Real number whose nth binary digit is 1 if n is prime and 0 if n is composite or 1
Not to be confused with Hardy–Littlewood's twin prime constant or Brun's twin prime constant .
The prime constant is the real number
ρ
{\displaystyle \rho }
whose
n
{\displaystyle n}
th binary digit is 1 if
n
{\displaystyle n}
is prime and 0 if
n
{\displaystyle n}
is composite or 1.
In other words,
ρ
{\displaystyle \rho }
is the number whose binary expansion corresponds to the indicator function of the set of prime numbers . That is,
ρ
=
∑
p
1
2
p
=
∑
n
=
1
∞
χ
P
(
n
)
2
n
{\displaystyle \rho =\sum _{p}{\frac {1}{2^{p}}}=\sum _{n=1}^{\infty }{\frac {\chi _{\mathbb {P} }(n)}{2^{n}}}}
where
p
{\displaystyle p}
indicates a prime and
χ
P
{\displaystyle \chi _{\mathbb {P} }}
is the characteristic function of the set
P
{\displaystyle \mathbb {P} }
of prime numbers.
The beginning of the decimal expansion of ρ is:
ρ
=
0.414682509851111660248109622
…
{\displaystyle \rho =0.414682509851111660248109622\ldots }
(sequence A051006 in the OEIS )
The beginning of the binary expansion is:
ρ
=
0.011010100010100010100010000
…
2
{\displaystyle \rho =0.011010100010100010100010000\ldots _{2}}
(sequence A010051 in the OEIS )
Irrationality
The number
ρ
{\displaystyle \rho }
is irrational .
Proof by contradiction
Suppose
ρ
{\displaystyle \rho }
were rational .
Denote the
k
{\displaystyle k}
th digit of the binary expansion of
ρ
{\displaystyle \rho }
by
r
k
{\displaystyle r_{k}}
. Then since
ρ
{\displaystyle \rho }
is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers
N
{\displaystyle N}
and
k
{\displaystyle k}
such that
r
n
=
r
n
+
i
k
{\displaystyle r_{n}=r_{n+ik}}
for all
n
>
N
{\displaystyle n>N}
and all
i
∈
N
{\displaystyle i\in \mathbb {N} }
.
Since there are an infinite number of primes, we may choose a prime
p
>
N
{\displaystyle p>N}
. By definition we see that
r
p
=
1
{\displaystyle r_{p}=1}
. As noted, we have
r
p
=
r
p
+
i
k
{\displaystyle r_{p}=r_{p+ik}}
for all
i
∈
N
{\displaystyle i\in \mathbb {N} }
. Now consider the case
i
=
p
{\displaystyle i=p}
. We have
r
p
+
i
⋅
k
=
r
p
+
p
⋅
k
=
r
p
(
k
+
1
)
=
0
{\displaystyle r_{p+i\cdot k}=r_{p+p\cdot k}=r_{p(k+1)}=0}
, since
p
(
k
+
1
)
{\displaystyle p(k+1)}
is composite because
k
+
1
≥
2
{\displaystyle k+1\geq 2}
. Since
r
p
≠
r
p
(
k
+
1
)
{\displaystyle r_{p}\neq r_{p(k+1)}}
we see that
ρ
{\displaystyle \rho }
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 .
External links
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