Does $\pi$ contain the combination $ 1234567890$?

Question

This question is related with Does Pi contain all possible number combinations?. More specifically, I want to know if $\pi$ contains $1234567890$. I checked this link https://www.facebook.com/notes/astronomy-and-astrophysics/what-is-the-exact-value-of-pi-%CF%80/176922585687811 and did not see it there. I think that $\pi$ does not contain $1234567890$. It is true or not. If it is true, how to prove it?

Answer 1

The nature of most real numbers is that, in any base, you can find any sequence of digits infinitely many times. The definition of "most" is technical, but rigorous.

We don't know if $\pi$ has this property, but we don't know it doesn't. It appears to have this property in base $10$, but we can't prove it, yet, and "appears" is always a bit of nonsense when we are saying, "We've checked the first $N$ examples out of infinity."

So, as Cameron commented, you are not going to find anybody here who is going to be able to prove that it doesn't occur, since, if we could, we'd have answered a long unresolved question.

If $\pi$ acted like a string of random digits, then you'd expect to have to check on the order of $10^{10}$ or $10$ billion digits before you found $1234567890$. If you tested $1$ trillion digits and still didn't find this sequence, I'd be shocked. But I don't know where you can download $1$ trillion digits of $\pi$...

In the first 1 billion digits of $\pi$, I found two instances of $123456789$, but no instances of $1234567890$.

Here's a simple example. In the first billion digits, there were $10049$ instances of $12345.$ There were $969$ instances of $123456$. There were $97$ instances of $1234567$. There were $9$ instances of $12345678$. And there were two instances of $123456789.$ If the digits of $\pi$ were random, we expect that approximately one tenth of the instances of $123456789$ in any sample would have next digit $0$.