From Wikipedia, the free encyclopedia
In mathematics, the Feller–Tornier constant C FT is the density of the set of all positive integers that have an even number of distinct prime factors raised to a power larger than one (ignoring any prime factors which appear only to the first power).[ 1]
It is named after William Feller (1906–1970) and Erhard Tornier (1894–1982)[ 2]
C
FT
=
1
2
+
(
1
2
∏
n
=
1
∞
(
1
−
2
p
n
2
)
)
=
1
2
(
1
+
∏
n
=
1
∞
(
1
−
2
p
n
2
)
)
=
1
2
(
1
+
1
ζ
(
2
)
∏
n
=
1
∞
(
1
−
1
p
n
2
−
1
)
)
=
1
2
+
3
π
2
∏
n
=
1
∞
(
1
−
1
p
n
2
−
1
)
=
0.66131704946
…
{\displaystyle {\begin{aligned}C_{\text{FT}}&={1 \over 2}+\left({1 \over 2}\prod _{n=1}^{\infty }\left(1-{2 \over p_{n}^{2}}\right)\right)\\[4pt]&={{1} \over {2}}\left(1+\prod _{n=1}^{\infty }\left(1-{{2} \over {p_{n}^{2}}}\right)\right)\\[4pt]&={1 \over 2}\left(1+{{1} \over {\zeta (2)}}\prod _{n=1}^{\infty }\left(1-{{1} \over {p_{n}^{2}-1}}\right)\right)\\[4pt]&={1 \over 2}+{{3} \over {\pi ^{2}}}\prod _{n=1}^{\infty }\left(1-{{1} \over {p_{n}^{2}-1}}\right)=0.66131704946\ldots \end{aligned}}}
(sequence A065493 in the OEIS )
The Big Omega function is given by
Ω
(
x
)
=
the number of prime factors of
x
counted by multiplicities
{\displaystyle \Omega (x)={\text{the number of prime factors of }}x{\text{ counted by multiplicities}}}
See also: Prime omega function .
The Iverson bracket is
[
P
]
=
{
1
if
P
is true,
0
if
P
is false.
{\displaystyle [P]={\begin{cases}1&{\text{if }}P{\text{ is true,}}\\0&{\text{if }}P{\text{ is false.}}\end{cases}}}
With these notations, we have
C
FT
=
lim
n
→
∞
∑
k
=
1
n
(
[
Ω
(
k
)
≡
0
mod
2
]
)
n
{\displaystyle C_{\text{FT}}=\lim _{n\to \infty }{\frac {\sum _{k=1}^{n}([\Omega (k)\equiv 0{\bmod {2}}])}{n}}}
Prime zeta function [ edit ]
The prime zeta function P is give by
P
(
s
)
=
∑
p
is prime
1
p
s
.
{\displaystyle P(s)=\sum _{p{\text{ is prime}}}{\frac {1}{p^{s}}}.}
The Feller–Tornier constant satisfies
C
FT
=
1
2
(
1
+
exp
(
−
∑
n
=
1
∞
2
n
P
(
2
n
)
n
)
)
.
{\displaystyle C_{\text{FT}}={1 \over 2}\left(1+\exp \left(-\sum _{n=1}^{\infty }{2^{n}P(2n) \over n}\right)\right).}