Totient summatory function

In number theory, the totient summatory function $\Phi (n)$ is a summatory function of Euler's totient function defined by:

$\Phi (n):=\sum _{k=1}^{n}\varphi (k),\quad n\in \mathbf {N}$

It is the number of coprime integer pairs {p, q}, 1 ≤ p ≤ q ≤ n.

The first few values are 0, 1, 2, 4, 6, 10, 12, 18, 22, 28, 32 (sequence A002088 in the OEIS). Values for powers of 10 at (sequence A064018 in the OEIS).

Properties

Using Möbius inversion to the totient function, we obtain

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

Φ(n) has the asymptotic expansion

$\Phi (n)\sim {\frac {1}{2\zeta (2)}}n^{2}+O\left(n\log n\right)={\frac {3}{\pi ^{2}}}n^{2}+O\left(n\log n\right),$

where ζ(2) is the Riemann zeta function for the value 2, which is ${\frac {\pi ^{2}}{6}}$ ¶.

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

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

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

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

where γ is the Euler–Mascheroni constant,

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

and

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

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

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

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

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

See also

• Arithmetic function

References

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

External links

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