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 $B$?