prime constant

The number $\rho$ defined by

$$\rho =\sum _p\frac{1}{2^p}$$

is known as the prime constant. It is simply the number whose binary expansion corresponds to the characteristic function of the set of prime numbers. That is, its $n$th binary digit is $1$ if $n$ is prime and $0$ if $n$ is composite.

The beginning of the decimal expansion of $\rho$ is:

$$\rho =0.414682509851111660248109622\mathrm{\dots }$$

The number $\rho$ is easily shown to be irrational. To see why, suppose it were rational. Denote the $k$th digit of the binary expansion of $\rho$ by $r_k$. Then, since $\rho$ is assumed rational, there must exist $N$, $k$ positive integers such that $r_n=r_{n+ik}$ for all $n>N$ and all $i\in \mathbb{N}$.

Since there are an infinite number of primes, we may choose a prime $p>N$. By definition we see that $r_p=1$. As noted, we have $r_p=r_{p+ik}$ for all $i\in \mathbb{N}$. Now consider the case $i=p$. We have $r_{p+i\cdot k}=r_{p+p\cdot k}=r_{p(k+1)}=0$, since $p(k+1)$ is composite because $k+1\ge 2$. Since $r_p\ne r_{p(k+1)}$ we see that $\rho$ is irrational.

The partial continued fractions of the prime constant can be found http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A051007here.