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Consider the number with binary or decimal expansion

 0.011010100010100010100...

that is, the $n$'th entry is $1$ iff $n$ is prime and zero else. This number is clearly irrational. Is it known whether it is transcendental?

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Related: math.stackexchange.com/questions/42231/… –  Dan Brumleve Jun 2 '11 at 1:45
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Wiki lists it as a "suspected transcendental" en.wikipedia.org/wiki/List_of_numbers#Suspected_transcendentals –  Dan Brumleve Jun 2 '11 at 2:02
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@Graham Enos: You mean of a transcendental number, in which case the answer is yes. By tradition, the first incommensurability proof involved $\sqrt{2}$, though some have argued it might have been the so-called golden number. –  André Nicolas Jun 2 '11 at 2:23
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@Dan, every number not known to be algebraic is suspected transcendental. –  Gerry Myerson Jun 2 '11 at 3:10
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1 Answer

Here, since I was "nominated" to answer, is what little I know.

The binary version is called the "prime constant" on some internet sites, but I am not aware of any substantial work on this number.

I initially found it by googling "0 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1", which brought me this link to the CRC concise encyclopedia of mathematics with references to OEIS. I could also have entered a similar search on oeis.org. I saw sequence A010051, the characteristic function of the primes. One of the cross references there is sequence A051006, also referenced in the encyclopedia article, which is the decimal expansion of the "prime constant", with that name given. Another Google search with name in hand brings up the Wikipedia article.

Both binary and decimal versions are irrational because the expansions do not repeat, but I can offer no useful comment on the question of transcendentality. As Dan Brumleve notes, a Wikipedia article claims that the binary version is suspected transcendental. As Gerry Myerson notes, "every number not known to be algebraic is suspected transcendental."

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thanks all (especially jonas m.)! much appreciated. cheers! numberman –  numberman Jun 2 '11 at 5:34
    
numberman, you can accept this answer by clicking the check mark. –  Dan Brumleve Jun 3 '11 at 3:54
    
The question has now been raised at MO, mathoverflow.net/questions/114905/… where some further references are given (the problem is still open). –  Gerry Myerson Nov 29 '12 at 22:18
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