Prime constant

Not to be confused with Hardy–Littlewood's twin prime constant or Brun's twin prime constant.

The prime constant is the real number $\rho$ whose $n$th binary digit is 1 if $n$ is prime and 0 if $n$ is composite or 1.

In other words, $\rho$ is the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is,

$\rho =\sum _{p}{\frac {1}{2^{p}}}=\sum _{n=1}^{\infty }{\frac {\chi _{\mathbb {P} }(n)}{2^{n}}}$

where $p$ indicates a prime and $\chi _{\mathbb {P} }$ is the characteristic function of the set $\mathbb {P}$ of prime numbers.

The beginning of the decimal expansion of ρ is: $\rho =0.414682509851111660248109622\ldots$ (sequence A051006 in the OEIS)

The beginning of the binary expansion is: $\rho =0.011010100010100010100010000\ldots _{2}$ (sequence A010051 in the OEIS)

Irrationality

The number $\rho$ is irrational.

Proof by contradiction

Suppose $\rho$ were rational.

Denote the $k$th digit of the binary expansion of $\rho$ by $r_{k}$. Then since $\rho$ is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers $N$ and $k$ 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 $k+1\geq 2$. Since $r_{p}\neq r_{p(k+1)}$ we see that $\rho$ is irrational.