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Let $n\ge 2\in\mathbb Z$. Suppose that a base-$n$-decimal $(0.a_1a_2a_3\cdots)_n$ represents $\sum_{k=1}^{\infty}\frac{a_k}{n^k}$ where $a_{i}\in\{0,1,\cdots,n-1\}\ (i=1,2,\cdots)$ is each digit number. Now, when we represent a rational number $0\lt p\lt 1$ as a base-$n$-decimal $(0.a_1a_2a_3\cdots)_n$, let's consider the following condition.

Condition : For every $0\le k\le n-1$, $$\lim_{N\to\infty}\frac{|\{i|1\le i\le N,a_{i}=k\}|}{N}=\frac 1n.$$

I've been interested in finding a rational number $0\lt p\lt 1$ which satisfies the condition for a $n$. For example, we can see $p=2/3$ satisfies the condition for $n=2$ because it is $(0.\overline{10})_2$. However, finding a $p$ which satisfies the condition for two or more $n$ seems difficult. We can see $p=21/26$ satisfies the condition for $n=2,3$ because it is $(0.1\overline{100111011000})_2$ and $(0.\overline{210})_3.$ However, I cannot find any $p$ for two $n$ other than $n=2,3$.

Then, here is my first question.

Question 1 : Does there exist a rational number $0\lt p\lt 1$ which satisfies the condition for two $n$ other than $n=2,3$ where the two $n$ are relatively prime?

In addition to this, the following questions seem very difficult.

Question 2 : Does there exist a rational number $0\lt p\lt 1$ which satisfies the condition for three $n$ where any two of the three $n$ are relatively prime?

Question 3 : Do there exist infinitely many rational numbers $0\lt p\lt 1$ which satisfy the condition for two $n$ where the two $n$ are relatively prime? How about the cases for three or more $n$?

Can anyone help?

Edit : As user level1807, who solved the question 1 and 2, points out, this is the question about the number which is simply normal to relatively prime bases. So, the question 3, which has not been solved yet, can be said as

Question 3 : Do there exist infinitely many rational numbers $0\lt p\lt 1$ which are simply normal to two relatively prime bases? How about the cases for three or more bases?

Update : I crossposted to MO.

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  • $\begingroup$ Is there some kind of backstory to this problem? Why is this interesting? $\endgroup$ – level1807 Jun 16 '14 at 14:20
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    $\begingroup$ @level1807: Well, sadly, no backstory. $\endgroup$ – mathlove Jun 16 '14 at 21:10
  • $\begingroup$ @mathlove I see. It does not even seem to me that a full investigation of the third question (i. e. proving there are infinitely many numbers) would be that hard, but without a proper motivation this is just not "fun" :) I say that as a mathematical physicist, mathematicians may disagree. $\endgroup$ – level1807 Jun 17 '14 at 4:03
  • $\begingroup$ @level1807: A full investigation using Mathematica? If so, I think you can't prove there are infinitely many rationals. Mathematica, I think, can't find 'infinitely' many rationals... To prove that, I think we need a completely new way. $\endgroup$ – mathlove Jun 17 '14 at 10:28
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    $\begingroup$ well, the thought is not very concrete anyways, sorry:) You are looking for rational numbers that are simply normal to several different bases. There may be some good ideas in this huge presentation on normal numbers. www-2.dc.uba.ar/ccr/talks/BsAs_normality.pdf $\endgroup$ – level1807 Jun 18 '14 at 13:37
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Basically, let $S$ be the set of all permutations of $\{0,1,\ldots,n-1\}$. Then for each $p\in S$ put

$$x=\sum_{m=0}^\infty \frac{1}{n^{m n}} \sum_{i=1}^{n} \frac{p_i}{n^i}.$$

You get $n!$ rational numbers that have the desired property.For example, for $n=4$, they are

$$\frac{9}{85},\frac{2}{17},\frac{13}{85},\frac{3}{17},\frac{18}{85},\frac{19}{85},\frac{5}{17},\frac{26}{85},\frac{33}{85},\frac{36}{85},\frac{38}{85},\frac{8}{17},\frac{9}{17},\frac{47}{85},\frac{49}{85},\frac{52}{85},\frac{59}{85},\frac{12}{17},\frac{66}{85},\frac{67}{85},\frac{14}{17},\frac{72}{85},\frac{15}{17},\frac{76}{85}$$

I believe this can be easily modified by constructing sequences of $2n$, $3n$ etc. digits. Since there are new "equi-distributed" periods of length $kn$ compared to $(k-1)n$ for each $k$, there are an infinite number of such numbers for each $n$. And of course adding some random digits before the periodic part won't change anything.

Edit: here is a formula for periods of length $Mn$. Again, $p\in S$ and $S$ is the set of all permutations of $M n$ digits ($M$ copies of each digit, $p=\{p_1,\ldots,p_{Mn}\}$).

$$x=\sum_{m=0}^\infty \frac{1}{n^{M m n}} \sum_{i=1}^{Mn} \frac{p_i}{n^i}.$$ For $n=2$ and $M=6$ one of the $924$ numbers given by this formula is $\frac{8}{26}$, which is $\frac{21}{26}-\frac{1}{2}$.

Edit2: okay, so by running a simple search in Mathematica I have been able to find a ton of examples of numbers with uniform digit distributions for $n=2$ and $n=5$ simultaneously. For example, $1/66$ is the "simplest": $$\frac{1}{66}=0.\overline{1111100000}_2 2^{-6}=0.\overline{0014213324}_5.$$ Another examples are $19/679$ for $n=3,4$; $9/142$ for $n=5,7$. For mysterious reasons there are no rationals with denominators up to $2000$ that meet the criterion for $n=4,5$. The first (with smallest denominator) number which meets the criterion for three bases $n=2,3,5$ is $809/2046$: $$\frac{809}{2046}=0.0\overline{1100101001}_2=0.1\overline{012000202022100112121102121012}_3=0.\overline{144203101322123012334420003441}_5.$$ The next for $n=2,3,5$ is $1237/2046$ (not surprisingly since these two sum up to 1). I would conjecture that this list is infinite as well (even after throwing out the ones with identical non-periodic part).

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  • $\begingroup$ Oh, so you are only interested in those special numbers? Somehow I didn't get that from reading your post. Okay, I guess I will try to run some searches. $\endgroup$ – level1807 Jun 16 '14 at 12:09
  • $\begingroup$ @barto it seems there are tons of such numbers for different pairs of relatively prime n-s. $\endgroup$ – level1807 Jun 16 '14 at 12:28
  • $\begingroup$ @level1807: Nice! You solved the question 1 and 2. Mathematica is great though it should not be able to solve the question 3... $\endgroup$ – mathlove Jun 16 '14 at 21:13

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