Does pi contain 1000 consecutive zeroes (in base 10)?

Question

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.

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

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.

Answer 5

Assuming that $\pi$ is a normal number, such a string should exist infinitely many times. However, under the normality assumption, decimal digits asymptotically follow an independent discrete uniform distribution in $\{0,1,\dots,9\}$. This means that two consecutive zeroes appear roughly speaking every $10^2$ digits (to be honest the set of all pairs of consecutive digits are not independent events, since two consecutive pairs have a common digit). A string of three consecutive zeroes appears every $1000$ digits. As an exemple, the string "00000" can be found once in the list of the first $10^5$ digits of $\pi$ in http://www.geom.uiuc.edu/~huberty/math5337/groupe/digits.html. So a string of 1000 consecutive zeros appears on average every $10^{1000}$ digits. At present only something like $10^{14}$ digits of $\pi$ are known, so I guess we are quite far to find such a string in the decimal expansion of $\pi$.