Prime constant
The prime constant is the real number [math]\displaystyle{ \rho }[/math] whose [math]\displaystyle{ n }[/math]th binary digit is 1 if [math]\displaystyle{ n }[/math] is prime and 0 if [math]\displaystyle{ n }[/math] is composite or 1.
In other words, [math]\displaystyle{ \rho }[/math] is the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is,
- [math]\displaystyle{ \rho = \sum_{p} \frac{1}{2^p} = \sum_{n=1}^\infty \frac{\chi_{\mathbb{P}}(n)}{2^n} }[/math]
where [math]\displaystyle{ p }[/math] indicates a prime and [math]\displaystyle{ \chi_{\mathbb{P}} }[/math] is the characteristic function of the set [math]\displaystyle{ \mathbb{P} }[/math] of prime numbers.
The beginning of the decimal expansion of ρ is: [math]\displaystyle{ \rho = 0.414682509851111660248109622\ldots }[/math] (sequence A051006 in the OEIS)
The beginning of the binary expansion is: [math]\displaystyle{ \rho = 0.011010100010100010100010000\ldots_2 }[/math] (sequence A010051 in the OEIS)
Irrationality
The number [math]\displaystyle{ \rho }[/math] can be shown to be irrational.[1] To see why, suppose it were rational.
Denote the [math]\displaystyle{ k }[/math]th digit of the binary expansion of [math]\displaystyle{ \rho }[/math] by [math]\displaystyle{ r_k }[/math]. Then since [math]\displaystyle{ \rho }[/math] is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers [math]\displaystyle{ N }[/math] and [math]\displaystyle{ k }[/math] such that [math]\displaystyle{ r_n = r_{n+ik} }[/math] for all [math]\displaystyle{ n \gt N }[/math] and all [math]\displaystyle{ i \in \mathbb{N} }[/math].
Since there are an infinite number of primes, we may choose a prime [math]\displaystyle{ p \gt N }[/math]. By definition we see that [math]\displaystyle{ r_p=1 }[/math]. As noted, we have [math]\displaystyle{ r_p=r_{p+ik} }[/math] for all [math]\displaystyle{ i \in \mathbb{N} }[/math]. Now consider the case [math]\displaystyle{ i=p }[/math]. We have [math]\displaystyle{ r_{p+i \cdot k}=r_{p+p \cdot k}=r_{p(k+1)}=0 }[/math], since [math]\displaystyle{ p(k+1) }[/math] is composite because [math]\displaystyle{ k+1 \geq 2 }[/math]. Since [math]\displaystyle{ r_p \neq r_{p(k+1)} }[/math] we see that [math]\displaystyle{ \rho }[/math] is irrational.
References
- ↑ Hardy, G. H. (2008). An introduction to the theory of numbers. E. M. Wright, D. R. Heath-Brown, Joseph H. Silverman (6th ed.). Oxford: Oxford University Press. ISBN 978-0-19-921985-8. OCLC 214305907. https://www.worldcat.org/oclc/214305907.
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