Totient summatory function

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

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

It is the number of ordered pairs of coprime integers (p,q), where 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 are 1, 32, 3044, 304192, 30397486, 3039650754, ... (sequence A064018 in the OEIS).

Applying Möbius inversion to the totient function yields

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

Φ(n) has the asymptotic expansion

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

where ζ(2) is the Riemann zeta function evaluated at 2, which is ${\displaystyle {\frac {\pi ^{2}}{6}}}$.

The summatory function of the reciprocal of the totient is

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

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,

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

and

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

The constant A = 1.943596... is sometimes known as Landau's totient constant. The sum ${\displaystyle \textstyle \sum _{k=1}^{\infty }1/(k\;\varphi (k))}$ converges to

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

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

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

• Arithmetic function

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

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