The prime constant is the real number \( \rho \) whose nth 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 kth 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 >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.
External links
Weisstein, Eric W. "Prime Constant". MathWorld.
vte
Irrational numbers
Chaitin's (Ω) Liouville Prime (ρ) Logarithm of 2 Gauss's (G) Twelfth root of 2 Apéry's (ζ(3)) Plastic (ρ) Square root of 2 Supergolden ratio (ψ)
Erdős–Borwein (E) Golden ratio (φ) Square root of 3 Square root of 5 Silver ratio (δS) Euler's (e) Pi (π)
Gold, square root of 2, and square root of 3 rectangles.png
Schizophrenic Transcendental Trigonometric
Undergraduate Texts in Mathematics
Graduate Studies in Mathematics
Hellenica World - Scientific Library
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