\[ g(n)= \begin{cases} 1, \quad \quad\quad \quad\quad\quad\quad\quad\quad \quad n = 2^0, 2^1, 2^1, 2^2, \ldots \\ \displaystyle \prod_{\text{odd prime }p} \left( \dfrac{p-1}2 \right)^{v_p (n) }, \quad \text{otherwise} \end{cases} \] Let \(g(n) \) be defined as above, where \(v_p(n) \) denotes the highest power of prime \(p\) that divides \(n\).

Let \(M\) be the set of all rational numbers \(\mu\) for which there is at least one positive integer n, such that \(\mu=\log _{ n }{ g(n) } \). In other words:

\(M=Q\cap \left\{ \log _{ n }{ g(n) } :\quad n\in \mathbb N \right\} \),

wher \(Q\) is the set of all rational numbers and \(N\) the set of all positive integers.

What is the cardinality of \(M\)?

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