# What is the largest number of consecutive repeating digits in pi?

As far as I can tell from googling, pi has been calculated to over 3 trillion decimal places. I'm curious whether a string of 100 2's has been found.

It has to happen, right? As I understand it, we should be able to find a string of 3 trillion consecutive 2's.

Am I wrong in my thinking?

## Answers and Replies

HallsofIvy
Homework Helper
We certainly might be able to find such a sequence but I can see no reason to think that there must be such a sequence. If, as many people believe, but I don't think has been proven, the digits of $\pi$ are "random" (every digit, every sequence of digits is equally likely to appear) it still does not follow that a given sequence will appear.

The notion that you're referring to is that of a normal number. A number is called normal if every finite sequence of digits occurs thesame number of times as every other finite sequence of digits. Thus in a normal number, there are as much numbers 2 as numbers 3, there are as much numbers 33 as 13.
So, in particular, in a normal number, you would expect that once in a time there would be a trillion 2's after eachother.

The thing is however, that pi hasn't yet been shown to be normal. So we don't really know if there is a sequence with a trillion 2's...

A number that IS normal is Champernowne's number, this is:

0.123456789101112131415161718192021222324...

Thus just concatenate all numbers. If you concatenate all PRIME numbers, then you also get a normal number:

0.23571113171923...

This is called the Copeland-Erdos constant.

However, as you might have guessed, showing that a number is normal is very hard...

Spinnor
jhae2.718
Gold Member
Out of curiosity, how would one go about proving a number is normal?

AlephZero
Homework Helper
If you concatenate all PRIME numbers, then you also get a normal number:

0.23571113171923...

This is called the Copeland-Erdos constant.

That seems counter-intuitive. Since every prime number (except two of them) ends in 1, 3, 7 or 9, one would feel the distribution was going to be biased - unless this effect washes out faster than something else, as the primes get bigger and contain more digits.

I suppose that just illustrate the fact that
However, as you might have guessed, showing that a number is normal is very hard...

Out of curiosity, how would one go about proving a number is normal?

A proof is found in http://www.mth.uea.ac.uk/~h720/teaching/dynamicalsystems/champernowne1933.pdf [Broken] but be warned, it's not exactly easy reading...

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jhae2.718
Gold Member
A proof is found in http://www.mth.uea.ac.uk/~h720/teaching/dynamicalsystems/champernowne1933.pdf [Broken] but be warned, it's not exactly easy reading...

Thanks.

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That seems counter-intuitive. Since every prime number (except two of them) ends in 1, 3, 7 or 9, one would feel the distribution was going to be biased - unless this effect washes out faster than something else, as the primes get bigger and contain more digits.

I suppose that just illustrate the fact that

I agree 100% with you, it is very counter-intuitive. In fact, if it wasn't proven, then I would have claimed that it wasn't true!

Some other counter-intuitive facts:
- almost all real numbers are normal (in the sense that the non-normal numbers have measure 0). Still it is very hard to come up with numbers which are normal.
- every real number is the product of two normal numbers
- multiplication by a nonzero rational number doesn't change the normalness of a number