I am defining a "somewhat normal number" as such:

A somewhat normal number is a number where every possible numerical string in base $10$ can be found in said number at least once, but the density of each $n$ digit string may not be equal to $10^{-n}$.

A somewhat normal number is a weaker version of a normal number, since it removes the requirement of the density of each $n$ digit string being equal to $10^{-n}$.

An example of a number that is somewhat normal but not normal would be $$0.\color{red}11\color{red}21\color{red}31\color{red}41\color{red}51\color{red}61\color{red}71\color{red}81\color{red}91\color{red}{10}11\color{red}{11}11\color{red}{12}11\cdots$$

i.e. every natural number (highlighted in red) with $1$s added to the end of each number is concatenated together (like in the Champernowne constant), and the number of $1$s is equal to the length of the natural number preceding it.

The number contains all of the natural numbers and therefore each possible numberical string can be found in this number. However there are $1$s than other digits in this number, and this number is therefore not normal.

Under this definition of "somewhat normal", can we say that $\pi$ is a somewhat normal number?

  • 2,615
  • 1
  • 14
  • 30
  • 1
    This is probably unknown. See https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations – C.Park Aug 02 '20 at 06:42
  • Showing that such occurs (even just once) is very difficult question and is not solved for most irrational numbers. Even for numbers much simpler than $\pi$ - say $\sqrt{2}$, are also left unsolved until recently. – C.Park Aug 02 '20 at 06:46
  • 1
    Why do you define "somewhat normal" when your question is "can we say that $\pi$ is a normal number?"? – David G. Stork Aug 02 '20 at 06:54
  • Oops, that’s a typo. Sorry. – Kyky Aug 02 '20 at 06:54
  • As a side note, this property is already known as [disjunctivty](https://en.wikipedia.org/wiki/Disjunctive_sequence). – Kyky Aug 02 '20 at 07:23

0 Answers0