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I didn't believe the screenshot, but this is real with ChatGPT 3.5 (the green avatar).

https://chat.openai.com/share/aadf6a17-f40e-4ea6-8f5e-236bc2...

ChatGPT 4 gives a more correct answer:

> Pi is an irrational number, meaning it has an infinite number of digits that don't repeat in a predictable pattern. So, there's no such thing as the "last" digits of pi.

https://chat.openai.com/share/6a7c0221-0aca-48af-8938-e4d98e...




I just tried it with ChatGPT 3.5:

> The decimal representation of pi is infinite and non-repeating, so it doesn't have "last 8 digits" in the way a finite number would. However, if you're looking for the first few digits of pi, they are 3.14159265. If you need more digits, you can find them using various sources or tools that provide the decimal expansion of pi.


You got more lucky than me because I tried a few times and I got the same answer with different digits every time. Also similar correct answers using ChatGPT 4 every time.


Better, but irrational numbers can have a predictable pattern like "0.110100100010000" ( E.g "0.1 10 100 1000 10000" )


I'd argue that every computable number has “a predictable pattern” in its digits.


I guess it depends.

A cryptographically secure pseudorandom number generator lets me pump out a stream of digits that's certainly computable, but unless you know my private key, you won't be able to predict it.

(Finding out the private key from the stream is 'computable', because the definition of computable is comfortable with running exponentially long brute force searches. But that's why 'computable' is not a useful definition in practice. You want something that captures 'tractable', not just 'possible on a Turing machine at all'.)


It a quite tautology. How do you define "predictability"?

The most intuitive answer is just computability.


That simple sequence, 1, 10, 100 etc, the digit is at least computable in O(1) time which is maybe a good way to look at predictability. It's a simple rule and no effort similar to computing every digit before it is required.


If you store the number of zeros as value z, computing z+1 is O(log n) in the long run for unbounded values of z.


If you define predictable as computable with finite memory, it is correct.

If you have finite memory, you have a finite number of states and will eventually return to a previous state. In your example, you eventually will run out of memory to track the number of consecutive zeros.




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