The motivation for this question comes from the novel Contact by Carl Sagan. Actually, I haven't read the book myself. However, I heard that one of the characters (possibly one of those aliens at the end) says that if humans compute enough digits of $\pi$, they will discover that after some point there is nothing but zeroes for a really long time. After this long string of zeroes, the digits are no longer random, and there is some secret message embedded in them. This was supposed to be a justification of why humans have 10 fingers and increasing computing power.

Anyway, apologies for the sidebar, but this all seemed rather dubious to me. So, I was wondering if it is known that $\pi$ does not contain 1000 consecutive zeroes in its base 10 expansion? Or perhaps it does? Of course, this question makes sense for any base and digit. Let's restrict ourselves to base 10. If $\pi$ does contain 1000 consecutive $k$'s, then we can instead ask if the number of consecutive $k$'s is bounded by a constant $b_k$.

According to the wikipedia page, it is not even known which digits occur infinitely often in $\pi$, although it is conjectured that $\pi$ is a normal number. So, it is theoretically possible that only two digits occur infinitely often, in which case $b_k$ certainly exist for at least 8 values of $k$.

Update. As Wadim Zudilin points out, the answer is conjectured to be yes. It in fact follows from the definition of a normal number (it helps to know the correct definition of things). I am guessing that a string of 1000 zeroes has not yet been observed in the over 1 trillion digits of $\pi$ thus computed, so I am adding the open problem tag to the question. Also, Douglas Zare has pointed out that in the novel, the actual culprit in question is a string of 0s and 1s arranged in a circle in the base 11 expansion of $\pi$. See here for more details.

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    $\begingroup$ The movie has excellent special effects and Jodie Foster. $\endgroup$
    – Will Jagy
    May 5, 2010 at 4:59
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    $\begingroup$ It's a conjecture: any possible pattern occurs in the decimal record of $\pi$ (so, $\pi$ is normal in base 10). $\endgroup$ May 5, 2010 at 5:16
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    $\begingroup$ In the book, it was a pattern of 0s and 1s arranged in a circle in the base 11 expansion of pi. If pi is normal, then this pattern should occur at some point, but in the novel this shows up surprisingly soon. en.wikipedia.org/wiki/Contact_(novel) $\endgroup$ May 5, 2010 at 5:32
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    $\begingroup$ If you want to find a given sequence of 1000 digits in pi (for example, 1000 zeros), then, given that it's a reasonable hypothesis that digits really do occur "randomly" (and hang the novel), you'll have to look through about 10^1000 digits before you find your pattern. So you can be pretty sure that we've not found 1000 zeroes yet! $\endgroup$ May 5, 2010 at 8:00
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    $\begingroup$ I assume you know about the Feynman point? en.wikipedia.org/wiki/Feynman_point $\endgroup$ May 5, 2010 at 8:08

4 Answers 4


Summing up what others have written, it is widely believed (but not proved) that every finite string of digits occurs in the decimal expansion of pi, and furthermore occurs, in the long run, "as often as it should," and furthermore that the analogous statement is true for expansion in base b for b = 2, 3, .... On the other hand, for all we are able to prove, pi in decimal could be all sixes and sevens (say) from some point on.

About the only thing we can prove is that it can't have a huge string of zeros too early. This comes from irrationality measures for pi which are inequalities of the form $|\pi-(p/q)|>q^{-9}$ (see, e.g., Masayoshi Hata, Rational approximations to $\pi$ and some other numbers, Acta Arith. 63 (1993), no. 4, 335-349, MR1218461 (94e:11082)), which tell us that such a string of zeros would result in an impossibly good rational approximation to pi.

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    $\begingroup$ As for the irrationality measure of $\pi$ Hata's longstanding record was recently beat: V.Kh. Salikhov, Russ. Math. Surv. 63, No. 3, 570-572 (2008); translation from Usp. Mat. Nauk 63, No. 3, 163-164 (2008). $\endgroup$ May 5, 2010 at 6:08
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    $\begingroup$ @Gerry: Thanks for reminding me of why the zeroes should be far away. $\endgroup$ May 6, 2010 at 11:53
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    $\begingroup$ matwbn.icm.edu.pl/ksiazki/aa/aa63/aa6344.pdf $\endgroup$
    – Charles
    May 21, 2010 at 4:05
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    $\begingroup$ @Charles, thanks for the link to the Hata paper. $\endgroup$ May 21, 2010 at 4:18
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    $\begingroup$ @Patrick, nope. $\endgroup$ Feb 26, 2018 at 8:40

A similar question (1 million consecutive 7 in the decimal expansion of pi) has been discussed by Timothy Gowers in a text published in 2006 (see Reuben Hersh: 18 unconventional essays).

His (quite classical) heuristic arguments in favor of yes were even used for a study on the influence of autority on persuasiveness in mathematics (See Matthew Inglis and Juan Pablo Mejia-Ramos, Cognition and Instruction Journal, Routledge, 2009).


Mahler's paper [1] shows that you can't have too many zeros too soon (or any other string of identical digits, or anything else that would give too good of a rational approximation). In this context the result is weak: there's no string of 1000 zeros starting after the 1000/41 st digit of pi... but it's easy enough to calculate 24 digits of pi as it is.

Even assuming that Salikhov's result holds for numbers this small, it won't exclude more than 150 digits (that is, no spans of 1000 zeros in the first 1150 decimal digits of pi).

[1]: K. Mahler, On the approximation of π, Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen Series A, Mathematical Sciences 56 (1953), pp. 30-42.

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    $\begingroup$ Thanks Charles. By 'after' do you mean 'before'? $\endgroup$
    – Tony Huynh
    May 21, 2010 at 15:26
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    $\begingroup$ Perhaps I was unclear. Mahler's result says that you can't have a string of 1000 zeros in the decimal expansion of pi appear just after the 24th digit, or before then -- that is, no such string in the first 1024 digits. (That this is 2^10 is coincidence; it's just 1000 (1 + 1/(42 - 1)) rounded down.) $\endgroup$
    – Charles
    May 21, 2010 at 20:46
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    $\begingroup$ The Mahler paper doesn't seem to be available online, but here's an MSN link: Mahler - On the approximation of $\pi$. $\endgroup$
    – LSpice
    Mar 7, 2018 at 23:14

I do not know what kind of effect this has on your question, but it might be a good thing to look at in this context. In 1995, a formula for the $n$-th digit of $\pi$ written in hexadecimal was found. See the wikipedia article, http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula.


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