Totient summatory function

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

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

It is the number of ordered pairs of coprime integers Template:Math, where Template:Math.

The first few values are 0, 1, 2, 4, 6, 10, 12, 18, 22, 28, 32, ... Template:OEIS. Values for powers of 10 are 1, 32, 3044, 304192, 30397486, 3039650754, ... Template:OEIS.

Properties

Applying Möbius inversion to the totient function yields

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

Template:Math has the asymptotic expansion

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

where Template:Math is the Riemann zeta function evaluated at 2, which is $\frac{{\pi }^2}{6}$.

Reciprocal totient summatory function

The summatory function of the reciprocal of the totient is

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

Edmund Landau showed in 1900 that this function has the asymptotic behaviorTemplate:Citation needed

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

where Template:Math is the Euler–Mascheroni constant,

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

and

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

The constant Template:Math is sometimes known as Landau's totient constant. The sum ${\sum }_{k=1}^{\mathrm{\infty }}1/(k\phi (k))$ converges to

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\phi (k)}=\zeta (2)\prod _{p\in \mathbb{P}}(1+\frac{1}{p^2(p-1)})=2.20386\dots .$

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

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

See also

• Arithmetic function

References

• Template:MathWorld

External links

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

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