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I am looking for pointers to any existing materials about the properties of this quantity.

For Euler's cototient, if a number $N$ is written as $2^a \cdot b$ with b odd then the cototient is $$\bar \phi (N) = 2^{a-1} (2b - \phi (b)) = 2^{a-1} (b + \bar \phi(b))$$

Since the quantity in parentheses is always even, this tells us that a cototient has at least as many factors of 2 as $N$ does. Treating $\bar \phi (N)$ as a given and rewriting $$2b - \phi(b) = b + \bar \phi (b) = \frac{\bar \phi (N)}{2^{a-1}}$$ I would like to determine which values of $a$ have solutions, and how many.

As a start, $2b-\phi (b)$ is always $2 \pmod 4$ if b has more than one factor, since $2b$ has exactly one factor of $2$ and $\phi$ gains at least one factor of two for each prime factor making it $0 \pmod 4$. Thus prime powers are the only possible solutions if $a$ is less than the exponent of $2$ in the factorization of $\bar \phi (N)$. If $a$ is greater than the exponent of $2$ then there are no solutions since $2b-\phi (b)$ is even.

Therefore most solutions will be for $a$ equal to the exponent of $2$ in the factorization of $\bar \phi (N)$, and the primary question becomes, how many solutions are there to $b + \bar \phi (b)= n$ for a given $n \equiv 2 \pmod 4$?

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