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Properties

Applying Möbius inversion to the totient function yields

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 ) = 3 π 2 n 2 + O ( n log ⁡ n ) , {\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 π 2 6 {\displaystyle {\frac {\pi ^{2}}{6}}} .¹

Reciprocal totient summatory function

The summatory function of the reciprocal of the totient is

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 ∈ 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\in \mathbb {P} }\left(1+{\frac {1}{p(p-1)}}\right),}

and

B = ∑ k = 1 ∞ μ ( k ) 2 log ⁡ k k φ ( k ) = A ∏ p ∈ 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\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 ∑ k = 1 ∞ 1 / ( k φ ( k ) ) {\displaystyle \textstyle \sum _{k=1}^{\infty }1/(k\;\varphi (k))} converges to

∑ k = 1 ∞ 1 k φ ( k ) = ζ ( 2 ) ∏ p ∈ P ( 1 + 1 p 2 ( p − 1 ) ) = 2.20386 … . {\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,² and its value is

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

See also

• Arithmetic function

• 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)

References

1. Weisstein, Eric W., "Riemann Zeta Function \zeta(2)", MathWorld /wiki/Eric_W._Weisstein ↩

2. OEIS: A065483 /wiki/On-Line_Encyclopedia_of_Integer_Sequences ↩