Define a number $q$ in binary notation whose $n$-th bit is $1$ for $n$ prime, and $0$ for $n$ composite. So its 2nd, 3rd, 5th, 7th, 11th, etc. bits are $1$, with all other bits $0$. Here is $q$ out to its $101$-st bit: $$.01101010001010001010001000001 010000010001010001000001000 001010000010001010000010001 000001000000010001$$ What is known about $q$? ($\approx 0.414683_{10}$). Has it been investigated? Does it have a name? It is irrational. But is it transcendental?
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3It breaks tables. I have a marvelous proof of this fact, but it is too large to contain in the body of your question. – Eric Tressler Nov 10 '15 at 01:48
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1http://math.stackexchange.com/questions/42697/what-is-the-name-of-this-number-is-it-transcendental – Dan Brumleve Nov 10 '15 at 01:58
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See also the answer linked in Dan's reference: Are these numbers irrational and/or transcendental?. – Joseph O'Rourke Nov 10 '15 at 02:04
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Your number is known as the prime constant. If we denote it by $\rho$, we have:
$$ \rho =\sum _{{p\in\mathbb{P}}}{\frac {1}{2^{p}}}=\sum _{{n=1}}^{\infty }{\frac {\chi _{{{\mathbb {P}}}}(n)}{2^{n}}}, $$
where $\mathbb{P}$ denotes the set of prime numbers and $\chi _{{{\mathbb {P}}}}$ is the characteristic function of prime numbers, i.e., the function such that for positive integer $n$:
$$ {\displaystyle \chi_\mathbb{P}(n):={\begin{cases}1&{\text{if }}n\in \mathbb{P},\\0&{\text{if }}n\notin \mathbb{P}.\end{cases}}} $$
The decimal expansion of $\rho$ begins with: \begin{align} \rho&=0.414682509851111660248109622\ldots \\ &=0.011010100010100010_2. \end{align}
and is included in the OEIS as sequence A051006.
Klangen
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