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In number theory , the totient summatory function
Φ
(
n
)
{\displaystyle \Phi (n)}
summatory function of Euler's totient function defined by:
Φ
(
n
)
:=
∑
k
=
1
n
φ
(
k
)
,
n
∈
N
{\displaystyle \Phi (n):=\sum _{k=1}^{n}\varphi (k),\quad n\in \mathbf {N} }
Properties
Using Möbius inversion to the totient function, we obtain
Φ
(
n
)
=
∑
k
=
1
n
k
∑
d
∣
k
μ
(
d
)
d
=
1
2
∑
k
=
1
n
μ
(
k
)
⌊
n
k
⌋
(
1
+
⌊
n
k
⌋
)
{\displaystyle \Phi (n)=\sum _{k=1}^{n}k\sum _{d\mid k}{\frac {\mu (d)}{d}}={\frac {1}{2}}\sum _{k=1}^{n}\mu (k)\left\lfloor {\frac {n}{k}}\right\rfloor \left(1+\left\lfloor {\frac {n}{k}}\right\rfloor \right)}
Φ(n ) has the asymptotic expansion
Φ
(
n
)
∼
1
2
ζ
(
2
)
n
2
+
O
(
n
log
n
)
,
{\displaystyle \Phi (n)\sim {\frac {1}{2\zeta (2)}}n^{2}+O\left(n\log n\right),}
where ζ(2) is the Riemann zeta function for the value 2.
Φ(n ) is the number of coprime integer pairs {p, q}, 1 ≤ p ≤ q ≤ n .
The summatory of reciprocal totient function
The summatory of reciprocal totient function is defined as
S
(
n
)
:=
∑
k
=
1
n
1
φ
(
k
)
{\displaystyle S(n):=\sum _{k=1}^{n}{\frac {1}{\varphi (k)}}}
Edmund Landau showed in 1900 that this function has the asymptotic behavior
S
(
n
)
∼
A
(
γ
+
log
n
)
+
B
+
O
(
log
n
n
)
{\displaystyle S(n)\sim A(\gamma +\log n)+B+O\left({\frac {\log n}{n}}\right)}
where γ is the Euler–Mascheroni constant ,
A
=
∑
k
=
1
∞
μ
(
k
)
2
k
φ
(
k
)
=
ζ
(
2
)
ζ
(
3
)
ζ
(
6
)
=
∏
p
(
1
+
1
p
(
p
−
1
)
)
{\displaystyle A=\sum _{k=1}^{\infty }{\frac {\mu (k)^{2}}{k\varphi (k)}}={\frac {\zeta (2)\zeta (3)}{\zeta (6)}}=\prod _{p}\left(1+{\frac {1}{p(p-1)}}\right)}
and
B
=
∑
k
=
1
∞
μ
(
k
)
2
log
k
k
φ
(
k
)
=
A
∏
p
(
log
p
p
2
−
p
+
1
)
.
{\displaystyle B=\sum _{k=1}^{\infty }{\frac {\mu (k)^{2}\log k}{k\,\varphi (k)}}=A\,\prod _{p}\left({\frac {\log p}{p^{2}-p+1}}\right).}
The constant A = 1.943596...Landau's totient constant . The sum
∑
k
=
1
∞
1
k
φ
(
k
)
{\displaystyle \textstyle \sum _{k=1}^{\infty }{\frac {1}{k\varphi (k)}}}
∑
k
=
1
∞
1
k
φ
(
k
)
=
ζ
(
2
)
∏
p
(
1
+
1
p
2
(
p
−
1
)
)
=
2.20386
…
{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k\varphi (k)}}=\zeta (2)\prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=2.20386\ldots }
In this case, the product over the primes in the right side is a constant known as totient summatory constant ,[ 1]
∏
p
(
1
+
1
p
2
(
p
−
1
)
)
=
1.339784
…
{\displaystyle \prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=1.339784\ldots }
See also
References
External links
