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A real number is said to be Decimal Alexandrian if its decimal representation contains every possible finite decimal sequence. It is a popular question whether $\pi$ is Decimal Alexandrian, or even in other bases. It is suspected to, but a proof has not been found yet. My question is whether any familiar number is. By familiar I mean Something like $0, 5, -8, \pi, e, \sqrt5, \phi, \gamma$, or a billion.

It is easy to generate a number just by picking its expansion, for example the number 0. 00 01 02 10 11 12 20 21 22 100 101 110 111 200 201 210 211 .... is Ternary Alexandrian. But are there any proofs that a well-known real is Alexandrian?

Daron
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(Added in response to comment) According to Wikipedia, no number that arises naturally from algebra or analysis is known to be normal. All known examples are contrived by concatenation of decimal (or other-base) digits.

The Champernowne constant 0.123456789101112131415161718192021222324...., is pretty well known. By construction, it has every finite string of decimal digits not prefixed by zeros. To see that it also has those prefixed by zeros, for example 000000000849102276, just note that such a string is a substring of the same string prefixed by 1: 100000000084912276.

John Bentin
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  • What I am really interested in is what a proof of normality/Alexandrianity/disjunctivity of a number defined in an algebraic/analytic manner -- without referring to its decimal expansion, say $\sqrt 5$, is -- would even look like, – Daron Dec 14 '13 at 00:41
  • @Daron: I added to my answer. – John Bentin Dec 14 '13 at 09:34