I read somewhere that any finite number sequence must be found in $\pi$. For example, $0998975645455$ must be somewhere in the digits of $\pi$.

The reason for this was that $\pi$ is irrational, meaning it's infinite and irregular.

But $\sum_{k=0}^n \frac{1}{10^{k!}}$ is transcendental and its digits are composed of $0$ and $1$ only.

So irrationality does not guarantee the property I mentioned.

Is it really true that one can find any number sequence from $\pi$?


marked as duplicate by Elliot G, user228113, Hayden, Joffan, Henning Makholm Jul 22 '16 at 0:12

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  • 5
    $\begingroup$ See math.stackexchange.com/questions/216343/… $\endgroup$ – Klint Qinami Jul 21 '16 at 23:55
  • $\begingroup$ For finite sequences here. It appears to be a very open problem. $\endgroup$ – user228113 Jul 21 '16 at 23:58
  • $\begingroup$ Irrationality is not enough as you correctly spotted. It is not actually proven that $\pi$ has this property, but it is believed by most to be true. See the linked post by @TnilkImaniq for a similar question with good answers. $\endgroup$ – Eff Jul 22 '16 at 0:07
  • $\begingroup$ So do normal numbers have the property? $\endgroup$ – Mike Park Jul 22 '16 at 0:11
  • $\begingroup$ @MikePark: Yes; the definition of a number being normal is that not only does every digit sequence appear in the decimal expansion; every digit sequence appears infinitely many times with the same (limiting) frequency we would expect to find it with in a string of random digits. $\endgroup$ – Henning Makholm Jul 22 '16 at 0:15

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