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I'm interested in this question apparently posed by John Nash, which I found in the book A Beautiful Mind.

If you make up a bunch of fractions of pi $3.141592\ldots$. If you start from the decimal point, take the first digit, and place decimal point to the left, you get $.1$

Then take the next 2 digits $.41$

Then take the next 3 digits $.592$

You get a sequence of fractions between $0$ and $1$.

What are the limit points of this set of numbers?

I would like to know what can be deduced by assuming $\pi$ is normal?

It seems to me that, if $\pi$ is normal, the sequence defined above must be non-convergent and therefore have at least two limit points (see the discussion here).

But I think by choosing carefully which subsequences to look for, you should be able to construct as many limit points as you like (again, if $\pi$ is normal). For example, you could look for subsequences bounded by $[0,1/3)$, $[1/3,2/3)$ and $[2/3,1]$ to find three distinct limit points.

Could anyone tell me if this reasoning is correct?

Henrik supports the community
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prdnr
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    It's still possible for a number to be normal if the first, second, fourth, seventh, eleventh, … digit are all the same, I think (not certain of that, I've just woken up). If that's true, then normality doesn't imply that you can look for subsequences bounded in the way you want; perhaps every member of the subsequence is in $[0, 1/3)$. Naturally that's not true for $\pi$, but it demonstrates that your argument will need to use some special property of $\pi$ that isn't normality. (Or it demonstrates that I'm wrong.) – Patrick Stevens Sep 10 '17 at 06:52
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    First of all: Welcome the maths.SE. At first (It also early here) I thought your argument was good. but then I realised the same thing @PatrickStevens points out, that normality doesn't rule out that the 1st, 2th, 4th, 7th,... digit can't all be the same, so you need something more. – Henrik supports the community Sep 10 '17 at 06:57
  • Thanks both for your comments. I see what you mean, looking again at the definition of a normal number it doesn't seem to rule this out. Do you think the argument for having at least two limit points holds - or is my assertion that the sequence must be non-convergent flawed? – prdnr Sep 10 '17 at 07:08
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    I suspect one can prove that $.11212312341234512345612345671234567812345678912345678910\dots$ is normal, and then the sequence converges. – Gerry Myerson Sep 10 '17 at 07:14
  • Any thoughts on the matter, prdnr? – Gerry Myerson Sep 12 '17 at 07:24
  • I think you need to change your sequence slightly when you reach 10, so that only one digit is added on each round. So that would be $.112123123412345123456123456712345678123456789123456789112345678910...$ for which the sequence definitely converges. I'm not 100% convinced it is normal, though. Do you think 9s are disfavored? – prdnr Sep 12 '17 at 09:37
  • I would think that you should insist on digit Autocorrelation=0 and a uniform or normal distribution of digit occurrences; to make a good question. Not that I am skilled enough to solve the problem but it might be approachable. – rrogers Sep 12 '17 at 20:14
  • I don't think 9s are disfavored. Are you familiar with https://en.wikipedia.org/wiki/Champernowne_constant ? – Gerry Myerson Sep 14 '17 at 09:08
  • Are you still here, prdnr? – Gerry Myerson Sep 15 '17 at 12:57
  • Still here. I have come across the Champernowne constant, but I thought this case might be different because there is so much repetition of the early digits. At least early on, the numbers are quite unbalanced. Of course, it might even out as you let it go to infinity. I've been trying some numerics, but haven't reached a conclusion. – prdnr Sep 16 '17 at 17:56
  • Counts at 189453389 digits are:
    0: 14398888
    1: 23894501
    2: 22783500
    3: 21672500
    4: 20561500
    5: 19450500
    6: 18339500
    7: 17228500
    8: 16117500
    9: 15006500     – prdnr Sep 16 '17 at 17:58

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