A number similar to Landau's Totient constant

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 $$B$$?

• 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. – KConrad Aug 6 at 1:47
• The decimal expansion, along with many references to the literature, is given at oeis.org/A005596 – Gerry Myerson Aug 6 at 3:17

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}