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.



Under what condition will this set contain a limit point of [0,1)?
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).


