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 simply 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 primes.

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 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 $k+1\geq 2$ . Since $r_{p}\neq r_{{p(k+1)}}$ we see that $\rho$ is irrational.

Notes

References

Weisstein, Eric W. "Prime Constant". MathWorld.

Irrational numbers

Prime ( $ρ$ ) Euler–Mascheroni ( $γ$ ) Logarithm of 2 Twelfth root of two (¹²√2) Apéry's ( $ζ (3)$ ) Sophomore's dream Plastic ( $ρ$ ) Square root of 2 (√2) Erdős–Borwein ( $E$ ) Golden ratio ( $φ$ ) Square root of 3 (√3) Square root of 5 (√5) Silver ratio ( $δ S$ ) Second Feigenbaum ( $α$ ) Euler's ( $e$ ) Pi ( $π$ ) First Feigenbaum ( $δ$ )

Schizophrenic Transcendental Trigonometric

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