Totient summatory function

Using Möbius inversion to the totient function, we obtain

${\displaystyle \Phi (n)=\sum _{k=1}^{n}k\sum _{d\mid k}{\frac {\mu (d)}{d}}={\frac {1}{2}}\sum _{k=1}^{n}\mu (k)\left\lfloor {\frac {n}{k}}\right\rfloor \left(1+\left\lfloor {\frac {n}{k}}\right\rfloor \right)}$

Φ(n) has the asymptotic expansion

${\displaystyle \Phi (n)\sim {\frac {1}{2\zeta (2)}}n^{2}+O\left(n\log n\right),}$

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 is defined as

${\displaystyle S(n):=\sum _{k=1}^{n}{\frac {1}{\varphi (k)}}}$

Edmund Landau showed in 1900 that this function has the asymptotic behavior

${\displaystyle S(n)\sim A(\gamma +\log n)+B+O\left({\frac {\log n}{n}}\right)}$

where γ is the Euler–Mascheroni constant,

${\displaystyle A=\sum _{k=1}^{\infty }{\frac {\mu (k)^{2}}{k\varphi (k)}}={\frac {\zeta (2)\zeta (3)}{\zeta (6)}}=\prod _{p}\left(1+{\frac {1}{p(p-1)}}\right)}$

and

${\displaystyle B=\sum _{k=1}^{\infty }{\frac {\mu (k)^{2}\log k}{k\,\varphi (k)}}=A\,\prod _{p}\left({\frac {\log p}{p^{2}-p+1}}\right).}$

The constant A = 1.943596... is sometimes known as Landau's totient constant. The sum ${\displaystyle \textstyle \sum _{k=1}^{\infty }{\frac {1}{k\varphi (k)}}}$ is convergent and equal to:

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k\varphi (k)}}=\zeta (2)\prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=2.20386\ldots }$

In this case, the product over the primes in the right side is a constant known as totient summatory constant,[1] and its value is:

${\displaystyle \prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=1.339784\ldots }$

• Arithmetic function

• Weisstein, Eric W. "Totient Summatory Function". MathWorld.

• Totient summatory function
• Decimal expansion of totient constant product(1 + 1/(p^2*(p-1))), p prime >= 2)