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Misdicorl | 1 year ago

This isn't true as you can build an infinite sequence that never repeats. An example sequence in binary is (the number of 0s between each 1 increases by 1 every time)

01001000100001...

discuss

order

poizan42|1 year ago

Exactly. A number with the property that every sequence occurs is called a rich or disjunctive number - a number can be rich in s specific bases or rich for all bases we don't know whether pi is any if that. A number where every sequence occurs equally often (scaled to the length of the sequence) is called a normal number, which is an even stronger property.

sponaugle|1 year ago

While Pi is not proven to be a disjunctive number, nor the stronger condition of being normal, it is generally believed to be normal. That being said we don't have a proof of Pi being normal, nor disjunctive.

I am not familiar with how a proof of that would be constructed, as clearly numerical or computational measurements could never be conclusive.

jfoutz|1 year ago

Starting from the fourth digit,

    01000100
But maybe I don’t understand your example

Misdicorl|1 year ago

You didn't understand the original claim which is that because Pi has an infinite decimal representation, every subsequence of it has a repeat

34679|1 year ago

Your example is governed by an arbitrary rule of your choosing. Pi is not.