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Let $T_1,T_2,....T_n$ be numbers such that $T_k= k$ no. of digits in decimal expansion of an irrational number, say $\alpha$, starting from $(\frac{k(k-1)}{2}+1)^{th}$ digit in the decimal expansion. e.g. for $\pi$, according to this definition, we have

$T_1=0.1, T_2=0.41, T_3=0.592$ and so on.

Question: Under what conditions regarding an irrational number $\alpha$ will the set of all such $T_k'$s will have a limit point of $[0,1)$ ?

$PS$: The same question for $\pi$ was asked by John Nash in one of the exams graded by him. The incident is given in the book "A beautiful mind". It was also given that this is an open problem. What I want to know is that whether there has been any work on this problem in general or in any specific cases.

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  • $\begingroup$ Probably the desired conclusion should be that all points of $[0,1]$ are limit points. Almost all (in the sense of Lebesgue measure) numbers have that property. And so $\pi$ surely has it as well. But, as noted, whether $\pi$ has it has no proof as yet. $\endgroup$
    – Gerald Edgar
    Commented Jan 4, 2012 at 13:56

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In the customary language a limit point means limit of a subsequence. If so, the condition just means that the sequence $T_k$ does not converge to $1$, and this reflects on the form of the decimal expansion of $\alpha$ quite in a simple way. Of course, it could be one of the billions of impossible, yet not very interesting questions, deciding whether some given real number like $\pi$ or $2^{\sqrt 2}$ is of this form.

A conjecture: maybe John Nash's question really had "limit point" in the sense of "limit of a subsequence" (and maybe, it also had the sequence $T_k$ starting from the $k$-th place, and of lenght $k(k-1)/2$, instead of the other way as quoted. If so, the question would be quite a fair for an examination, the answer being no, just because $\pi$ is irrational). Unfortunately, I do not have a copy of "A beautiful mind" here, to check the exact statement of John Nash's question as reported by the author of the book, and her sources. But after all I do trust John Nash as an examinator (more than Sylvia Nazar as historian, I have to say).

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  • $\begingroup$ Thanks, I will check A beautiful mind for the precise version. I remember it to be just like this, but I could well be wrong. $\endgroup$
    – nb1
    Commented Jan 4, 2012 at 11:47
  • $\begingroup$ Taking $k$ digits starting from the $(k(k-1)/2+1)$th one simply means that you chop off successive pieces of the decimal expansion of lengths 1, 2, ... . If it were the other way round, there would be no explanation for the pretty much arbitrary choice of $k(k-1)/2$. $\endgroup$
    – Emil Jeřábek
    Commented Nov 25, 2013 at 14:39

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