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Famously, it is not known whether $\pi$ is a normal number. Indeed, there are far weaker statements that are not known, such as the statement that there are infinitely many 7s in the decimal expansion of $\pi$. I'd like to have some idea of where the boundary lies between what we know and what we do not know. For example, I would guess that it is not even known not to be the case that all decimal digits from some point on are 0s and 1s. Am I right about this?

A partial answer to this question was given by Timothy Chow in a discussion of another question: Is pi a good random number generator?. He pointed out that some very weak facts can be deduced from known results about how well $\pi$ can be approximated by rationals. I suppose I could ask whether that is essentially the only technique we have. Could it be, for instance (as far as what is proved is concerned -- obviously it isn't actually the case) that the digits of $\pi$ are all 0s and 1s from some point on and that there is a constant $C$ such that the number of 1s in the first $n$ digits is never more than $C\sqrt{n}$?

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    $\begingroup$ just imagine the newspapers when some day there is a correct proof that $\pi$ ends with $0$s and $1$s ... $\endgroup$
    – Martin Brandenburg
    Jan 12, 2011 at 16:39
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    $\begingroup$ If I remember well, we have estimates of the irrationality measure of $\pi$ (not larger than $8.0161...$, after M. Hata), and this tells us that not everything is possible. For instance $\pi$ cannot ressemble a Liouville number. $\endgroup$
    – Denis Serre
    Jan 12, 2011 at 16:48
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    $\begingroup$ We would need bounds on π's irrationality measure to say that, not just estimates of π's irrationality measure. $\endgroup$
    – user5810
    Jan 12, 2011 at 16:57
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    $\begingroup$ A mildly related question involves this function: $$ f(n) = \begin{cases} 1 & \text{if there is a a sequence of }n\text{ consecutive 7s in the decimal expansion of }\pi, \\ 0 & \text{otherwise}, \end{cases} $$ I was slightly startled to read in a book by Hartley Rogers that this function is computable. The proof that there is an algorithm that, when given $n$, returns $f(n)$, is short, simple, and somewhat thought-provoking. $\endgroup$
    – Michael Hardy
    Jan 12, 2011 at 20:21
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    $\begingroup$ Wadim Zudilin commented in an answer to this question, that Hata's irrationality measure for $\pi$ was recently beaten: V.Kh. Salikhov, Russ. Math. Surv. 63, No. 3, 570-572 (2008); translation from Usp. Mat. Nauk 63, No. 3, 163-164 (2008). $\endgroup$
    – Tony Huynh
    Jan 12, 2011 at 21:22

2 Answers 2

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I believe that Gerry's answer to the earlier question on MO is exhaustive enough for this question as well. The normality of $\pi$ (which would imply the irrationality measure 2) is not known yet, and even the questions like "all decimal digits from some point on are 0s and 1s" cannot be shown unconditionally.

The best known estimate for the irrationality measure of $\pi$, due to V. Salikhov [Russ. Math. Surv. 63:3 (2008), 570--572] (see also MR2483171 (2010b:11082)) reads $|\pi-p/q| > q^{-7.6063\dots}$ for all integers $p$ and $q>q_0$. The improvement of Hata's record is achieved by constructing rational approximations to $\pi$ alone (Hata's result is in fact about $\mathbb Q$-linear independence of $1$, $\log 2$ and $\pi$).

An interesting approach of attacking the normality of numbers like $\pi$, as well as problems like 0s--1s, is related to the so-called BBP formulas. However all such formulas for $\pi$ correspond to its binary (also hexadecimal) expansions. The last piece of news is a base 10 formula for $1/\pi^2$ due to G. Almkvist and J. Guillera (arXiv:1009.5202 [math.NT]), $$ \sum_{n=0}^\infty\frac{(6n)!}{n!^6} (532n^2+126n+9)\frac1{10^{6n}}=\frac{375}{4\pi^2}, $$ which is not of BBP type but close to it in a certain sense.

A nice exposition of The Life of $\pi$ is J. Borwein's public lecture given at the 2010 Australian Math Society meeting.

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    $\begingroup$ Not being able to prove the normality of $\pi$ is compatible with being able to prove unconditionally that it is not the case that all digits from some point on are 0s and 1s. However, it seems that even this extreme non-normality is hard to rule out rigorously. $\endgroup$
    – gowers
    Jan 12, 2011 at 23:41
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    $\begingroup$ I see from the answer you refer to that this kind of fact is indeed unknown. Indeed, it seems that it really is the case that there's nothing one can say (about the decimal expansion) that doesn't follow from results about rational approximation. $\endgroup$
    – gowers
    Jan 12, 2011 at 23:50
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    $\begingroup$ The inequality in the 5th row should be in the other way. $\endgroup$
    – GH from MO
    Jan 13, 2011 at 0:18
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    $\begingroup$ There is a plausible conjecture about dynamical behaviour of BBP sum terms; it's in the "Exp. Math." book(s) of BB and also in an article in Exp. Math. (journal). The conjecture would imply the normality in the corresponding base. I can't find any advantage (=a simplified version of the conjecture) to single out the 0-1 problem... A good news: if one considers the binary expansion of $\pi$ instead, then eventually it ends with 0s and 1s only. :-) $\endgroup$
    – Wadim Zudilin
    Jan 13, 2011 at 13:09
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    $\begingroup$ "The normality of $\pi$ (which would imply the irrationality measure 2)..." Am I missing something? I don't think normality implies irrationality measure 2, as witnessed by Champernowne's constant. $\endgroup$
    – Wojowu
    Jan 28, 2016 at 15:25
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I think the closest we know is Furstenberg's result that $10^m3^n\pi$ is dense modulo $1$ (say). On the other hand, this is true for any irrational in place of $\pi$. A simple proof appears here. See also this generalization.

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