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Any series $\displaystyle \sum_{k=0}^{\infty}a_k2^{-k}$, where $a_k\in\{0,1\}$, converges to some $x\in[0,2]$ and since the sequence $a_n$ is unique for each $x\in[0,2]$ there is an bijection between $\text{func}(\mathbb{N},\{0,1\})$ and $[0,2]$.

Is there any known sequence $a_n\in\text{func}(\mathbb{N},\{0,1\})$ (or real number $x\in[0,2]$) such that for any $n\in \mathbb{N}$ there is an $m>n$ such that the binary pattern $a_0a_1\cdots a_m$ corresponds to a prime? That is, does it exists such a sequence so that $\displaystyle \sum_{k=0}^{m}a_k2^{m-k}$ is a prime for infinitely many m?

The Mersenne primes correspond to $x=2$.


It seems like orangeskid gave me the answer in comments but I didn't understand it first (or second, or third):

From results about disjunctive sequences it is clear that any binary string $a_0a_1\cdots a_m$ corresponding to a prime can be extended to a bigger string corresponding to a prime. Therefore it exists $x\in[0,2]$ that is related to the primes in this way.

And from Dirichlet's theorem it follows that any $x\in[0,2]$ can be arbitrarily well approximated by a sequence $a_0a_1\cdots a_m$ and then be completed to a prime string as above, repeatedly.

$\therefore$ The set of all prime related numbers in $[0,2]$ is dense.


So I reverse the question, are there non trivial examples of

$\displaystyle \sum_{k=0}^{\infty}a_k2^{-k}\in[0,2]$ for which there is a maximal prime $\displaystyle \sum_{k=0}^{m}a_k2^{m-k}$?

Lehs
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  • It's conjectured that there are infinitely many Mersenne primes. It may not be simple to check whether there are any such sequences at all. Starting from the other end is of course no problem, using Dirichlet's theorem on primes in arithmetic progressions. A non- issue: the decomposition is unique providing you take the infinite one for diadic rationals in $[0,2]$. Sounds like a nice conjecture! Any tests? – orangeskid Sep 16 '14 at 17:36
  • @orangeskid: I have made a big integer program that testing decimal sequences and have tested a lot of patterns inclusive $\pi$ but the result seems very similar to the Mersenne case: a few rather small then a big gap to the a big one, so big it seems pointless to seek for a next coming... – Lehs Sep 16 '14 at 18:36
  • @Lebs: Oh, it is not hard to check that there exist prime numbers starting with any given sequence of digits http://math.stackexchange.com/questions/60825/proof-that-there-are-infinitely-many-prime-numbers-starting-with-a-given-digit-s so yes, such sequences definitely exist. Testing does seem difficult indeed. Providing an explicit such sequence seems hard. – orangeskid Sep 16 '14 at 18:44
  • @orangeskid: I don't know what to conjecture. Maybe there exists a biggest pattern match for every pattern? Or, some of the patterns, I don't know which, has infinitely many matches and some don't. Or, every pattern has infinitely many matches...? – Lehs Sep 16 '14 at 18:45
  • @orangeskid: but I want perfect matches: all digits from the start of the pattern to a selected stop should correspond to a prime. – Lehs Sep 16 '14 at 18:49
  • @Lebs: You want infinitely many perfect matches. Here is how to construct in principle such an infinite sequence: Start with any finite pattern of $0$ and $1$. Find a prime number that starts with this pattern. Take the pattern of this prime. Find another longer prime pattern that starts with this pattern. Take the pattern of that prime. Find another longer prime pattern that starts with this pattern. Continue this to infinity. Still, not an explicit enough form. – orangeskid Sep 16 '14 at 18:53
  • @orangeskid: For an infinite pattern I want a certain start of it to be a prime. Just don't be a prime that starts with the pattern. For example: some finite subsequences of 111111111... correspond to primes (Mersenne) but it isn't known if there are infinitely many perfect matches. – Lehs Sep 16 '14 at 19:08
  • @orangeskid: Thanks, I finally understood. – Lehs Sep 17 '14 at 20:34
  • @Lebs: No worries. It looks like an infinite pattern like that can be computed by a machine. Of course, it does get very complicated. – orangeskid Sep 19 '14 at 01:03

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