Hacker News Reader

Comment by univacky | Hacker News Reader

Do we know that every possible finite sequence exists in pi?

https://en.wikipedia.org/wiki/Normal_number

Pi is not proven to be normal, but is suspected.

Note that normality of a number in a given base is a sufficient condition for existence of all finite sequences in its expansion in this base, but not necessary.

It might turn out that Pi is not normal say in base 2, but all possible finite sequences of 0s and 1s occur in it, just with some of them being slightly more frequent than others of the same length.

From that article, it seems that correct term is "disjunctive sequence".

IIRC, since there are an infinite number of digits in Pi, and there are no infinitely repeating finite sequences in Pi, there are therefore an infinite number of finite sequences in Pi.

Granted, the index might be larger to represent than the actual sequence you're indexing, but...

That doesn't imply that every finite sequence is in pi.

I agree. There are an infinite number of finite sequences that are not my social security number and do not contain my social security number. An infinite list is still infinite though it lack any specific finite sequence.

"An infinite number of finite sequences" does not mean "all finite sequences". For example, the number 1.10100100010000100001... contains an infinite number of finite sequences, but does not contain, for example, any sequences with a 2, or even the sequence '10101'.

You cannot bend the rules in this way and still equate it to the question at hand - you're introducing characters outside of base2 in your example. If you stuck within the realm of base2, '2' in its binary form certainly appears in the pattern.

Not only that, but Pi is normal (all digits distributed evenly), and your sequence clearly is not.

I understand what you are saying - indeed there are infinite finite sequences - but it just does not apply here.

You can trivially modify their example to be 1.02003000400005... cycling through the digits and the argument still holds.

As another commenter pointed out, pi is not proven to be normal, but is suspected to be.

They already showed 21 (in binary, 10101) doesn't appear anywhere, even as a binary pattern.

The example isn't base two. It's base 10. It just happens to be a number that only has 0s and 1s. Like eleven. Or 1000.

To phrase it another way, there are infinitely many sizes of Infinity. Just because you have one infinite sequence, does not mean it is the same size as another Infinity and as such may include or exclude elements of the other Infinity.

Pi is not known to be normal.

> there are therefore an infinite number of finite sequences in Pi

Sure, but this doesn't answer the question: Do we know that every possible finite sequence is present? You can have the former without the latter being true. For example, a sequence of the form "1011011101111011111..."[0], while infinite, never contains the finite sequence "1234".

[0] https://oeis.org/A094946

Not sure that counts as proof.

We do not.

Reply on news.ycombinator.com