The real number given by the absolutely convergent series

$$\displaystyle A = \sum_{k=1}^\infty \frac{|\mu(k)|}{k \phi(k)}$$

is known as Landau's Totient Constant. It can be explicitly evaluated to be $\frac{\zeta(2)\zeta(3)}{\zeta(6)}$. Indeed, we see that $A$ can be expanded into an Euler product

$$\displaystyle A = \prod_p \left(1 + \frac{1}{p(p-1)} \right).$$

We then have

\begin{align*} 1 + \frac{1}{p(p-1)} & = \frac{p^2 - p + 1}{p(p-1)} \\ & = \frac{p^3 + 1}{p(p^2 - 1)} \\ & = \frac{(p^3 - 1)(p^3 + 1)}{p(p^2 - 1)(p^3 - 1)} \\ & = \frac{p^6 - 1}{p(p^2 - 1)(p^3 - 1)} \\ & = \frac{1 - p^{-6}}{(1 - p^{-2})(1 - p^{-3})}, \end{align*} and from here we see that

$$\displaystyle \prod_p \left(1 + \frac{1}{p(p-1)} \right) = \prod_p \left(\frac{1 - p^{-6}}{(1 - p^{-2})(1 - p^{-3})} \right) = \frac{\zeta(2) \zeta(3)}{\zeta(6)}$$

as desired.

I am looking to evaluate the related number

$$\displaystyle B = \sum_{k=1}^\infty \frac{\mu(k)}{k \phi(k)} = \prod_p \left(1 - \frac{1}{p(p-1)} \right),$$

where there is no absolute value in the numerator. The "complete the cube" trick used above obviously will not work, as one gets the quadratic polynomial $x^2 - x - 1$ instead in the numerator. Is there a nice expression for $B$? In lieu of that, is there a reasonable interpretation for $B$?

  • 4
    $\begingroup$ That number has a "well-known" interpretation: it is the proportionality factor in the Artin primitive root conjecture. I suspect there is no tidy formula for this number. $\endgroup$
    – KConrad
    Commented Aug 6, 2020 at 1:47
  • 1
    $\begingroup$ The decimal expansion, along with many references to the literature, is given at oeis.org/A005596 $\endgroup$ Commented Aug 6, 2020 at 3:17

1 Answer 1


This constant is known as Artin's Constant. The book Finch S. R. Mathematical constants (section 2.4) gives the following information.

A rapidly convergent expression for Artin's constant is as follows [12-18]. Define Lucas' sequence as $$ l_{0}=2, \quad l_{1}=1, \quad l_{n}=l_{n-1}+l_{n-2} \quad \text { for } n \geq 2 $$ and observe that $l_{n}=\varphi^{n}+(1-\varphi)^{n},$ where $\varphi$ is the Golden mean [1.2]. Then $$ \begin{aligned} C_{\text {Artin }} &=\prod_{n \geq 2} \zeta(n)^{-\frac{1}{n} \sum_{k \mid n} l_{k} \cdot \mu\left(\frac{n}{k}\right)} \\ &=\zeta(2)^{-1} \zeta(3)^{-1} \zeta(4)^{-1} \zeta(5)^{-2} \zeta(6)^{-2} \zeta(7)^{-4} \zeta(8)^{-5} \zeta(9)^{-8} \cdots. \end{aligned} $$


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