Funny/interesting magic numbers

[chrisname]

They're the same thing, just in a different numeric base.

[hamsterman]

That's an interesting point Veltas made (well, this probably isn't what he meant, but whatever.. )
This thread assumes that since pi is infinite and not periodic, it contains all possible combinations of bits, however a decimal expression of pi (a string containing chars '0' to '9') too is infinite and not periodic, however it does not contain all possible combinations of bits (for example, there is no 11111111 in it). Therefore the properties of being infinite and non periodic are not sufficient to prove that a sequence contains all finite combinations of its elements.
So does pi have some other property to prove this, or Seraphimsan just wrong?

^{edit: yeah, there actually are plenty of 00000000s..}

[ne555]

@chrisname: No sir, I'm an oscillating one.
x \leftarrow x - \tan(x) that converges to \pi if x starts close enough.

@hamsterman: I thought that it was an open problem that pi is normal in every base (or normal at any base). However I think you are cheating when you change the representation of the symbols.
It will be like say a '1' indicates the start of a number, after that the number of '0' count to n+1. There will be no way to obtain two '1' next to each other.
I guess there is a confusion between "all finite combinations of its elements" and "all possible combinations of bits"

since pi is infinite and not periodic, it contains all possible combinations of bits its elements,

Nope. By instance: 110100100010000... it is infinite an no periodic (with elements '1' and '0'). The condition is necessary, but no sufficient.

[hamsterman]

Thanks for making that clear. I hadn't heard of normal numbers before. I'll look into them..