# Do the digits of $\pi$ contain every possible finite-length digit sequence? [duplicate]

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Prove there are no hidden messages in Pi

This is not a practical problem. I am asking out of curiosity. Any links/references are most welcome.

Say, we write the digits of $\pi$ in base $10$. Does this sequence of digits contain every possible finite length digit sequence? What about $e$, $\sqrt{2}$ or some other commonly known irrational numbers?

Is this property of numbers independent of base? If a number has this property when written in base $10$, will it also have it in base $2$, $3$ and all other bases?

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## marked as duplicate by Asaf Karagila, Douglas S. Stones, Raskolnikov, jspecter, Hans LundmarkJan 5 '12 at 14:05

According to Wikipedia, "It is not even known whether all digits occur infinitely often in the decimal expansions of those constants." –  joriki Jan 5 '12 at 11:54
@joriki Thanks for the pointer. I did not know the term normal number. Wikipedia says: "infinite sequence of digits in every base $b$ is distributed uniformly." Do these numbers also have the property I describe? My intuition says yes, but I am not entirely sure. –  Szabolcs Jan 5 '12 at 11:58
@Asaf: That question is a bit different; though JDH's answer refers to normality and disjunctivity, it's only implicit that these are widely believed to hold, and no information on what's know about this is given, so this question isn't really answered by that answer. –  joriki Jan 5 '12 at 12:01
@Szabolcs: That's a somewhat misleading quote taken out of context. If you look at that entire sentence, it answers your question. –  joriki Jan 5 '12 at 12:03
@joriki Sorry, you're right. I didn't pay attention. Alright, so I consider this question answered by mentioning the keyword I need to search for: normal numbers. I'd accept that if anyone posted it ... –  Szabolcs Jan 5 '12 at 12:07

According to the Wikipedia article on normal numbers, "It is not even known whether all digits occur infinitely often in the decimal expansions of those constants."

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It seems very little indeed is known (in terms of rigorous proofs) on this topic. But I thought I'd add some ideas for what I think seems true and logical and might hopefully eventually be proven.

I do think those numbers are normal in every base. But more than that as well. Normality is only one aspect of randomness and I think these numbers exhibit practically all of them.

I think the answer to your last question is no. I would hypothesise for example that you can "construct" a number which is normal in every base except for 2 where "11" never occurs (obviously not including bases of powers of 2). As restrictive as that is, there's still heaps of room for irrational, fancy numbers. In fact, I believe you could arbitrarily pick any "admissible" set of disallowed sequences in any bases and still make a number which is normal otherwise. I'd say uncountably many such numbers would exist. Randomness is the norm.

Basically, what I'm saying is that the bases are very independent of each-other and you can do whatever you like to each one without really affecting the others at all.

Another relevant example is this. Most would believe that numbers like $\pi$, $e$ and $\sqrt{2}$ probably don't have "patterns" in their base expansions and I'd agree. We know at least that irrational numbers never repeat indefinitely in any base. Now consider $0.1001000010000001000000001\dots$ in some chosen base with $1$'s in the positions $1,4,9,16,\dots$ It is clearly irrational (and probably transcendental). But the expansion has a simple pattern. Moreover I'm confident it's actually normal in every other base. Obviously, we could use any pattern we liked and the idea is the same.

SUMMARY: A pattern in any one base is absolute chaos in every other base with the one and only exception being rational numbers.

Now please shoot me down with some counter-examples!

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What you say is probably true for coprime bases, but certainly not in general -- a number that has no $1$ in its expansion in base $4$ also has no $1$ in its hexadecimal expansion, and a number that has no $3$, $6$ or $7$ in its octal expansion has no $7$, E or F in its hexadecimal expansion. –  joriki Jan 5 '12 at 20:14
@joriki Yes of course, I tried to get that across and hopefully it goes without saying to some degree. Although, I'm pretty sure they needn't be relatively prime. In fact, as long as they're not rational powers of each-other, I think it will still work! I'm curious about your second example, how did you find that out? I can see how a three digit pattern would affect a four digit pattern... –  user826788 Jan 6 '12 at 8:36
For the hexadecimal expansion to have $7$, E or F, there must be three $1$s in a row in the binary expansion; then there must be at least two $1$s in a row in the binary expansion of one of the octal digits; and $3$, $6$ and $7$ are the only such octal digits. –  joriki Jan 6 '12 at 9:32