^{n}\dfrac{r_{k}(n)}{3^{k}}+O(1) $ where the "error term" is an integer possibly less or equal to $ 1 $ in absolute value.

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  • $\begingroup$ The error term is an integer and has size potentially larger than log n . If you can prove it is bounded above by a constant, I and many others would like to see the idea. Gerhard "It Is Of Order N" Paseman, 2018.03.28. $\endgroup$ Mar 28, 2018 at 22:39
  • $\begingroup$ Good point. The problem comes from the $ r_{k}(n) $ that are too big compared to $ 3^{k} $, I'd say those whose absolute value exceeds $ 3^{k/2} $ . $\endgroup$ Mar 28, 2018 at 23:09
  • $\begingroup$ Let $ S^{+}(n) $ (resp. $ S^{-}(n) )$ be the sequence of positive (resp. negative) $ r_{k}(n) $ whose absolute value is at least $ 3^{k/2} $ sorted in decreasing (resp. increasing( order( $ s^{+}(n) $ and $ s^{-}(n) $ their respective number of elements. Let $ \delta(n) : = $\endgroup$ Mar 28, 2018 at 23:28
  • $\begingroup$ $ s^{+}(n)-s^{-}(n) $ . If $ \delta(n)>0 $ (resp. $ <0 $ ) let's replace the first $ \vert\delta(n) \vert $ elements $ r_{i}(n) $ of $ S^{+}(n) $ (resp. $ S^{-}(n) $ ) with $ r_{i}(n)-3^{i} $ (resp. with $ r_{i}(n)+3^{i} $ ). This should reduce the magnitude of the error term. $\endgroup$ Mar 28, 2018 at 23:43

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